 
129

CALCULUS

Book 1

## by

## Peter J. Ponzo

## TABLE OF CONTENTS

EXAMPLE PROBLEMS 5

ASSORTED PROBLEMS 5

LECTURE 0 8

SOME BASICS 8

NUMBERS ... and INFINITY 8

EQ \f(5,0) 8

∞ is NOT a number 8

INEQUALITIES 8

FUNCTIONS 9

vertical line test: 9

functions and their domain 10

ABSOLUTE VALUES 10

To plot y = | f(x) | 11

SOME TRIG IDENTITIES 11

the RADIAN measure of an angle 11

SOME TRIG GRAPHS 12

SOME GEOMETRY 13

LOGARITHMS and EXPONENTIALS 13

exponential functions 13

logarithmic function 13

ODDS 'n' ENDS 14

geometric series 14

SIGMA NOTATION 15

MAPLE 15

LECTURE 1 15

LIMITS 15

LIMIT RULES 17

ONE SIDED LIMITS 19

LECTURE 2 20

INFINITE LIMITS 20

ASYMPTOTES 21

CONTINUOUS FUNCTIONS 22

LECTURE 3 24

TECHNIQUES FOR EVALUATING LIMITS WHEN THE \ 24

The form EQ \f(∞,∞) 24

The form ∞ - ∞ 24

The form EQ \f(0,0) 24

Reduce the given limit to one you know 24

to make tea 24

the SQUEEZE THEOREM 26

the graph of y = f(x) sin x 27

LECTURE 4 28

the DERIVATIVE 28

the DIFFERENTIAL 30

more on DERIVATIVES 31

DIFFERENTIATION RULES 31

the CHAIN RULE 32

HIGHER DERIVATIVES 32

concave up 33

concave down 33

the Logistic Equation 34

velocity 34

acceleration 34

LECTURE 5 34

IMPLICIT DIFFERENTIATION & TRANSCENDENTAL FUNCTIONS 34

IMPLICIT DIFFERENTIATION 34

the slope at a point (x,y) 35

greatest integer function 37

trig, exponential and log functions 37

weird limits 37

the TRIG FUNCTIONS and their derivatives 38

the EXPONENTIAL and LOG functions 38

ln x 38

LECTURE 6 39

INVERSE FUNCTIONS 39

horizontal line test 40

TESTING TO SEE IF A FUNCTION HAS AN INVERSE 40

Examples of Inverses 41

the Derivative of an Exponential Function 41

LOGARITHMIC DIFFERENTIATION 42

About Exponential Growth 43

About the number e 43

ODDS 'n' ENDS ON CURVE SKETCHING 45

Even and Odd Functions 45

Quick&Dirty Curve Sketching 46

LECTURE 7 47

MORE ON INVERSE FUNCTIONS 47

the INVERSE TANGENT 49

the INVERSE SINE 49

restricting the domain 52

Check the dimensions 52

the limit of EQ \f(sin A,A) 54

y = sin x with x in DEGREES 54

LECTURE 8 55

ABSOLUTE MAXIMUM AND MINIMUM 55

closed interval 55

critical point 56

RELATIVE MAXIMA and MINIMA 58

First Derivative Test 58

give it a name and use it! 61

Snell's law 61

LECTURE 9 63

RELATED RATE PROBLEMS 63

LECTURE 10 66

The TANGENT LINE APPROXIMATION 66

Rule of 72 67

POLYNOMIAL APPROXIMATIONS 68

quadratic approximation 68

cubic approximation 68

quartic approximation 68

best 72

LECTURE 11 72

NEWTON'S METHOD for finding roots 72

a computer algebra system 73

The error goes to zero! 74

What is the annual rate of return from this mutual fund? 76

a computer spread sheet 76

DIFFICULTIES WITH NEWTON'S METHOD 77

Pick a reasonable value for x1 78

LECTURE 12 78

L'HÔPITAL'S RULE 78

the EQ \f(0,0) 78

Interpretation of a Limiting Value 80

LECTURE 13 81

POLAR COORDINATES 81

a distance and a direction 81

polar curves 83

y2 = f(x) 85

INTERSECTION OF POLAR CURVES 86

LECTURE 14 87

the AREA UNDER A CURVE 87

the SUM of rectangles 88

the DEFINITE INTEGRAL 88

a Riemann SUM 90

PROPERTIES of the DEFINITE INTEGRAL 90

THE FUNDAMENTAL THEOREM 91

the \ 92

an ANTIDERIVATIVE 92

constant of integration 93

LECTURE 15 94

DEFINITE INTEGRATION 94

negative 94

EQ \i( , ,umming) 95

47,000,000 elemental rectangles 97

The error in area 97

LECTURE 16 99

AREAS IN POLAR COORDINATES 99

AREA SWEPT OUT BY THE RADIUS 100

this \ 101

check it for reasonableness 102

LECTURE 17 102

TECHNIQUES OF INTEGRATION 102

THE METHOD OF SUBSTITUTION 102

next 104

integration is an ART 104

Heaviside calculus 104

Shift Theorem 105

INTEGRATION BY PARTS 105

who's u and who's v 106

the Ponzo function 108

the lower limit 108

the upper limit 108

LECTURE 18 109

VOLUMES 109

cut the solid into many very thin slices 110

Volume of a cylinder 111

Volume of a cone 111

VOLUMES OF SOLIDS OF REVOLUTION 112

the volume of a sphere 112

the volume of a torus 112

make a reasonable diagram 114

the centre of area 114

THE THEOREM OF PAPPUS 114

the CENTROID 114

The centroid of a triangle 115

LECTURE 19 115

Volumes of solids of revolution using horizontal rectangles 115

a cylindrical shell 116

the volume of a \ 116

guess who's the student? 116

distance travelled 117

whatzits per doodle 117

The total work 118

digging a well 118

cost of manufacturing 118

a reasonable approximation 118

AVERAGE VALUE OF A FUNCTION 119

an average temperature 119

the \ 119

average velocity 120

A PARADOX 120

LECTURE 20 121

IMPROPER INTEGRALS 121

DEFINITION of an IMPROPER INTEGRAL 121

f(x) must approach zero very rapidly 123

Another Kind of Improper Integral 123

f(x) must get small enough fast enough 124

SOLUTIONS TO \ 124

EXAMPLE PROBLEMS

(done in the text)

• A lake is stocked with 1000 fish. It is found that N(t), the number of fish after t years, increases so that its rate of increase is governed by the equation: = k N (10,000 - N) (called the "Logistic Equation"). Show that the rate of change, , is a maximum when the population is 5,000 fish.

• The cost of manufacturing an item is $100 even if no items are manufactured ... and the cost decreases with each item; for x items the cost per item is 100 - .1x (i.e. the cost decreases by $0.1 for each item). Graph C(x), the cost of producing x items.

• An orchard contains 240 apple trees, each tree producing 30 bushels of apples. For each additional tree planted, the yield per tree decreases by 1/12 bushel (due to overcrowding). Sketch N(x), the total apple production as a function of x, the number of additional trees planted.

• A conical drinking cup is formed from a circular piece of paper by removing a sector and joining the edges. If the radius of the piece of paper is 10 cm., what should be the angle q so as to yield a cup of maximum volume?

• A man can run 10 times faster than he can swim. He begins in the water at a point P, swims to shore, then runs to Q. Describe his path so the total time is a minimum.

• A certain amount of money is left in the bank to accumulate interest (compounded at i% per year). If you want to double your money in n years, what should the interest rate be?

• In drilling a well, the cost per metre depends upon the type of sand, gravel or rock which must be excavated. Suppose the cost is C(x) _dollars/metre_ at a depth x _metres_. (As x changes, the type of material changes, hence the cost changes.) Express, as a definite integral., the cost in digging a well of H _metres_.

• The cost of manufacturing an item depends upon the number of items manufactured. For the first few items the cost is high and the profit low, but as we produce more items the cost decreases hence the profits (when we sell the items) increase. Suppose the profit for the nth item is p(n) dollars per item. Express, as a definite integral, the profit in producing (and selling) K items.

• You invest $10,000 in a mutual fund, then, 5 months later you put an additional $15,000 into the fund, then, 3 months later put in an additional $5,000. At the end of a year, your investments (totalling $30,000) have grown to $31,470. What is the annual rate of return from this mutual fund?

ASSORTED PROBLEMS

(which you'll be able to solve by the end of this course)

1. Evaluate:

(a)

(b) f'(2) if f(x) = (x-2) sin _using the limit definition of derivative!_

(c) f'(- ) if f(t) = | sin t | (you needn't use the definition ... unless you want to)

2. Compute the area bounded by the curves y = and x + 3y = 7. (Include a sketch, and since this is a complicated evaluation with lots of room for errors, _check for reasonableness_.)

3. Evaluate

4. Use Newton's Method to find approximations to the roots of the following equations, correct to five decimal places. In each case, make a reasonable plot of the function in order to obtain an the initial "guess", x1.

(a) x3 \- x2 \+ x - 22 = 0 (b) x _ln_ x = 6

5. Use l'Hopital's Rule to evaluate the following limits:

(a) (b)

6: Express each of the following in terms of one or more definite integrals using both HORIZONTAL and VERTICAL rectangles. **DO NOT EVALUATE** but include a sketch:

(a) The area bounded by y = 0, y = x, y = _ln_ x, and y = 1.

(b) The volume when the region described by 0 ≤ y ≤ 1 - x2, x ≥ 0, is revolved about the y-axis.

7. Evaluate each of the following limits (or explain why the limit does not exist):

(a) (b)

8. Evaluate:

(a) cos (b) ... and simplify your answer

9. Calculate:

(a) the average value of f(x) = on 0 ≤ x ≤ 2 (b)

10. Evaluate: (a) (b)

11. The function f(x) = , with domain x≥ 0, has an inverse g(x).

(a) Test f(x) to verify that it does have an inverse when x ≥ 0

(b) Calculate g(2) (c) Calculate g'(2)

12. The numbers π and e are both near "3". Which of eπ or πe is larger?

13. Determine the shortest distance from the origin to the curve y = (3 - x2)/2. Include a sketch!

14. Find if : (a) y = (b) y = x cos ( _ln_ x)

15. Calculate the areas described (and include a sketch):

(a) below y = and above y =

(b) enclosed by y = , y = 0, x = -4 and x = -2.

16. Evaluate if y =

17. (a) If y = ax, determine by first taking the _ln_ of each side.

(b) If y = xx, determine by first taking the _ln_ of each side.

18. Determine an approximate value to 641/3 by using a linear approximation.

19. Evaluate

20. Sketch the graph of f(x) = showing where f(x) is increasing, decreasing, asymptotes and all critical points.

21. Find the area bounded by y = and y = tan x, between x = 0 and the first intersection of these two curves to the right of x = 0. (Include a reasonable sketch.)

22. Calculate the area for each of the following regions (bounded by _polar_ curves):

(a) inside r = sin q

(b) inside the cardioid r = 1 + sin q and outside the circle r = 1

(c) inside the smaller loop of the limaçon r = 1 - 2 sin q

(d) inside the lemniscate r2 = cos 2q

23. Aluminum pop cans to hold 300 mL are made in the shape of right circular cylinders. Find the dimensions which minimize the amount of aluminum used.

24. A rectangular field next to an ocean is to be fenced on three sides with 1000 m of fencing. (The fourth side, being the shoreline, is not fenced.) Determine the dimensions of the field so the area is as large as possible.

25. A man 2m tall walks at 3 m/s directly away from a streetlight that is 8 m high. How fast is the length of his shadow changing?

26. The hour and minute hands of a clock are 3 cm and 4 cm long respectively. How fast are their tips approaching each other at 3 o'clock?

27. Obtain a cubic polynomial approximation to cos x at x = and use it to approximate cos 57˚

28. A picture 2 m tall hangs on a vertical wall, the lower edge being 1 m above your eyes. How far from the wall should you stand in order to obtain the "best view" of the picture? (i.e the angle subtended by the picture, at your eye, should be a maximum.)

29. Show that the curves xy = and x2 \- y2 = 1 intersect so that, at the point(s) of intersection, their tangent lines are perpendicular.

30. A tangent line, drawn to the curve + = 1, has x- and y-intercepts at P and Q. Show that the sum of the intercepts is a constant, independent of where the tangent line is drawn.

31. Determine where the polar curves r = q and r = cos q intersect, for q > 0. (You'll need to use Newton's method!)

32. Prove that π (1 - x) is always larger than 2x _ln_ .

Hint: consider f(x) = π (1 - x) - 2x _ln_ . Is it always positive?

33. Evaluate the following integrals:

(a) (b) (c) (d)

(e) (f) (g) (h)

LECTURE 0

This is an introduction to the ideas of the calculus. Many students will have had a previous calculus course and been exposed to the idea of a limit, differentiation "rules" (product, quotient, chain rule, implicit differentiation, etc.), derivative of trig, exponential and log functions, optimization and related rate problems, some curve sketching, Riemann sums and the calculation of area. So, nearly all of this course should be vaguely familiar to many students - but we won't count on it! In fact, we'll start from the beginning, assuming very little previous knowledge of the calculus ... so it'll be a review for many.

SOME BASICS

**NUMBERS ... and INFINITY** :

Once upon a time there were only the _positive_ integers: 1, 2, 3, 4, ... and we could add them or multiply them and we would again get positive integers: 17 + 21 = 38, 7 x 8 = 56.

When we subtract however, we can get _negative_ integers: 17 - 21 = -4, so we add these negative integers to our collection of numbers. Now we have _all_ the integers (including 0) and when we add, subtract or multiply numbers in this collection we get other numbers in the same collection.

When we divide, however, we can get _fractions_ which are _not_ in our collection of positive and negative integers ... so we add them to our collection to get the set of "rational numbers". For example, is a (rational) number which, when multiplied by the _integer_ 3 yields the _integer_ 5.

What, then, is ? If it's a number, then it should yield **5** when multiplied by 0. But every number yields **0** when multiplied by 0 ... so is NOT a number. **Remember this!** (It may be a cauliflower, but it's NOT a number!)

We will have occasion to "let x approach infinity", which we write: x->∞. This simply means that x is allowed to increase without bound; it becomes larger than ANY number.

Is ∞ a number? No. It's a convenient symbol which we will use to indicate that a quantity (like x) is increasing so as to exceed (eventually) every number.

Consider the limiting value of the ratio as x->∞. (We will have more to say about "limits" later.) The numerator becomes infinite, as does the denominator. Can we write the limiting value as ... and is this equal to 1? No. In fact, the limiting value of is the number 2 (as x->∞). Similarly we can't say that ∞ - ∞ is 0 (as in the limiting value of x2 \- as x->∞). Since ∞ is NOT a number it's not surprising that it doesn't behave like a number!

Remember:

0 x ∞ ≠ 0 and ∞ - ∞ ≠ 0 and ≠ 1 and ≠ ∞ and ≠ 1

PS **:**

S: Why isn't = 1? I can accept ≠ 1 (since ∞ isn't a number then division isn't an operation one can perform on ∞) but 0 is a number.

P: Let' see. When multiplied by 0, must yield 0. That's certainly true of the number 5 (that is, 5 multiplied by 0 does yield 0), so maybe = 5. But then any number yields 0 when multiplied by 0. So maybe can be any number. It certainly can't be any specific number like 5. Let's say it's indeterminate.

S: Fair enough. One last thing. Any number multiplied by 0 gives 0, right?

P: Right.

S: Then surely ∞ multiplied by 0 must give 0.

P: And a brown cow multiplied by 0? Does it give 0? Remember, ∞ is NOT a number.

S: Okay, okay. But are there any more numbers? I mean, if subtracting gave us the negative numbers and division the rational numbers, why don't we do something else and get more numbers? Are there any more?

P: Sure. Besides adding, subtracting, multiplying and dividing we can take roots, say the square root. That'd give us more numbers. For example, is not a rational number. It's irrational.

S: And will taking roots give us all the numbers?

P: No. There are numbers like π which can't be obtained by taking the root of a rational number.

S: And what do you call these numbers?

P: They're irrational, just like , but they're also called transcendental numbers.

S: I've got a headache. Can we just keep going'?

**INEQUALITIES** :

We will denote the set of numbers in the interval from -1 to 7, including both -1 and 7, as [-1,7] ... or sometimes we'll use: -1 ≤ x ≤ 7. If -1 is included but 7 is _not_ , we'll use the notation [-1,7) ... or sometimes -1 ≤ x < 7. If neither -1 nor 7 is included we'll use (-1,7) or perhaps -1 < x < 7.

Intervals of the type a ≤ x ≤ b (which include both end-points) are called CLOSED intervals. Intervals of the type a < x < b which include neither end-point are called OPEN intervals. The interval -1 < x ≤ 7 is neither open nor closed.

We are often required to "solve" inequalities like: ≥ 1. Here, for example, we must find the values of x which make greater than, or equal to, 1. If we simply multiply both sides of the inequality by (3x+1), just as we would with an equality (i.e. an "equation") we would get 4x ≥ 3x+1 hence x ≥ 1 (subtracting 3x from each side) and we would conclude that only those values of x greater than (or equal to) 1 will make ≥ 1. It is a surprise, then, to find that x = - 10 _also_ makes ≥ 1. (Try it!)

The rule is this: when multiplying an inequality by a number (or an expression, like 3x+1) we must change the _direction_ of the inequality if the number (or expression) is negative.

Examples:

• Although 7 > 5, after multiplying both sides by -3, we get -21 ≤ -15 (and the direction of the inequality has _changed_ ). However, multiplying by 2 (a positive number) we get 14 ≥ 10 (and we don't change the direction of the inequality).

• If ≥ 1, then 4x ≥ 3x+1 provided 3x+1 > 0. Hence, when we found that x ≥ 1, we were only finding x-values satisfying 3x+1 > 0 (i.e. x > -1/3).To find _all_ solutions we proceed as follows:

(i) We look for solutions satisfying 3x+1>0 (i.e. x > -1/3). Then ≥ 1 is true provided 4x ≥ 3x+1, that is, provided x ≥ 1. We now have all solutions satisfying 3x+1 > 0, namely all x-values satisfying x ≥ 1. But x > -1/3 _and_ x ≥ 1 means all x-values satisfying x ≥ 1.

(ii) Now we look for solutions satisfying 3x+1<0 (i.e. x < -1/3). Multiplying the inequality by (3x+1) will change its direction: 4x ≤ 3x+1, hence x ≤ 1. We now have, as solutions, all x-values satisfying x < -1/3 _and_

x ≤ 1 ... hence all x-values satisfying x < -1/3 (and that includes the solution x = -10 mentioned above!)

(ii) Finally, then, any value of x which satisfies either **x ≥ 1** or **x < -1/3** will satisfy ≥ 1

Of course, a picture is worth a thousand words, so here's the graph of y = . Note that y _is_ greater than or equal to 1 when x <\- and then again when x ≥ 1. |

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**FUNCTIONS** :

If a quantity, which we'll call "y", depends upon another quantity (we'll call it "x") so that y assumes a single, unique value for each value of x (chosen from some set of values), then we say that "y is a function of x" and write y = f(x). The requirement of a "single, unique value" is important; if we don't have that, we don't have a "function". Graphically, this means a function must satisfy the _vertical line test_ : every vertical line (in the domain) must cross the graph of y = f(x) only once. If it crosses more than once, then there are two or more y-values and the graph is NOT that of a _function_.

Examples:

• A person's height is a _function_ of her age: height = f(age)

• The volume of a sphere is a _function_ of its radius: V = f(r)

The set of values from which x is chosen is called the _domain_ of the function. The set of possible values for y is called the _range_ of the function.

Note:

It's common, when writing y = f(x), to call "x" the _independent variable_ , "y" the _dependent variable_ and "f" the _function_. The notation f(x) really means the _value_ of the function "f" when the independent variable has the value x. In keeping with this notation we will sometimes refer to the function "f" (although we will sometimes, for the sake of clarity, refer to the function "f(x)"... unless confusing the _value_ with the _function_ makes things less clear!)

For example we might say "Consider the function x2" when we _really_ mean "Consider the function which _squares_ " or perhaps "Consider the function which takes a number x and generates the number x2". The _function_ is really the operation (like "squaring"), not the result of applying this operation.

Further, although we will often use "f" as a label for our function ... hence we'll refer to f(x) ... we will also use g(x) and h(x) etc. Curious how the labels for functions are often chosen from the middle of the alphabet: "f", "g", "h" ... constants are chosen from the beginning: "a", "b", "c" ... and variables from the end: "x", "y", "z". In fact, if we're talking about temperature T as a function of time t, we might just write T(t), and if we're talking about pressure P as a function of volume V we might write P(V) ... and so on. In particular, it is common practice to write x(t) for the position of an object at time t and v(t) for its velocity.

Let's talk a little about functions and their domain:

Examples:

• y is a function of x according to the rule: y = , where the _domain_ is the set of numbers -1 ≤ x ≤ 1 (else y will be the square root of a negative number) and the _range_ is the set: 0 ≤ y ≤ 1. (For each x in the _domain_ , y will lie in this _range_.). Sometimes the _domain_ isn't specified explicitly, but is understood to be whatever x-values will provide a real value for y. In this example, we would identify the domain as -1 ≤ x ≤ 1 without being told!

• y = x2 +4x - 7. What is the _domain_? Since every real number x provides a real value for y, the _domain_ is the entire real line: -∞ < x < ∞ (unless otherwise specified). The _range_ is harder to identify. It's the totality of possible values of y, and happens to be -11 ≤ y < ∞. (Can you prove this?)

• Suppose the value of y is related to the x-value according to x2 \+ y2 = 25. In this example, if we write

y2 = 25 - x2 and recognize that y2, hence 25 - x2, must be positive (or, at least, non-negative) then we must have x2 ≤ 25 so x must be restricted to the interval -5 ≤ x ≤ 5. However, for each x in this interval we can't guarantee that the relation defines a single, unique y-value! For example, if x = 3, then y2 = 25 - 32 = 16 and the y-value could be 4 or -4. Hence, the relation x2 \+ y2 = 25 does NOT define y as a "function" of x. We could, of course, "solve for y" and obtain the possible "solutions": y = or perhaps y = - and either one will define a function with _domain_ -1 ≤ x ≤ 1 ... but the original relation did NOT define a function.

**ABSOLUTE VALUES** :

We use the notation | N | to mean the _absolute value_ of the number (or expression) N. It's easy for numbers: | 7 | = 7 and | -7 | = 7. In general, | N | = N if N is positive, and | N | = -N is N is negative, and | 0 | = 0.

More formally, we define:

| N | =

and note that N can be a number (such as -7 or ) or an expression (such as x2 \- 3 or sin x).

Examples:

• | x+1| = i.e

• | 4-x2| = i.e

• |x| + |y| = x + y if x ≥ 0 and y ≥ 0 (i.e. in the first

quadrant of the x-y plane, including the positive x- and

y-axes), whereas |x| + |y| = x - y if x ≥ 0 but y < 0

(the fourth quadrant) and |x| + |y| = -x - y in the

third quadrant and, finally, |x| + |y| = -x \+ y in the

second quadrant.

The graph of |x| + |y| = 1 is shown at the right. |

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• To plot y = | f(x) |, just plot y = f(x) and reflect the negative parts in the x-axis (i.e. replace negative values of f(x) by - f(x))

Final Note: Now, with both absolute values and inequalities covered, we can consider expressions like

| x2 \- 4 | < 2, which is the same as - 2 < x2 \- 4 < 2 or 2 < x2 < 6 hence x lies in the interval < x < OR in - < x < \- . This manipulation is important when we come to consider expressions like

**| x - a | < h** (which is the same as a - h < x < a + h) or the expression **| f(x) - L | <** e (which is the same as L - e < f(x) < L + e).

• Reminder: | N | < 3 means -3 < N < 3 and

| x+5 | < 3 means -3 < x+5 < 3 (hence -8 < x < -2)

etc. etc. etc.

**SOME TRIG IDENTITIES** **and other trig stuff** * :

• sin2A + cos2A = 1 for any number (or angle) A.

• sin (A+B) = sin A cos B + cos A sin B and sin (A-B) = sin A cos B - cos A sin B

• cos (A+B) = cos A cos B - sin A sin B and cos (A-B) = cos A cos B + sin A sin B

• sin A - sin B = 2 cos sin

• sin2 A = and cos2 A =

Unless otherwise specified, if we write sin A , cos A, etc., and if you wish to

_consider A an angle, then assume it's the RADIAN measure of an angle_ _!_

• sin A and cos A are defined for real numbers "A" as follows (and you'll note that the definition has little to do with angles!)

  |

Given any positive number A, begin at the point (1,0) on the unit circle and move counter-clockwise a distance equal to A, reaching a point we can call P. The x-coordinate of P is cos A and the y-coordinate of P is sin A (and _that's_ the definition of the sine and cosine of a number A). If A is negative, then move clockwise.

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We can now introduce an angle "A". Note that the angle at the centre of the circle has the radian measure A ... since and since our radius is "1", the arclength and the central angle are equal ... provided the central angle is measured in RADIANS! We're all familiar with which is the above formula with the central angle a complete revolution: 2π, in RADIANS! We'd never write circumference = 360 r, where the central angle is 360 when measured in degrees, so we restrict ourselves to RADIANS when we consider angles (unless, of course, we indicate otherwise).

  | Anyway, because of the definition we can easily identify a number of points on the unit circle ... like (1,0) and (0,1) and (-1,0) and (0,-1) ... hence the value of sin A and cos A for various numbers A:

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PS:

S: Can't I write cos 45 = ?

P: Sure. Just be sure you indicate that the angle is in degrees, like cos 45˚ = , else somebody will measure off 45 units along the circle and take the x-coordinate of the resultant point as cos 45 ... and it won't be .

S: You gotta be kiddin'. Everybody knows cos 45 and sin 90 etc. are in degrees.

P: And if I write cos 47?

S: No, I mean the standard angles like 30, 45, 60, 90 and so on ... not 47.

P: Wait till we get into course. We'll approximate cos 47, knowing cos 45, and we'll all be confused if we don't agree on this.

S: Okay ... radians it is. But how about the good ol' sin A = . Has that gone?

P: No, that still works. We take some angle A and stare at the coordinates of the point P.

See the triangle? Notice that = sin A. (Of course, the hypotenuse is "1" and the "opposite" is just the y-coordinate which, by definition, is sin A).

S: Seems like cheating. Your hypotenuse is always "1". Is that necessary?

P: Let's enlarge the whole diagram ... blow it up ... like a photographic enlargement, by a factor R. Then we have the following diagram. See? The triangle has hypotenuse R and = = sin A, again, and of course, = = cos A. Happy?

S: No. |

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**SOME TRIG GRAPHS** **to remember:**

S: How about the special angles, 45˚ and 30˚ and so on.

P: There's nothing special about them, except that most people remember the 6 trig functions for 30˚, 45˚ and 60˚ from the triangles shown at the right ===>>>

Then sin 45˚ (or sin ) = and cos 60˚ (or cos ) = and cos 30˚ (or cos ) = and, of course, knowing these triangles you can read off the other trig functions as well, like tan 60˚ (or tan ) = and |

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so on. These will come in handy when we have to sketch the graph of curves involving angles.

**SOME GEOMETRY** **:**

All straight lines have an equation of the form: Ax + By = C.

The equation of a line through the points (x1,y1) and (x2,y2) is: = .

The equation of a line through (a,b) with slope m is: = m (the _point-slope_ form).

If m1 and m2 are the slopes of two lines, then they will be perpendicular if m1 m2 = -1.

**LOGARITHMS and EXPONENTIALS** :

The functions 2x, 5x, πx or ax (for any positive number "a") are called _exponential_ functions, and "a" is called the _base_. There are RULES:

(A)

If ay = x, then y is called the _logarithm_ of x to the base a, written: y = loga x, a _logarithmic_ function (or simply a _log_ function). There are RULES:

(B)

Every entry in (B) actually follows from the corresponding entry in (A). For example, assume the logs in (B) are to the base "a". Then alog x + log y = alog x alog y = xy = alog xy hence we get the first entry in (B), where we've used the fact that alog N = N if the log-base is "a".

Notice that x = ay is an _exponential_ function of y, and if we solve for y we get y = loga x, a _log_ function of x. In fact, x = ay and y = loga x are two different ways of writing the _same_ relation between x and y. (i.e. if x = ay then y = loga x AND if y = loga x then x = ay.)

Hence we may write or even .

Examples:

• log3 37 = 7 and log5 5 = 1 and logπ1 = 0

• 3log37 = 7 and 56 = 26log25 and π0 = 1

PS:

S: That's confusing ... isn't it? I was never very good at logs.

P: Everybody loves exponentials. Nobody loves logs. Everybody knows the RULES (A). Nobody knows (B). Well ... some don't, and the examples are even more unfriendly, right?

S: Right! So should I memorize (B)?

P: Yes ... and remember: a "log" is an "exponent" and all properties of logs follow from the rules for exponents ... the log-rules just look different, but they're really familiar things.

S: Sure, sure.

MORE ON LOGS:

If p = logaN, then ap = N. Now we take logs of each side, to the base b: logb(ap) = logbN. Using a magic property of logs (to any base) we get: p logb a = logbN, hence p = . Remembering who "p" is, we

have: . So, we can easily change logs from one base to another. If you have a table of logs to the base 10 and desperately need logs to the base 2, just use log2N = . The above relation has a cousin: let N = b and recall that logb b = 1. Then we get: which will be useful when we consider log functions later on.

PS:

S: Wait just one minute. Do you really expect me to remember all this .. all this ...

P: No. But it's here so you can look it up when you need it ... and someday you may need it.

S: So I only have to know that there are umpteen weird relations and I should know where to look to find them?

P: Yes.

S: You said log properties are familiar things, but I wouldn't call logab = 1/logba a familiar thing.

P: Pay attention: if x = logab then ax = b, and if y = logba then by = a and we want to show that x = 1/y, right? Okay, if ax = b then raise both sides to the power y and get (ax)y = by , but by = a, so axy = a and that means that xy = 1. See? It's just a property of exponents. Right?

S: zzzzzz

**ODDS 'n' ENDS** ... mostly ENDS:

• A geometric series has the form: a + a r + a r2 \+ a r3 \+ ... + a rn-1 (there are n terms here ... count 'em!).

The sum of the above series is a . If | r | < 1, then = since rn->0.

That is, the sum of the "infinite" geometric series is a + a r + a r2 \+ a r3 \+ ... = provided | r | < 1!!

• Polynomials are functions like 1 + x - 3x2 or x3 \- 5x4. They all have the form A + Bx + Cx2 \+ Dx3 \+ ... and so on, ending after a finite number of such terms. "A", "B", etc. are constants. Further, if p(x) and q(x) are polynomials, then is called a _rational_ function.

PS:

S: Whoops. Has that anything to do with rational numbers?

P: It's the same kind of definition. If p and q are integers, then is a rational number. If p and q are polynomials, then is a rational function. See? The polynomial functions play the role of the integers.

S: Sure, sure.

• SIGMA NOTATION:

We will have occasion to refer to a series, say a1 \+ a2 \+ a3 \+ ... \+ an, and it gets tiring to have to write out several terms (as we have just done) every time we want to identify the series. For this reason we may use SIGMA notation: each term of the series has the form ak where k = 1 or 2 or 3 ... or n, and the series is the sum of such terms, so we write where means "sum all such terms from k = 1 to k = n". For example:

1 + 2 + 3 + ... + 100 = and = e7 \+ e8 \+ e9 \+ e10 \+ e11 \+ e12 and

f(1+h) + f(1+2h) + f(1+3h) + ... + f(1+nh) = and = sin 4π + sin 5π = 0

and = .

PS: The Greek letter is called SIGMA ... it's upper case ... lower case sigma is s.

• In these lectures we will, from time to time, need to do some algebra and arithmetic with lots of digits of accuracy (just to illustrate some point we're making; we won't expect you to do these calculations). Then we'll use a computer program called **MAPLE** (except we'll call it   'cause it looks nicer).   is a so-called **C** omputer **A** lgebra **S** ystem (or CAS) which knows many of the techniques we'll learn in this course and, in particular,   can do arithmetic with great (infinite?) precision.

LECTURE 1

LIMITS

PS.

P: What's the limit of the sequence of numbers: 4.9, 4.99, 4.999, 4.9999, etc. etc. ?

S: It's 5.

P: Why 5?

S: Because the numbers get closer and closer to 5.

P: They also get closer and closer to 17 ... so why isn't the limit 17?

S: Well ... they get closer to 5 than to 17.

P: But one of the numbers is exactly 4.99, so maybe that's the limit. After all, how much closer can you get?

S: Oh, they have to get closer and closer without actually reaching the limit ... so that's why it's 5 and not 4.99 ... I think.

P: How about the following sequence: 4.9, 5, 4.99, 5, 4.999, 5, 4.9999, 5, etc. etc. where every second number is exactly 5. Now what's the limit? It's still 5, right? Yet the numbers actually reach 5 from time to time. Clearly we need a definition of what we mean by "the limit is 5" so that only one number satisfies the definition and that one number is 5 ... and not 4.99 or 17.

S: If you say so ... but what good is all this?

P: We want to be able to describe, in a reasonably precise manner, the various features of a graph such as:

S: You're kidding. Surely these functions never occur in a real problem, right?

P: Well, we want to establish a kind of vocabulary so we can talk about the features ... with a terminology that has some precise meaning. Of course, if we can analyze the above weird function then something like ==>>

is really easy.

S: What does this function represent?

P: It could be the concentration of a drug in the bloodstream as a function of time, or the amount of energy radiated by a hot body as a function of the wavelength of the emitted light, or some probability distribution or maybe ... |

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S: Okay, okay, but what would one want to know about such a function?

P: Where the maximum occurs or what the limiting value of y is when x becomes infinite or how y behaves for small values of x or ..

S: Okay, let's go.

Consider the function f(x) = . Notice that f(x) has no value at x =1.5 since f(1.5) gives which is _not_ a number. Nevertheless, f(x) does have a _limiting_ value of 6. (We make this claim without even defining what we mean by _limiting value_! We'll provide this definition below in such a way that the mathematical definition agrees with our common sense notion of _limiting value_. First, then, we need to think about this common sense notion.) To see that f(x) does, indeed, have a limit of 6, we compute f(x) for various values of x approaching 1.5, and we also include the _error_ , namely |f(x) \- 6|, the absolute value of f(x) - 6 (to see how well we're doing in achieving the _limit_ of 6).

Table 1 Table 2

x f(x) error

1.3 5.6 0.4

1.4 5.8 0.2

1.49 5.98 0.02

1.499 5.998 0.002

1.4999 5.9998 0.0002 | x f(x) error

1.7 6.4 0.4

1.6 6.2 0.2

1.51 6.02 0.02

1.501 6.002 0.002

1.5001 6.0002 0.0002

---|---

In spite of the fact that f(1.5) doesn't exist, it seems clear from the table that the limit (as x approaches 3/2) is 6. (We write this as: f(x) = 6.) Indeed, by making x sufficiently close to 1.5, we can make the _error_ , namely |f(x) - 6|, as small as we please. For example, suppose we want the error to be less than, say, .001 (meaning we want the values of f(x) to lie in 5.999 < f(x) < 6.001). We can achieve this simply by making x sufficiently close to the number 1.5 (for example, we can restrict x to lie in the interval 1.4999 < x < 1.5001).

For this particular function and this particular x-value, we can say:

Since the error, |f(x) - 6|, can be made as small as we please simply by restricting x to lie in some sufficiently small interval about the number 1.5, then the _limit_ of f(x) is 6, as x approaches 1.5 and this is indicated by the notation: f(x) = 6.

S: That's confusing. I mean, it seems easy enough to do ... this business of making something small.

P: Not just small but as small as we please, and remember that we should be able to do this without prescribing the value of f(x) or even the value of x but only by controlling how far x is from the number 1.5, and it's precisely this property we want for our "limit" definition, and ...

S: So why does that make 6 the limit and not, say, 7?

We can't make the error, |f(x) - **7** |, as small as we please by restricting x to lie in some small interval about 1.5, because when x is close to 1.5 the values of f(x) are close to 6, NOT close to **7** ... so **7** is NOT the limit. In fact, 6 is the ONLY number that satisfies this definition!

S: I think you have a problem there. I can make x really close to 1.5 by making it EQUAL to 1.5, then is your "error" small? I mean, how small is |f(1.5) - 6|? I make it | - 6 | which isn't even a number, let alone a small number .

P: Very good! You've put your finger on a problem with our definition ... so lets modify it.

Alas, no matter what interval we choose about 1.5, x = 1.5 will be in that interval ... and there's no way we can make the error small for x = 1.5 since f(x) doesn't even have a value! To fix this, we modify our definition to read:

We say that the _limit_ of f(x) is 6, as x approaches if the error, |f(x) - 6|, can be made as small as we please simply by restricting x to lie in some sufficiently small interval about the number 1.5, with the exception of x =1.5 itself!

S: Don't you find that awkward? I mean, "by restricting x to lie in some sufficiently small interval about the number 1.5, with the exception of x =1.5 itself" sound like mumbo-jumbo.

P: Okay, let's modify it ... again. We have to find something to replace that phrase which, I admit, is rather awkward, but I'm not sure you'll like the modification.

We can improve upon the wording (at the expense, perhaps, of making it less understandable!) by replacing the phrase "by restricting x to lie in some sufficiently small interval about the number 1.5, with the exception of x =1.5 itself" with the phrase "by choosing a sufficiently small number, h, and restricting x to lie in the interval, 0 < |x - 1.5| < h". Note that |x - 1.5| < h means that 1.5 - h < x < 1.5 + h, so we indeed have restricted x to some interval about 1.5 ... and we can make the interval small by choosing a small h ... and 0 < |x - 1.5| means that

x ≠ 1.5 ... so we've now got a reasonable definition of _limit_.

We'll generalize this notion of _limit_ to other functions, and other x-values:

We say that f(x) = L if we can make the error, |f(x) - L|, as small as we please, simply by choosing a sufficiently small number, h, and restricting x to lie in the interval, 0 < |x - a| < h.

S: You're right. I don't like the mods you've made. Can we go back to the original?

P: No. We'll leave it as it is. Don't you see how pretty it is? You just write 0 < |x - 1.5 | < .0001 and you've said that x is not equal to 1.5, but it's very close. Now pay attention.

Now let's return to the function f(x) which we can write as = = 2x+3 provided 2x-3≠0. But if we're considering the limit as x->3/2, then x is close to _but different from_ 3/2, so the division of numerator and denominator by 2x-3 is valid ... and it's clear why f(x)->6 as x->3/2 (since f(x) is identical to 2x+3 for every x-value _except_ x = 3/2 ... so the limit of f(x) is the same as the limit of 2x+3, namely 6).

Often, we can avoid using the above definition to find a limit (assuming a limit exists). In fact, from the definition, one can prove some nice rules (but we'll omit the proofs).

LIMIT RULES

Suppose f(x) = L and g(x) = M. Then we have the following rules:

SUM RULE: = L + M DIFFERENCE RULE: = L - M

PRODUCT RULE: f(x) g(x) = L M QUOTIENT RULE: =

**Example:** If f(x) = , calculate f(x).

**Solution:** As x->2 we have lim f(x) = lim = = = = - 4.

Note that the _value_ of this function, namely f(2) (which we get by direct substitution), is also -4.

It's important to notice, however, that we did NOT evaluate the _limit_ by substituting x = 2. Instead we used the various LIMIT RULES. If the limit turns out to be f(2) it's because we have a CONTINUOUS function (see below for a discussion of "continuous" functions).

PS:

S: Hold on! It looks to me like you just plugged in x = 2 to get the limit, right? And that gives you the value, right? And that makes the limit equal to the value, right?

P: Wrong. I used the various LIMIT RULES. It's just happens that the result is -4, the same as the value. But that's only true for certain functions ... called "continuous" functions. It's not always true that the limit and the value are equal.

S: Example?

P: Okay, let's see ... I can invent a function which has no value but does have a limit. Want to see it?

S: Sure.

P: It's f(x) = which has no value at x = but has a limit of 6. Like it?

S: Yeah, it's great. But how about a function which actually does have a value. Then it's the same as the limit, right?

P: Wrong. I'll invent a function defined for every real number x, and it will then have a value at x = 1.5, but this function will also have a limit as x->1.5 and it'll be different. Want to see it?

S: Sure.

P: It's f(x) = for all x ≠ 1.5 and f(1.5) = 47, or I could write it as: f(x) = . Like it?

S: Hey! I don't mean a double-barrelled function! I mean ...

P: Okay, here's another one: f(x) = x2 \+ + + + + ... Like it?

S: No. I assume it goes on forever. I don't know anything about functions that go on forever. I mean ...

P: Not true. You know how to find the sum of such an infinite series as this. Look at it. It's a geometric series and the common ratio is ...

S: Wait, I'll do it. The common ratio is ... uh, I divide the second term by the first and I get ... uh, . Right?

P: Sure, but you should also divide the third by the second and the fourth by the ...

S: I know that, but you said it was geometric. Anyway, I can only add an infinite geometric series if the common ratio is less than 1 and ... uh, well, I guess is less than 1. Terrific. Then it adds up to and that's and that's ... uh, 1 + x2. So what?

P: So find f(x).

S: I guess it's = 1. So that's the same as f(0), right? I mean, f(x) = 1 + x2 so f(0) = 1.

P: Wrong! When x ≠ 0, then your common ratio is less than 1 so you can add the infinite series using , BUT when x = 0 your common ratio is exactly 1 so you CAN'T use this formula.

S: Then how do I get f(0)? Wait! I just plug it in! f(0) = 0 + 0 \+ 0 + ... which I guess is 0, right?

P: Right! So for this function the limit is "1" but the value is "0". Nice, eh?

S: No. Anyway, I read, somewhere, the following definition of "limit":

f(x) means:

for any e > 0, a d can be found such that | f(x) - L | < e whenever 0 < | x - a | < d

P: Yes, it's the same definition as ours. When we say we can "make the error as small as we please" we really mean "smaller than any number e > 0", and when we choose an "h" and "restrict x to lie in the 0 < |x - a | < h", it's the same as finding a "d" and restricting x to lie in the interval 0 < | x - a | < d. See? It's the same. Don't let the Greek letters fool you.

S: Sure, sure. So why even use this definition?

P: Suppose f(x) = 47. Then, by restricting x to a sufficiently small interval about "3", say 0 < | x - 3 | < d, we can make the error | f(x) - 47 | < .1, say. Hence we can force f(x) to lie in the interval 46.9 < f(x) < 47.1, and if we want the error to be even smaller, say | f(x) - 47 | < .001, then we can select a smaller value for d, and ...

S: Wait. You've already said all this. I asked why the e - d definition is any better than our earlier definition.

P: I guess it's because we've given a name to the error, namely e which could be .1 or .001, etc.

S: I still don't see what good all this is. Is there any useful application of this stuff ... something I'd understand?

P: Well ... suppose you were building a box with all sides of equal length and it was to have a volume of 8 m3 with an error of, say, 0.1 m3 (which is e by the way). Then each side would be about 2 metres (since 23 = 8). Now, the big question: how accurately must you cut the sides so that the error in the volume is less than 0.1 m3? We write V(x) = x3 and insist that the error in volume, | V(x) - 8 |, is less than 0.1 by restricting x to lie in some interval about x = 2, say 2 - h < x < 2 + h. The answer to the question above is the value of "h" ... so how small must h be?

S: Let me do it! I'd want V(x) to lie between 8 - 0.1 = 7.9 and 8 + 0.1 = 8.1 ('cause that'd make the error less than 0.1 m3). That means I'd want 7.9 < x3 < 8.1, so I'd take the cube root and find that 1.9992 < x < 2.00083, so I'd have to cut the sides with an error less than ... uh, about .0008 m, right?

P: Right. You can guarantee an volume error less than 0.1 m3 by making the side length lie in 2 - h < x < 2 + h, with h = .0008 (i.e. x must lie in 1.9992 < x < 2.0008). Of course, if somebody wanted the volume error even smaller, less than e = 0.001 m3 for example, then you'd have to choose an even smaller value of h.

S: That doesn't sound like calculus to me ... just common sense and a little arithmetic.

P: Here's the calculus: the reason you can guarantee that an "h" exists for every specified error (which we can call "e") is because the function V(x) = x3 is a continuous function of x, hence has a value and a limit at x = 2 and they're the same. If this weren't the case then you couldn't guarantee an arbitrarily small error.

S: For example? I mean, are there really problems like that ? I mean real-world problems, not mathematical problems.

P: Sure. Suppose the cost of postage is determined by the weight of the parcel according to the prescription: cost = $10.00 for parcels under 1 kg, $15.00 for parcels from 1 kg to less than 2 kg, $20.00 for parcels from 2 kg to less than 3 kg, etc. etc.

If x is the weight and C(x) the cost, the graph looks like this ==>

Note that C(x) =15 , and because this limit exists we can guarantee the cost to be near $15 (in fact, in this problem, exactly $15) by restricting the weight to lie in some sufficiently small interval about 1.5 kg. However, |

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C(x) doesn't exist so it's NOT possible to guarantee a cost near $15 or near $20 (or near any amount L) by restricting the weight to lie in some interval about 2 kg. In fact, a wee bit under 2 kg and the cost is $15 and a wee bit over 2 kg and the cost jumps to ...

S: Okay ... I got it ... calculus is wonderful.

P: One more thing. Most students are wary of double-barrelled functions defined like:

f(x) = . In fact, there is a feeling that only mathematicians could love such functions. But the cost of postage is given by C(x) = and we could add to this array for parcels where x≥3 kg. Can't you just see such a sign hanging in the post-office?

S: No.

**ONE SIDED LIMITS** :

The graph of y = f(x) = is shown. Note that f(x) has no value at x = 1.5 (which we indicate with an "open circle"), however, as x approaches 1.5 either _from the left_ (i.e through values such as 1.49, 1.499, 1.4999, etc. as in Table 1 above) or _from the right_ (through values such as 1.51, 1.501, 1.5001, etc. as in Table 2), the values of f(x) clearly have the limit 6. |

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Now consider the graph of another function which we'll call g(x). This function _does_ have a value at x = 1.5, namely g(1.5) = 6 (indicated by the "closed circle"), however, if x approaches 1.5 _from the right_ , the values of g(x) have a limit of 7. We indicate this by writing g(x) = 7 where the notation x->1.5+ means x is approaching 1.5 through values more positive than 1.5. |

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Further, if x approaches 1.5 _from the left_ the limiting value of g(x) is 6, and we can write g(x) = 6 where x->1.5- means x is approaching 1.5 through values more negative than 1.5.

The question we pose is: does g(x) have a limit as x approaches 1.5?

The answer will be "yes" only if our definition of _limit_ is satisfied. If, for example, we suspect that

g(x) = 6, then we must be able to make the error, |g(x) - 6|, as small as we please by restricting x to lie in some interval about 1.5 such as: 0 < |x - 1.5| < h. This is clearly impossible since an x-value just slightly larger than 1.5 will give g(x) a value near 7 so the error is already larger than 1.0, hence g(x) = 6 is NOT true.

Similarly, g(x) = 7 isn't true either (since values of x close to 1.5 but slightly smaller will give an error larger than 1.0). Indeed, there is NO number L such that |g(x) - L| can be made as small as we please by restricting x to lie in an interval about 1.5, hence g(x) = L is NOT true and we conclude that g(x) doesn't exist (meaning _no_ number L will satisfy our definition of _limit_ ).

In this example, g(x) has a _right-sided limit_ , g(x) = 7, and a _left-sided_ limit, g(x) = 6, but it doesn't have a limit: g(x). On the other hand, for the function f(x) = we have:

f(x) = f(x) = f(x) = 6.

That is, it has _right-_ and _left-sided_ limits and they're identical and equal to the _limit_. We make note of this:

f(x) = L provided f(x) = L and f(x) = L .

Also,

if f(x) = L and f(x) = L then f(x) = L.

In words:

If f(x) = L exists, then both left- and right-sided limits will be equal to L.

Also, if both left- and right-sided limits exist and are equal to L, then L is also the limit of f(x).

Sometimes it's necessary to evaluate both _left-_ and the _right-_ limits in order to see if the limit exists.

**Example:** Does f(x) = have a limit as x approaches 3?

(Or, to put it differently, does exist?)

Note: When we see an absolute value sign we desperately want to get rid of it by using the fact that

|m| = m if m ≥ 0 whereas

|m| = - m if m < 0

(where m is any expression, such as x2-9).

**Solution:** To rid ourselves of the absolute value sign we need to know whether x2 \- 9 is positive or negative, so we first consider the limit as x approaches 3 _from the left_ (i.e. x->3-). Then, for x slightly smaller than 9 (such as 8.9 or 8.99 etc.), x2-9 < 0 so |x2 \- 9| = - (x2-9). (Recall the definition of absolute value!)

Hence: =

= = = - 6.

On the other hand, if we approach _from the right_

(so that x2 \- 9 > 0 and |x2 \- 9| = x2 \- 9) we have: |

---|---

= = = = 6.

Since these limits are different, we conclude that does NOT exist.

In fact, to the left of x = 3, the graph of y = is the same as the graph of y = - (x + 3) and, to the right of x = 3, the same as y = x + 3 as shown in the diagram. Clearly, left- and right-handed limits are NOT equal.

PS.

S: If g(x) has a limit as x approaches 3 from the left, and it also has a limit as x approaches 3 from the right, then it has a limit no matter how x approaches 3 (they ain't no other way to approach 3, is there?) So how come you say there's no limit as x approaches 3?

P: It's that neat definition that does it. In order to have a limit (according to our definition) there must be some number L such that the error |g(x)-L| can be made as small as we please simply by forcing x to lie in some small interval about 3, such as 3-h<x<3+h (except for x=3 itself, of course). But it's impossible to find such a number L and such an interval because if x is in the right part of the interval, 3 < x < 3+h, then g(x) can be made as close as we please to 6 ... so clearly L (whoever he is) must be VERY close to 6. On the other hand if x is in the left half of this interval, 3-h<x<3, the values of g(x) can be made as close as we please to -6 hence L must be VERY close to -6 as well. But there's no number L which is simultaneously VERY close to 6 and VERY close to -6 ... so there's no L, hence there's no limit. See?

S: Not really.

P: Don't worry about it ... it's not something we'll spend much time on. One of these days you'll wake up and run naked thru' the streets shouting eureka! ... and all will be clear.

S: (He's kidding ... right?)

LECTURE 2

INFINITE LIMITS, ASYMPTOTES and CONTINUOUS FUNCTIONS

**INFINITE LIMITS** :

We want to be able to describe the features of the above graph at x = -1 and x = 5. To do this we first consider the limit: . For x in some small interval about x = 0 (say -.001 < x < .001, excluding x = 0 itself) the values of are larger than = 1,000,000. Further, as we make the interval even smaller, the values of become even larger. In fact, we can make the values of this function larger than any number you care to mention \- and certainly larger than L for _any_ number L - hence the the limit cannot be L for _any_ L - hence there is no limit. i.e. does not exist.

Now consider sin . This limit won't exist either, but for quite a different reason.

For x in some small interval about x = 0 (say -.001 < x < .001, excluding x = 0 itself) the values of sin will be between -1 and +1 (that's because the sine function _always_ lies between -1 and +1). Furthermore, no matter how small we make the interval we can always find x-values in the interval such that is a multiple of π and that will make sin = 0 (since the sine of a multiple of π is 0) ... so, if sin did exist and is equal to some number L, then L must clearly be VERY close to 0. But, no matter how small the interval we can also find x-values such that is an odd multiple of and that will make sin = ±1 (since the sine of an odd multiple of is either +1 or -1). Conclusion? sin takes on values -1 _and_ 0 _and_ +1 in our interval (no matter what interval we choose!) so the limit L (if there is such a number) must be close to -1 _and_ close to 0 _and_ close to +1, all at the same time - clearly there is no such number - hence there is no limit. i.e.sin does not exist.

PS.

S: Wait a minute. You say: we can always find an x-value such that is a multiple of π ... what does that mean?

P: Pick an interval about x = 0.

S: How about -.01 < x < .01?

P: Good. Now watch me find an x-value in your interval which will make sin = 0. That means I have to find an x such that is a multiple of π ('cause that'll make sin = 0). Let's see, if I choose = 1000π (that's a multiple of π, right?) that'll mean x = and since π is roughly 3 then x is roughly 1/3000 or roughly .0003 so my x certainly lies in your interval. And believe me, if you choose any other interval about x = 0 I'll still be able to find x-values which will make sin = 0. Not only that, for any interval you choose I can also find an x-value such that sin = 1. Not only that ...

S: Wait, wait ... let's see you find an x in -.0001 < x < .0001 which makes sin = 1.

P: Okay, I'll need to be \+ a multiple of 2π ('cause that'll make sin = 1) and I'll choose a huge multiple of 2π (so my x lies in your interval) so let's try = + 100,000(2π) (which makes sin = 1) and we'll see if it's in your interval ... π is roughly 3 so is roughly 1.5 + 600,000 which is roughly 600000 so x is roughly which is roughly .0000017 so my x does indeed lie in your interval, and if you choose a smaller interval I'll just choose a bigger multiple of 2π and my x would still lie ...

S: Let's forget the whole thing.

P: Wait. Don't you see? You can't possibly make the values of sin as close as you please to some number L by restricting x to lie in some interval about x = 0. Why? Because I can find x-values which make sin equal to 1 and other x-values which make sin equal to 0 (to pick two convenient values of the sine function, though I could pick others ...), so L (whoever she is) must be VERY close to 1 and VERY close to 0 all at the same time ... and there is no such number. In fact, since sin takes on every value between -1 and +1 in every interval about x=0, the limit L must be close to every number between -1 and +1, all at the same time ... and that's impossible ... so there is no limit L. Understand?

S: zzzzz

Neither of the limits and sin exist (because there is no number, L, which satisfies our definition of a "limit"). Yet, the first fails to exist for a particular reason, namely: the values of become larger than any number when x is restricted to smaller and smaller intervals about x = 0 (excluding, of course, x = 0 itself). In order to indicate this particular reason for nonexistence of a limit we write: = ∞. Writing this doesn't mean that we now have a limiting value for . It simply means that the limit fails to exist because becomes larger than any number as x approaches 0, or, to put it differently, we say: approaches infinity as x approaches zero. In general:

f(x) = ∞

means that the values of f(x) can be made arbitrarily large by restricting x to lie in some sufficiently small interval about x = a, excluding x = a itself.

or, in sexier words,

f(x) = ∞

means that the values of f(x) can be made arbitrarily large by choosing a sufficiently small number h and restricting x to lie in the interval 0 < |x - a | < h.

It should be clear what we meaning by: = -∞

We also have (think about these!):

and and and

Note that we use ∞ to mean +∞ (the + is understood, just like 5 means +5).

**ASYMPTOTES** :

If any of the following are true, then we say that the graph of y = f(x) has a

VERTICAL ASYMPTOTE, namely: x = a.

f(x) = ∞ f(x) = -∞ f(x) = -∞ f(x) = ∞

**Examples:** Evaluate the following limits (or explain why they don't exist)

(a) (b)

(c) 21/x (d) 21/x

**Solutions:** (a) = -∞ whereas = ∞, and they're different, so doesn't exist. (If both left- and right-limits had been, say, -∞, the limit still wouldn't exist but we could at least use the phrase: "limit = -∞". This would be the case with = -∞)

(b) = -∞ since the numerator has a limit of 3 (a positive number) whereas the denominator approaches 0 through negative values (and is a large negative number).

(c) 21/x = ∞ since, as x approaches 0 through positive values, 21/x takes on values which are "2 raised to a large positive number "... hence the answer "∞".

(d) 21/x = 0 since, as x approaches 0 through negative values, 21/x takes on values which are "2 raised to a large negative number" (for example 2-100 = ) hence the answer 0.

We've talked about limits where x approaches a number and y = f(x) becomes infinite. Now let's talk about limits where x becomes infinite and y approaches a number.

We write = - and we mean that, as x becomes arbitrarily large (and positive), the values of y = get arbitrarily close to the number - . Graphically, it means that the graph of y = approaches the line y = - as x becomes infinite (as in the diagram). |

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Examples:

= 0 = - ∞

(meaning that y = approaches the line y=0) (meaning that y = heads south!)

If f(x) = a or f(x) = a

then we say that the graph of y = f(x) has a HORIZONTAL ASYMPTOTE, namely: y = a.

**CONTINUOUS FUNCTIONS** :

As x approaches some number, say 5, maybe f(x) has a _limit_ ... and maybe is doesn't. Also, maybe f(x) has a _value_ at x=5 ... and maybe it doesn't. Even if f(x) had a limit _and_ a value, maybe they're different numbers! The nicest functions are ones that have a limit and a value and they're the same ... and these functions are called CONTINUOUS functions.

If (1) f(x) = L and

(2) f(a) exists and

(3) f(x) = f(a)

then f(x) is said to be CONTINUOUS at x = a.

PS:

S: So why are they the "nicest" functions?

P: Because the graph of such a function has no breaks. At every point on the graph of a continuous function there is a left-limit, a right-limit and a value and they're all the same! The graph could have points or sharp corners, but it has no breaks. However, if a function is discontinuous at a point (say x = 3) then the graph could go in one direction as you approach 3 from the left and in another direction as you approach from the right ... and the actual value of the function (hence the point on the curve at x = 3) might be neither of these!

**Example:** Is f(x) = continuous at x = 1.5? at x=2?

**Solution:** Since f(x) doesn't have a value at x = 1.5, it's not continuous there. (It doesn't matter whether it has a limit!) But, at x = 2, f(2) = = 7 exists _and_ , as x->2, we have lim =

= = =7 and they're EQUAL, so f(x) _is_ continuous at x=2. (Notice that we didn't get the limit by substituting x = 2 since that gives the _value_ , not the _limit_ and they may or may not be the same!)

Example:

For the function graphed at the right, write limit statements for :

x->∞, x->-∞, x->0+, x->0-,

x->2-, and x->2+.

Where is the function discontinuous?

Why? |

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**Solution:** f(x) = -1, f(x) = -1, f(x) = ∞, f(x) = ∞,

f(x) = 0 and f(x) = 2. The function has NO limit at x = 0 and x = 2, hence is discontinuous there. (Note that f(x) has a value at x = 2, namely f(2) = 0, but since it has no limit is cannot be continuous there.)

NOW, having established all the "limit terminology" we'll need, stare at the following graph (of a fictitious function f(x)) & verify the comments beneath.

LECTURE 3

MORE on LIMITS

**TECHNIQUES FOR EVALUATING LIMITS WHEN THE "RULES" DON'T APPLY** **:**

In what follows, we consider certain limits to which we cannot apply the various limit rules (for one reason or another), yet, by using some tricky manipulation we can often put the expression into a form where we CAN apply the rules.

• **The form**

= = = - where we divided numerator and denominator by the largest power of x (i.e. x2) so that every term then approaches a constant. This is a useful technique when you have . Note that we can only use the LIMIT RULE when all limits exist. In our example, neither limit exists ( _until_ we divide numerator and denominator by x2).

• **The form** ∞ - ∞

= where we multiplied both numerator and denominator by + to "rationalize the numerator". We then get = 0 (since the denominator becomes infinite while the numerator does not). Note that the LIMIT RULE which says lim (f - g) = lim (f) - lim (g) doesn't apply since neither of the limits lim (f) & lim (g) exist.

• **The form**

= where we "rationalize the numerator"and =

• **Reduce the given limit to one you know**

Given that = 1, then we can write = 1 where _anything_ can be plugged into { }. For example: = 1 or, equivalently, = 1 (since x->-∞ will make 2x->0). Of course, in a _real_ problem, nature isn't so accommodating; you're more likely to see this problem as: 2-x sin 2x and it's up to you to change it to the form .

P: This technique of reducing a problem to one you've already solved is very useful, especially for a mathematician. Did I tell you the story of the mathematician and the engineer?

S: Don't tell me.

P: They were both given an empty kettle and asked to make tea. Both did exactly the same thing: fill the kettle with water, boil the water, put tea bags into the teapot and pour in the boiling water. Then, a new problem. They were given a kettle of boiling water and asked to make tea. The mathematician first emptied the kettle thereby reducing the problem to one he'd already solved.

S: I don't like tea.

Examples:

Evaluate each of the following limits (or explain why the limit doesn't exist):

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

**Remember** :

(i) | N | has the value N when N ≥ 0, and the value -N when N < 0. This will enable you to eliminate the absolute value sign.

(ii) The expression x->∞ means x->+∞ (i.e. the "+" is understood).

(iii) The notation " f(x) = ∞ " means the limit doesn't exist for a particular reason: _the values of f(x) become arbitrarily large (and positive) for x near "a"._

Use this notation when appropriate, don't just say "limit doesn't exist".

Solutions:

(a) =

= = -(5+x) -> -10.

(b) = = = = 10.

(c) doesn't exist since left- and right-limits differ (as shown above).

(d) = = = - 4.

(e) = = = -1 (dividing num. and denom. by x3)

(f) = = -∞ (since cos2x->1 and sin x ->0-).

(g) = = ->

where, in the last step, we've divided the numerator by x and the denominator by .

(h) === 1.

(i) = -> as h->0.

S: I'll tell you something; I can't always tell when something has a limit and when it doesn't. I mean, how should I know when to use some trick if I just looked at the expression and I think it already has a limit ... without doing any work?

P: I don't know what you're talking about. Give me an example.

S: Well, like maybe ... uh, the limit of - when x goes to ∞. It's 0 and that's obvious so why would I do some magic like rationalizing something?

P: That's obvious? Remarkable! It's not obvious to me! I see two huge numbers and when I subtract ...

S: You see ∞ - ∞ and that's ... uh, wait ... it isn't 0, is it?

P: No. In fact, 0 happens to be the correct limit, but that's pure luck: - has a limit of ∞ whereas - x has a limit of -∞. But let me test you on your ability to recognize expressions where you have to do some work to get the limit ... because it's NOT obvious. In each of the following, x->∞. What's the limit?

(a) (b) (c) (d) 10-x2

S: In (a), it looks like so I have to do some work. In (b) it's so I have to do some work. In (c) ...

P: No! In (a) you have a huge number divided by a huge number and you can't tell (without doing some work) whether that's small or large. But, in (b), you have a huge number divided by a number which is almost 1. Surely you know what that is. Just think of 1,000,000 divided by 1. It's huge!

S: Okay, so the limit is ∞. In (c) I get ... uh, sin in the numerator and that sin 0 and that's 0 so the answer is zero and I didn't have to do any work. In (d) ...

P: Well ... I wish you wouldn't say = 0, it's the limit that's 0, but in (c) you have to look at the denominator too. If the denominator approached 0 you'd be in trouble because the form isn't one where you can write down the answer. In fact ...

S: But in (c) the denominator isn't approaching 0 ... it's approaching ∞ and that makes the fraction even smaller so it gets to 0 even faster. In (d) I get 10-∞ and I have to do some work. How'm I doin' boss?

P: Terrible! If you don't know anything about 10 raised to a large negative number, try an example! What about 10-1000. Is it large? Small?

S: It's ... uh, and I'd say that's pretty small. So I guess 10-x2 = 0, right?

P: Remember this:and where "large number" could be large and positive or large and negative, like - 1,000,000.

Examples:

Assuming = 1, evaluate each of the following:

(a) t sin (b) (c) cot x

Remember:

Given = 1, we may conclude that = 1, where m is an expression which approaches zero.

Solutions:

(a) t sin = = = 1 where q = -> 0 as t->∞.

(b) = == -1 where t = z-π .

(c) cot x =cos x = (-1)(1) using result from (b).

Example:

Sketch the graph of a function which satisfies _all_ of the following:

f(x) = -∞, f(x) = 0, f(x) = 1, f(x) = 1

Solution:

The graph of f(x) could look like _any_ of the following:

Examples:

A function which has no _value_ at x = a, but has a _limit_ as x->a, may be made continuous (at x = a) by _defining_ the value to be the limit. For example, f(x) = has no value at x = 5, yet

f(x) = 10, so the "redefined" function f(x) =

is continuous .

For each of the following, f(a) is undefined. Determine whether it is possible to define f(a) so as to make the function continuous there:

(a) f(x) = , a = 0 (b) f(x) = , a = 3

(c) f(x) = e-1/|x| , a = 0 (d) f(x) = , a = 1

Solutions:

(a) Define f(0) = = 1.

(b) Define f(3) = = = .

(c) Define f(0) = e-1/|x| = 0

(d) = = -1 and = 1 differ, so f(x) doesn't exist, so f(x) cannot be made continuous by redefining f(1).

**MORE ON LIMITS** :

PS:

S: Since sin x->0 as x->0, then f(x) sin x->0 as x->0, for any function f(x). Right?

P: Wrong. If f(x)->∞ as x->0, then the product f(x) sin x could approach anything, depending upon the function f(x). For example, if f(x) = , then the product approaches 1 and if f(x) = , then the product approaches .

S: What if f(x) doesn't approach ∞? Then surely f(x) sin x->0 as x->0. Right?

P: Sure, if f(x) doesn't get too large.

S: Let's see you prove that.

P: Okay. Suppose |f(x)| ≤ 1000 for all x. Then |f(x) sin x| ≤ 1000 | sin x |. Further, |f(x) sin x| is an absolute value so it's never negative, hence 0 ≤ |f(x) sin x| as well. We then have:

0 ≤ | f(x) sin x | ≤ 1000 | sin x |

Now we let x->0. Since |f(x) sin x| is stuck between two functions, both of which have a limit of 0 (namely g(x) = 0 and h(x) = 1000 | sin x | ), then |f(x) sin x| also has a limit of 0.

S: Is that some kind of theorem?

P: Yes. Here it is:

the SQUEEZE THEOREM

If g(x) ≤ f(x) ≤ h(x) and g(x) = h(x) = L, then f(x) = L as well.

In words: If f(x) lies between two other functions, and they have the same limit, then f(x) also has this limit.

This theorem is often used to show that some limit of the form f(x) g(x) is 0, when one factor approaches 0 (say, lim (f) = 0) but the other factor doesn't have a limit ... but _is_ bounded. In that case you can't use the LIMIT RULE which says: lim (f g) = lim (f) lim (g) because this RULE requires that both limits, lim (f) and lim (g), exist. (It would be nice if you _could_ use the RULE; you'd say lim (f g) = lim (f) lim (g) = 0 lim (g) = 0, but if there _is_ no number "lim(g)", then what's the meaning of "0 lim (g) = 0"? It's like multiplying zero by a yellow rose. Is it zero?)

**Example:** Evaluate x sin . (Note that sin doesn't _have_ a limit ... and don't confuse this problem with x sin which can be written = 1 which is the same as with m = , hence its limit is 1.)

**Solution:** Since | sin | ≤ 1, then 0 ≤ |x sin | ≤ | x | and since both sides of this inequality have the same limit, namely 0, then |x sin |->0 as well. In this example, it's tempting to write: -1 ≤ sin ≤ 1 so that

-x ≤ x sin ≤ x and then use the SQUEEZE theorem. Unfortunately, x sin ≤ x isn't even true for negative x. Try x = -π for example. We'd get -π sin (-π) ≤ -π which says 0 ≤ -π (which certainly isn't true). Moral? If you'd like to prove that some limit is zero, you can prove that its absolute value has a limit of zero.

In fact, while we're at it, let's say something about y = f(x) sin . Since sin oscillates between -1 and +1, then y = f(x) sin oscillates between y = - f(x) and y = f(x) and, in fact, touches the curve y = f(x) whenever sin has the value +1 and it touches y = - f(x) whenever sin = -1.

  | The same is true of the graph of y = f(x) sin x (or, for that matter, y = f(x) cos x). The graph oscillates between y = f(x) and y = - f(x). It's even more interesting if f(x) changes slowly and sin x changes rapidly. We could accomplish this by considering

f(t) sin wt where w is a very large number (hence

sin wt oscillates rapidly). Then f(t) sin wt oscillates so rapidly between y = f(t) and y = - f(t) that it almost fills the space between. Indeed, if f(t) is the voltage produced by a microphone while recording Beethoven's Fifth

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  |

Symphony (and that's why I changed the name of the variable to t = time!), and if we modify the amplitude of a second sinusoidal voltage, sin wt, so it becomes

f(t) sin wt, we've got _amplitude modulation_ (AM for short) and we can send it up a big antenna and you can receive it on your car antenna and your AM radio can delete the sin wt and recover the f(t) (i.e. _demodulate_ the radio signal) and send f(t) to your speaker and you can relax with Beethoven. (In fact, each radio station has its own distinct w which you can "tune in".)

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PS:

S: If I have to evaluate, say, , can't I just say it's = ∞?

P: If we agree that the expression is shorthand for the limit of a ratio where the numerator has a limit of 3 and the denominator has a limit of 0, then it's okay ... sort of. However, most profs get very nervous when students use this notation. They're likely to cross-multiply and get 3 = 0 x ∞ which is meaningless. Besides (and this is important), the denominator could approach 0 through negative values, such as would be the case for where the limit is -∞ (and you'd be tempted to still write = ∞, and get the wrong answer). Also, for the case , you'd still get but now the left-limit is = -∞ whereas the right-limit is = ∞. So you have to be very careful.

S: Can I write = = -∞ and = = ∞ where I show the "sign of zero" ... or at least how I approach 0 ... so I get the right sign?

P: Sure, as long as you indicate the convention you're using so everybody who reads what you write will understand what you're saying. For example, it's often convenient to write

= = ∞ (since the two negatives give a positive result).

Using this shorthand notation indicates that the graph of y = has a vertical asymptote and that the curve goes due north as x approaches 1 from the left. |

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S: So now that we have all this stuff about limits, what good is it?

P: The good part is yet to come. Calculus comes in two flavours: differential calculus and integral calculus. Differential calculus is about rates of change and the slope of tangent lines to curves ... the DERIVATIVE. Integral calculus is about areas under curves and breaking down a complicated problem into simpler ones and summing ... the DEFINITE INTEGRAL. Both the derivative and the integral are defined in terms of a limit, hence we began our study with LIMITS (... unfortunately, it's perhaps the most difficult part of our study!). Next we'll turn to the differential calculus, and the DERIVATIVE ... but first a few problems.

Problems:

1. Evaluate each of the following (or explain why the limit doesn't exist):

(a) (b)

(c) (d)

2. Determine the points of discontinuity (if any) for the function f(x) =

Solutions:

1. (a) For x > 4, = = x+4->8 as x->4+

For x < 4, = = - (x+4)->-8 as x->4-

Hence doesn't exist (... left and right limits don't agree).

(b) = = = -4

where |2x2-1| = -(2x2-1) since 2x2-1 < 0 when x is near 0 (and, of course, 2x2+1 > 0).

(c) = = = -2.

(d) = = ->- **3** as x->∞.

2. Since f(x) is defined differently for x > 0 and x < 0, we consider both left- and right-limits:

f(x) = = 0 and f(x) = = 0

Since they're the same, this function _does_ have a limit: f(x) = 0. However, it's not equal to f(0) = 1, so the function is NOT continuous at x = 0. For every other value of x, say x = a > 0, we would get

f(x) = x2 = = (a) (a) = a2, using a LIMIT RULE. Since this is equal to f(a) = a2 (for a > 0) , the function is continuous for every x > 0. Similarly we could prove it continuous for every x = a < 0. Hence x = 0 is the only discontinuity.

**Comments:** If we've just found a limit, say = 1, then we've actually shown that sin x and x have very nearly the same value when x is small (since their ratio is nearly "1"). For example, choosing x = .0123 we find that sin (.0123) = .01229969 (and if we use a calculator for this evaluation, we make sure it's in _radian_ mode, else sin x and x are NOT close in value for small "x"). In fact, = 1 says that the two curves, y = sin x and y = x, are very nearly coincident for x small. It may not be such a surprise, then, to learn that y = x is the _tangent line_ to the curve y = sin x, at x = 0 ... which brings us to tangent lines.

LECTURE 4

the DERIVATIVE and its RULES

**the DERIVATIVE** :

Our problem is to determine the slope of the tangent line to the curve y = f(x), at x = a. To do this we consider two points on the curve; one at x = a (and y = f(a)) and a second point at x = a + h (and y = f(a+h)). The slope of the line joining these two points is . On the other hand, we could take the second point as x (and y = f(x)) in which case the slope of the line joining the two points would be . Regardless of what we call the second point, the slope of the tangent line at x = a will be obtained by taking the limit of this slope as h->0, or as x->a. |

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f'(a) = =

This limit is called the derivative of f(x) with respect to x, at the place x = a.

**Example:** Determine the derivative of f(x) = at x = 2.

**Solution:** f'(2) = = =

= = .

We could also use the second form for the derivative:

f'(2) = = = = = .

Of course, we could also use or since it matters little what names we give to the variables. If you understand the diagram, and the fact that we take one point as x = a and some second point (giving it _any_ name), and let the second point approach the first, then the limiting value of the slope of the line joining the two points is the derivative of f(x) at x = a (that is, f'(a)).

If the limit which defines f'(a) exists, we say f(x) is differentiable at x = a.

The above limit definition (take your pick which one) gives the derivative of f(x) at some point "a". If we want the derivative at a variable point "x", we just replace "a" with "x" in the definition:

S: That's confusing, I mean ...

P: Pay attention. Suppose we want to compute the derivative of f(x) at, say, x = 6. Okay, (6, f(6)) is the point in question. Now pick another point.

S: Okay, I pick x = 7.

P: Sorry, I mean pick a name for another point. We have to let this second point approach the first and we can't do that if it's stuck at 7. See? Okay, pick a name.

S: Sam.

P: Hmmm. The point with x = Sam has y = f(Sam). Now what's the slope between these two points on y = f(x)?

S: It's ... uh, , right?

P: Good! Now let Sam->6 and you've got ... what?

S: Huh? I don't understand the question.

P: You've got and that's f'(6), the derivative of f(x) at x = 6. Nice, eh? And now you can see that it makes little difference what we call the second point so long as it's approaching the first. I'll give you yet another notation, maybe more appealing.

The change (or _increment_ ) in x, which we call "h" above, is sometimes called Dx ( a convenient notation which indicates that it's a small increment if the variable "x"). The corresponding change in y, namely

f(x+Dx) - f(x), is often denoted by Dy. This makes the definition of the derivative take the form:

which is familiar to many and is the reason for the notation for the derivative of y with respect to x (i.e. becomes , in the limit)* . If y = f(x) we use or y' or f'(x) or sometimes just f' to represent the derivative.

Using the limit-definition of the derivative we can show that xn = n xn-1 where n is any integer. We illustrate with n = 5.

If f(x) = x5, then = =

= t4 \+ t3x + t2x2 \+ tx3 \+ x4 -> x4 \+ x4 \+ x4 \+ x4 \+ x4 = 5 x4 as t->x

In fact,works for any exponent, whether integer or not.

For example, = x1/2 = (1/2) x-1/2 = .

the DIFFERENTIAL

We've already mentioned that, after having evaluated some limit such as = 1, we can interpret this to mean that sin x and x are very close in value when x is small. The same holds for the limits which yield the derivative. Let's see how that works:

We have y = x3 and have just computed = 3x2, so we know that is very nearly 3x2 when Dx is small. What does that mean? It means that when x changes from x to x+Dx, the corresponding change in y = x3, namely Dy = (x+Dx)3 \- x3, is such that ≈ 3x2 and that means we'd expect Dy ≈ 3x2 Dx .

Is that useful? Yes. If x is the length of the side of a cubical box, then y = x3 is its volume. Now suppose the box is to be built with x = 2 _metres_ (so the volume is 23 = 8 _metres_ 3) and there are small errors in building the box and the sides are made to within .01 _metres_ . What about the volume? We put x = 2 and Dx = .01 (because this is the possible error, or change, in x) and get an estimate of the change in volume, namely Dy ≈ 3x2 Dx = 3 (2)2 (.01) = .12 _metres_ 3.

S: I don't need calculus for that! I'd just say (2.01)3 \- 23 and that's the change in volume, and that's .121 m3 and that'd be exact ... no estimate, no approximation.

P: Of course I was giving you a simple problem, just to illustrate the idea of using the derivative to estimate small changes. I could give you a tougher problem. Here it is:

**Example:** When x = 49, y = has the value 7. Estimate the change in y when x changes to 47.

**Solution:** For y = x1/2, then = x-1/2 = so we can estimate changes in y using ≈ or

Dy ≈ Dx and if we put x = 49 and Dx = -2 (because x changes from 49 to 47, a change of -2), we get

Dy ≈ = - . Conclusion? is approximately: 7 - ≈ 6.857

In general we have the following prescription for estimating changes in the value of a function when the variable changes by some small amount:

We have y = f(x) and have just computed = f'(x), so we know that is very nearly f'(x) when Dx is small. What does that mean? That means that when x changes from x to x+Dx, the corresponding change in y = f(x), namely Dy = f(x+Dx) - f(x), is such that ≈ f'(x) and that means we'd expect Dy ≈ f'(x) Dx .

Dy ≈ f'(x) Dx is called the DIFFERENTIAL of y = f(x),

and it depends upon x and upon Dx, the DIFFERENTIAL of x,

and it gives an estimate of the change in y when x changes by Dx.

**Example:** The distance travelled after a time t _hours_ is x(t) = 100 + t3 _metres_. Estimate the distance travelled from t = 2 _hours_ to t = 2.1 _hours_.

**Solution:** Since ≈ = 3t2, we compute the "differential of x", namely Dx = Dt = 3t2 Dt. Substituting t = 2 and Dt = .1 we get Dx = 3 (2)2(.1) = 1.2 _metres_ as our estimate of the distance travelled.

P: Do you see anything interesting here?

S: Nope.

P: If I told you that the velocity was = 3t2 = 3(2)2 = 12 m/hr at the end of 2 hours, then I asked how far would you would travel in the next tenth of an hour, what would you say?

S: Uh ... (12)(.1) = 1.2 metres.

P: Right, and that's just what the differential is doing for us. It gives quite a reasonable approximation if you don't change things too much.

S: But .12 metres is exact, isn't it?

P: No, because the velocity is changing, even during the time from t = 2 to t = 2.1 hours. In fact, the velocity is increasing so the distance would actually be somewhat larger. To be exact, the distance would be (100 + (2.1)3) - (100 + 23) = 1.261 metres.

more on DERIVATIVES

S: I notice that you keep saying "the derivative with respect to x". Why the "with respect to"?

P: If y depends upon x and x changes by Dx and the corresponding change in y is Dy, then the limit of is the derivative of y with respect to x. Now if y depended upon somebody called u, and u changed by Du, then

would be ...

S: Don't tell me ... the derivative of y with respect to u.

P: Right! And one other thing. If y is measured in hectares and x is measured in degrees Celsius, then = is measured in hectares per degree Celsius. See? And if u were measured in volts, then = would be measured in hectares per volt. See? The "with respect to" is important ... although we'll often omit this phrase if it's obvious who we're differentiating with respect to. See?

S: I guess so.

P: And one more thing. Since the derivative is a "rate of change", it tells how rapidly y changes when x changes. That is, if = 10, say, it means that y is increasing 10 times more rapidly than x is. If = - 1/3 then it means that y is decreasing as x increases (because is negative) and it's decreasing 1/3 as rapidly as x increases ... and ...

S: Yeah, I got it.

P: And one more thing. Don't get tied to any particular set of labels. The independent variable could be called y or z or V or even x, and we could use the notations or or or . See?

S: Can we keep going?

P: Not yet -- we should recognize when things increase and decrease and how the derivative gives us this info and ...

S: A picture is worth a thousand words, remember?

P: Okay, here's the graph of some invented function f(x). For a while it increases, then it decreases, then it increases again and so on. These increases and decreases are reflected in the value of the derivative, f'(x), so if we plot f'(x) versus x we'll find that when f(x) is increasing, f'(x) will be positive and when f(x) is decreasing f'(x) will be negative. See how it goes?

S: Yeah ... and it looks like ... uh ... what's happening at x = .8 and x = 1.5 and x = 2.2 or thereabouts?

P: When f(x) stops going up and starts coming down, it has a horizontal tangent so its derivative f'(x) is zero. That happens at about x = .8 and again at about x = 2.2 and at x = 1.5 the derivative f'(x) is again zero because f(x) again has a horizonatl tangent.

S: Hmmm ... does this kind of thing ever show up in real problems?

P: Yes ... trust me. |

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DIFFERENTIATION RULES

If f(x) and g(x) are both differentiable at x, then:

SUM: = f'(x) + g'(x) DIFFERENCE: = f'(x) - g'(x)

PRODUCT: = f(x) g'(x) + f'(x) g(x)

QUOTIENT: =

with the proviso that, in the QUOTIENT rule, g(x) ≠ 0 (else the expression in undefined!).

**Examples** :

• Suppose we knew that sin x = cos x and cos x = - sin x. Then

tan x = = = = = sec2x

(using the QUOTIENT rule).

• We can also use the PRODUCT rule to compute: x3x5 = x3(5x4) \+ (3x2)x5 = 8x7 which (fortunately) agrees with xn = n xn-1 when n = 8.

the CHAIN RULE:

One of the most useful rules (all of which can be derived from the definition of the derivative!) is:

the CHAIN RULE

If y = f(u) and u = g(x), then:

=

or

= f'(u) g'(x)

**Examples** :

• If y = sin x3, we can write this relation as y = sin u and u = x3 so that

= == (cos u) (3x2) = 3x2 cos x2. Hence sin x3 = 3x2 cos x2.

• Suppose that y = sin. We can write y = sin u, u = , v = cos w and w = x3. Then

= and all are easy differentiations! We get:

= = - .

**Example:** Find the equation of the tangent line to the curve y = 2x sin x at the point (,π).

**Solution:** = 2x (cos x) + 2sin x (using the PRODUCT rule) is the slope of the tangent line at _any_ place x. For x = we get = 2 cos + 2 sin = 0 + 2 = 2. We use the _point-slope_ form of the equation of a straight line to get: = 2 or y = 2x as the tangent line. (Note that this tangent line passes through the origin. Try this problem: "Find a point on y = 2x sin x such that the tangent line passes through the origin." Pretend you don't know the answer!)

**Example:** Find the normal line to the curve y = 2x sin x at the point (,π).

**Solution:** Since the tangent line has slope 2, the normal line (which is perpendicular to the tangent line) has slope - and passes through the same point (,π). Its equation is then:

= - or x + 2y = .

**HIGHER DERIVATIVES** :

Starting with some differentiable function y = f(x) we generate a second function, f'(x), by differentiating: = f'(x) which, as we've said, can be interpreted as the slope of the tangent line to the curve y = f(x) at the place x. We can also differentiate again to obtain the _second derivative_ : = f'(x) which is written using the notation: = f''(x). If f'(x) is increasing, then

_its_ derivative will be positive, i.e. > 0. If f'(x) is decreasing, then < 0. Geometrically speaking, if > 0 then the tangent line has an increasing slope so the curve is _concave up_. If < 0 then the tangent line has an decreasing slope so the curve is _concave down_. |

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 | For example, if s is the distance travelled after a time of t _hours_ (measured in _kilometres_ , say), then v = is the velocity (in _kilometres/hour_ ) so v > 0 when s is increasing and v < 0 when s is decreasing and so on. In addition, the second derivative of s, the acceleration a = = , tells when v is increasing or decreasing so when a > 0 the velocity is increasing and so on.

To the left are three graphs. One of them is s(t), the distance travelled (plotted versus t), one of them is the velocity v(t) = and one is the acceleration, a = .

Which is which?

P: Are you listening? Which is which?

S: I'd say that ... uh, one is the slope of the other and ... uh ... I don't understand the question.

P: The middle one is s(t) versus t and its slope is given by the first graph and you'll notice that at the point marked with a big dot the distance is constant for awhile so the velocity is zero. Further, the first graph decreases then increases so its derivative must first be negative then positive and that's what happens in the last graph so that must be , the acceleration.

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Finally, you may notice that, since a = is the 2nd derivative of s(t) it gives the concavity of s(t), so when the middle graph is concave down, the acceleration is negative. Nice, eh?

S: Why would anybody be interested in the slope of the tangent line to some curve, or, for that matter, whether it's concave up or down? I can't get too excited about acceleration. Is that why we're studying calculus?

P: No. Calculus involves derivatives and these occur everywhere ... the universe unfolds according to equations involving derivatives ... and that's why we study derivatives.

S: For example?

P: We've already mentioned that if s is the distance travelled by an object, then v = is its velocity and = is its acceleration. Newton's law of universal gravitation looks like = - , again involving derivatives. The spot which paints the picture on your TV screen moves across the screen according to + k(s2-1)+ s = 0. See? Derivatives! A hot object at temperature T cools according to = - k (T - c). More derivatives. The volume of a sphere is V = r3 and the surface area is . If C(x) is the cost of producing x items then is the marginal cost per item and if R(x) is the revenue then is the marginal revenue. The consumer price index ...

S: zzzzz

**Example:** A lake is stocked with 1000 fish. It is found that N(t), the number of fish after t years, increases so that its rate of increase is governed by the equation: = k N (10,000 - N) (called the "Logistic Equation"). Show that the rate of change, , is a maximum when the population is 5,000 fish.

**Solution:** We determine how rapidly increases by taking its derivative:

= = k N (- ) + k where we've differentiated k N (10,000 - N) as a product. Hence

= k . Note, first, that = k N (10,000 - N) is positive when 0 < N < 10,000. Now, when N < 5,000 we have > 0 so is increasing. When N > 5,000 we have < 0 so is decreasing. Hence the maximum rate of increase occurs when a population of 5,000 is reached.

Note: If the fish population obeyed the Logistic Equation, = k N (10,000 - N), then N(t) increases until

N = 10,000 which is the eventual population of the lake (called the "carrying capacity" of the lake). Normally one wouldn't know the eventual population so we'd have to assume an equation = k N ( P - N) where the eventual population is some _unknown_ constant P. Then we'd drop 1000 fish in the lake and each year we'd measure the population, N(t). The rate of increase will increase at first, then decrease. When this happens the population is the carrying capacity. If the population at that time is, say, 15,000 then we can expect the eventual fish population to be 30,000.

(A nice problem: Assuming = k N ( P - N), where k and P are constants, show that the rate of change, , is a maximum when the population is fish.)

**Example:** An object moves in a straight line so that s, its distance (in metres) after a time t (seconds), is given by the relation: s = t sin(πt). Determine its velocity and acceleration at time t = 1.

**Solution:** We need to find the velocity and acceleration . First, = πt cos(πt) + sin(πt) using the PRODUCT rule, then = - π2t sin(πt) +2π cos(πt) and, at t = 1, we get = - π and = - 2π.

(Note that < 0 means that s is decreasing; the object is getting closer. Also, < 0 means that is decreasing; the object is slowing down.)

LECTURE 5

IMPLICIT DIFFERENTIATION & TRANSCENDENTAL FUNCTIONS

Some Notes:

• If f'(x) = 0 over some interval, then f(x) = constant over that interval (i.e. the graph of y = f(x) is a horizontal line). This means that y is NOT changing when x changes.

• Note the interpretation of the derivative as a _rate of change_.

This interpretation was used by Newton*, who imagined x moving along a line at constant speed, and y = f(x) moving along another, parallel line. Then the derivative indicated how much faster (or slower) y was moving. In fact, it's _this_ interpretation of the derivative which makes the differential calculus such an important tool (and _not_ the interpretation as the slope of the tangent line to a curve ... as if anybody were interested in the slope!).

• If y = f(u) depends upon u and u = g(x) depends upon x thentells how rapidly y changes when u changes ("y runs 6 times faster than u"), and tells how rapidly u changes with x ("u runs 7 times faster than x"), and then gives how rapidly y changes with x ... "y runs (6)(7) = 42 times faster than x" ... and now the CHAIN rule seems almost trivial!

**IMPLICIT DIFFERENTIATION** :

Sometimes we're given a relation between x and y which, although is _does_ define y as a function of x, doesn't give y explicitly in terms of x. For example, consider the relation y3 \+ y5 = x. For x = 1 the y-value must be found from y3 \+ y5 = 1 and there is only one y-value which satisfies this equation. (Can you prove this?) Similarly, for any other value of x there will be a single value of y which satisfies the equation. (Can you prove this?) Consequently, y _is_ a function of x (even though we can't usually find the y-value ... except for certain peculiar x-values such as x = 0 in which case y = 0, or x = 2 in which case y = 1). However, remarkably, we _can_ find as follows:

Regarding y as a function of x we differentiate the relation y3 \+ y5 = x with respect to x and get

= x or, using the CHAIN rule we get = 1 hence (3y2 \+ 5y4) = 1,

so = . If we know a point on the curve, such as (2,1), then the slope of the tangent line at that point is = = .

**Example:** Find at the point (1,1) if xy = sin .

**Solution:** First we check that (1,1) _does_ lie on the curve! It does.

Next we differentiate the entire equation, regarding y as a function of x: = sin .

This gives: x + y = cos . Now substitute the given point, x = 1, y = 1 and get:

\+ 1 = 0 so = -1. (Note: the tangent line would be = -1 or y = -x + 2)

**Example:** Assuming xn = n xn-1 holds for integers n, prove that it holds for fractions as well.

**Solution:** Let y = xp/q, then yq = xp and we differentiate _implicitly_ : yq = xp gives

q yq-1 = p xp-1 (where we need only use the differentiation formula for integers p and q). Then

= = = xp/q - 1 ... and the formula works for any rational number!

PS:

S: What about the rest of the numbers, like and π which aren't rational numbers?

Is xπ still equal to π xπ-1 ?

P: Yes ... but it isn't easy to prove.

S: One other thing. You said that the standard equation of a circle, x2 \+ y2 = a2 (where "a" is a constant), doesn't define y as a function of x ... since there are usually two y-values for each x-value. Right?

P: Right.

S: Then if I go right ahead and take of everything, that is, I differentiate implicitly, even though y is NOT a function, I get 2x \+ 2y = 0 so = - . Is that wrong?

P: Actually, it's okay ... sort of. Since you can extract a function from this relation, y = for example, then

= - gives the slope of the tangent line at a point (x,y) on the graph of this extracted function. See?

S: No.

P: Well, let's take y = for example. Then = 1/2 = -1/2(-2x) = - which, of course, is - . If we take another function out of x2 \+ y2 = a2, say y = - , then = = - again.

S: Are they the only two functions you can "extract" from x2 \+ y2 = a2?

P: No. We can also define y = when -a ≤ x ≤ 0 and

y = - when 0< x≤ a. This is a function, with a single defined value for each x in the domain -a ≤ x ≤ a; its graph is shown in the diagram. (By the way, it doesn't have a derivative at x = 0).

Whether we compute for negative x (using y = ) or for positive x (using y = - ) we'd get the result = - .

So the we get from implicit differentiation actually gives us the correct slope no matter what function we extract from the relation (provided exists!). That's because = - doesn't give the slope at some x-value (where there may be two y-values satisfying the relation), it gives the slope at a point (x,y).

S: Wait a minute. You said it doesn't have a derivative at x = 0? Can you prove that? |

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P: Remember the definition! In order to have a derivative, the limit must exist.

In particular the right-hand limit must exist and that's:

= = -∞.

In fact, it's clear from the graph that as x approaches 0 from the right, the chord joining the two points (one being the point (0,a)) becomes vertical and its slope becomes large and negative, approaching -∞.

S: Do I have to know this for the final exam?

P: No. |

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Examples:

For each of the following:

sketch a graph of the function indicating asymptotes and discontinuities, and

write "limit statements" for x->-∞, x->∞, and

if f(x) has a discontinuity at x = a, write _two_ limit statements for x->a- and x->a+.

(a) f(x) = (b) f(x) = (c) f(x) =

Solutions:

(a) f(x) = = has discontinuities at x = 0 and x = - 2

f(x) = 0, f(x) = 0, f(x) = = -∞,

f(x) = ∞, f(x) = ∞ and f(x) = -∞ |

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(b) f(x) = = =

has discontinuities at x = 0 and x = - 2.

Since f(x) = thenf(x) = -1, f(x) = -1

f(x) = -∞, f(x) = ∞, f(x) = -2, f(x) = -2 |

  | (c) f(x) = has a

discontinuity at x = 0.

f(x) = 0 and

f(x) = 0

(Note: 0 ≤ | f(x) | ≤ , then we use the SQUEEZE theorem to get the limit.)

f(x) =f(x) = 1

P: Notice anything interesting about the graph of y = sin x ?

S: Nope.

P: It oscillates between y = and y = - . Remember? Since sin x oscillates between +1 and -1, then y = f(x) sin x oscillates between y = f(x) and y = - f(x) and in this case ...

S: Yeah, I remember.

Examples:

The _greatest integer function_ [[x]] is defined as "the greatest integer not exceeding x".

(For example, [[-5.7]] = -6, [[-5]] = -5, [[-4.8]] = -5, [[4.8]] **=** 4 )

(a) Sketch a graph of f(x) = [[x]] for -4 ≤ x ≤ 4 and identify points of discontinuity.

(b) Sketch a graph of f(x) = [[x + ]] for -4 ≤ x ≤ 4 and identify points of discontinuity (and note that f(x) rounds any number "x" to the nearest integer!)

Solutions:

(a) For f(x) = [[x]], points of discontinuity occur at every integer.

(b) For g(x) = [[x + ]], points of discontinuity occur halfway between successive integers.

P: Remember the sign in the post office ... about the cost of postage? It was C(x) = .

S: Nope.

P: Well, look at your notes. Anyway, the cost could also be written C(x) = 5 + 5[[x+1]]. Do you see that?

S: Nope.

P: Can you imagine such a sign ... hanging in a post office?

S: Nope.

P: Your cerebral prowess leaves something to be desired.

TRANSCENDENTAL FUNCTIONS: trig, exponential and log functions

Many functions we deal with can be evaluated for every x (in the domain of the function) using a $4.95 calculator (that can only add, subtract, multiply, divide and extract roots). These are called _algebraic_ functions. (e.g. , and a+bx+cx2+dx3.) Any function which is not _algebraic_ is called _transcendental_. These include sin x, tan x (and the other four trig functions) and 10x (and all the exponential functions ... with any base), log2 x (and all log functions, with any base) and many others. Unfortunately, the RULES for differentiation (product, quotient, Chain Rule, etc.) are of little help if we wish to differentiate a transcendental function ... unless, of course, we can express a transcendental function in terms of algebraic functions, using products quotients or composite functions .... but then they wouldn't be transcendental! Hence we must resort to the definition of the derivative ... and since the derivative is defined in terms of a limit, we can expect to find some weird limits which require some ingenuity to evaluate.

• To deduce sin x = cos x requires evaluating the limit:

=

= cos x , hence reduces to the evaluation of a weird limit: = 1.

(We won't prove this, but we'll use it ... as we've already done.)

the TRIG FUNCTIONS and their derivatives

Beginning with we can generate the derivatives of the other five trig functions (without resorting to the definition of the derivative!), because they are all related by simple _algebraic_ equations. For example, sin2 x + cos2x = 1 so if we differentiate implicitly (and use sin x = cos x) we get:

2 sin x (cos x) + 2 cos x cos x = 0 hence we can solve for . Similarly, tan x = so tan x = = hence . Also,

sec x = so sec x = -1 = - (cos x)-2 (- sin x) = which can also be written

. Finally, we "leave it as an exercise" to verify that and .

Note that all the _co_ -functions (cosine, cosecant and cotangent) have the (-) sign ... and that makes it somewhat easier to remember.

the EXPONENTIAL and LOG functions

• To find 2x requires the limit: and ultimately the weird limit .

• To determine log3 x requires, ultimately, evaluation of another weird limit, namely:

1/t = e ≈ 2.71828 (... which is how "e", the base of the "natural" logarithms arises).

Let's start with f(x) = loga x (for some positive base "a").

Then f'(x) = = loga = logax/h

where we've used two magic properties of logarithms: log A - log B = log and n log P = log Pn. In the above, we must assume that x > 0 (since loga x isn't defined for x ≤ 0) and that means we won't then get into trouble with the factor . Now, for any positive x, as h->0, then t = ->0 as well. Hence our weird limit is 1/t , just as we said. The values of (1+t)1/t, as t->0, approach a number slightly less than 3 (approximately 2.7182818) and this number is called "e".

or, equivalently,

Finally, then, we have

It's now clear that the world's greatest choice for a base for our logarithms is precisely the number e, since we'd get: loge x = logee and since logee = 1 we'd have . In fact, since this is the natural choice for a base, logarithms to the base e ≈ 2.7182818 are called _natural logarithms_ and are written _ln_ x (pronounced "lawn x" or "ell en x").

What could be simpler! Imagine the frustration of the pre- _ln_ mathematicians who could find functions whose derivatives were xn, for every number n ... namely ... with the single exception of n = -1.

Anyway, from now on we'll assume our logs are _natural_ logs unless otherwise specified.

**Examples:** Find for each of the following:

(a) y = x _ln_ x (b) y = _ln_ cos x (c) y = _ln_ (sec x + tan x)

Solutions:

(a) = x _ln_ x + _ln_ x = x + _ln_ x = 1 + _ln_ x (using the PRODUCT rule for differentiation).

(b) Let u = cos x so y = _ln_ u and (using the Chain Rule) = = = = - tan x.

(c) Let u = sec x+tan x so y = _ln_ u and = = = = sec x.

Okay, we now have the derivative of logax. To find the derivative of ax we could try to evaluate the limit

= ax ... but it wouldn't be easy! Instead we digress to consider INVERSE functions which will make it easier to find ax once loga x is known.

PS:

S: Wait wait wait. You said that log3 x isn't defined if x is negative.

P: ... or zero.

S: Okay, log3 x isn't defined for x negative or zero. Why not?

P: Remember the definition of the logarithm to the base 3? If y = log3 x, then x =3y. But 3y is never negative or zero, so x can't be negative or zero.

S: There must be an easier way ...

P: Let's go over logs one more time. Suppose we want to find a number p such that 5p = 47. You can see that 52 = 25 and 53 = 125 so there is some magical number between 2 and 3 such that 5p is 47. Whatever that number, it's called ...

S: The log of 5 to the base 47?

P: No! p is the log of 47 to the base 5, written log5 47. Remember the significance of the word "base": if 5p = 47 then "5" is the base. NOW, if there were such things as the log of a negative number, say p = log5(-47), then 5p = -47. But that's impossible, right? You can't raise 5 to any power and get a negative number, so there are no logs of negative numbers.

S: Or logs of zero, right?

P: Right! Now let's talk about Inverse Functions. Pick a number between 1 and 10. Now multiply it by 3. Now subtract 7. Now add 7. Now divide by 3. You get the number you started with, right?

S: Amazing! How'd you do that?

P: You pick a number x, then form f(x) = 3x-7. That is, you apply the function f(x). Then you undo what f(x) did to x, by adding 7 and dividing by 3. That is, you apply another function g(x) = . See? You first apply the function

f(x) = 3x - 7, then you apply the function g and to get g(f) = = = x and the latter function undoes what f did, and you get your "x" back again. The second function, "g", is called the INVERSE of "f". For the function "f" you "multiply by 3 then subtract 7": f(x) = 3x - 7. For the function "g" you "add 7 then divide by 3": g(x) = ... and g(f(x)) = x.

S: Why did you insist I take a number from 1 to 10?

P: I didn't think you could multiply big numbers by 3.

LECTURE 6

**INVERSE FUNCTIONS** :

If g(x) is the INVERSE of f(x), then g(f(x)) = x. Of course, in order that g(f(x)) make sense, x must lie in the domain of f (because we've got an f(x) in there!), and f(x) must lie in the domain of g (since we're evaluating "g" at the place "f(x)"). How to find g, the inverse of a given function f?

Examples:

• To determine the inverse of y = f(x) = 3x \- 7, do the following:

(1) Write x = 3y - 7 (2) Solve for y = = g(x), the inverse of f(x).

Note: g(f(x)) = = x.

• To determine the inverse of y = f(x) = x3:

(1) Write x = y3 (2) Solve for y = x1/3= g(x), the inverse of f(x).

Note: g(f(x)) = f(x)1/3 = x.

• To determine the inverse of y = f(x) = :

(1) Write x = (2) Solve for y = = g(x), the inverse of f(x).

Note: g(f(x)) = = x.

• To determine the inverse of y = f(x)

(1) Write x = f(y) (2) Solve for y = g(x), the inverse of f(x).

**Note** : The first step is to interchange x and y: from y = f(x), you write x = f(y). This changes the graph of y = f(x) by reflecting it in the line y = x. ("Interchanging" replaces every point (x,y) by (y,x) which is the _mirror image_ in the line y = x.) The second step, solving x = f(y) for y in terms of x, doesn't change anything in the graph! In fact, x = f(y) and y = g(x) are two forms of the _same_ relation between x and y ... and they have the _same_ graph. This is handy. If you don't like the looks of a function g(x), as in y = g(x), then you can write the relation as x = f(y) where the inverse function, "f", may be nicer to look at (and, in particular, to differentiate!).

The last step, "solve for y", may be impossible ... because not all functions _have_ inverses. |

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Consider y = x2. To find the inverse we interchange x and y, writing x = y2. Then we solve for y = ?!%#*, but there are TWO values of y, namely y = ± , so x = y2 doesn't define y as a _function_ of x. That's because f(x) = x2 doesn't _have_ an inverse function. We might have anticipated this by considering the graph of y = x2. When we interchange x and y, hence reflect the curve in the line y = x, we don't get the graph of a function. (A function must have ONE y-value for each x-value in its domain, and x = y2 doesn't!) |

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Remember the _vertical line test_ for a function? Every vertical line (in its domain) must cross the graph only once. In order for a function to have an inverse, it must also satisfy a _horizontal line test_ : every horizontal line (in its range) must cross the graph only once. y = x2 satisfies the first test (so it's a function) but not the second (so it has no inverse). Remember: to find the inverse of a function f(x) we must solve x = f(y) for y (or, what is equivalent, |

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solve u = f(v) for v ... or w = f(q) for q ... don't get tied to any particular labels!). To see if this is possible, we needn't go so far as interchanging x and y. We can simply ask: "Given y, is it possible to solve y = f(x) for x?" (i.e. can we solve the original equation, y = f(x), for x, if y is given?) When y = 3 is given, it's like drawing a horizontal line y = 3 and trying to find the x that satisfies 3 = f(x). If there _were_ one solution (and ONLY one) then there is a unique x for y = 3 (and the horizontal line test is satisfied). (That's NOT the case with 3 = f(x) = x2.)

However, if there _were_ a unique x for all values of y in, say, c ≤ y ≤ d, then y = f(x) would define x as a _function_ of y (as well as defining y as a function of x!).

What is this _function_? It's x = g(y) where "g" is the INVERSE of "f".

If f(x) has an inverseg(x)then y = f(x)when solved for xgives x = g(y)

**TESTING TO SEE IF A FUNCTION** _HAS_ **AN INVERSE** :

If the "horizontal line test" is satisfied for some function y = f(x) (on a domain a ≤ x ≤ b), then f(x) has an inverse. But the horizontal line test WILL be satisfied if the function f(x) is always increasing or always decreasing (i.e so-called "monotonic" functions)... and that can be used to test a function for an inverse: check that f'(x) > 0 or f'(x) < 0 on the domain.

**More Examples of Inverses** **:**

• What is g(x), the inverse of f(x) = ? (on x > 0 where we can check that f'(x) > 0)

Write y = , interchange x and y so x = , then solve for y = g(x) = - 2.

The graphs of y = and y = - 2 are reflections of each other in the line y = x.

• What is the inverse of f(x) = e2x ? (on -∞ < x < ∞ where f'(x) > 0)

Write y = e2x , then x = e2y, then solve for y = _ln_ x.

The graphs of y = e2x and y = _ln_ x are reflections of each other, in the line y = x.

PS:

S: If "g" is the inverse of "f", then what's the inverse of "g"?

P: We can see the answer graphically. Start with y = f(x), reflect it in the line y = x and get its inverse, y = g(x). Now start with y = g(x), reflect in y = x and get its inverse ...

S: y = f(x), right? But is that a proof?

P: Here's a proof: since g(f(x)) = x (that's what makes "g" the inverse of "f"), then apply the operation "f" to each side and get f(g(f(x))) = f(x), or, to put different labels on things, f(g(u)) = u (where we've replaced f(x) by u). Now stare at

f(g(u)) = u and realize that this makes "f" the inverse of "g": whatever "g" does to u, "f" undoes it, recovering u again.

S: You were about to say something clever about the derivative of ax, remember? You said knowing loga x makes it easier to find ax ... then you digressed into inverses.

**the Derivative of an Exponential Function** :

If we write y = ax and want to determine we can differentiate directly (but we'd have to use the definition of the derivative) or we can write this same relation as x = loga y and find by _implicit_ differentiation. Since we've already found the derivative of the log function (by using the definition) the latter scheme is easier.

y = ax fi x = loga y fi x = loga y fi 1 = logae

logay = and we _know_ logay). Now we solve for = . As is usual with implicit differentiation, the answer has y's in it. But y = ax so we get, finally: = .

To recap (about the exponential and log functions): and .

These involve logs to the base "a" but can be changed to "natural" logs as follows:

If p = logae then ap = e so (taking _ln_ of each side) _ln_ ap = _ln_ e or p _ln_ a = 1, so p = . In other words,

. That's not something special about logs to the base e. In fact: . We then have:

ax = ax ln a and loga x =

In particular, if we choose as base the number "e" (noting that _ln_ e = 1):

ex = ex and ln x =

Note that f(x) = ex is a function whose derivative is the _same_ function! (Can you think of any others? How about f(x) = π ex or f(x) = -47 ex? Clearly f(x) = C ex has this nice property, for _any_ constant C ... and these are the only functions with the property that f'(x) = f(x).)

Above is a reasonably accurate (computer-plotted) graph of y = ex and y = _ln_ x. They are inverses one of the other and knowledge of one (say y = ex) will tell you everything you want to know about the other. For example, it's clear that e0 = **1** , so . Also, e1 = **e** so .

Also, _ln_ e-3 = -3 _ln_ e = -3. Further, since , then .

Note that (as mentioned on earlier occasions) the logarithmic function is defined _only_ for positive x.

**LOGARITHMIC DIFFERENTIATION** :

Once in a while one has to differentiate exponential functions with variable exponents and variable bases. If the base is a constant, say f(x) = **2** x2, it's easy:

Let u = x2 then f(x) = 2u and = = = (2u _ln_ 2) (2x) = 2x 2x2 _ln_ 2.

If the exponent is constant it's also easy. For f(x) = (sin x5)3, = 3 (sin x5)2 cos(x5) (5x4) = 15 x4 sin2x5 cosx5.

If both base and exponent are variables, it's NOT so easy ... unless we use "logarithmic differentiation":

LOGARITHMIC DIFFERENTIATION: Take the _ln_ of each side ... then differentiate

Because _ln_ f(x)g(x) = g(x) _ln_ f(x) the result of taking logs is to eliminate variable exponents!

**Example** : Determine if y = xx.

**Solution** : First takes logs of each side ("natural" logs, of course) : _ln_ y = _ln_ xx = x _ln_ x. Now find using "implicit differentiation": _ln_ y = x _ln_ x gives = x () + _ln_ x = 1 + _ln_ x. Now solve for

= y (1 + _ln_ x) = xx (1 + _ln_ x). (Note that the answer is definitely NOT given by xx = x (xx-1) because the rule xp = p xp-1 only works when the exponent is a constant!)

**Note:** Taking logs before differentiation is a good scheme in many instances.

**Example** : Determine if y = .

**Solution:** Here _logarithmic differentiation_ is also useful. Take the _ln_ of each side, then find implicitly.

_ln_ y = _ln_ = 10 _ln_ (x+1) \+ 3 _ln_ x - 2 _ln_ (x-1) so = + - hence

=

**Example:** Use the differential to estimate the percentage change in volume of a cubical box if there is a 1% error in the side length.

**Solution:** If V = x3, where x is the side length, then we write ≈ = 3x2 so DV ≈ 3x2 Dx. This gives an estimate of the change in volume when the side changes from x to x+Dx. To get a percetage change, we divide by V = x3 giving ≈ = 3 hence the % change in volume is three times the % change in side length.

S: Huh?

P: That's what we mean by percentage change, isn't it? We divide the change in volume by the original volume ... and we can multiply by 100 to get the % change. But let me show you something slick, using logarithmic differentiation (in case you were wondering why this example is stuck in with logarithmic differentiation!).

Since V = x3 then _ln_ V = _ln_ x3 = 3 _ln_ x and now we differentiate and get = 3 and we use ≈ and get = 3 so that ≈ 3 . The fractional change in V is 3 times that in x (and we could multiply by 100 to get % changes). In fact, if we're interested in estimates of "fractional changes" (or percentage changes) in some function y = f(x) when x changes by Dx, we should :

(1) take logs................................................ _ln_ y = _ln_ f(x)

(2) differentiate............................................ =

(3) replace by the approximation ........... ≈

(4) solve for the fractional change ............. ≈ Dx

What we'd find is that the % change in the volume of a cubical box (measured in _metres_ 3 ) is **3** times the % change in side length, and the % change in the area of a square (measured in _metres_ 2 ) is **2** times the % change in side length. Try it!

**About Exponential Growth** **:**

One often hears the expression "Holy cow! It grows exponentially!" (or some such phrase). Indeed, the growth of y = ex (or y = 2x or y = 10x) is explosive, although it wouldn't seem so from the above diagram. However, if we plot y = ex for -100 ≤ x ≤ 100 the graph looks very much like the negative x-axis together with the positive y-axis! For negative x it vanishes almost immediately (to zero). For positive x it explodes (to infinity). Among other things, this explains the "miracle of compound interest". If you stick $1000 in a bank which pays 12% interest each year, then you'll have 1000 (1.12) or $1120 after a year and 1000 (1.12)(1.12) = 1000 (1.12)2 = $1254.40 after 2 years and, after n years, 1000 (1.12)n dollars. The amount of money is an exponential function of "n": M = 1000 an where a = 1.12 and, for a lifetime (where n = 75 years, say), the $1000 will grow to about $5 million dollars! (moral: start saving early and retire a millionaire.)

PS:

P: That reminds me of the story of how the inventor of the game of chess was rewarded (or was it checkers?). The king asked what she wanted and she said a grain of wheat on the first square, 2 on the second square, 4 on the next and so on, the number doubling with each square. The king laughed at the trivial request, but only for a moment. He pulled out his math notes and quickly calculated that, on the last square alone, there would be 263 grains of wheat and that was more than all the wheat in the kingdom. (263 is roughly 10,000,000,000,000,000,000. How do I know? Because 210 = 1024, or roughly 103, and I can write 263 = (210)6.3 which is then roughly (103)6.3 which is roughly 1019.)

S: If the exponential function looks like the negative x-axis together with the positive y-axis, that means the log function must look like the ... uh, let's see ... the negative y-axis together with the positive x-axis. Right?

P: Right.

S: That means the log function grows very slowly.

P: You got it. |

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**About the number e** :

The base of natural logarithms, the number e, is one of the most important numbers in mathematics ... perhaps the most important after the number π*. It occurs in a variety of situations:

• Radioactive substances "decay". (For example, radium eventually "decays", turning into lead). The amount of the substance, S(t), after a time "t", satisfies an equation like S(t) = A e-bt where A and b are constants.

• The current in an electric circuit is often described by I(t) = A e-bt.

• Populations sometimes grow according to an equation N(t) = A ebt.

• The so-called "normal probability distribution" is described by e-b(x-c)2 (the infamous "bell curve").

• The amount of a substance entering into a chemical reaction is often described by .

• If $A is left in the bank to accumulate interest at the rate of i% per year, the amount of money after n years is A ( 1 + )n. If the interest is calculated monthly at a rate %, the amount after n years is A ( 1 + )12n. If calculated daily at a rate %, the amount would be A ( 1 + )365n (if there are 365 days per year). If the interest is calculated T times each year at a rate % (T=12 means "monthly" and T=365 means "yearly") then the amount after n years is A ( 1 + )Tn . The big question: how much money would you have after n years if the interest is calculated continuously (meaning T->∞). The answer is A ( 1 + )Tn. To evaluate this limit we set N = , or T = and note that N->∞ as T->∞, so our limit can be written:

A ( 1 + )inN/100 or A in/100

and we recognize the limit N = e. Hence the amount after n years is A ein/100 dollars.

For example, $1000 left in the bank for n = 1 year at 10% per annum compounded continuously will grow to $1000 e.1 = $1105.17 compared to $1100.00 if it were compounded annually.

Examples:

Plot the graph of y = x2 e-x, showing where it's increasing or decreasing.

Solution:

  | Note first that y = 0 _only_ at x = 0 (and is positive for all other x-values). Further,

= x2 (-e-x) + 2x e-x = x (2 - x) e-x

which is negative for x < 0 and again for x > 2 (hence the function is decreasing there) and is positive for 0 < x < 2 (so the function is increasing there).

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Note, too, that x2 e-x = ∞ (since both x2 and e-x become infinite). Finally, to complete the picture, we should compute x2 e-x. Unfortunately, this is a problem where one factor becomes infinite and the other becomes zero. Nevertheless, if we write y = and have faith in the explosive growth of ex, it's not hard to accept the fact that x2 e-x = 0. We'll show this later when we consider limits which have the form or . In fact, ex grows so rapidly (with increasing x) that = 0 ... and you could substitute _any_ power of x and get the same "0" limit. (No matter how hard x1000 tries to get to infinity, ex drags the fraction to zero.) On the other hand, the function _ln_ x grows so slowly that = ∞ for any positive power p, no matter how small.

On the other hand, 2x is also an exponential function as is 10x and πx ... and they all grow very rapidly.

PS:

S: I thought you said that "e" was important. Is it more important than 2 or 10?

P: Well, if you want to know the truth, "e" usually occurs in the form of a function eax, for some constant "a". However, eax can always be written 2bx where "b" is some constant ... so, in a sense, that makes "2" just as important as "e", doesn't it?

S: How's that? Is eax = 2bx? Is that what you're saying?

P: Sure. Let's do an example. I'll find "b" so that e3x = 2bx. First I take the ln of each side and get 3x = ln 2bx = bx ln 2 (using that important log-property: log AB = B log A) and that means that b = . See?

S: I really don't see why all the fuss about "e". Let's do something else, can we?

**Example** : The cost of manufacturing an item is $100 even if no items are manufactured ... and the cost decreases with each item; for x items the cost per item is 100 - .1x (i.e. the cost decreases by $0.1 for each item). Graph C(x), the cost of producing x items.

**Solution** : For x items the cost is C(x) = x (100 - .1 x) = 100 x - .1 x2 which is positive for 0 < x < 1000. (For x > 1000 the cost is negative and makes no sense ... so we only consider x < 1000.) Further, C'(x) = 100 - .2 x is first positive, for x<500 (the total cost increases) then negative, for x>500 (C(x) is decreasing). Finally, C''(x) = -.2 is negative so that C'(x) is decreasing (meaning that the graph is concave downward).

It's clear that the maximum cost occurs for 500 items, and this technique of determining the maximum (or minimum) value of a function by investigating its derivative is one we'll use later on. |

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S: Are you saying the cost of producing 1000 items is zero?

P: That's what's called a mathematical idealization. Mathematicians do it all the time. When a mathematician shows you the solution to a problem you can be sure that she's made some assumptions and you have to be clever enough to argue with the assumptions. Assuming the cost per item decreases by $0.1 per item is, of course, ridiculous. So pay attention and argue with my assumptions! BUT, having made the assumptions, the math just carries on to the bitter end and knows little of nonsensical conclusions.

**Example** : An orchard contains 240 apple trees, each tree producing 30 bushels of apples. For each additional tree planted, the yield per tree decreases by 1/12 bushel (due to overcrowding). Sketch N(x), the total apple production as a function of x, the number of additional trees planted.

**Solution** : The total production is ( _number of trees_ )x( _bushels per tree_ ). For x additional trees, the _number of trees_ is 240 + x and the _bushels per tree_ is 30 - x/12. The total production is:

N(x) = (240 + x)(30 - ) = 7200 + 10 x - and N'(x) = 10 - is first positive (until x = 60) meaning that the total production is increasing, then N'(x) < 0 (for x > 60) so the production decreases. Note, too, that

N''(x) = - < 0 so the curve is concave down. (It's clear that x = 60 more trees should be planted to yield the maximum number of apples from the orchard.) |

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**Example:** Repeat the above if the number of bushels per tree is 10.

All other values are the same.

**Solution:** The total production is

N(x) = (240 + x)(10 - ) = 7200 - 10 x - and

N'(x) = -10 - is decreasing for _every_ choice of x ≥ 0. Again, N''(x) < 0 so the curve is concave down. (Conclusion? don't add any more trees!) |

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**ODDS 'n' ENDS ON CURVE SKETCHING** :

**Even and Odd Functions** **:**

For simple functions like f(x) = x2 or f(x) = x3 it is sufficient to sketch a function which passes through the origin and is increasing with increasing x. However, certain properties make sketching a little easier.

Some functions, like f(x) = x2, have the same value for -x as for +x; that is, f(-x) = f(x). Such functions are called EVEN functions. Some examples: cos x, 1+x2 \- x4, x sin x and e|x|.

(I)

Other functions, like f(x) = x3, assume the opposite sign when x is replaced by -x. These are the ODD functions. Some examples: sin x, x - x3 \+ x5, x cos x and sin x3.

(II)

When graphing these functions, it's only necessary to graph the function for x ≥ 0 since the graph for x < 0 can be obtained via the above EVEN or ODD symmetry.

If neither of (I) or (II) holds, then the function is neither even nor odd. Some examples: 1 + x, sin (x+1), ex and, of course _ln_ x (which only has values for x > 0).

**Quick &Dirty Curve Sketching** **:**

If a picture is worth a thousand words then we should spend some time in sketching the graph of functions: y = f(x). We first find the places where f(x) changes from positive to negative (by crossing the x-axis, or perhaps jumping discontinuously across the x-axis), then the places where f'(x) is positive or negative (to see where the graph is increasing or decreasing) then, if we're not exhausted, we find where f''(x) is positive or negative (to see where the graph is concave up or concave down), then take note of any horizontal asymptotes (where x->∞) or vertical asymptotes (where y->∞). It's a lot of work!

**Example** : Sketch the graph of y = .

**Solution:** Note that y = 0 when x = 0, and y < 0 when x < 0, and y > 0 when x > 0; the graph lies in the first and third quadrants. In fact, f(-x) = = - = - f(x) so the function if ODD.

Now consider = = . It's positive when x2 < 1 (i.e. in the interval -1 < x < 1) and negative outside this interval.

We can also calculate = and find that y''>0 (hence is concave upward) when x lies in

-< x < 0 and again in x > . Elsewhere it's concave downward.

There are no vertical asymptotes (since f(x) is never infinite for any value of x) but (dividing numerator and denominator by the highest power of x) = = 0 so there is a horizontal asymptote, namely y = 0 (the x-axis). |

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NOW ... if we're not interested in the details but want an quick&dirty picture of y = , we can sketch it for very small x and for very large x.

For | x | << 1 (meaning x is _very_ small), we can neglect the x2 in the denominator (compared to the "1") so, approximately, y ≈ = x, and we sketch

y = x (for x very small). |

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Next, for large x, we neglect the "1" compared with the x2 and get the approximation y ≈ = which we sketch for very large x (i.e. |x| >> 1). Then we just join these pieces (praying that nothing too wild happens in between small and large x).

**Example:** Sketch y = 1 - x + x2.

For small x, y is very nearly the straight line y = 1 - x (neglecting the x2). In fact, this is the tangent line at the origin!

Next, for very large x, y is nearly the parabola y = x2. |

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**Example:** Sketch y = x + .

  | For small x we neglect the x term and get:

y = .

For large x we neglect the term and get:

y = x (approximately).

These two curves, y = and y = x, are easy to sketch (for small and large x respectively) ... then we join them (and pray).

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S: In the last two examples, how do you know that y = 1 - x + x2 and y = x + don't cross the x-axis?

P: I said this was quick&dirty ... especially dirty. But, of course, we could check if 1 - x + x2 = 0 for any x-value, or if x \+ 1/x = 0. In both cases, there are no roots of these equations ... but if we had to do too much work the method wouldn't be quick.

S: Then the graph of N = 7200 - 10 x - begins (for x near zero) just like the straight line y = 7200 - 10x then decreases like y = \- ?

P: Right.

S: And C = 7200 + 10 x - starts out like the line y = 7200 + 10 x?

P: Mmm

S: And y = looks like y = = for small x and like ... uh, y = = -x for large x, right?

P: Mmm

S: And what about y = ? It looks like y = for small x, and y = ≈ when x is large and negative, and y = (x + ) sin looks like y = x sin when x is large and that looks like which looks like y = 1. And what about y = ...

P: zzzzz

LECTURE 7

MORE ON INVERSE FUNCTIONS

  |

Recall that a function y = f(x) has an inverse on the domain a ≤ x ≤ b provided it's monotonic there (i.e. f(x) is always increasing or always decreasing on a ≤ x ≤ b). Further, the inverse (we'll call it g(x)), when graphed, is just the reflection of y = f(x) in the line y = x. Note, too, that if the range of f(x) is c ≤ y ≤ d, then this becomes the _domain_ of the g(x), and the domain of f(x), namely a ≤ x ≤ b, becomes the _range_ of g(x). (All of this can be seen from a typical graph of y = f(x) and its reflection/inverse, y = g(x).)

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**Example** : Verify that f(x) = has an inverse on the domain 0 ≤ x < ∞, then find the inverse g(x) and its domain and range.

**Solution:** First we compute f'(x) = = which exists and is negative for all x in 0 ≤ x < ∞, hence f(x) has an inverse there (because it's monotonically decreasing for all x in its domain). Note, too, that 0 ≤ < 1 for 0 ≤ x < ∞; the range of y = f(x) is then 0 ≤ y < 1. We then have the domain of g(x) as 0 ≤ x < 1 and its range as 0 ≤ x < ∞ (and we have these even before we find g!). To find g we first write y = f(x) = then interchange x and y, writing x = . Now we solve for y: cross-multiply |

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to get x + xy = y so x = y - xy = (1 - x) y hence y = which is the inverse: g(x) = .

PS:

S: Hold on! Are you sure that y = g(x) is the reflection of y = f(x)? Shouldn't you check?

P: No need to (unless I've made a mistake ... and I never make a mistake). Nevertheless, here's the graph of both. See? A reflection ===>>>

S: But why did you restrict the domain of f(x) to

0 ≤ x < ∞?

P: Because f(x) would be an increasing function on this domain and I could guarantee it had an inverse (and that's why I didn't need to check the graphs of each to see this ... I knew it would be okay).

S: Are you saying that no other domain would do?

P: No, just that the one I picked would do.

S: I'd like to see you pick another domain. |

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P: Okay, let's sketch the graph of y = f(x) (a picture is worth a thousand words, right?)

I note that, for x very small, y = is much like y = x (neglecting the x in the denominator). Also, I can see that y = 0 when x = 0 and there's a vertical asymptote at x = -1 and a horizontal asymptote y = 1 since = 1 (which I could also expect because y is much like = 1 when x is large ... neglecting the 1 in the denominator). Now look at the graph. See? It's increasing for 0 ≤ x < ∞?

S: Are you kidding? It's increasing everywhere!

P: Well, that's true ...

S: So you could have picked any domain for f(x) and you'd still have an inverse. Go ahead, try the horizontal line test on y = . It never intersects more than once! |

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P: Remember we're looking for the inverse of a function f(x) on some domain. On any domain which includes x = -1, f(x) = isn't a function ... it doesn't have a single, unique value at x = -1 ... it doesn't have any value at all! We have to start with a function, one which satisfies the vertical line test: every vertical line in the domain must intersect the graph precisely once. If I choose a domain which includes x = -1 it doesn't intersect at all at x = -1. See?

S: Sounds like cheating to me. What does g(x) look like if you use the whole graph y = f(x)?

P: Let's reflect. It looks like this ===== >>>>>>>

S: That's a perfectly good function, isn't it? Try your line tests on it! Is the math so dumb it won't let me find an inverse for the whole f(x)? I could understand if there were two y-values for each x, in g(x) I mean ... but just look at the graph, I mean ...

P: Okay, okay. We could have looked for an inverse for the function f(x) = on the entire real line

\- ∞ < x < ∞ with the exception of x = -1. Happy?

S: I don't know why you made such a fuss ...

P: Watch what can happen if we're not VERY careful. Let's consider another function which is defined everywhere except at certain isolated points (like f(x) above where the isolated point is x = -1). And we'll also pick a |

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function which increases everywhere .

S: Except at those isolated points, right? I can hardly wait.

P: I'll pick y = tan x. Like it?

S: I don't even remember what it looks like.

P: Like this ====>>>

And it does increase everywhere, except at those points where it has no value (or, if you like, a vertical asymptote), like x = - and x = and so on. So do you think it has an inverse on the entire real line: - ∞ < x < ∞ ? (except at the points of discontinuity, of course).

S: No, because it doesn't satisfy the horizontal line test.

P: That's good! In fact it doesn't have an inverse, unless we restrict the domain to ...

S: Let me do it ... uh, to - < x < , right?

P: Right! And the range is -∞ < y < ∞, so the domain and range of the inverse will be ... |

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S: The domain is - ∞ < x < ∞ and range is - < y < ... I can see that. And I can also tell you what g(x) looks like ... I'm using g(x) to represent the inverse of f(x) = tan x ... uh, is there a special name for this inverse?

P: Yes, it's called "arctan x".

S: Okay, y = arctan x looks like this ====>>>

Now, let's see you find if y = arctan x. |

**the DERIVATIVE of ARCTAN X, the INVERSE TANGENT** **:**

If y = arctan x and we wish to determine we can rewrite this relation as x = tan y. (Remember, if f(x) and g(x) are inverses, then y = f(x) is the _same_ relation as x = g(y). If you "solve y = f(x) for x" you get the inverse

x = g(y) and, if you "solve x = g(y) for y" you'll again get the inverse y = f(x).) Now, to find it's easier to find it implicitly from x = tan y by taking the derivative of both sides: x = tan y or 1 = sec2y hence = and, as usual when using implicit differentiation, we get the derivative with y's in it. To obtain in terms of x alone we need to find sec2y in terms of x, knowing that tan y = x. But sec2y = 1 + tan2y and tan y = x hence

sec2y = 1 + x2. Finally, then:

Note that the derivative is always positive (indicating that the function arctan x is increasing, as it is!) and the derivative = the slope of a tangent line, approaches 0 as x->∞ (i.e. = 0) as it should (judging from the graph of y = arctan x). Further, we can use the Chain Rule to show that:

**the DERIVATIVE of ARCSIN X, the INVERSE SINE** **:**

If we attempt to define the inverse of sin x we run into the same difficulty as with y = tan x: the horizontal line test isn't satisfied. However we can restrict the domain as we did with the tangent function so that this test _is_ satisfied. The most natural selection is shown below.

Hence we consider y = sin x with domain - ≤ x ≤ and range -1 ≤ y ≤ 1. The inverse sine, denoted y = arcsin x, will then have domain -1 ≤ x ≤ 1 and range - ≤ y≤ . Again, to determine from y = arcsin x, we first rewrite this relation as x = sin y and differentiate implicitly: x = sin y or 1 = cos y so that

= . Once again we should eliminate y by finding cos y in terms of x, knowing that sin y = x. We have cos2y = 1 - sin2y = 1 - x2 so that cos y = ± and we have an ambiguity in sign!#$%*? If we choose

cos y = - then = - would be negative. However, it's clear that y = arcsin x has a positive slope, so we choose instead cos y = and get:

and, in general

PS:

S: Hold on. Isn't the mathematics smart enough to pick out the sign for us? I mean, our inverse has a positive so why does the math give us a choice?

P: We restricted the domain of the sine function to - ≤ x ≤ , remember? With this choice of domain our inverse will have a positive . However ...

S: I got it! Had we chosen, say 0 ≤ x ≤ π, then our inverse would have positive and negative slopes.

P: No! Choosing a domain 0 ≤ x ≤ π would give us a piece of the sine function that doesn't satisfy the horizontal line test ... so it wouldn't even have an inverse.

S: Okay ... suppose we'd chosen a domain, say, ≤ x ≤ . Then we would have an inverse and the derivative of this inverse would have the negative ... I see that now. One other thing; why the strange choice of names: "arctan" and "arcsin"?

P: I almost hate to mention this, but some people like to call them tan-1x and sin-1x because that's a common notation in mathematics for "inverses". For example, the "inverse" of the number "2" is written 2-1 and the inverse of the matrix A is written A-1 so it seems natural to call f-1(x) the inverse of the function f(x). I used to use the notation tan-1x and sin-1x myself, then I started to use a computer algebra system (that actually does calculus) and to enter these functions at the computer keyboard you have to type arctan x and arcsin x, not tan-1x or sin-1x (which isn't so easy on a keyboard). Besides, there's enough in calculus that's confusing so it's better not to get students even more confused by using sin-1x which some will want to rewrite as and that's quite wrong, of course.

S: Hold on ... did you say a computer that actually does calculus ? If computers can do all this stuff then why am I?

P: You have a problem in economics or biology or kinesiology or physics and you turn it into a mathematical problem with equations and functions which need to be differentiated, etc., and then you can give the computer the task of performing the differentiations, etc. ... but you have to do the first part.

S: You never explained why the funny name "arcsine".

P: Remember the definition of sin A for any number A? You measure off an arc of length A, on the unit circle, and the y-coordinate of the resulting point is sin A, that is, y = sin A. Now suppose you were given the y-coordinate (i.e. the value of the sine), and had to find the arc length A. Using our notation, A = arcsin y, so you'd ...

S: Don't tell me ... you'd be finding the "arc" of the "sine" or the arcsine for short. Cute. One last thing ... will we be finding inverses for the other four trig functions?

P: We could, if you're really interested.

S: Will I have to know it for the final exam?

P: No.

S: Then I'm not really interested.

**Examples:** Evaluate

(a) arcsin (b) sin (c) cos

Solutions:

(a) Let A = arcsin so sin A = - . Then A lies in the range of the arcsine function, namely - ≤ A ≤ . We choose an angle in this range whose sine is - ; clearly A is negative and is recognized as A = - .

(We use the convenient sin A = with _opposite_ = -1 and _hypotenuse_ = 2, correctly placing the angle A in the fourth quadrant. In degrees, A = - 60˚) |

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(b) Let A = arctan so tan A = - and A lies in the range of the arctan function, namely - < A < . Here we must choose the angle in the fourth quadrant. It's not one of the familiar angles, but we only need to know its sine, namely: sin A = sin= - = - . |

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(c) Let A = arcsin so sin A = and A lies in the range of the arcsine function, namely - ≤ A ≤ . We choose the angle in this range whose sine is ; clearly A is positive, lies in the first quadrant and has a cosine of: cos A = cos= . |

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**Problem:** Which of the following is the graph of:

(a) y = sin (arcsin x)? (b) y = arcsin (sin x)?

(c) y = tan (arctan x)? (d) y = arctan (tan x)?

Solution:

PS:

S: Hold your horses! You said that if g(x) is the inverse of f(x), then g(f(x)) = x. So shouldn't we have sin(arcsin x) = x and arcsin(sin x) = x and arctan(tan x) = x and ...

P: But we defined arcsin x by first restricting the domain of sin x to - π/2 ≤ x ≤ π/2, remember? Else sin x wouldn't even have an inverse. In the above problem I'm asking what the graph of arcsin(sin x) looks like without restricting x to lie in - π/2 ≤ x ≤ π/2. For any number x (i.e -∞ < x < ∞) we certainly have a value for sin x and this value lies in -1 ≤ sin x ≤ 1 and this is precisely the domain of the arcsine function so we can certainly compute arcsin(sin x) ... even without restricting x. If we did restrict x to lie in the interval - π/2 ≤ x ≤ π/2 then we would have arcsin(sin x) = x. You just have to look at the graph of arcsin(sin x) to see that.

S: What you mean is, if x is unrestricted then arcsin x is NOT the inverse of sin x so we shouldn't expect arcsin(sin x) to be equal to x.

P: Precisely!

S: Then why didn't you say that? Anyway, how did you pick out the correct graphs?

P: First, I know that sin(arcsin x) isn't even defined unless x lies in - 1 ≤ x ≤ 1, the domain of the arcsine function. That gives me the first graph which only lies in -1 ≤ x ≤ 1. All the others are defined for all values of x in - ∞ < x ≤ ∞ ... except arctan(tan x) where tan x doesn't even have a value at odd multiples of π/2. In fact, as x goes from just under π/2 to just over π/2, tan x jumps discontinuously from -∞ to ∞ so arctan would jump discontinuously from arctan(-∞) = - π/2 to arctan ∞ = π/2 (remember the graph of y = arctan x?). That gives me the next graph. Next, as x goes from -∞ to ∞, sin x just oscillates continuously from -1 to +1 to -1 etc. so arcsin(sin x) would oscillate from arcsin(-1) to arcsin(1) to arcsin(-1) etc., that is, from - π/2 to π/2 to -π/2 etc. That gives me the third graph. Finally, as x goes from -∞ to ∞, arctan x goes continuously from - π/2 to π/2 (remember the graph of y = arctan x?) so tan(arctan x) goes continuously from - ∞ to ∞ (just like x did!).

S: But how did you know they would look exactly like the given graphs?

P: I didn't (except for x in - π/2 ≤ x ≤ π/2, of course, where they would all be the same as y = x). But the problem is worded so that these were exactly the graphs of four functions ... so I just had to identify which goes with which.

S: Sounds like cheating to me.

P: Not at all! If I said one of the following is the formula for the volume of a sphere of radius r, then you'd be able to pick which one even if you didn't know the formula. After all, it's one of the given formulas.

S: What formulas?

P: V = πr2, V = 4πr2, V = 4πr4 and V = r3. Which is the volume of a sphere of radius r?

S: I haven't the foggiest ... wait, it's πr2. I remember that one, vaguely.

P: No, if r is measured in metres then πr2 is measured in metres2, so it's the area of something, not the volume of anything! See? Check the dimensions.

S: Okay, it's V = r3 metre3, a volume. So what about V = 4π r3? It's measured in cubic metres. What's it the volume of?

P: Three spheres.

S: Very funny.

Examples:

1. (a) Calculate if xy = arctan x.

(b) Prove that f(x) = ex \- _ln_ x has an inverse on x ≥ 1. If the inverse is called g, compute g'(e).

(c) Calculate g'(-1), if g(x) = arcsin(x+1).

(d) Calculate h'(1), if h(x) = (x-1) tan (ex-1).

(e) Calculate arcsin(tan(arcsin( - ))).

(f) Calculate at (0,0), if arctan y = l _n_ (sec x + tan x).

2. (a) Calculate at x = , if y = l _n_ (arcsin x).

(b) Determine the equation of the tangent line to _ln_ (x2 \+ y2) = arctan at the point (1,0).

Solutions:

1.

(a) xy = arctan x gives x + y = . Now solve for = .

(b) f'(x) = ex \- ≥ e - 1 when x ≥ 1, so f'(x) > 0 hence f has an inverse on x ≥ 1. From y = f(x) = ex \- _ln_ x we interchange x and y and write x = f(y) = ey \- _ln_ y which, when solved for y gives y = g(x), the inverse of f(x). Hence x = or 1 = so = g'(x) = . To determine g'(e) we need to know y when x = e. But x = ey \- _ln_ y so, when x = e, we have e = ey \- _ln_ y which has the obvious solution y = 1 (since e1 \- _ln_ 1 = e - 0 = e). Substituting y = 1 into gives g'(e) = .

(c) g(x) = arcsin(x+1) gives g'(x) = = . Substitute x = -1 and get g'(-1) = 1.

(d) h(x) = (x-1) tan (ex-1) gives h'(x) = (x-1) + tan(ex-1) and substituting x = 1 gives h'(-1) = 0 + tan(e0) = tan(1).

It's just as easy to use the definition: h'(-1)==tan(ex-1) = tan(1).

(e) Let A = arcsin(- ) so sin A = - and A lies in the fourth quadrant. (In fact, A = - .) Then tan(arcsin(- )) = tan A = - and finally, the problem:

arcsin(tan(arcsin( - ))) = arcsin(tan A) = arcsin= ?.

As above, we can let B = arcsinso that

sin B = - and note that B also lies in the fourth quadrant. |

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Unfortunately B is not one of the garden-variety angles (0˚, 30˚, 45˚, 60˚, 90˚, etc. i.e. 0,π/6, π/4, π/6, π/2, etc.), so we'd need a calculator to find the angle B. Punch in and then ask for the arcsin and get the number 35.26438965 which is in the first quadrant, so we take its negative and get the answer - 35.26438965˚ which is in degrees (!?&%) so we either put our calculator in _radian mode_ and do it all again ... or simply multiply by π/180 to convert to radians. We get B = = -.6154797084 (using π = 3.1415926535, or some such approximation to this number).

PS:

S: Couldn't I just leave the answer as B = - 35.2644˚? (My calculator only has 6 digits.) Besides, I can never remember whether I should multiply by 180/π or π/180.

P: Sure ... just don't forget to indicate that it's in degrees. But remember: π/180 is π radians per 180 degrees and is therefore measured in radians per degree so multiplying by degrees gives radians. Try it on 180˚. You'd get

π/180 (180˚) = π radians. See?

S: Then what'd I get if I multiplied, say 35˚, by 180/π?

P: A very big angle ... and marks taken off.

S: I never could understand why the big fuss about radians. Just why do you insist upon using radians instead of degrees?

P: Remember the definition of sin A and cos A? We measure off a distance A along the circumference of a circle and the coordinates of the terminal point are (cos A, sin A) ... and the central angle just happens to be precisely A provided its measured in radian! But that's not all. When we come to differentiate f(x) = sin x we have to resort to the definition of the derivative and get

= so we get: cos x , hence it reduces to the evaluation of a weird limit:

the limit of as A->0, where A = . Note that is an EVEN function of A (since

= ) so we need only compute ... the left-limit will be the same.

To evaluate this limit we consider the diagram which defines the angle "A" (for some small positive A) and use the Squeeze Theorem. To do this we need to find something both smaller and larger than , each of which has the same limit, then will also have this limit. The lengths of various lines and arcs are shown. The piece of the circumference has length A (as per definition). Also a side of the right-triangle within the circle has length sin A (it is, after all, the y-coordinates of the point which defines the sine function). Finally we construct a tangent line at this point and extend it to intersect the x-axis. |

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This line segment has length tan A. (Remember tan A = and adjacent = 1 so opposite = tan A.) Now compare the three lengths; we have sin A < A < tan A or sin A < A < or since A is small and positive then sin A > 0 and we can divide through by sin A (and not change the direction of the inequality!) to get: 1 < < . Now take the limit as A->0+ and get 1 ≤≤ 1 (since cos A->1 as A->0). That does it!

S: What about ? You got = 1.

P: But = = = 1.

S: And just where did you use the fact that A is in radians?

P: If the central angle is A, in RADIANS, then the arclength is also A ... but ONLY if A is in radians. If you want to consider the limit of when the central angle is A degrees, then the limit will NOT be "1". In fact, let's suppose the central angle is A degrees. Then what's the arclength?

S: It'd be ... uh, let's see, there's a formula: a = r q where "a" is the arclength and "q" is the central angle in radians and r is the radius. Here r = 1 and our central angle, in radians, is q = A (where A is in degrees). So far so good?

P: Yes, good ... keep going ... and I'm glad to see you multiplied by and not .

S: Okay, the arclength would then be a = q = A, if A is in degrees. So what would that make the limit?

P: You're doing fine. Keep going.

S: Let's see ... where are those three lengths ... yeah, I got it: sin A < A < tan A, except I'd have to put the correct arclength in the middle, so it'd become sin A˚ < A˚ < tan A˚ (and I've even put in the "degrees" symbol, just to make you happy). Now I'd want in the middle so I'd have to multiply through by and that'd

give me: < < and now I can let A go to 0˚ and put on the squeeze and get:

≤ ≤ so the limit is ... am I right? Wait ... the limit of is , right?

P: Right! And if x is measured in degrees, what's sin x ?

S: Huh?

P: We started all this by trying to differentiate sin x, remember? The derivative of sin x is

= = cos x = cos x provided "x" is measured in RADIANS. BUT, if "x" is in degrees, what's sin x ?

S: I'd say ... uh, sin x = cos x. Does that make sense?

P: Sure. We can interpret the derivative of a function as the slope of the tangent line to the graph of that function. So what you've shown is that y = sin x, when plotted against x in DEGREES, has a slope at the place x which is cos x. Here's a plot of y = sin x, with x in RADIANS:

It's not too surprising that the slope of the tangent line, at x = 0, is 1. Now here's a graph of y = sin x with x in DEGREES. What's the slope at x = 0?

S: Don't tell me! It's small ... it's ... no, that's big ... it's , right?

P: Right. So do you still want to work in degrees? Remember, you'd have to use sin x = cos x, etc.etc.

S: Okay ... radians are great ... so when do we do something useful with all this stuff?

LECTURE 8

OPTIMIZATION PROBLEMS

**ABSOLUTE MAXIMUM AND MINIMUM** **VALUES OF A FUNCTION** :

Having to find the value of x which maximizes some function, f(x), is a problem which occurs frequently. The function f(x) could be a profit (which we'd want to maximize) or a temperature (so we'd be finding the hottest spot) or the strength of a beam (so we'd be finding the strongest beam) or the yield of apples from an orchard. Sometimes f(x) represents a loss (which we'd want to minimize) or expenses (which we'd also want to minimize) or perhaps the coldest spot in the lake (to see if the water will freeze). Calculus helps.

Below we show several functions defined on some closed interval of the form a ≤ x ≤ b. Each function (except one) has both an absolute maximum and an absolute minimum value on [a,b] ... and we can see where these occur.

For h(x), the absolute min and max values occur at x = c and x = d respectively (where h'(x) = 0).

For g(x), the absolute min occurs at x = c where g'(x) = 0 but the absolute max occurs at x = a.

For f(x) the absolute maximum and minimum both occur at the endpoints x = a and x = b respectively.

For F(x), the absolute minimum occurs at x = b but there is no absolute max.

To see this we should define _precisely_ what we mean by "the absolute maximum value of a function on an interval".

The absolute maximum value of f(x) on an interval, is f(x0) where

(i) x0 is in the interval and

(ii) f(x0) ≥ f(x) for every x in the interval.

The absolute minimum value of f(x) on an interval, is f(x0) where

(i) x0 is in the interval and

(ii) f(x0) ≤ f(x) for every x in the interval.

Note that the maximum value must actually _be_ a value of the function! For the function F(x) depicted in the diagram above, the limit as x->a+ exists, but it isn't the absolute maximum value because it is NOT a value achieved for _any_ x in the interval. On the other hand, F(b) is the absolute minimum because F(b) ≤ F(x) for any x in the interval and is actually achieved for x = b. Whereas all other functions depicted have both absolute maximum and absolute minimum values, F(x) does not. Note that F(x) isn't continuous for every x in the interval whereas the other functions _are_ continuous.

Does that mean that _continuity_ guarantees the presence of an absolute maximum and minimum? Look at the following functions:

For y = 1 - | x - 1|, we have y = 1 - = x when 0 ≤ x < 1 ... since | x - 1 | = - (x-1) ... and we also have y = 1 - (x - 1) = 2 - x when 1 ≤ x ≤ 2. This function is made up of two lines, it's _continuous_ for all x in the interval [0,2] and has both an absolute maximum and an absolute minimum (which occur at x = 1 and x = 0 or 2 respectively). The next function _isn't_ continuous for all x in [0,2] and fails to have an absolute maximum, although it does have an absolute minimum (which occurs at x = 0, x = 1 and x = 2). It looks like continuity will guarantee an absolute maximum and minimum ... but look again:

The first function is f(x) = on the interval 0 < x ≤ 1 and although it IS continuous for every x in this interval, it _still_ doesn't have an absolute maximum (although it does have a minimum at x = 1). Perhaps that's because it has a vertical asymptote (i.e. becomes infinite)? No, because the second function, namely f(x) = x on the interval 0 < x < 1 also has no abs. max (neither does it have an abs. min) ... and it's never infinite.

So what are the criteria which guarantee that some f(x) will have both an absolute maximum and an absolute minimum on some interval?

If f(x) is continuous on the closed interval a ≤ x ≤ b, then it has both an

absolute maximum and an absolute minimum on that interval.

The thing is, f(x) must be continuous and the interval must be closed (i.e. include both end-points). Look at all of the above graphs and convince yourself that this does indeed happen. This doesn't preclude a discontinuous function from having an absolute maximum and/or an absolute minimum ... even on an interval which is not closed. It's just that we can't guarantee it. More than that, if a function doesn't have an abs. max or an abs. min we _can_ guarantee that the function is either discontinuous somewhere in the interval and/or the interval isn't closed.

Suppose we _do_ have a continuous function on a closed interval. Just where should we look to find the absolute extrema? A study of the various graphs shown above (and below) leads us to consider places in the interval where either f'(x) = 0 or where f'(x) doesn't exist.

Hence, we define a _critical_ point:

If (i) f'(x0) = 0, or

(ii) f'(x0) does not exist,

then x0 is called a critical point of the function.

Then:

If f(x) is continuous on the closed interval a ≤ x ≤ b, then the

absolute maximum and absolute minimum will occur either

(i) at a critical point in a < x < b, or

(ii) at an end-point: x = a or x = b.

That means that we need only find the critical points interior to the interval and evaluate f(x) at each, then evaluate f(x) at the end-points, then pick the largest and smallest of all these values. BUT, this procedure is guaranteed to succeed only if f(x) is continuous on an interval, and the interval is closed (as the following examples show).

Examples:

min & max min & max no max, but min occurs at

at end-point and occur at each min occurs critical point,

a critical point. end-point. at an end-point. but no max.

**Examples** :

For f(x) = 1 - | x - 1 |, the only critical point in 0 < x < 1 is x = 1 (because f'(1) doesn't exist). Hence the absolute maximum and absolute minimum values are the largest and smallest of the values f(0), f(1) and f(2), hence the abs. max is f(1) = 1 and the abs. min is f(0) = f(2) = 0.

For h(x), the abs. max and abs. min are the largest and smallest (respectively) of the numbers f(a), f(b), f(c) and f(d) (noting that x = c and x = d are critical points in a < x < b because f'(x) = 0 there).

For g(x), the abs. max and abs. min are the largest and smallest (respectively) of the numbers g(a), g(b), and g(c) (noting that x = c is a critical point in a < x < b because g'(x) = 0 there).

**Example:** Determine the absolute extrema of f(x) = x - 2 sin x on 0 ≤ x ≤ 2π.

**Solution:** f'(x) = 1 - 2 cos x = 0 where cos x = . On 0 ≤ x ≤ 2π this occurs at x = and x = . Hence the absolute maximum and minimum values of f(x) are the max and min of the numbers: f(0), f( ), f( ) and f(2π). Hence we calculate: f(0) = 0, f( ) = - 2 sin = - ª -.685, f( ) = - 2 sin

= + ª 6.968 and f(2π) = 2π ª 6.283, hence, the abs. min is - and the abs. max is + .

**Example:** A conical drinking cup is formed from a circular piece of

paper by removing a sector and joining the edges. If the radius of the piece of paper is 10 cm., what should be the angle q so as to yield a cup of maximum volume?

Note: volume of a cone =

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**Solution:** From the diagram, the volume of the cone

is V = π r2 h. But 102 = r2+h2 (that's Pythagorus talking) so

r2 = 100 - h2 hence the volume is V(h) = π (100-h2)h and we wish to find the absolute maximum of this continuous function for h in some interval ... hopefully a _closed_ interval. Clearly the smallest h is 0, but that means we don't cut any sector out of the circular paper, so q = 0, and we don't have a cone at all. On the other hand if we cut out |

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_all_ the paper, so q = 2π, then although h is a maximum (h = 10) we again don't have a cone. It seems we have an open interval to consider, namely 0 < h < 10. Nevertheless, we continue: V'(h) = π (100 - 3h2) which is first positive, for 0 < h < 10/, then negative, for 10/< h < 10 (so V(h) then decreases) and the maximum clearly occurs at the critical point, h = 10/, where V'(h) = 0. Since we were asked for q, we must find the relationship between h and the angle q.

Note that the circumference of the circle (on the cone) is 2πr and this equals the remaining circumference on the paper circle, namely: 2π(radius) - arc length = 2π(10)-10q = 20π - 10q ... using arc length=(radius)(angle at centre). Hence, 2πr = 20π-10q so q = 2π - r but r = = = 10 so q = 2π - 2π .

Finally, we really _could_ have considered V(h) on the closed interval 0 ≤ h ≤ 10 (even though the "cone" would have zero volume at the end-points: h = 0 and h = 10). However, we pretended that this was an open-interval problem so we could talk about ....

**RELATIVE MAXIMA and MINIMA** **:**

The previous example illustrates how we might obtain max and min values of f(x) on intervals which are NOT closed: use the derivative to determine where f(x) is increasing or decreasing. Indeed, if f'(x) > 0 just to the left of x = c (meaning that f(x) is increasing in value) and f'(x) < 0 just to the right of x = c (so f(x) is decreasing), then we might suspect that f(c) is larger than nearby values of f(x). That should make f(c) a RELATIVE MAXIMUM. Of course, there's always the possibility that f(x) is discontinuous at x = c and |

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the actual value, f(c), is different from the left- and right-hand limits. Then f(x) wouldn't have a derivative at x = c, but that would still make x = c a _critical point_ , so we again look to the critical points for RELATIVE MAXIMA and RELATIVE MINIMA.

For the function shown at the right (which is the previous function, now graphed over _all_ of (a,b]), f'(c) = 0 so x = c is a critical point (hence a candidate for a relative max or min) so we check the sign of f'(x) just to the left and right of x = c and find that f(x) first increases then decreases. Although this makes f(c) a relative maximum, it's clearly not the absolute maximum on the interval shown, since f(b) is. For the function depicted, there are two relative minima and three relative maxima in the interval. (Can you find them?) |

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For the previous function, there is no absolute minimum at x = a because f(x) doesn't have a value there (indicated by the open circle). For a similar reason, there isn't a relative minimum at x = a either.

It's time we defined relative extrema:

f(c) is a relative maximum if f(c) ≥ f(x) for all x in the domain which are sufficiently near c

If you're standing on the top of a mountain you're at a _relative_ maximum (of elevation). To be at an _absolute_ maximum, the mountain has to be Mount Everest.

Now, we consider the following "First Derivative Test" for a relative maximum:

If f'(x) > 0 immediately to the left of x = c and f'(x) < 0 immediately to the right of x = c, then f(c) is a relative maximum **??**

Note the **??** Is this test valid? Can we think of a counterexample? (i.e. a function which is increasing just to the left of x = c, decreasing just to the right, yet f(c) is NOT a relative maximum?)

On the right is the graph of a function, f(x), almost identical to the one portrayed previously, yet f(c) is no longer a relative maximum. Indeed, f(c) is now a relative (and absolute) minimum! Hence, if f(x) increases until x = c then decreases after x = c, it's not enough to guarantee a maximum (of either flavour). We need continuity at x = c. (That is not to say that f(c) _cannot_ be a relative or absolute maximum if f(x) is discontinuous at x = c, it's just that we can't guarantee it!) |

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So we modify our test:

First Derivative Test for a Maximum

If f(x) is continuous at x = c and f'(x) > 0 immediately to the left of x = c and f'(x) < 0 immediately to the right of x = c, then f(c) is a relative maximum.

First Derivative Test for a Minimum

If f(x) is continuous at x = c and f'(x) < 0 immediately to the left of x = c and f'(x) > 0 immediately to the right of x = c, then f(c) is a relative maximum.

This theorem must be taken with a grain of salt. i.e. it's true, except it doesn't include some important cases. For example, if a relative maximum occurs at x = a (the left end-point of our domain), then it's pretty hard to insist that f'(x) > 0 to the _left_ of x = a. Similarly, for the previous graph, f(b) wouldn't satisfy the requirements of this test even though it certainly is a relative maximum. Clearly, for end-points, we need _continuity_ and the appropriate sign for f'(x) on one side of the end-point. It just makes sense to say: "If f(x) increases to the value f(b) at the right end-point, then f(b) is a relative maximum for f(x) on the interval".

Note, too, that if f(c) is a relative maximum, we needn't have

f'(c) = 0. It may be that f'(c) doesn't exist as is the case for f(x) = 1 - |x-1| at x = 1. However, if f(c) is a relative maximum, then either f'(c) = 0 or it doesn't exist. In either case it's a _critical_ point.

We now have a scheme for identifying relative maxima and absolute maxima. To reiterate: |

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If it's required to find extrema (either relative or absolute) for f(x) on some interval, then we first determine the critical points of f(x) _interior_ to the interval. (We'll consider the end-points separately.) If f(x) is continuous at such an interior critical point then we apply the First Derivative Test to see if it's a relative maximum or minimum. We also evaluate f(x) at each critical point. Then we evaluate f(x) at the end-points. If f(x) is continuous throughout the closed interval [a,b], the absolute maximum and minimum are simply the largest and smallest values of f(x) evaluated at the end-points and the interior critical points.

S: What if f(x) isn't continuous throughout ... or what if f(x) isn't even continuous at some critical point? Then what? Your theorems seem pretty weak to me.

P: Actually, these theorems are only for the very lazy who like to turn some crank and have the answers pop out. The best way to determine the absolute and relative extrema is just as we've been doing with our graphs. We just look ... and see. After all, a picture is worth ..

S: Yeah, I know, a thousand words.

P: Here's a nice problem where the graph tells all. Take a length of wire, of length, say, 1 metre. Cut it into two pieces. With one piece, form a circle. With the other, form a square. How much of the wire should form the circle and how much the square if you want the maximum total area? Okay, suppose we cut a length x metres for the circle and the rest, 1-x, for the square. The area of the square is 2 and the area of the circle is πr2 where r is the radius. Since the circumference |

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is x, then x = 2πr so r = so the area of the circle is π2. The total area (which we wish to maximize) is:

A(x) = π2 + 2 and we now graph A(x) by determining where it's increasing and decreasing. We have A'(x) = - = which is negative at first (when x is small) then positive (when x is larger than ).

The graph then looks like this ====>>>

The absolute minimum obviously occurs at x = and the absolute maximum occurs at one of the end-points. To see which, we evaluate A(0) = and A(1) = . The latter is larger, so the maximum area is achieved when x = 1; ALL of the wire is used to form the circle. |

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S: But A(x) is continuous on 0 ≤ x ≤ 1 (which, by the way, is a closed interval) so you can find the absolute maximum by evaluating A(x) at the end-points and at the interior critical point, right?

P: Right.

S: Then you didn't need to graph A(x) at all ... or determine where it was increasing or decreasing. Just pick the biggest of A(0), A() and A(1). Am I right?

P: Yes ... and see how much you've learned already? I'll bet if you did this problem last month, you'd set A'(x) = 0, find that x = and conclude that that gave the maximum area ... just because I asked for the maximum. If I had asked for the minimum area, you'd do the same thing. Or maybe you'd apply some kind of test, or ..

S: Certainly not. And now, can we continue?

P: Just one last thing. If you had used the second piece of wire for an equilateral triangle, or perhaps some other figure, do you know what the solution would be? I'll tell you. It would STILL be x = 1; ALL of the wire should be used for the circle. Do you know why? I'll tell you. The circle provides the maximum possible area for a given perimeter ... so you shouldn't waste any wire on anything else. In fact, for a given surface area, what solid provides the maximum possible volume? I'll tell you: a sphere. That's why soap bubbles are spherical. In fact, every soap bubble has taken this math course and ...

S: Please, let's keep going.

**Example:** For each of the following, determine where the function is increasing or decreasing, locate _critical points_ , determine the location of _relative maxima_ and _relative minima_ and use all of this information to sketch the graph of y = f(x).

(a) f(x) = x5 \- 5x

(b) f(x) = x ex (note: to help in graphing, use x ex = 0)

Solutions:

(a) f'(x) = 5(x4 \- 1) ≤ 0 (i.e. f(x) is decreasing) where x4 ≤ 1, i.e. where -1 ≤ x ≤ 1. Elsewhere,

f'(x) > 0 (and f(x) is increasing). Critical points occur where f'(x) = 5(x4 \- 1) = 0, hence at x = ± 1. Since f'(x) is first positive then negative as x crosses -1, f(-1) is a relative maximum. Similarly, f'(x) is first negative then positive as x crosses +1, so f(1) is a relative minimum. (Also, for x small, f(x) behaves like y = - 5x.)

(b) f'(x) = (x+1) ex ≤ 0 when x ≤ -1 (so f(x) is decreasing there) and f'(x) > 0 elsewhere (so f(x) is increasing). The only critical point occurs at x = -1 (where f'(x) = 0). Also, f(x) < 0 when x < 0 and f(x) > 0 when x > 0, so the graph lies in the first and third quadrants. Finally, since x ex = 0, y = 0 is a horizontal asymptote.

S: How did you know that x ex = 0?

P: I write it as and use the explosive growth of the exponential function (since the exponent approaches +∞) to drag the fraction to zero as x->-∞.

S: Is that a proof?

P: No ... but later we'll discuss a method for evaluating such a limit.

In the "wire problem", we wanted to cut the wire in an "optimum" manner (to maximize the total area). When we chose the variable "x", we did so knowing that when we found x it would determine precisely how to cut the wire. That's important. If we want to maximize a quantity Q, then choose a variable, call it x (or some convenient name), which will determine _precisely_ how to maximize Q ... then express Q in terms of x; for each x there is a Q-value. Then find the x-value which maximizes (or minimizes) Q(x).

For example, if the problem were to determine the isosceles triangle of maximum area which can be inscribed in a given circle of radius R, we would NOT pick as variable the base length r. Why not? Because even if we knew the "optimum" value of r it still wouldn't tell us what triangle to use! In fact, there are TWO triangles for each value of r. (See diagram ===>>>). Moral? Pick a variable (such as the height or the angle at the vertex) which DOES provide a unique triangle. |

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**Example** : A man can run 10 times faster than he can swim. He begins in the water at a point P (see diagram), swims to shore, then runs to Q (the cottage). Describe his path so the total time is a minimum.

Note: distances "a", "b" and "c" are given. |

**Solution:** Since the total time is to be minimized, we let it be called T. Then we pick some variable which describes a (unique!) path from P to Q and express T in terms of this variable. Then, minimize T.

There are several possible variables to describe a particular path from P to Q: the distance x or the distance y or the angle q or the angle f. (If any of these is known, then the precise path will be known, hence the time T). Let's pick x. Then the total time, T(x), is made up of the time spent in the water + the time on land. We calculate each: |

We use time-in-water = = where we let u = speed-in-water (since we weren't given this quantity, we simply give it a name and use it!) Note that the distance travelled in the water is calculated from Pythagorus' theorem.

Also, time-on-land = = where v = speed-on-land. Hence, the total time is:

T = + . As is often the case, our function depends upon two variables, x and y, which are related. We must eliminate one of them so T is a function of a single variable. Since we chose to express T in terms of x, we'll eliminate y. There must be a relation between x and y ... and there is. It's x + y = c (which was given). Hence, y = c - x so we have, finally, T(x) = + and we remind ourselves that a, b,c, u and v are constants which we assume are known. Our problem is to minimize T(x) for x in some interval ... but what interval? It's clear that x lies in 0 ≤ x ≤ c (since the swimmer would head for shore, somewhere between P and Q) hence we have the problem of a continuous function T(x) on a closed interval [0,c]. We find the critical points:

T'(x) = - = 0 and we must now solve this for x. If it should turn out that x lies in (0,c), then it's an interior critical point and we evaluate T(x) there, as well as at x = 0 and x = c (the end-points). The smallest of these three numbers is the minimum time. Note how convenient to have the theorem which says we don't need any derivative test if T(x) is continuous and the interval is closed ... just pick the smallest of these values! Without this we'd have to analyze the terrible expression T'(x) = - to see where it's positive and where it's negative. Or, worse yet, we might be tempted to use the "second derivative test" which requires finding !! It will turn out (of course!) that T'(x) = 0 at one point in the interval 0 < x < c and T'(x) < 0 to the left of this point and T'(x) > 0 to the right, so this point provides the minimum.

Let's get back to the equation which gives the critical point: = or

= which, although it looks messy, is actually a very nice way to express the condition for a minimum time because each of and is recognized (!after considerable staring at the expressions!) as the sine of one of the angles q and f !! The condition for a minimum can then be written

and, for our problem, the ratio of speeds is = so we didn't have to know each speed after all, just their ratio! This magnificent condition for the _minimum time_ is called Snell's law and is known to all beams of light ... because it's precisely how light travels from one medium to another! When light moves from glass to air (where it travels faster) it is _refracted_ at the boundary between the two media (just like the path of our swimmer ... because the speed of light is greater in air than in glass). Having taken this math course, the light beam changes course at the boundary so it takes the minimum time to travel from one point (in glass) to another (in air).

PS:

S: You haven't finished the problem! So what's the optimum path? So what's x?

P: Well, I need to solve = = .

S: Go ahead. I'll wait.

P: Okay ... I'll let r = = , the ratio of speeds (so I only have to do this once and when I'm finished you can stick in any ratio you like). Then I square both sides and get an equation to solve for x, namely ... uh, I think it's a quartic equation: (1-r2) x4 \- 2c (1-r2) x3 +(b2+c2(1-r2)-a2) x2 \+ 2ca2r2 x - a2c2r2 = 0. See? Once you know all the numbers a, b, c and r ... you just solve this equation for x.

S: You gotta be kiddin' ... just solve for x? Besides, how do I know you didn't make a mistake. Maybe your quartic equation is wrong.

P: Well, we can check it ... sort of. If x is measured in metres, then every term must have the same dimensions (else I did make a mistake). Since r is the ratio of speeds, it has no dimensions (i.e. is dimensionless.) Now look at each term. (1-r2)x4 has dimensions m4, and since a, b and c are also in metres, then 2c (1-r2) x3 and (b2+c2(1-r2)-a2) x2 and 2ca2r2 x and finally a2c2r2 are also in m4. So that checks out. Now let's assume r = 1 (so the speed is the same in water and on land) and b = a. Then we know the solution: the path should be a straight line from P to Q and x = must be the solution. Our equation becomes: 2ca2x - a2c2 = 0 and x = is indeed the solution! So that checks out. Hence I have great faith in my equation ... I don't think I made a mistake.

S: Okay, solve it if a = 1, b = 2, c = 3 and r = .

P: The equation becomes: 99 x4 \- 594 x3 \+ 1290 x2 \+ 6x - 9 = 0 and I'll use Newton's method to find x. Aaah, wait, you don't know Newton's method, do you? You'll just have to wait for a lecture or three until we get to that topic.

S: Sure, sure. But tell me, what's this about checking the dimensions? You've mentioned this before.

P: It's a neat idea. If somebody gives you a formula for something, you can check that it's dimensionally correct. If not, then the formula is wrong. Suppose somebody says the area of a circle is πr3. Then, for r in metres, this would give cubic metres for the area (because of the r3) so it can't be a correct formula since area is measured in square metres. Suppose somebody says the volume of a torus (that's a donut) is V = πa2b2 where "a" and "b" are

certain lengths, then it's an incorrect formula because it gives m4, not m3 as it should for a volume. Suppose Einstein had said E = mc3. Then you could check and discover that the dimensions of energy E are not the same as the dimensions of mc3, so Einstein was wrong. In fact you could point out to him that he's better off with E = mc2 because that, at least, is dimensionally correct. You see? A = πr2 is the area of something and C = 2πr is the length of something and V = r3 is the volume of something, etc. so you would never say the area of a circle is 2πr because it's not even an area!

S: But what if I said the area of a circle is A = 2πr2? It has the correct dimensions, so then what? |

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P: Then checking the dimensions wouldn't help. But, sometimes, you can check particular cases too, as we did with our equation above. We knew the answer when r = 1 and a = b. It's the same for, say, the volume of a torus. If I said the volume was V = 2 then it not only has the correct dimensions (metres3 if a and b are in metres) but it also gives V = 0 when a = b (so, for example, V = 3 would definitely be wrong).

S: Is it the correct formula for a torus?

P: Yes, I think so, but wait until we get to volumes later on in the course; we'll compute the volume from scratch.

S: Newton's method comes later ... and volumes comes later ... and I thought we were almost finished. Okay, let's go!

Example:

A page is to have 100 cm.2 of printed text, with 2 cm. borders.

Find the dimensions of the page of smallest area.

Solution:

We must minimize the area of the page, so we give it a name: call it A. Now we pick some variables which will determine the dimensions of the page: say x = width and y = height. Then A = xy must be a minimum. As before, we have two variables, but they're related since the width of the text is (x-4) and its height is (y-4) and its area is given as (x-4) (y-4) = 100. Now we can find y = 4 + and substitute to get A(x) = xy = 4x + and _that's_ |

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the function we wish to minimize. Unfortunately, we don't have a closed interval because x can be any length greater than 4 cm. (If x = 4.0001, then we have little width for our text, but y = 4 + is enormous so the page is a mile high!) Okay, we'll find A'(x) and see when it's increasing and when it's decreasing:

A'(x) = 4 - which is VERY negative when x is slightly greater than 4 (example: x = 4.0001) and positive when x is very large (since the second term is small). Hence A(x) is first decreasing, until A'(x) = 4 - = 0, then increasing. The minimum occurs when (x-4)2 = 100 or x - 4 = 10, hence x = 14 cm. (and y = 4 + = 4 + 10 = 14 cm. as well).

If we wanted to sketch A(x) (although it's not necessary, since we have the answer) we could note that A(x) looks like when x is near |

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4 (neglecting the first term, 4x), so A(x) has a vertical asymptote ... and A(x) looks like 4x when x is large (neglecting the second term).

LECTURE 9

RELATED RATE PROBLEMS

Many times we know the rate of change of some quantity and wish to know the rate of change of a related quantity; that's a "related rates" problem. For example, if the radius of a balloon is increasing at 10 m/second, how rapidly is the volume increasing? We know and we wish to know . Identifying these rates of change is the first step. The second step is to find the relation between V and r: V = r3. Then we differentiate both sides with respect to t and get = 4π r2 , hence we can compute for any radius r, since = 10 is known. A typical problem is worded: "If the radius changes at the rate 10 m/s then find the rate of change of volume when the radius is 2 m." See? Both and r are given and we just plug them into = 4π r2 to get .

Example:

Sawdust is falling onto a pile at the rate of K metre3/second. If the pile maintains the shape of a right circular cone with its height equal to the diameter of the base, how fast is the height increasing when the pile is H metres high?

(Note: your answer will be in terms of numbers K and H.) |

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**Solution:** We identify what rate of change is requested: where h is the height (in meters). The rate of change which is given is = K m3/s. (V is the volume of the cone.) Next, we find the relation between V and h.

The volume of the conical pile (when the height is h and the radius is r)

is π r2 h (depending upon two variables). But r = , so the volume is V(h) = h3 metres3 and is changing at the rate:

= = h2 = K m3/sec. (given). Hence = m/sec. and, when h = H, we have = m/sec. (We can also check that the dimensions of this quantity are: = m/s which is okay for .)

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Example:

Water leaks from a conical tank at the rate _A_ m3/minute. If the tank is _H_ metres high and _R_ metres in radius (across the top), how rapidly is the depth of water changing when the tank is half full? (Express your answer in terms of _A_ , H and R .)

**Solution** :

Again we note that = _A_ m3/min is given and is required and the relation is again V = π r2h. But the relation between r and h is obtained |

by similar triangles: = so r = h hence V = π h3 (which has the dimensions m3 ... which is comforting). Now = π h2 = _A_ m3/min. (given) so we need to find h when the container is half full,

substitute into π h2 = _A_ and solve for . We'll interpret "half-full" to mean half the volume of the full container. But when h = H, the volume is πR2H so we want to know h when V = π h3 = πR2H. Solving we find that h3 = |

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so h = Using this h we get = m/min.

(and is in metres/minute ... which is a nice check).

PS:

S: But isn't h supposed to be decreasing? I mean, why doesn't the math give < 0?

P: Good point ... and my fault. I should have put = -A m3/min because the volume is decreasing. Of course, I can just assume that A is negative, then < 0.

S: Hmmm. Do I have to know the volume of a cone is πr2h?

P: You can look it up, if you have time (like on an assignment). On an exam, I might give you the formula.

S: Whew!

P: But, later on in this course, we'll actually derive this formula.

S: I can hardly wait.

**Example:** A rod of length 25 cm. rotates at 1000 r.p.m. (revolutions per minute) **.** Attached to the arm is a second rod of length 100 cm. with one end, P, constrained to slide horizontally along the x-axis. How rapidly is P moving at any given time? |

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**Solution** : We're given the rate of change = 1000 (2π) radians per minute (since each revolution is 2π radians). We're asked to find the rate of change: . The relation between q and x is given by the "cosine law" for triangles: 1002 = 252 \+ x2 \- 2 (x)(25) cos q = 252 \+ x2 \- 50 x cos q. Now differentiate with respect to t to get:

0 = 2x - 50 (- x sin q \+ cos q). Since is given, we can compute for any given position (meaning any x and q which satisfies 1002 = 252 \+ x2 \- 50 x cos q) from . For example, when q = 0 or π we have = 0. Similarly when q = we have = - 25 cm/sec. ... and so on for any x and q.

PS:

S: Can you check your formula with that dimensional stuff?

P: Not easily, because there are numbers in there which have dimensions. It would have been better to assume the rod lengths to be "a" and "b" (rather than 25 and 100 ) and then we'd get: which is dimensionally correct since it's: (metres) which is metres/second.

S: Whoa! I make it (metres)(radians) per second.

P: Well, radians don't really have dimensions. You just have to think of any valid formula involving an angle in radians, like a = r q (the length of arc of a circle of radius r subtending an angle q at the centre) then you'd see that q = which is dimensionless ... so we don't count it. Just count length and mass and time or anything constructed with these dimensions, like force or energy or whatever.

S: I think I'll forget about this dimensional stuff.

P: You may be interested to know that the problem we just solved has to do with a piston in a car engine. The crankshaft rotates at and P slides up and down the cylinder and ...

S: Not interested.

**Example:** A 10 m ladder leans against a vertical wall. The bottom of the ladder is pulled away from the wall at 3 m/s. How quickly is the top of the ladder sliding down the wall when the bottom is 6 m from the base of the wall?

**Solution** : Here we know = 3 m/s and we wish to determine . Pythagorus gives the relation between x and y, namely: x2 \+ y2 = 102. Differentiating this relation with respect to t gives 2 x + 2 y = 0. Now substitute the known quantities and solve for . The known quantities are x = 6, = 3 and y = = 8 so = - = - m/s. (It's negative indicating that y is decreasing.) |

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**Remember** : For related rate problems, one variable, say x, has a known rate of change. Another variable, say y, is related to x via some equation y = f(x). Then = = f'(x) gives the rate of change of y for any given x and . Sometimes the relation between x and y is in the form F(x,y) = C where C is some constant. (That is, y is given _implicitly_ in terms of x, such as x2 \+ y2 = 102 in the previous example). Then differentiate _implicitly_ to find which will (usually) be in terms of x, y and . To find , enough information must be given to compute x, y and ... hence .

**Example** : A car moves south-west along a highway described by the curve y = x2 (where the x-axis points east-west and the y-axis north-south). Its headlights illuminate a fence which lies along the x-axis. Investigate the speed with which the beam of light moves along the fence.

**Solution:** We need a reasonable diagram (we always need a reasonable diagram!) The headlight beam is tangent to the curve y = x2 and will strike the fence (i.e. will intersect the x-axis) at the x-intercept of the tangent line. So we'll pick some point of the curve (and we'll know it's on the curve if y = x2), then we'll find the equation of the tangent line at this point, then we'll find the x-intercept, then we'll find of this x-intercept. That's our method of attack.

Let the point on the curve be (x1,y1) where y1 = x12 ... and it's _this_ |

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relation which puts the point on the curve. (Remember, we have to tell the mathematics, somehow, that the point is on the curve and this is how we do it.) Then the tangent line equation is = slope of tangent line = 2x1 (since = 2x if y = x2). The tangent line intersects the x-axis when y = 0 so we solve for x = x1 \- . Plug in y1 = x12 and get the x-intercept as x = Anyway, if we take of the relation x = we get = so the speed with which the light travels along the fence is always half the speed with which the car moves west. There's nothing more we can do with this problem.

PS:

S: Wait a minute. Nothing more? Suppose the car is moving at 100 km/hour? Then how fast is the light moving?

P: That's a nice problem. Unfortunately, we have to wait until we get to "parametric equations" before we can tackle it. Remind me when we get there and we'll come back to this problem. Wait, I'm not sure we even cover parametric equations ... which would really be a shame.

S: Yeah ... a shame.

Example:

A lighthouse is located on an island 2 km from a straight shoreline. The light rotates at 3 revolutions per minute. How fast is the spot of light (at P) moving along the shoreline when P is 4 km from the lighthouse?

Solution:

x = 2 tan q hence = 2 sec2q

and = 3 x 2π = 6π radians/minute.

When P is 4 km from the lighthouse, sec q = = 2

Hence = 2 (22) 6π = km/min.

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LECTURE 10

the LINEAR (TANGENT LINE) and OTHER POLYNOMIAL APPROXIMATIONS

The TANGENT LINE APPROXIMATION

Recall that the derivative, f'(a), gives the slope of the tangent line to the curve y = f(x) at the place x = a. Further, at x = a, y = f(a) so we have a point (a,f(a)) and a slope f'(a) so we can calculate the equation of the tangent line: = f'(a) or (to put it into a better form)

as if you didn't know! Is this useful? When would we _really_ want the equation of a tangent line? In fact, the tangent line and the curve itself are very nearly the same curve so long as we don't stray too far from the point of tangency ... and this is the basis for using the tangent line to approximate f(x) for x-values near x = a.

**Note:** To use y = f(a) + f'(a) (x - a) to approximate f(x), we clearly need to compute f(a) and f'(a)! Hence, we need to choose an "a" where these _can_ be computed with relative ease.

**Example:** Compute an approximate value for .

**Solution:** Consider y = . Then, for x = 49, y = 7. Since 47 isn't too far from 49 we find the tangent line to f(x) = at x = 49 and use _it_ to compute f(47) ... approximately. The tangent line is y = f(49) + f'(49) (x - 49) and since f'(x) = then f'(49) = = and our tangent line is y = 7 + . For x-values near x = 49 we conclude that ≈ 7 + ... that is, and 7 + have approximately the same value. In particular, ≈ 7 + = 7 - = 6.8571

Above is a reasonably accurate (computer) plot of y = and its tangent line at x = 49. A few things seem clear: (1) the tangent line is indeed a good approximation near x = 49, and (2) it gets worse as we move away from the point of tangency, and (3) the tangent line will give an approximation to which is too large (since the tangent line lies _above_ the curve). In fact, = 6.8557 (to 4 decimal places). Also, (4) using the tangent line approximation (sometimes called the linear approximation because y = f(x) is approximated by a linear equation) means we move not along the curve y = f(x) but along the tangent line.

S: That looks awfully familiar! I mean, ≈ 7 - .

P: It should. We got that before, using the differential. In fact, y = f(a) + f'(a) (x - a) says that y changes by Dy = y - f(a) when x changes by Dx = x - a and if we move along the tangent line , the tangent line equation says that Dy = f'(a) Dx and that's the differential! It's just a change in notation, that's all. When we used the differential we were saying that y = f(x) changes at the rate f'(a) when x = a so a small change in x, namely Dx, would give a change in y of about f'(a) Dx which isn't exact because f'(x) isn't constant, but it's good enough for small changes in x. If the velocity is 12 metres/hour then, even if the velocity changes we'd expect to go about (12) (.1) = 1.2 metres in the next .1 hours. Nice, eh, how everything hangs together?

**Example:** Show that 1 + is an approximation to when x is near "1".

**Solution:** The tangent line approximation has the form f(a) + f'(a) (x-a) so we take f(x) = and a = 1 and get f'(x) = so that f(1) = 1 and f'(1) = . Then f(a) + f'(a) (x-a) gives 1 + . (For example, ≈ 1 + = 1.04 whereas the exact root is = 1.03923 to five decimal places.)

**Example:** Use the tangent line approximation to compute sin 47˚, approximately.

**Solution:** Fortunately, 47˚ is close to 45˚ and we can compute f(x) = sin x and f'(x) = cos x at x = 45˚. The tangent line to y = sin x is y = f(45˚) + f'(45˚) ( x - ) = + . Hence, for x near ,

sin x ≈ + . Now we'd like to put x = 47˚ but we need to convert to radians. It's easier, however, to recognize that x - is just 2 degrees (i.e. 47˚ - 45˚) which, in radians, is 2 = . Hence our approximate value for sin 47˚ is + = .7318 (while the exact value, to 4 decimal places, is .7314). Since the tangent lies above the sine curve (at x = π/4), it's not surprising that the approximation is too large. Below is a computer plot of the situation.

PS:

S: Two things: first, you used degrees instead of radians and you told me that ...

P: Hold on. I used radians where it mattered ... in the expression ) where using degrees would be disastrous. But when I plug in the values of sin for example, I can plug in the values of sin 45˚, right? They are the same numbers, right?

S: Okay, okay, but you also said the tangent line lies above the sine curve ... so the approximation will be too large. That's easy for you to say, but how am I supposed to know what lies above what? Am I supposed to plot the sine curve? Then I'd need a calculator and if I had a calculator why would I be interested in an approximate value for sin 47˚? I'd just punch it up on the calculator and ...

P: Let me ask you a question ... just to see if you've been paying attention. What is it about a curve, y = f(x), that makes it lie below its tangent line? Think about it.

S: I haven't the foggiest idea. Wait ... uh, the curve curves down. I mean, if y = f(x) is concave down then it'll be below its tangent line. Right?

P: Right on! And if it's concave up, it's above its tangent line. Now, the big question: how can you tell if a curve is concave up or down?

S: I give up.

P: The second derivative! If f''(x) > 0 then y = f(x) is concave up and lies above its tangent line ... and the linear (tangent line) approximation will be too small. If f''(x) < 0 then y = f(x) is concave down and lies below its tangent line ... and the linear approximation will be too large. For f(x) = sin x, f''(x) = - sin x which is negative near so it's concave down and our approximation will be too large. In the previous example we had f(x) = = x1/2, so f'(x) = x-1/2 and f''(i) = - x-3/2 which is also negative near x = 49 (or near any positive x for that matter) so the tangent line will always give an approximation which is too large.

S: I still don't understand why we'd want to use a tangent line approximation. Everybody owns a calculator. Why not use it and get the exact value?

P: Here's a nice use of the tangent line approximation:

**Example:** A certain amount of money is left in the bank to accumulate interest (compounded at i% per year). If you want to double your money in n years, what should the interest rate be?

**Solution:** If the amount of money is $A, then after n years it will have grown to $A (1 + )n. In order to double, we need (1 + )n = 2, or, taking _ln_ of each side, we need n _ln_ ( 1 + ) = _ln_ 2 hence

_ln_ ( 1 + ) = . Now _ln_ is a fancy function so it's not easy to solve this equation for i, but since is small we can approximate f(i) = _ln_ ( 1 + ) near i = 0 by its tangent line: f(i) ≈ f(0) + f'(0) i = _ln_ 1 + i (since f'(i) = = when i = 0). Also, _ln_ 1 = 0 so we have ≈ . Further _ln_ 2 ≈ .69 (roughly) so we get i ≈ . For example, to double in 10 years you'd have to get = 6.9% interest, approximately.

PS:

S: How accurate is that answer? I mean 6.9% ... how accurate is it?

P: The correct answer is 7.2 % (to one decimal place) ... and it's for that reason that people in the financial world use the "Rule of 72" : .

S: Well, we could have guessed that our approximation would be low because we're using the tangent line for ln ( 1 + ) and the ln curve is concave down as I recall ... I could differentiate ln ( 1 + ) twice to check, but I won't ... so the tangent line is above the curve ... so ... hey! Why isn't the approximation too large. The tangent line is above the curve!

P: This problem is different from the others. Previously we were given x and trying to evaluate y = f(x). (i.e we were given x = 47 and trying to compute ... or sin 47˚). In this problem we were given y = f(x) = ln ( 1 + ) ... namely ... and we were trying to find x! That's different! See?

S: No.

P: Let's sketch y = ln ( 1 + ) near x = 0 ... and its tangent line at x = 0. Now, given a y-value of , what's the x-value? The x-value on the tangent line is smaller than the x-value on the curve ... and that's because the tangent line is above the curve. See? The picture is worth ...

S: That's confusing.

P: Then don't think about it. |

---|---

S: Just one more thing. You call the tangent line approximation a "linear" approximation. Are there other approximations which aren't "linear"?

P: I'm glad you asked that question ...

POLYNOMIAL APPROXIMATIONS

The thing about the approximation y = f(a) + f'(a) (x - a) is that it has the same value as f(x) and the same derivative as f(x), at x = a. In fact, if we wanted a linear approximation, meaning a straight line approximation,

y = A + Bx and we wanted y to have the value f(a) when x = a we'd need A + Ba = f(a). Also, if we wanted y to have the same derivative at x = a then we'd want = B = f'(a). This gives us two equations to solve for A and B, namely and . The solution is A = f(a) - a f'(a) and B = f'(a) and the linear approximation becomes y = f(a) - a f'(a) + f'(a) x or y = f(a) + f'(a) (x - a) which is (no surprise!) the tangent line. However, when we derive it that way it's natural to ask what quadratic approximation is "best" in the sense that it matches f(x) in value and first derivative and second derivative. So we consider y = A + Bx + Cx2 and find the constants A, B and C by requiring that, at x = a, , and . After solving these 3 equations in 3 unknowns and substituting we get the quadratic approximation:

Actually, the procedure is easier if we started off by assuming a quadratic approximation in the form:

y = a0 \+ a1 (x - a) + a2 (x - a)2 (which is still a quadratic ... because the highest power of x is x2 ... but in a form which simplifies the calculations). Then, at x = a, we have and and and equating these to f(a), f'(a) and f''(a) respectively gives, immediately, a0 = f(a), a1 = f'(a) and a2 = f''(a), hence the required quadratic approximation, as above.

Note that the quadratic approximation is, in some sense, the "best" parabolic approximation to the curve whereas the tangent line is the "best" straight line approximation ... at least at x = a. We could also determine the "best" cubic approximation or quartic approximation, etc. etc. In fact, since a cubic y = A + Bx + Cx2 \+ Dx3 ... or, better, y = a0 \+ a1 (x - a) + a2 (x - a)2 \+ a3 (x - a)3 ... has precisely **4** constants, we can insist upon **4** conditions (yielding 4 equations to solve for the **4** unknown constants). The **4** conditions are that, at x = a (meaning any specified value of x), the values of y, , and must agree with f(a), f'(a), f''(a) and f'''(a). We'd get the same values for a0, a1 and a2 as we did for the quadratic approximation ... and a3 would be f'''(a).

As with the linear approximation, we usually pick "a" at a place where f(a), f'(a), f''(a) and f'''(a) aren't difficult to compute.

**Example:** Determine a quadratic approximation to at x = 49.

**Solution:** Note that a = 49 is someplace where we _can_ easily evaluate f(x) = x1/2, f'(x) = x-1/2 and

f''(x) = - x-3/2. We get f(49) = , f'(49) = - , f''(49) = so our quadratic approximation (at x = 49) becomes y = 7 + - 2 = 7 + - 2 . To see how good it is, we compute an approximation to as 7 + - 2 = 6.85561 (to 5 decimal places). The exact value (also to 5 decimal places) is = 6.85566 and the linear, tangent line approximation gives 6.85714 (to 5 decimal places). This is illustrated in the graph below:

S: The tangent line approximation is just the first two terms of the quadratic approximation. Is that always the case? I mean, to get the quadratic approximation we just added another term. To get the cubic approximation, do we add yet another term?

P: Sure, and remember the format of these polynomial approximations:

POLYNOMIAL APPROXIMATIONS

linear: y = f(a) + f'(a) (x - a)
quadratic: y = f(a) + f'(a) (x - a) + f''(a) (x-a)2

cubic: y = f(a) + f'(a) (x - a) + f''(a) (x-a)2 + f'''(a) (x-a)3

S: Hold on. I see how it goes ... the fourth degree approximation would get another term ... uh, something with

f ''''(a) (x-a)4 ... but what's the number out front? There's then comes ... then what?

P: Write y = a0 \+ a1 (x - a) + a2 (x - a)2 \+ a3 (x \- a)3 \+ a4 (x - a)4. Then differentiate four times and put x = a and notice that everything differentiates to zero except the term a4 (x - a)4 which becomes (4)(3)(2)(1) a4 so = 4! a4 and this must equal f ''''(a) hence a4 = f ''''(a). The number out front, as you put it, is = . Now it's easy to see what's the polynomial approximation of degree 5 or 6 or whatever.

**Example:** Determine an approximate value for sin 47˚ using a quadratic approximation.

**Solution:** We consider f(x) = sin x and determine the quadratic approximation at x = 45˚ (since we know all about the sine function and its derivatives at this x = a). We have f(x) = sin x = and f '(x) = cos x = and

f ''(x) = - sin x = - . Our quadratic approximation is then

y = f(a) + f'(a) (x - a) + f''(a) (x-a)2 with a = and this gives: y = + - 2 (where we were careful to substitute a = 45˚ in radians in (x - a) and (x -a)2 etc.) Now, to get an approximation to sin 47˚ we need to express x - in radians; that is, 2˚ in radians which is 2 = so finally we get:

sin 47˚ ≈ + - 2 = .731359 (to 6 decimal places). To 6 dec. places, the exact value is .731354 and the linear approximation, + , is .731789 (so we can appreciate the improvement in going to the quadratic approximation, and, as you'd expect, the cubic approximation is even better).

PS:

P: Did you notice anything interesting about the formulas for the linear and quadratic approximations?

S: Nope.

P: Let me write them out again: y = f(a) + f'(a) (x - a) versus y = f(a) + f'(a) (x - a) + f''(a) (x - a)2. Remember? If the curve y = f(x) lies below its tangent line, at x = a, then the linear approximation is too large ... and we can predict that by looking at the sign of the second derivative, f''(a). Now do you see?

S: Uh ... not really.

P: When f''(a) < 0 then y = f(x) is concave down and the curve lies below its tangent line so the linear approximation is too large ... so we should really subtract something from the linear approximation ... and the quadratic approximation does just that! It adds f''(a) (x-a)2 to the linear approximation which, if f''(a) < 0, really means it's subtracting something. Neat, eh?

S: Aaah ... mathematics is wonderful. Let's do some more ... something really useful ... if there is anything useful about this stuff ... my calculator can do all this ... I don't know why we're studying this ...

**Example:** Solve the equation _ln_ (1 + ) = for i, using a quadratic approximation for the logarithm.

**Solution:** We have f(i) = _ln_ (1 \+ ) so f'(i) = = and f''(i) = - and, evaluating at i = 0 (we choose this value since it's easy to compute f(0), f'(0), etc.) we have f(0) = _ln_ 1 = 0, f'(0) = and f''(0) = - hence our quadratic approximation is y = f(0) + f'(0) (i - 0) + f''(0) (i - 0)2 = i + i2 = - 2. Then we solve - 2 = which is a quadratic equation to solve for i. (This is no surprise since we're using a quadratic approximation for f(x)!) To solve, it's easier to let

x = then solve for x. We rearrange the equation, rewriting it in the form: x2 \- 2 x + = 0 and use the magic formula for the roots of a quadratic: to get x = 1 ± = 1.9281 or .0719 and we pick the smaller root and conclude that ≈ .0719 so i = 7.19% (which gives the "Rule of 72").

  | If we look at the graph of y = _ln_ (1 \+ x) and the quadratic (i.e. parabolic) approximation, we see that it's excellent near x = 0 and becomes worse as we move away from the point of approximation. Further, the quadratic approximation is too small for x > 0 and too large for x < 0 (although it's hard to tell from the diagram).

Also, the curve y = _ln_ (1 + x) has a vertical asymptote at x = -1 (since _ln_ (1 + x) -> -∞) and we wouldn't expect the parabola y = x - to be any good as an approximation anywhere near x = -1. Even a cubic or quartic approximation won't have a vertical asymptote at x = -1 hence can't provide good approximations there.

PS:

S: Why did you discard the root 1.9281?

P: Instead of solving ln (1 + x) = which has

---|---

exactly one solution for x (hence for i) ... because it's an increasing function ... we solved x - = and this equation has two solutions (as most quadratic equations do!). In fact, looking at the graph above we see that, although y = ln (1 + x) has the value only once, y = x - has this value twice ... since the parabola goes up then comes back down. It's the first one that we want ... the one nearest to the point of approximation ... the one nearest x = 0. One thing I should mention: although the graphs were plotted on a computer and show the beautiful quadratic approximation, the horizontal dotted line is NOT y = ≈ .069 (since I wanted to illustrate the two intersections with the quadratic, and it was clearer with a line something like y = .4). Anyway, as you can plainly see, the picture is worth ... well ... you know.

**Problem:** Verify each of the polynomial approximations illustrated above.

Note: For the first, f(x) = _ln_ x, the polynomial is constructed about x = 1. Why can't we find a polynomial (of the type we've been constructing) about x = 0?

PS:

S: You've been calling these the "best" quadratic and the "best" cubic approximations, etc. Aren't they the "best"? And if so, why do you put "best" in quotes. It's as though you're not convinced ... or not sure, or something.

P: Okay, let's consider something simple, say, the "best" linear approximation to f(x) = x3 where we'll take the approximation about x = 0. Then f(0) = 0 and f'(0) = 0 as well and our "best" linear approximation is

y = f(0) + f'(0) x = 0 (using f(a) + f'(a) (x - a) with a = 0). Hence our "best" straight-line approximation is the x-axis itself! If we look at the graph, it's easy to see that when we require our "best" line to match the value f(0) and the derivative f'(0), it gives a pretty poor approximation. In fact, y = x looks quite a bit better (as a straight-line, "linear" approximation). On the interval -1 ≤ x ≤ 1, for example, our "best" (tangent line) approximation, y = 0, has a maximum error of "1" whereas the maximum error using the approximation y = x is only ... uh, let's see ... the error is the difference between y = x and y = x3 and that's | x - x3| and, on 0≤x≤1 we can delete the absolute value sign because x - x3 ≥ 0, so this error has a maximum value, on 0 ≤ x ≤ 1 either at a critical point in 0 < x < 1, or at the end-points x = 0 or x = 1. The critical points are where = 1 - 3x2 = 0, meaning x = so the maximum is - ≈ .38 which is better than "1". See?

S: See what?

P: The so-called "best" linear approximation for f(x) = x3 (using the method we've been describing ... that is, matching the value and derivatives of f(x) at x = 0, and so on) doesn't do as well as y = x. In fact, there are even better approximations than y = x . Of course, it all depends upon what one means by "better", doesn't it?

S: If you say so.

P: You might consider the line y = mx and try to vary the slope "m" and see which gives the smallest maximum error. It's another meaning for the word "best" ... and it's fun.

S: If you say so. |

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LECTURE 11

NEWTON'S METHOD for finding roots of f(x) = 0

In earlier lectures we've run into a variety of problems which require finding roots of equations of the form: f(x) = 0. For example, we've considered 99 x4 \- 594 x3 \+ 1290 x2 \+ 6x - 9 = 0, and which is a root of the equation x2 \- 47 = 0, and _ln_ (1 + x) = which can also be written in the form f(x) = 0 with

f(x) = _ln_ (1 + x) - . Now we use the linear tangent line approximation in a special way, called Newton's Method, to find such roots to as many decimal places as we wish.

First note that the curve y = f(x) crosses the x-axis at an x-value which satisfies f(x) = 0, so finding a root is the same as finding an x-intercept for such a curve. Suppose we sketch, or plot, y = f(x) and find that it changes sign between, say x = a and x = b. That is, f(a) has a different sign than f(b); for example, in the diagram, f(a) < 0 and f(b) > 0. Then there is a root lying in a < x < b. Now we pick any number x1 in a ≤ x ≤ b (it could be either "a" or "b", or a number in between as shown in the diagram) and construct the tangent line to y = f(x) at x = x1. This tangent line will intersect the x-axis at some x-value, say x2, and this will be closer to the root |

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of f(x) = 0 than was x1. In other words, if x1 is an approximation to the root, then x2 is a better approximation.

So what's x2?

The tangent line at x = x1 is: y = f(x1) + f'(x1) (x - x1) and this crosses the x-axis at x = x2 where

y = 0 = f(x1) + f'(x1) (x2 \- x1) and, solving, we get: . If x2 is better than x1 (as an approximation), then we can repeat the procedure, finding the tangent line at x = x2 and determining x3, where _it_ crosses the x-axis. There's no need to repeat the calculation; it's similar to that found above: . We can now repeatedly use the same prescription, generating a sequence of number x2, x3, x4, ..., xn, ... which get closer and closer to the root of f(x) = 0 ... or, to use a more apt notation: xn = c

  | From the diagram at the left you can watch the sequence of approximations marching toward the root of f(x) = 0, at x = c. Just remember that x1 is some initial approximation which we generate by guessing or by plotting or by divine insight. The rest of the approximations come from Newton's formula:

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**Example:** Compute , using Newton's Method.

**Solution:** We invent the function f(x) = x2 \- 47 so that the positive root of f(x) = 0 will give us . Note that we don't invent the function f(x) = x - since we can't evaluate f(x) if we don't know !! (We want to "evaluate" f(x) and f'(x) using only addition, subtraction, multiplication and division; think of doing all this on a $3.95 calculator which _doesn't_ have a square root button.) Then, to compute the Newton _iterates_ (that's what the x1, x2, etc. are called) we guess at a root, say x1 = 7 (which seems a reasonable guess) and plug this x-value into x - = x - = where we've simplified the expression somewhat. Then

x2 = = = 6.857142858 and now we plug this x2 into the same expression giving

x3 = = = 6.855654762 and we now we plug this x3 into the same expression giving

x4 = = = 6.855654601 and we now we plug this x4 into the same expression giving

x5 = = = 6.855654601 and if we continue we'll get x6 = 6.855654601 as well. We're finished. To 9 decimal places, _that's_ .

Of course, if we carried more decimal places we could go and on, getting better and better approximations ... and many more decimal places ... and the error would approach zero. In fact, it's very instructive to actually watch the error go to zero. To do this we'll begin again with x1 = 7 but now we'll carry 75 digits.

The numbers which follow were computed using  : a computer algebra system from U. of Waterloo. (My $4.95 calculator can't do 75 digits). When we have need of lots of digits, we'll let   do it.

In what follows, we first ask for 75 digits (Digits:=75;) then ask   to **evalf** (meaning **eval** uate as a **f** loating point, or decimal, number) the exact square root (and we'll call it "root"), then we'll define

y = (x + 1/x)/2 and we'll "iterate", computing x1, x2, etc. and each time we'll compute an error = x2 \- root, etc.

_______________________________________________________________________________

• Digits:=75;

Digits := 75

_______________________________________________________________________________

• root:=evalf(sqrt(47));

root := 6.85565460040104412493587144908484896046064346100132627548510818567851711514

_______________________________________________________________________________

• y:=.5*(x+47/x);

y := .5 x + 23.5 1/x

_______________________________________________________________________________

• x2:=subs(x=7,y);

x2 := 6.85714285714285714285714285714285714285714285714285714285714285714285714286

_______________________________________________________________________________

• error:=x2-root;

error := .00148825674181301792127140805800818239649939614153086737203467146434002772

_______________________________________________________________________________

• x3:=subs(x=x2,y);

x3 := 6.85565476190476190476190476190476190476190476190476190476190476190476190476

_______________________________________________________________________________

• error:=x3-root;

-6

error := .16150371777982603331281991294430126130090343562927679657622624478962*10

_______________________________________________________________________________

• x4:=subs(x=x3,y);

x4 := 6.85565460040104602726699535902922054430628339896844346594726451466195334222

_______________________________________________________________________________

• error:=x4-root;

-14

error := .190233112390994437158384563993796711719046215632898343622708*10

_______________________________________________________________________________

• x5:=subs(x=x4,y);

x5 := 6.85565460040104412493587144908511289322242413572750032746808991073007441445

_______________________________________________________________________________

• error:=x5-root;

-30

error := .26393276178067472617405198298172505155729931*10

_______________________________________________________________________________

• x6:=subs(x=x5,y);

x6 := 6.85565460040104412493587144908484896046064346100132627548510819075903143518

_______________________________________________________________________________

• error:=x6-root;

-62

error := .508051432004*10

_______________________________________________________________________________

• x7:=subs(x=x6,y);

x7 := 6.85565460040104412493587144908484896046064346100132627548510818567851711514

_______________________________________________________________________________

• error:=x7-root;

error := 0

_______________________________________________________________________________

Did you see? The error goes to zero! (Well ... at least "zero", to 75 digits). Remarkable! The errors are something like 10-3, 10-7, 10-15, 10-31, 10-62 then something less than 10-75 (in fact, probably about 10-120). In fact, it's this remarkable rapidity with which Newton's Method gives roots that endears it to many. Once an iterate finds itself near a root, the remaining iterates march to the root with unerring accuracy. Each successive error is roughly the _square_ of the preceding error!

**Example:** Compute a root of 99 x4 \- 594 x3 \+ 1290 x2 \+ 6x - 9 = 0 (to 5 decimal places).

Solution:

This problem arose (earlier) in connection with finding the path, from P to Q, which takes the minimum time. Point Q is on land and P is in water and swimming speed is 1/10 of running speed.

For f(x) = 99 x4 \- 594 x3 \+ 1290 x2 \+ 6x - 9 we have the iteration scheme based upon:

x - = x - . Come to think of it, maybe I'll let   do it. We'll define y = f(x), ask |

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  to compute y', construct the iteration equation x - y/y', substitute various values of x into y ... until we find two values where y changes sign, then we'll pick one of these two x-values as x1 ... and iterate, until the iterates are identical to, say, 6 digits. Note: in the following excerpt from a session with  , the curious command "evalf" means evaluate as a floating point (i.e. decimal) number.

_________________________________________________________________________________

• Digits:=6;

Digits := 6

_______________________________________________________________________________

• y:=99*x^4 - 594*x^3 + 1290*x^2 + 6*x - 9;

4 3 2

y := 99 x - 594 x + 1290 x \+ 6 x - 9

_______________________________________________________________________________

• Dy:=diff(y,x);

3 2

Dy := 396 x - 1782 x + 2580 x + 6

_______________________________________________________________________________

• iterate:=x - y/Dy;

4 3 2

99 x - 594 x + 1290 x + 6 x - 9

iterate := x - \------------------------------------

3 2

396 x - 1782 x + 2580 x + 6

_______________________________________________________________________________

• subs(x=0,y);

-9

_______________________________________________________________________________

• subs(x=1,y);

792

_______________________________________________________________________________

• x1:=0;

x1 := 0

_______________________________________________________________________________

• x2:=evalf(subs(x=x1,iterate));

x2 := 1.50000

_______________________________________________________________________________

• x3:=evalf(subs(x=x2,iterate));

x3 := .33712

_______________________________________________________________________________

• x4:=evalf(subs(x=x3,iterate));

x4 := .165492

_______________________________________________________________________________

• x5:=evalf(subs(x=x4,iterate));

x5 := .101483

_______________________________________________________________________________

• x6:=evalf(subs(x=x5,iterate));

x6 := .0843389

_______________________________________________________________________________

• x7:=evalf(subs(x=x6,iterate));

x7 := .0827736

_______________________________________________________________________________

• x8:=evalf(subs(x=x7,iterate));

x8 := .0827600

_______________________________________________________________________________

• x9:=evalf(subs(x=x8,iterate));

x9 := .0827601

_______________________________________________________________________________

• x10:=evalf(subs(x=x9,iterate));

x10 := .0827601

_______________________________________________________________________________

We conclude that a root of f(x) = 99 x4 \- 594 x3 \+ 1290 x2 \+ 6x - 9 = 0 (to 5 decimal places) is .08276 ...

PS:

S: Wait! How come it took so long ... and you only asked for 6 digits?

P: Remember, we're approximating y = f(x) by its tangent line and actually finding where this tangent line intersects the x-axis. Maybe the tangent line isn't a good approximation to y = f(x), for this particular f(x).

S: How could we check that?

P: Well, if the slope of f(x) doesn't change too rapidly near x = .0827601, then we'd expect the tangent line to be pretty good. After all, the tangent line has a constant slope but y = f(x) doesn't. When we wanted to compute y = x2 \- 47 near x = , the Newton iterates quickly got us to a root. For y = x2 \- 47 , = 2x and = 2 which isn't so large so isn't changing too rapidly. Let's check y = 99 x4 \- 594 x3 \+ 1290 x2 \+ 6x - 9 near the root we just found. gives:

_______________________________________________________________________________

• y:=99*x^4 - 594*x^3 + 1290*x^2 + 6*x - 9;

4 3 2

y := 99 x - 594 x + 1290 x \+ 6 x - 9

_______________________________________________________________________________

• Dy:=diff(y,x);

3 2

Dy := 396 x - 1782 x + 2580 x + 6

_______________________________________________________________________________

• D2y:=diff(Dy,x);

2

D2y := 1188 x - 3564 x + 2580

_______________________________________________________________________________

• evalf(subs(x=.0827601,D2y));

2293.179894

_______________________________________________________________________________

so the second derivative is quite large (and the curve bends rapidly away from its tangent line ... so the tangent line isn't a very good approximation near x = .0827601).

S: Are there any other roots of f(x) = 99 x4 \- 594 x3 \+ 1290 x2 \+ 6x - 9 = 0?

P: I don't know. Shall we plot it and see? We can just evaluate f(x) for a whole bunch of x-values and see if it changes sign anywhere, then we can use Newton's method again, starting with an initial guess, x1, nearby and ..

S: Let's forget it ... except ... I almost hate to ask, but will I ever really have to find the root of 47 or where to swim to get to the cottage or ...

P: Okay, that's a fair question: "When would anyone ever need Newton's Method, outside of a calculus course?"

**Example:** You invest $10,000 in a mutual fund, then, 5 months later you put an additional $15,000 into the fund, then, 3 months later put in an additional $5,000. At the end of a year, your investments (totalling $30,000) have grown to $31,470. What is the annual rate of return from this mutual fund?

**Solution:** Let the monthly rate of return be i . For example, if i =.01 it means a 1% return on your investment per month so each dollar will grow to (1.01)n after n months. The first $10K invested has been in the fund for 12 months, so will grow to 10(1+i)12 (measured in kilobucks). The next $15K grows to 15(1+i)7, having been in the fund for 7 months. The last $5 will grow to 5(1+i)4, having been in the fund for 4 months. The total value of your investments, after 12 months, is 10(1+i)12 \+ 15(1+i)7 \+ 5(1+i)4 = 31.47 kilobucks. We must solve this equation for i.

We let 1+i = x and rewrite the equation in the standard format: and use the iteration scheme based upon x - = x - . If we guess at a monthly return of 1% (we need an initial guess!), then we can use i = .01 hence x1 = 1+i = 1.01 is our initial guess. Iterating, we get ...

S: Wait! How many iterations will it take? Guess!

P: Well, it depends upon how rapidly the slope of f(x) changes and that depends upon f''(x) and

f''(x) = 1320x10+630x5+60x2 and, for x = 1.01 that gives ... did you bring your calculator?

S: Just stick in x = 1, that's close enough ... and that gives f''= 1320+630+60 = 2010 and that's pretty big, right?

P: You're getting very clever. Yes, it's large, so I'd say about 6 or 7 iterations. Of course, it depends upon how close x1 is to the root of f(x) = 0. Anyway, let's proceed ... in fact, let's use a computer spread sheet where I've programmed:

x2 = x1 \- , x3 = x2 \- and so on.

In what follows, I type in the numbers shown in boldface and the spreadsheet does the rest. I also have the spreadsheet compute (for each iteration x1, x2, etc.) the equivalent Annual Rate, as a percentage , namely 100(x12 \- 1):

S: Aha! You only needed two iterations!

P: Amazing! Now you can see how clever Newton is ... uh, was. However, we can also estimate the root of f(x) = 0 by using a quadratic approximation for f(x). Want to try it?

S: Why not.

We'll use a quadratic approximation for f(x) =10x12 \+ 15x7 \+ 5x4 \- 31.47, about x = 1 (since we expect x = 1+i to be near 1).

We have f(x) ≈ f(1) + f'(1) (x-1) + f''(1) = -1.47 + 245 (x-1) + 1005 (x-1)2. Instead of solving

f(x) = 0, we solve the quadratic equation 1005 (x-1)2 \+ 245 (x-1) - 1.47 = 0 or putting x - 1 = i, we solve

1005 i2 \+ 245 i - 1.47 = 0. Using the world famous formula i = gives i = .00586 where we took the positive square root because we clearly want i positive! This compares quite favorably with the value obtained via Newton's method.

S: Favorably? It's right on the button!

P: Well, to 5 decimal places. To 7 decimal places the exact root (using Newton's method) is i = .0058570 whereas the quadratic approximation gives i = .0058592 which is pretty good, eh what?

S: Yeah. One other thing ... does Newton's method always work?

P: I'm glad you asked that question:

**DIFFICULTIES WITH NEWTON'S METHOD** **:**

If we begin our iterations with x1 far from a root of f(x) = 0, we can't guarantee that the iterates x2, x3, etc. will march to the root. Since we have a nice geometrical picture of what the method is doing (i.e. repeatedly moving along tangent lines to the x-axis) we can see what happens in various cases:

  |

Suppose f(x) has two roots, x = p and x = q, and we begin at x = x1, too far to the right of q (which, we assume, is the root we want). It's possible that the tangent line at x = x1 will intersect the x-axis nearer to x = p and the iterates will have a limit of p (instead of q). That's not too bad ... at least we get a root.

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  |

Worse still, we might pick an x1 where the tangent line is horizontal ... then it _never_ intersects the x-axis! (In this case, x2 = x1 \- involves a division by f'(x1) which is zero.)

  |

Also, we might pick x1 so that the tangent line intersects the x-axis (at x2) too far to the right of q and the next tangent line takes us a mile or two along the negative x-axis ... and who knows what f(x) looks like there?

  |

Also, we might find that x3 is identical to x1 and the iterates just repeat: x1, x2, x1, x2, etc. etc. not having any limiting value at all.

As you might imagine, there are other weird things that can happen.

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The moral? Pick a reasonable value for x1 before you start iterating!

Before we leave Newton's method, let's look once again at the iterative procedure for finding a square root. To find we invent f(x) = x2 \- N and, to solve f(x) = 0, we iterate using x - = x - = and some reasonable first iterate, x1. (i.e. we use xn+1 = for n = 1, 2, 3, 4, ... etc.). But look at how clever this scheme is! If x1 is smaller than , then will be larger than so Newton's method picks, as x2, the average of x1 and ! In fact, at each stage of our iterations, xn and will be on either side of so we actually have the root in some interval which (hopefully) gets smaller and smaller. Very nice. This technique makes such sense that it's used by people who have never heard of Newton!

**Examples:** (a) Solve x3 \- x2 \+ x + 22 = 0.

(b) Solve x _ln_ x = 6

**Solutions:** (a) Using xn+1 = xn \- with f(x) = x3-x2+x+22 we get: xn+1 = xn \- . The only root is slightly

larger than 3, so we use x1=3 and get x2=3.045789, x3=3.044724, x4=3.044723, x5=3.044723 and we conclude that the root is 3.04472 to five dec. places.

(b) With f(x) = x _ln_ x - 6 we get:

xn+1 = xn \- . There is one root between 4 and 5. If we use x1=4.3 we get x2=4.189351, x3=4.188760, x4=4.188760 and we

conclude that the root is 4.18876 to five dec. places. |

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LECTURE 12

L'HÔPITAL'S RULE for evaluating limits of the form

In an earlier lecture we investigated various LIMIT RULES (such as lim (f + g) = lim(f) + lim(g) and so on) which allowed us to avoid computing limits by resorting to the _definition_ of "limit". Often these rules were of no value, such as would be the case for the rule: , in the case where = 0. However, if

f = L (a number different from zero) and g = 0 we agreed to use the notation = ∞ (perhaps +∞ or -∞ depending upon the sign of L and whether g(x) = 0 from the right or from the left). One of the most irritating limits is the case where g(x) = 0 AND f = 0 ... then we say that the ratio has the indeterminate form . It's this form which we want to consider now.

To begin, let's consider f(x) = x2 \- 4 and g(x) = x3 \- 8 and which has the form. (I know we can factor both numerator and denominator, cancel the factor (x - 2) and then use the limit rule ... but we won't because we want to generate another technique.) We rewrite the ratio in the form which doesn't change anything because f(2) = 0 and g(2) = 0 and we've divided both numerator and denominator by (x - 2) which isn't zero if x->2 (because we want x near 2, but not equal to 2). Now note that = f'(2) and

= g'(2) so NOW the numerator and denominator _have_ limits different from zero (i.e. we've eliminated the dreaded form), and we get = = []x=2 = , where the convenient notation [ ]x=2 means "evaluated at x = 2".

This prescription for evaluating limits of the indeterminate form is called:

l'HÔPITAL'S RULE for

If f(x) = 0 and g(x) = 0 then = provided this limit exists

Of course, in order for l'Hopital's rule* to be valid, the functions f(x) and g(x) must _have_ derivatives near x = a (else the limit of their ratio has no meaning).

Note: The form is precisely the form of the difference ratio that defines the derivative: .

If we apply l'Hopital's rule we get = = = f'(a) as expected!

**Examples:** Evaluate the following limits

(a) (b) (c) x2e-x

Solutions:

(a) = = -1 (where we differentiated numerator and denominator).

(b) == = where the first application of l'Hopital's rule yielded a limit still in the form , so we used l'Hopital again!

(c) x2e-x = =

= x3 e-x (rearranging a bit) and we wind up with an expression which is _worse_ than the one we started with! If we continue in this manner, the limits will NOT get easier to evaluate ... they'll get worse ... and that brings us to the second form, , which, miraculously, also succumbs to l'Hopital's rule:

l'HÔPITAL'S RULE for

If f(x) = ∞ and g(x) = ∞ then = provided this limit exists

As you might imagine, the functions must actually _have_ derivatives in order to use this rule.

Now we write x2e-x = = =

= 0.

P: Do you recognize x2e-x = 0?

S: Nope.

P: We talked about it when I mentioned the explosive growth of ex. In fact, I think I said that has a limit of zero because no matter how hard x1000 tries to drag the fraction to ∞, the ex in the denominator ...

S: ... drags it to zero. Yeah, I remember now.

P: Well, now we can see that: = =

= = ... etc. etc. = = = 0.

See? We keep differentiating numerator and denominator until we no longer have the form , and ex just waits patiently until x1000 is differentiated to a constant, then it drags the fraction to zero.

S: Are you going to prove l'Hopital's rule?

P: No, just trust me; it works.

**Interpretation of a Limiting Value** **:**

Having evaluated a limit, say f(x) = L, then we know that | f(x) - L | is very small when x is close to "a". In fact, that sometimes says something quite interesting about f(x) in cases where f(x) = , a ratio of functions (which is what we've been considering here). If -> L, then is "close to" L (when x is near "a").

Consider, for example, the by-now-familiar = 1. It says that is very nearly "1" when x is close to zero, so y = sin x and y = x have nearly the same values for small values of x. (We pointed out, earlier, that choosing x = .0123 we find that sin (.0123) = .01229969). In a sense, we're comparing the values of sin x and x.

We can do the same for = -1 so that the values of 1 - ex are roughly the negative of the values of x (when x is small, of course). That is: 1 - ex ≈ -x for small x, hence ex ≈ 1 + x.

Similarly = tells us that 1 - cos x ≈ hence cos x ≈ 1 - when x is small.

Similarly, we might ask: _If sin x is "close to" x (when x is small), then how about sin x - x? What's_ _it_ _"close to"?_ We might try to compare sin x - x with, say, x2 by evaluating =

= = 0, which says that sin x - x is _very much smaller_ than x2 (when x is very small) since the ratio of sin x - x and x2 approaches zero. Let's then compare sin x - x with something smaller than x2 ... say x3 (which is certainly smaller than x2 when x is small). We have = = = = - . We now have that sin x - x is "very close" to - x3 (for small x), or, sin x ≈ x - and that's the cubic approximation we considered earlier. In fact, 1 - is the quadratic approximation to cos x (at x = 0) and 1 + x is the linear (tangent line) approximation to ex (at x = 0).

S: Why are you always writing this is "close to" that? Why the quotes? Are they close, or aren't they?

P: Good question! Now I have one for you: how do you measure "closeness". I mean, if I gave you two numbers, what would you do to see if one was "close to" the other?

S: I'd just look at them ... what else?

P: No. Give me a prescription which I could place in a "Manual on Closeness" ... some algorithm or procedure which anyone could follow. You can't say "when you have two numbers you just look at them". I see you don't know what I mean. Okay, I have in my pocket two numbers. Their difference is .00001 and I'd like to know it they're "close" in value.

S: Of course they are.

P: The two numbers happen to be .00002 and .00001, the first being 100% larger than the second, yet their difference is .00001, so do you still think these two numbers are "close"?

S: Huh?

P: The distance to the sun is 150,000,000 km and to an asteroid is 149,500,000 km. Are these two distances "close"?

S: I'd say so.

P: Yet their difference is 500,000 which is a pretty large. Funny eh? .00002 is NOT "close to" .00001, yet their difference is only .00001, while 150,000,000 is "close to" 149,500,000 even though their difference is 500,000. The thing which makes one number "close to" another is that their ratio should be close to 1, not their difference close to 0. It's a natural way to compare, just as we're doing when we compare p(x) to q(x) via the ratio . The ratio is 2 (not very close to 1) whereas = 1.003 so the latter numbers are close to one another ... in this special sense of the phrase "close to".

S: Okay, so when you say cos x is "close to" 1 - then you're really saying that their ratio is close to 1, right?

P: Right.

S: Hah! Gotcha! Their difference is also close to zero! See? cos x - (1 - ) has a limit of 0 as x->0, so their difference is close to 0!

P: That's just an accident. I mean, it doesn't always happen that way. Sometimes ->1 yet p(x) - q(x) doesn't ->0. See? One has a tendency to say ≈ 1 hence p ≈ q hence p - q ≈ 0 and that could be wrong!

S: Prove it.

P: Let's see ... uh, yes ... let's try it with p(x) = + and q(x) = . Then = = 1+x ->1 as x->0 hence p is "close to" q when x is small (in our special meaning of the phrase "close to") so we might think that p(x) - q(x) ->0, but we'd be wrong since p(x) - q(x) = which certainly doesn't approach zero as x->0! It's like the sun and the asteroid; because their ratio is very close to 1 doesn't mean their difference is small. Got it?

S: Got it. However, before you go on, tell me why you said, earlier, that ... to use your exact words:

= tells us that 1 - cos x ≈ hence cos x ≈ 1 \- when x is small.

You just said you couldn't do this, didn't you?

P: You're right. Mia culpa. I apologize ... BUT the statement is still true, even though the reasoning is fallacious.

S: Can I do that on an exam? I mean, get the right result with the wrong reasoning?

P: Sure ... but you won't get any marks for it. But hold on, let me do it properly so you can see why the final statement is true. First, cos x ≈ 1 - means that cos x - -> 0 (as x->0), so that's what we must prove, starting with the known result = which we got using l'Hopital's rule. Now = means (according to the definition of "limit") that, for any choice of error e we can make < e by restricting x to lie in some small interval about x = 0, say 0 < | x | < h. Hence - e < \- < e and we can reorganize this to read:

- ex2 < cos x - < ex2. Now let x->0 and get (using the ol' SQUEEZE theorem):

≤ ≤ and since the outside limits are 0, then

= 0 as well.

S: Do I have to know this for the final exam?

P: No. I just thought you might be interested.

S: Wrong!

**Example:** Evaluate .

**Solution:** We must reorganize the expression - , so it has the form , so we rewrite it as . Using l'Hopital's rule twice we get:

= = = = 0.

Note: this says that not only is sin x ≈ x but also ≈ .

An interesting question: If p(x) - q(x) -> 0, does - -> 0? Answer? Sometimes, but not always. See if you can find a simple example where, say, = 0, yet ≠ 0

LECTURE 13

POLAR COORDINATES

We usually describe the location of points on a plane by giving the distance left-right and distance up-down from some origin. Sometimes we think of it as distance east-west and north-south. Perhaps we think of latitude and longitude. In any case, they are "rectangular" or "Cartesian" coordinates for the point ... but they aren't the only way to describe the location of a point and, indeed, sometimes rectangular coordinates are a terrible choice.

For example, if you're in a desert and were describing how to get somewhere you wouldn't say "go 3 km east then 4 km north" (using rectangular coordinates). You're more likely to say something like "Go 5 miles north-east", giving a distance and a direction. These are POLAR COORDINATES. The distance from the origin is called r and the direction is specified by the angle q measured (in RADIANS!) counterclockwise from the positive x-direction as shown ===>>>

It's clear from the diagram that there is a simple relationship between the rectangular coordinates of the point P, namely (x,y), and the polar coordinates (r,q). These are: |

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x = r cos q, y = r sin q and r2 = x2 \+ y2, tan q =

Before we continue with polar coordinates, let's digress to consider yet _another_ coordinate system (just so you don't think these are the only ones). Whereas (x,y) provides 2 distances and (r,q) gives a distance and an angle, we might consider a coordinate system which describes the location of a point in the plane using two angles. We pick two origins (why not?) and give two angles q1 and q2 as shown.

There are major problems with this (q1, q2) coordinate system. If q1 > q2 , then the two rays emanating from the origins don't intersect anywhere. Further, what are the coordinates of the points P1, P2 and P3 shown? They are P1(π,π) and P2(0,π) and |

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P3(0,0) but they aren't the only points with these coordinates! In fact every point between the two origins has the same coordinates, namely (0,π). Because of this we won't spend any more time on this coordinate system!!

Back to polar coordinates:

In identifying a point (r,q) it is convenient to think of "polar" graph paper. Note that "rectangular" graph paper has the curves x = constant and y = constant drawn for you (such as x = -2, x = -1, x = 0 etc.) whereas polar graph paper has the curves r = constant and q = constant drawn ==>>

Until you are accustomed to thinking in polar coordinates, it is convenient to convert polar equations (such as r = 2 or q = π/4) into rectangular coordinates: r = 2 becomes x2 \+ y2 = 22, a circle of radius 2 and centre the origin. q = π/4 becomes tan q = 1 = y/x so y = x (a line through the origin with slope 1). |

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The diagram illustrates various points ===>

If the circles shown are described by r = 1, r = 2, etc. then p1 has polar coordinates (4,π/4) and p2(3,π/6), p3(4,5π/6), p4(4,5π/4), p5(3,4π/3) and p6(2,7π/4). However, every point in the plane has a whole host of polar coordinates! The point p6 can also be described by r = 2 and q = - π/4 (with negative q corresponding to clockwise measurements of the angle). We also have p1(4,9π/4) where the angle q corresponds to a complete revolution of 2π plus an additional π/4. Indeed, p1 can be described by r = 2 and q = 2πn + π/4 for any integer "n". But that's not all. If we regard negative q as meaning "move opposite to the positive direction" then we can also consider negative |

values for r, also meaning "move opposite to the positive direction". The positive direction is in the direction of q so a negative r means move in the direction opposite to this. That gives to p1 (for example) the polar coordinates r = - 4 and q = π/4 + π = 5π/4. In spite of the plethora of polar coordinates for each point in the plane, we usually identify a point with a positive r and an angle (in RADIANS!) in the interval 0 ≤ q < 2π.

Now for some more interesting polar curves:

Whereas the equation y = x (in rectangular coordinates) describes a straight line through the origin with slope 1, r = q describes quite a different curve in polar coordinates. As the q increases, r increases and the points with polar coordinates satisfying r = q lie on a spiral. Although the computer-plotted graph below shows r = q for positive values of q (hence of r), we can also plot r = q for negative q as well. See what it'll look like? Just replace the polar coordinates of every point by their negatives. For example, (-2π, -2π) and (-π/2, -π/2) also satisfy r = q.

Consider the polar curve r = 2 cos q. We'll convert to rectangular coordinates, hoping to recognize the curve. To do this we want to create terms r2 and/or r cos q and/or r sin q and/or tan q which we'll replace by x2+y2, x, y and y/x respectively. Multiplying by r gives r2 = 2 r cos q hence x2+y2 = 2x is the equivalent rectangular equation and this can be rewritten (x - 1)2 \+ y2 = 1, a circle of radius "1" with centre at (1,0).

After a while it gets tiring having to convert to rectangular coordinates ... and often we don't recognize the rectangular equation anyway, so we should get accustomed to sketching directly in polar coordinates.

In order to sketch a polar curve r = f(q) it's convenient to first sketch this relation as though r and q were rectangular coordinates (or we could make a table of values, but a picture is worth a thousand tables).

In the above example, we've identified 4 points at q = 0, q = a (where r is a local maximum), q = b (where r is a local minimum) and q = c ... then we show these points in polar coordinates, noting where r is decreasing and increasing.

But something wonderful happens if f(q) is negative, as shown in the example below:

Again we sketch r = f(q) and identify a few points of interest at q = 0, a, b, c, d and e. Note that, when

q = b, r has decreased to zero (so the distance from the origin, in _polar coordinates_ , is zero and the polar curve goes through the origin). From q = b to q = d, r = f(q) is negative and although the q-direction is (roughly) into the second quadrant, the negative r-values indicate that we move opposite to that direction, placing the points (roughly) in the fourth quadrant. In particular, at q = c (which points roughly north-west, in polar coordinates), r has its most negative value ... so we move south-east.

**Examples:** The polar curves, r = are _conic sections_ (hyperbolas, parabolas, circles, ellipses), depending upon the value of "a". For each of the following a-values, plot the curve and identify:

(a) a = 0 (b) a = .5 (c) a = 1 (d) a = 2

Solutions:

(a) a = 0 is easiest. The polar equation is r = 1, a circle.

(b) For a = .5, r is always positive and goes from a minimum value of = to a

maximum of = 2 (when q = π). It's the ellipse.

(c) For a = 1, r = has its minimum of at q = 0 but increases to ∞ as q->π (the denominator becoming zero). Or, to put it differently, r = ∞ and q = π gives the only infinity. This is the parabola.

(d) For a = 2, r = not only becomes infinite but does so twice ... and r is also negative.

From q=0 to q=, r increases from its initial value of , becoming infinite as q-> .

From q=to q=, r <0 (since 1+2 cos q<0) and the polar curve lies in the fourth, then the first quadrant. From q=to q=2π, r is again positive and returns from infinity (when q = ) to .

The curve is a hyperbola and it has two branches, the right-most branch being traversed while r is negative and q points into the second and third quadrants.

S: That's hard, isn't it? I mean, do you really expect ...

P: Don't worry. I only hope you can follow the arguments I gave, and could sketch these curves if you had enough time, but I won't expect you to reproduce them on an exam.

S: Whew!

P: But they are nice, aren't they? In fact, these polar equations describe the conic sections with a focus at the origin and that's different than the usual rectangular equations. In fact, if you wanted to describe the orbit of heavenly objects as they moved about the sun then you'd likely pick the sun as focus and you'd get one of these polar equations for the orbit of planets or comets. Nice, eh?

S: Wonderful ...

**Examples:** Sketch r2 = sin 2q

**Solution:** Note that sin 2q is negative when π < 2q < 2π (since that puts 2q in the third or fourth quadrant where the sine function is negative) ... and again when 3π < 2q < 4π. Hence, when < q < π and again when

< q < 2π, there is no curve!! That's because r2 = sin 2q cannot be negative. For q-values in between (starting at q = 0), r2 increases to a maximum when 2q = then decreases to zero when 2q = π (i.e. q = ). Then we come to the sector where there is no curve, then we start again with q = π where r increases to a maximum of 1 when 2q = and decreases again to zero for 2q = 4π, or q = 2π ... and that takes us through one complete circuit of the origin and if we continue, the curve repeats.

This polar curve is called a LEMNISCATE*.

S: You say r2 can't be negative ... so there's no curve there. That's weird, isn't it?

P: You've seen it before, you just don't recognize it. In fact, if you sketch y2 = 4 - x2 in rectangular coordinates you'd say that, for x2 > 4, there is no curve because y2 can't be negative ... hence the curve lies only in x2≤ 4.

S: I'd never say that! I don't even recognize y2 = 4 \- x2!

P: How about x2 \+ y2 = 4?

S: Aah, that's different. Anyway, why didn't you sketch r2 = sin 2q in rectangular coordinates first, as you suggested we do?

P: Okay, I'll do it that way. First I sketch r = sin 2q (which is easy) then I take ± its square root to get the graph of

r2 = sin 2q (but only where sin 2q is positive, of course).

S: How's that again!?

P: It's how you can sketch y2 = f(x). Just sketch y = f(x), then throw away all the negative pieces of f(x) (because there's no curve there, remember?), then take ± with what's left. That's y2 = f(x).

Then you get those cute little loops wherever f(x) is positive.

The method is something like sketching y = | f(x) |. We first sketch y = f(x), then reflect all the negative pieces of f(x) in the x-axis, i.e. replace them with - f(x). Remember?

S: But the curve is different, right? I mean, y = doesn't look like y = f(x), does it?

P: No, but when f(x) increases or decreases, so does and that's enough to sketch y = ... uh, except for one other thing which I almost hate to mention.

S: Go ahead. Nothing'll scare me now.

P: Well, where f(x) is zero, the graph of y = usually has a vertical tangent (i.e the derivative is infinite). That's because, for y = , we have = 1/2= -1/2 f'(x) = which becomes infinite as f(x)->0 (unless, of course f'(x)->0 as well ... in which case we've got a form and needn't be infinite).

S: That does scare me.

**Examples** : Sketch each of the following:

(a) r = 4 + 3 cos q

(b) r = 4 + 4 cos q

(c) r = 4 + 5 cos q

Solutions:

  | All are plotted on the same graph. Note that r = 4 + 3 cos q has max and min r-values of 7 and 1 and they occur at q = 0 and q = π (respectively) while r = 4 + 4 cos q has max and min r-values of 8 and 0, the latter being at the origin (and occurs for q = π). Finally, the last curve has negative r-values which occur whenever

4 + 5 cos q < 0, hence when q lies between two particular angles in the second and third quadrant. For q in this interval, although the q-direction is west of the y-axis, the negative r-value places the point east of the y-axis ... generating a small loop with right-most point r = -1, q = π.

All are called LIMAÇONS (with the form r = a \+ b cos q) although the middle curve, where b = a, is more commonly known as a CARDIOID. When b > a there's an inner loop.

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**INTERSECTION OF POLAR CURVES** :

A problem we'll meet shortly involves finding the points of intersections of two polar curves, say r = f(q) and r = g(q). The procedure is much the same as for rectangular coordinates: solve these 2 equations in 2 unknowns: r and q. Because there are an infinite variety of ways to describe a single point in polar coordinates (i.e. (r,q) and (r,q+2π) and (-r,q+π) etc. etc.) it's best to sketch the polar curve and anticipate the location of points of intersection. That way you'll know when you have them all!

**Example:** Find the point(s) of intersection of r = cos q and r = 1 - cos q.

**Solution:** A sketch indicates two points of intersection: one in the first and one in the fourth quadrant. To find them, set cos q = 1 - cos q so cos q = hence q = and q = or, perhaps this one's simpler to describe as q = - . In any case r = cos = so the |

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two points are (, ±). Note that r = 1 - cos q is a cardioid with its minimum r-value (namely r = 0) occurring at q = 0 and its maximum (r = 2) at q = π. Also, r = cos q is a circle as we've already seen.

PS:

S: I haven't seen ... have I?

P: Well, I didn't sketch it last time. Pay attention: r = cos q has its maximum (r = 1) at q = 0, then r decreases as q increases until, when q = , we have r = 0 (so the curve is now at the origin) and after that, q points into the second quadrant but now r is negative so we move opposite to the q-direction and get the curve going into the fourth quadrant (the lower half of the circle) until q = π and r = -1 which actually gives the right-most point on this circle even though q points due west (because r is negative, so the curve is east) and now q continues into the third quadrant but r is still negative so the curve is traced out in the first quadrant, again(!), retracing the upper half of the circle, then finally q points into the fourth quadrant and r becomes positive again and we go in the q-direction and retrace the lower half of the circle (again!) so we stand back and notice that when q goes from 0 to 2π the curve r = cos q is actually traced out twice! Got it?

S: zzzzz

P: Wake up ... here's one for you. Sketch the FOLIUM:

(x2+y2)2 = x2 y, but I'd suggest changing first to polars. And while you're at it, tell me the maximum value of r (i.e the maximum distance from the origin). And while you're at it, notice how many times the curve is traced when q goes from 0 to 2π. Got it? Okay, I'll do it myself. Put x2 \+ y2= r2 and x = r cos q and y = r sin q and get r4 = r3 cos2q sin q so we cancel r3 from each side, checking to see if |

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we've thrown any points away and noticing that the origin has r = 0 but we're not throwing this point away because it's satisfies the equation that's left, namely r = cos2q sin q (in fact we get to the origin every time cos q = 0 or sin q = 0) and now we imagine q increasing from q = 0 (where we begin, at the origin) so r increases to some maximum value (and we'll find out where that occurs in a minute) then r decreases again to 0 when q = π/2 ('cause cos q = 0) then q points into the second quadrant and cos q goes negative but r is still positive 'cause we've got a cos2q in the equation, but soon we get to q = π where sin q = 0 so we go to the origin again and for the rest of the time q points into the third and fourth quadrants but r is negative (because of the sin q) so the curve is traced out in the first and second quadrants (again!) until, finally, q = 2π and we re-arrive where we began, at the origin, ready to retrace the curve for a third time should q decide to continue. Got it?

S: You forgot to find the maximum r-value.

P: Aah yes, well, we must maximize the continuous function r = cos2q sin q on the closed interval 0 ≤ q ≤ π ... got that? a continuous function on a closed interval ... so we find the critical points within this interval ... where

= cos2q (cos q) + 2 cos q (-sin q) sin q = 0 and that means that cos3q \- 2 cos q sin2q = 0 or, let's see, I can factor this so it reads: cos q (cos2q \- 2 sin2q) = 0 hence either cos q = 0 (meaning q = ) or cos2q \- 2 sin2q = 0 which I can rewrite, putting cos2q = 1 - sin2q, and I get 1 - 3 sin2q = 0 so sin q = and there are two q-values in 0 < q < π which have this sine and each yields a maximum for r (one's in the first and one in the second quadrant, at the points labelled p and q). See?

S: You still haven't found the maximum r-value ... and you have to check the end-points to find the maximum ... and you forgot that 1 - 3 sin2q = 0 means sin q = ±, so you forgot that too.

P: Okay, the end-points ... q = 0 and q = π each give r = 0 (certainly not the maximum). Also, I only need to consider q's in the interval 0 < q < π, where sin q > 0 (so I ignore sin q = \- 1/), but I guess I should find the maximum r-value. I substitute the appropriate q into r = cos2q sin q = (1 \- sin2q) sin q and get rmax = (1 - ) = .

S: You can't leave it in that form ... rationalize the denominator. I learned that in kindergarten.

P: Go back to sleep.

LECTURE 14

AREAS ENCLOSED BY CURVES ... and RIEMANN SUMS

**the AREA UNDER A CURVE** **:**

Now we begin the study of the so-called INTEGRAL CALCULUS. Whereas we've been studying DIFFERENTIAL CALCULUS which deals with _differentiation_ ... arising from considerations involving the rate of change of functions and the slopes of tangent lines ... NOW we consider _integration_ arising from considerations involving the area enclosed by curves.

In the _differential calculus_ we stood close to a function and investigated its local behaviour ... how it changed when the independent variable changed by some small amount. In the _integral calculus_ we stand back and look at the global behaviour ... involving large changes in the independent variables ... and, as we shall see, this is most easily accomplished by summing the effects of small changes!

To begin with, we consider the problem of finding the area beneath a curve y = f(x), from x = a to x = b, meaning the area enclosed by the curve y = f(x) and the three lines x = a, x = b and y = 0 (the x-axis). Later we'll generalize to area enclosed on all sides by curves.

In fact, to simplify even further, we'll consider the area under y = mx from x = 0 to x = b. Since it's a triangle and we have a formula for that, namely , then we can check the method we're about to describe.

The first step is to divide the interval from x = 0 to x = b into n smaller "subintervals" (where n is some large number, like 1,000,000). For the interval 0 ≤ x ≤ b, the n subintervals will have length h = and the x-coordinates of the points of subdivision are 0, h, 2h, 3h, .... etc. until b = nh (which is obvious, since h = b/n).

The next step is to erect a rectangle on each subinterval, choosing as height some convenient value of y in each subinterval. (The width of every rectangle is the same: h.) For now, we'll choose as height the value of y at the right-end of each subinterval, meaning mh, m(2h), m(3h) ... etc. (since y = mx gives the height for any x).

The next step is to add all the areas of the rectangles (which is easy because we have a formula for that). For our choice of heights, we get a SUM of terms each being (width)(height) for each rectangle:

.

Finally, we let n∞. If the SUM of areas of all the rectangles has a limiting value as n∞, then that's the area under the curve. For our problem, we need to find which seems a formidable task ... but isn't. Remember, m is a constant (the slope of the given straight line) and we can put . The SUM is then m 2 (1 + 2 + 3 + ... + n) and there's a magic formula for the sum of the integers from 1 to n, namely . Hence we can write our SUM of rectangular areas as and find its limit as n∞ which is easy since = 1. Finally, then, SUM = = = and (voila!), the technique we've described works for our triangle! In fact, had we taken as height not the value of y at the right-end of each interval but, say, the value at the left-end, we'd get a SUM = m 2 (0 + 1 + 2 + 3 + ... + (n-1))

= b2 which still has the correct limit: . Indeed (remarkably) any y-value in each subinterval could have been chosen as height. The limiting value of the SUM of rectangles would still be ... and that gives us great confidence in this technique. In fact, had we taken some other y-value (rather than the right-y or the left-y which happen to be the maximum-y and the minimum-y in each subinterval) we'd get a SUM of rectangular areas which would be less than the first SUM we obtained, and greater than the second ... and since these two SUMs have the same limit, any other SUM would too (remember the SQUEEZE theorem?).

Now we generalize to an arbitrary y = f(x) on a ≤ x ≤ b ... except we must insist that f(x) be continuous on this closed interval and it must also be positive. (Later we'll see what happens when these conditions aren't met.)

  | First we subdivide the interval a ≤ x ≤ b into n subintervals, each of width .

Then we pick some convenient y-value (i.e an f(x)-value) in each subinterval. (In the diagram, we've chosen the value of f(x) at the right-end of each subinterval as we did before.)

Then we construct n rectangles each with width h and height equal to the chosen y-value.

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Then we SUM the areas of all rectangles. For our chosen heights, we'd get:

.

Finally, we compute which we may write (using the sigma-notation for sums):

and we'd have to compute this limit, realizing that "a" and "b" are constants and h = . If the limit exists (regardless of which y-values we chose in each subinterval for our heights) then it's the required AREA under y = f(x) from x = a to x = b.

Since h is an increment in x, it's sometimes convenient to use the notation h = x (as we mentioned when we considered derivatives, writing f'(x) = ). Then our prescription for finding the AREA becomes: .

This limit is called the DEFINITE INTEGRAL of f(x) from "a" to "b" and is written:

f(a+kx) x =

Note: If f(x) is measures in units of _watts_ and x is measured in _seconds_ , then the SUM f(a+kx) x and hence the definite integral are measured in _watt-seconds_. If f(x) is in _metres_ and x in _degrees_ , then the definite integral is measured in _metre-degrees_. This is sometimes useful, especially if f(x) is a rate of change, like _metres per second_ and x is measured in _seconds_ , then f(x) dx, and hence , is measured in _metres_. This almost suggests the type of applications one might expect of the definite integral. In fact, let's talk more about these _units_ so we get a feel for what the definite integral is doing for us:

• If is measured in _kilometres_ and x in _hours_ then f(x) is in _kilometres_ _per_ _hour_ ... a rate of change, so the total number of _kilometres_ is obtained by summing the incremental changes in _kilometres_ as the _hours_ go by, from x = a to x = b.

• If is measured in _dollars_ and x in _kilograms_ , then f(x) is in _dollars_ _per kilogram_... a rate of change, so the total number of _dollars_ is obtained by summing the incremental changes in _dollars_ as the _kilograms_ change, from x = a to x = b.

• If is measured in _hectares_ and x in _months_ , then f(x) is in _hectares per month_ ... a rate of change, so the total number of _hectares_ is obtained by summing the incremental changes in _hectares_ as the _month_ go by, from x = a to x = b.

Although we interpret as an AREA, it's only so we have some convenient geometrical picture (worth a thousand words?) just as we interpret as the slope of some tangent line when, in fact, derivatives are almost never used to determine "slopes". Besides, if f(x) is measured in _metres/second_ (a velocity) and x is measured in _seconds_ , then f(x) dx (hence the "sum" ) is measured in _metres_... a strange unit perhaps, for AREA! In fact, this so-called AREA actually gives the total distance travelled by an object whose velocity at time "x" is "f(x)".

**Example:** Compute the total area of n rectangles under y = sin x from x = 0 to x = π.

**Solution:** We'll subdivide the interval [0,π] into n subintervals of length and evaluate sin x at x = h, 2h, 3h, ..., nh. This will give the heights of the rectangles. The widths are all the same, namely h. We get:

SUM = h sin(h) + h sin(2h) + h sin(3h) + ... \+ h sin(nh) or, substituting h = :

. Now we'll evaluate, to 20 decimal places, for various values of n ... but we'll ask   to do it! We'll: (1) ask for 20 digits, then (2) define h = π/n (  calls π Pi), then (3) define our SUM, then (4) substitute various values of n, evaluating each time as a floating point (decimal) number (using the   command "evalf"). If we're lucky we'll see the SUM approach some limiting value which is the AREA under the graph of y = sin x from x = 0 to x = π:

• Digits:=20;

Digits := 20

•h:=Pi/n;

h := Pi/n

• SUM:=sum(h*sin(i*h),i=1..n):

• evalf(subs(n=10,SUM));

1.9835235375094545037

• evalf(subs(n=100,SUM));

1.9998355038874435049

• evalf(subs(n=1000,SUM));

1.9999983550656618233

• evalf(subs(n=5000,SUM));

1.9999999342026151656

The limit seems to be the number 2, so that's AREA under the graph of y = sin x from x = 0 to x = π.

PS:

P: So let's see you evaluate the definite integral: .

S: Huh?

P: But we've just done that one: = .

S: Do you expect me to do all this? I mean, am I expected to find the area by adding up all those rectangles?

P: There's an easier way which we'll get to soon. But remember when we defined the derivative? It was

f'(x) =and we actually computed a few derivatives using this definition ... but usually there were easier ways, like the product rule and quotient rule and chain rule. Here, too, we'll seldom have to resort to the definition above. Anyway, how do you like the notation? We start with a SUM, which we write f(a+kx) x, then we take the limit and get . See the similarity? It's like starting with the slope of a line joining two points on a curve, , then taking the limit as x0 and getting . Nice notation, eh?

In fact I'm told that the symbol is an old German symbol for s as in sum... so you can think of as elemental rectangular areas like "f(x) dx". Don't you see? That's the area of a rectangle with height f(x) and width dx. Nice, eh? This is quite nice because you can easily construct the definite integral which represents an area without going through the hassle of finding a Riemann SUM and then ...

S: A who? |

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P: Oh, I forgot to mention, the SUMs we've been talking about are called Riemann* Sums. That is, f(a+kx) x is called a **RIEMANN SUM** and its limit is the **DEFINITE INTEGRAL**. When we found the area under y = mx we first got a Riemann sum: m 2 (1 + 2 + 3 \+ ... + n). The big problem is not to find the Riemann sum, that's just adding up areas of rectangles ... the big problem is finding its limit as n∞.

**SOME TERMINOLOGY and PROPERTIES of the DEFINITE INTEGRAL** **:**

In we call "a" the LOWER LIMIT and "b" the UPPER LIMIT and f(x) the INTEGRAND.

Because of the way in which the definite integral was defined we have the following properties (which really require a proof ... but we'll just accept them without proof).

(A) = +

(B) = -

(C) += (D) = k

(E) = 0 (F) = -

The geometrical meaning of each of the above properties is clear, I think.

(A) The area under y = f(x) + g(x) is the area under y = f(x) PLUS the area under y = g(x).

(B) The area under y = f(x) - g(x) is the area under y = f(x) MINUS the area under y = g(x).

(C) The area under y = f(x) from "a "to "b" PLUS the area from "b" to "c" is the area from "a" to "c".

(D) If the function is multiplied by a constant k, its area is too.

(E) The area from x = a to x = a is zero. (What else?).

(F) Aaah, this one is not so obvious. If a < b (as usual) then the Riemann sum for subdivides the interval into subintervals of width h = which is negative, so the Riemann sum has the same terms as the Riemann sum for ... except for the sign of each term (because h < 0). Hence the definite integrals are the negative, one of the other. If that's confusing, then just look at (C) and put b = a. You'd get + = = 0 because of (E) ... and the result is just (F). And if _that's_ confusing, then just accept it.

**Example:** Find a Riemann sum whose limit (as n∞) is .

(Practice the mathspeak: the _lower limit_ is 1, the _upper limi_ t is 3 and the _integrand_ is ex and we're looking at the integral of ex from 1 to 3 ... NOT the integral "from 3 to 1" ... we've got to get the jargon right!)

**Solution:** The interval is 1 ≤ x ≤ 3 and we subdivide it into n parts of width h = = . The points of subdivision are then 1, 1+h, 1+2h, 1+3h, ..., 1+nh (the last of which, of course, is "3"). On each subinterval we construct a rectangle whose height is the value of y = ex at, say, the left-end. The heights are e1, e1+h, e1+2h, ..., e1+(n-1)h. The areas are each (height)(width), namely: e1 h, e1+h h, e1+2h h, ...., e1+(n-1)h h and the Riemann sum is given by: . The limit of this SUM, as n∞, is : the area under y = ex from x = 1 to x = 3.

PS:

S: Can you find this limit?

P: No problem. Normally I can't, but this one is easy. I factor out h e and rewrite the Riemann SUM as

h e (1 + eh \+ e2h \+ ... + e(n-1)h) and recognize the sum in the brackets as a geometric series for which I have a formula. It sums to ...

S: How do you recognize the sum as a geometric series? I don't.

P: A geometric series has the form a + ar + ar2 \+ ar3 \+ ... where each term, divided by the preceding term, gives the same number "r". In my series, each term divided by the preceding term gives the same number, namely eh ... so it's a geometric series. The sum of n terms of such a series is a so my Riemann sum adds up to

h e = h e . I now just have to find the limit of this as n∞.

S: Good luck.

P: Remember that h = so nh = 2 so enh = e2 and my SUM can then be written e (e2 \- 1) . Instead of replacing h by and letting n∞, I'll leave the h in there and take the limit as h0. Our limit is just: e (e2 \- 1) and this limit has the wonderful form so we can use l'Hopital's rule and get: e (e2 \- 1) = e (e2 \- 1)

= e (e2 \- 1) = e3 \- e (where, to use l'Hopital, we differentiated numerator and denominator with respect to h).

S: Are you saying that the area under y = ex is e3 \- e?

P: From x = 1 to x = 3 ... yes.

S: Amazing.

P: But there's an easier way. Pay attention:

**THE FUNDAMENTAL THEOREM** **:**

In order to avoid finding limits of Riemann sums, we do the following (and whoever thought of this was a genius!). To evaluate we make the upper limit variable, calling it "t". We now have a function of t, namely A(t) = . The first thing you'd like to do with such a weird and brand-new kind of function is differentiate it! Unfortunately, none of the "differentiation rules" help ... so we must resort to the definition of the derivative. (Imagine the first mathematician who did this. He/she read a book on the Newton/Leibniz differential calculus and decided to use it on this area function, A(t). Imagine the surprise when ... but we get ahead of ourselves.)

Consider . The numerator is our AREA from x = a to x = t+h MINUS the area from x = a to x = t. Using properties of the definite integral we can rewrite this as:

, the area from x = t to x = t+h. The area of this narrow strip lies between the area of two rectangles, each of width "h": one has height equal to the minimum value of f(x) in [t,t+h] and the other has height equal to the maximum value of f(x) in [t,t+h]. This sounds reminiscent of the SQUEEZE theorem (!) which now comes to our rescue. We write:

h f(x)minimum ≤ ≤ h f(x)maximum

which, when divided through by h gives

f(x)minimum ≤ ≤ f(x)maximum

and now we can use the SQUEEZE theorem. Let h0 and recognize that the two "outside" limits are equal ... equal to what?

P: Did you hear? Equal to what?

S: I wasn't listening. Could you repeat the question?

P: Never mind. Think of the maximum value of f(x) in t ≤ x ≤ t+h. It occurs somewhere in that interval, say at x = t1. Think also of the minimum value of f(x) in the same interval. Suppose it occurs at x = t2. Then the above inequality reads : f(t1) ≤ ≤ f(t2). Okay, now let h0. What happens to t1 and t2?

S: I give up.

P: Think! They both lie in this interval from t to t+h, and that interval is shrinking to zero! They're both being squeezed to the left as h0. They're both approaching ... what?

S: I give up. I really wasn't listening. Besides, I haven't understood much of what you said ...

P: Pay attention! Both t1 and t2 are approaching x = t, the left-end of the interval. Then f(t1) and f(t2) are both approaching f(t) ... provided f(x) is continuous!! Remember I said we'll assume f(x) is continuous? Well, here's why. As h0 the left- and right-sides of the inequality f(t1) ≤ ≤ f(t2) have the same limit, namely f(t). Hence, the SQUEEZE theorem tell us that also has this limit. In other words

A'(t) = = = f(t)

and we've managed to differentiate this area function we've invented.

S: But this is confusing, isn't it? And you don't have any pictures! You always said a picture is worth a thousand ...

P: Okay, okay, here's a picture.

The shaded area is A(t+h) \- A(t) = , the area under the curve from x = t to x = t+h. See? It's greater than the area of the rectangle whose height is the minimum value of f(x) and less than the area of the rectangle whose height is the maximum value of f(x). That means that A(t+h) - A(t) lies between h fmin and h fmax. That means that lies between fmin and fmax. Now, as h0, the minimum and maximum of f(x) (in the interval t ≤ x ≤ t+h) approach f(t). That makes the limit of |

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equal to f(t) as well, because of the SQUEEZE theorem. That means we've managed to differentiate this rather weird function we've invented, and since A(t) = gave us A'(t) = f(t) , we now see the general rule:

. In fact, if you remember the "differential", we would say = f(t) and replace by and get ≈ f(t) so A ≈ f(t) t and we'd see that the tiny change in area is approximately f(t) t which is the area of a certain rectangle. Nice, eh?

S: Okay, if you say so.

P: But we're not finished. Our goal is to evaluate without having to resort to limits of Riemann sums. To see how the rest goes, let's do an example.

**Example:** Evaluate

**Solution:** First we introduce the "area function" A(t) = . Remember that, eventually, we want A(3). Now we differentiate A(t) and get: A'(t) = et, using the general rule, obtained from our investigation above:

. Now that we know the derivative of A(t), can we find A(t) itself ... hence A(3)? Yes, because A(t) must be one of the functions et \+ C (for some value of the constant C). These are the _only_ functions which yield et when differentiated. But if A(t) = et \+ C, what's C? We know one other thing about A(t) = which we haven't used: A(1) = = 0, hence A(1) = 0 = e1 \+ C and we now know that C = \- e1. Finally, then, A(t) = et \- e1. When t = 3 we get A(3) = e3 \- e (as we did before, when we used the Riemann sum definition of the definite integral).

The general procedure to evaluate is as follows:

(1) Write A(t) = (2) Then A'(t) = f(t) (the integrand, evaluated at the upper limit).

(3) Then, if F(t) is any function whose derivative is f(t), we must have A(t) = F(t) + C.

(4) We also know that A(a) = = 0 and that gives 0 = F(a) + C, hence C = - F(a).

(5) Finally then, A(t) = = F(t) - F(a) and, for t = b we get:

= F(b) - F(a) where F(t) is any function whose derivative is f(t).

Some Terminology:

If F(x) = f(x), then we call F(x) an ANTIDERIVATIVE of f(x).

Examples:

• ex is an antiderivative of ex (because ex = ex).

• cos x is an antiderivative of - sin x (because cos x = - sin x)

• _ln_ x is an antiderivative of ... but only if x > 0.

• arctan x is an antiderivative of

• _ln_ sec x is an antiderivative of tan x (because _ln_ sec x = = tan x) ... but only if sec x>0.

• _ln_ (sec x + tan x) is an antiderivative of sec x (check this out!) ... but only if sec x + tan x >0.

S: You keep saying "but only if something > 0". What's that all about?

P: Remember, the logarithm of (something) is defined only for positive something ... and ln is, of course, a logarithm.

S: Suppose you wanted an antiderivative of, say , when x < 0?

P: Okay, let's consider y = ln (-x) because now -x is positive if x < 0. Let u = -x and get y = ln u so = = = = . See? You still get . That means that ln x = when x > 0 and ln (-x) = when x < 0. Nice, eh? We've got an antiderivative for for both positive and negative x. Of course we wouldn't expect an antiderivative which gives for x = 0 because doesn't even have a value at x = 0.

S: Does that mean we've gotta know in advance whether x is positive or negative ... else we won't know which to use? I mean, ln x or ln (-x) ... which one do we use?

P: What we need is a function, call him SAM, which is "x" when x > 0 and is "-x" when x < 0, then we can say that an antiderivative is ln SAM. Can you think of such a function?

S: You gotta be kiddin'. I'm sorry I asked.

P: How about | x |. Remember the absolute value function? It has precisely the right properties. Hence ln | x | is an antiderivative of for all x ≠ 0.

**Note:** Every function has an infinite number of antiderivatives. If F(x) = f(x), then = f(x) as well so F(x) + C is also an antiderivative, regardless of the choice of constant C. To get the most general antiderivative, pick any one ... and add an arbitrary constant.

**Examples:** ex \+ C is the most general antiderivative of ex .

cos x + C is the most general antiderivative of - sin x .

_ln_ | x | + C is the most general antiderivative of .

arctan x + C is the most general antiderivative of .

Usually, antiderivatives are called INDEFINITE INTEGRALS and denoted by: without any limits. In other words:

denotes an ANTIDERIVATIVE (or INDEFINITE INTEGRAL) of f(x)

A few examples are in order (and can be easily verified by differentiating the right-side):

= + C = _ln_ | x | + C = arctan + C

= arcsin + C = _ln_ |sec x | + C = _ln_ | sec x + tan x | + C

Note: REMEMBER THE ABOVE INTEGRALS !!

**Note:** Remember that is the of elements and each has the dimensions of which, if "a" and "x" are metres, has the dimensions of so it's not surprising that this (indefinite) integral is arctan which also has this dimension (because of the factor ... not because of the arctan which is an angle in RADIANS hence has no dimensions). Note too that should be dimensionless ... which it is! These observations will help determine when the appears in the evaluation of the integral and when it doesn't.

The rule for evaluating DEFINITE integrals can now be restated as:

the FUNDAMENTAL THEOREM

= F(b) - F(a) = [F(x)]where F(x) is any antiderivative of f(x)

and the notation [F(x)]means F(b) - F(a)

Important observation: after having found an antiderivative F(x), then evaluating at the upper and lower limits and subtracting, the result is some number ... without any x's!

**Examples:** Evaluate the following definite integrals:

(a) (b) (c)

Solutions:

(a) = [- cos x]= (- cos π) - (- cos 0) = 1 + 1 = 2

Note, too, that = [- cos t]= (- cos π) - (- cos 0) = 1 + 1 = 2 and = [- cos z]= 2.

What name you give to the variable under the is irrelevant.

S: Hold on! Why do you pick - cos x as the antiderivative. I thought you said that - cos x + C was the antiderivative of sin x? See? C! The constant of integration.

P: Right. Suppose I pick - cos x + C, then I'd evaluate it at the upper limit, at the lower limit, then subtract. I'd get - = (- cos π) - (- cos 0) = 1 + 1 = 2. See? The C has cancelled out. Moral?

S: Forget the "C".

P: ... when evaluating definite integrals, but never when evaluating indefinite integrals!

(b) = [arctan x]= (arctan 1) - arctan 0 = π/4 - 0 = π/4

S: How do you know that arctan 1 = π/4?

P: If y = arctan x then x = tan y (because they're inverses, remember?). Hence, if we want to know y = arctan 1, then write this as 1 = tan y (everybody loves the tangent, nobody loves the arctangent, so we rewrite the relation to get it in terms of the tangent function)... so what's y? That is, what angle (in RADIANS!) has a tangent of 1? It's π/4, so that's y. See?

S: No. There are millions of angles whose tangent is 1. Am I right?

P: Already you've forgotten about our inverse trig functions! Remember! We had to restrict the domain so as to guarantee the existence of an inverse. That means that, if y = arctan x, then y must be chosen from the range - π/2 < y < π/2, and now there's only one angle there which has a tangent of 1 ... and it's π/4.

S: Okay, okay. I'm sorry I asked.

(c) = [ ]= ( ) - ( ) = 0

The last is perhaps surprising. Does it _really_ say that the area under y = x3, from x = -1 to x = 1, is zero?

We'll review a bit in the next lecture ... and see what this means.

LECTURE 15

MORE ON DEFINITE INTEGRATION

We first considered the problem of finding the area enclosed by y = f(x), x = a, x = b and y = 0 (the x-axis), where f(x) was both continuous and positive in this interval. We then subdivided the area into the SUM of a multitude (namely "n") of tiny rectangles (because we have a formula for the area of rectangles). Each rectangle had a width h = and a height equal to some convenient value of f(x) in each subinterval. Letting the number of rectangles become infinite, we arrived at the AREA (the limit of the Riemann SUM) which we denoted by . Now, what happens if f(x) is always negative on [a,b]? We repeat the same procedure, but now the heights of the rectangles are all negative and we used (height)(width) to compute the area of each ... and (height) < 0 ... so we actually get the negative of the areas. The limit of the Riemann SUM as n∞ yields not the area bounded by y = f(x), x = a, x = b and y = 0 but the negative of that area. (Note that we now avoid using the phrase "the area beneath y = f(x)" because this curve lies below the x-axis ... so the area below _it_ is pretty large!)

Now back to . We may break this up into two integrals (using one of the properties of definite integrals) and write: = + where the second integral is indeed the area "beneath" y = x3 from x = 0 to x = 1 because y = x3 is positive. However, on -1 ≤ x ≤ 0, y = x3 is negative so the evaluation of this integral gives the negative of the area enclosed by the the curve, the x-axis and x = -1, x = 0. The integral actually gives - , which cancels = .

In other words, whenever y = f(x) lies below the x-axis, the definite integral yields the negative of the area so it's quite possible to have these "negative" areas cancel with "positive" areas to give zero. |

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**Example:** = [ - cos x ] = (- cos 2π) - (cos 0) = -1 +1 = 0 which seems clear since the "area" from 0 to π is positive while the "area" from π to 2π is negative ... and they cancel.

In the diagram on the right we plot the graph of some invented function f(x). Note that the "area" between the curve and the x-axis is sometimes positive, sometimes negative. We also plot the graph of which is zero when x = 0, then increases until x = P (at x ≈ .8) because f(x) is positive so the "area" is increasing. After x = P, f(x) goes negative so the "area", namely , decreases, until x = Q (x ≈ 1.5) after which f(x) becomes positive again so the "area" increases again ... and so on. In fact, after x = R, (x ≈ 1.2) f(x) becomes so negative that the negative area cancels completely the positive area accumulated so far, and goes negative.

All this makes sense even if we didn't have any calculus or any fundamental theorem. However, we DO have the fundamental theorem and it says that = f(x) so the derivative of the area function A(x) = is actually f(x) and so it's clear that the graph of A(x) will have a horizontal tangent wherever f(x) = 0 ... and that's precisely what happens at P, Q and R! |

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Under what circumstances would you expect = 0? Of course there must be "negative area" exactly cancelling "positive area" ... but can you predict that, just by inspecting f(x)?

Remember the definition of an ODD function? If you change the sign of x, you change the sign of f(x). That is, f(-x) = - f(x). Under these circumstances the values of the function on -L ≤ x ≤ 0 are just the negative of the values on 0 ≤ x ≤ L ... and the "areas" cancel. |

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How about when f(x) is an EVEN function? Now the values of f(x) on -L ≤ x ≤ 0 are exactly the same as those on 0 ≤ x ≤ L so the "areas" are as well. That is, = . Hence we |

can write = 2 ... which is sometimes very convenient.

**Examples:** = 0 (because f(x) = x cos x is ODD) and

= 0 (because f(x) = x2 sin x is ODD) and

= 2 = 2 where we recognize as the area under y = from x = 0 to x = 1, namely the area of a quarter-circle of radius "1", i.e. π 12. (Note that y = means that x2 \+ y2 = 1 ... so we _do_ have a circle).

Let's calculate some areas:

**Examples:** Calculate the area (in the first quadrant) enclosed by the curves y = , y = x.

**Solution:** First we sketch the two functions so we can see the area:

Then we find the points of intersection (y = = x has solutions: (0,0) and (1,1)).

Then we compute the area under y = and subtract the area under y = x.

We get - = [ x3/2]- [ ]= - = .

Finally, we check to see if this is reasonable! Note that the area lies within a square of area "1", so it should be less than this (which it is). In fact, it's clearly less than the area of half this square (since our area lies within the upper, triangular half) ... and it is. So we have some faith in our answer.

Note, however, that we could have used a property of definite integrals to combine the two integrals and write our area as:

= [ x3/2 \- ]= - = .

Since it's convenient to think of as "elemental" rectangular areas of height f(x) and width dx (reflecting the origins of the definite integral as the limit of a Riemann SUM), then we can also think of

as elemental areas of height (- x ) and width dx . We'll do this when convenient.

Indeed, if the required area is enclosed by curves y = f(x) and y = g(x) as shown, then we imagine subdividing the area into elemental rectangular areas located at the place x (somewhere between x = a and x = b), with width dx and height f(x) \- g(x). The required area is the the of these (beginning with rectangles at x = a and ending with x = b), yielding the definite integral: |

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**Example:** Compute the first-quadrant area enclosed by y = x, y = 1 and y = sin x.

**Solution:** First we sketch the area.

Then we find the points of intersection: y = 1 = sin x at (π/2,1) and y = x = 1 at (1,1) and, of course y = x and y = sin x meet at the origin. We label these points on our sketch.

Then we imagine the area subdivided into elemental rectangular areas located at x, of width dx, and height ... well, that's a problem. When x lies in [0,1], the rectangle rises from y = sin x to y = x so has height (x - sin x) and area (x \- sin x) dx. However, for x in [1,π/2] the rectangle rises from y = sin x and ends on y = 1 hence |

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has area (1 - sin x) dx. We'll need to find the areas separately: +

= [ + cos x]+ [ x + cos x]= - + - (1 + cos 1) = .

Finally, we check for _reasonableness_. Our area is certainly less than the area of the triangle shown and this triangle has area

(1/2) (base) (height) = (1/2) (π/2 - 1) (1) = .

Is < ? Yes, so we have some faith in our answer.

S: Some faith? What does that mean?

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P: You'd be surprised how many students will get an answer like 12,473 then go on to the next question, thinking they're finished.

S: Really? You're kidding. I'd never do that.

P: We'll see.

We've seen that the definite integral (which gives the required area) can be generated by imagining a subdivision into elemental rectangles, determining the area of these rectangles, then from the first rectangle (at x = a, say) to the last (at x = b, say).

Let's review this technique:

Suppose we want the area enclosed by y = f(x) and y = g(x). We first sketch the area, then find the points of intersection by solving y = f(x) and y = g(x) (two equations in two unknowns). Suppose the intersections are at (a,c) and (b,d), so the area we seek lies between x = a and x = b, y = c and y = d, as shown. Then we subdivide into elemental rectangular areas located at the place x, of width dx and height f(x) - g(x) (the upper y-value _minus_ the lower y-value), hence the area (f(x) - g(x)) dx. Now we these rectangular

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areas from x = a to x = b via the definite integral: .

S: That's not right, you know. I mean, the height isn't really f(x) - g(x). That's only the height at x, not at x + dx. So how come it gives the right answer ... or does it?

P: Remember that we got by taking the limit of a Riemann SUM ... not by this gimmick of imagining elemental rectangles and . We do this "elemental rectangle" business in order to get the definite integral which represents the area without always having to go back and construct a Riemann SUM ... but I can if you'd like.

S: Sure. Go ahead. I've got lots of time.

P: Okay, we subdivide [a,b] into n subintervals of width h = . The x-coordinates of the points of subdivision are x=a, a+h, a+2h, a+3h, ..., a+nh (which is also "b"). On each subinterval we pick a convenient x-value and compute f(x) - g(x) at that value ... and construct a rectangle with this height, and width h. For example, suppose that we choose x1 in the first subinterval, x2 in the second and so on. Then f(x1) - g(x1) is the height of the first rectangle, f(x2) - g(x2) the height of the second, etc. Now we construct a Riemann SUM which is the sum of the areas of all n rectangles:

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h + h + ...+ h = . Finally we let n∞ (so that h = 0 as well) and define the limit as . See? That's the area we're after ... but we could have gotten to this definite integral by imagining a bunch of "elemental" rectangles of area dx and from x = a to x = b.

S: Actually, the area of your rectangles don't look much like the area we're after. I guess the definite integral is only a good approximation, right?

P: No, it's exact. I just chose a few subdivisions in my diagram so you can actually see the individual rectangles. Here, let me show you what it looks like with 47,000,000 elemental rectangles. See? The elemental rectangles really give a good approximation ..

S: Aha! An approximation! Didn't I just say ... ?

P: Wait, it's only an approximation if you take 100 or 47,000,000 rectangles. The error gets smaller as the number of rectangles goes up. In the limit, as n∞, the error goes to zero ... and the definite integral is exactly the area. Anyway, what's your favourite number?

S: Huh? ... seven, I guess.

P: Then let's look carefully at the 7th subinterval which goes from x = a + 6h to x = a + 7h and we'll pick a convenient x-value in this subinterval and called it x7 and construct a rectangle with height f(x7) - g(x7) and width h. It doesn't look like a very good approximation to the actual area between f(x) and g(x), but as h0 and this rectangle becomes skinny and the number of such rectangles becomes infinite ... voila! The SUM of all of them has the correct limit!

S: That's amazing, isn't it? I mean, all those errors ... millions of them ... and you add them up and they go to zero. You'd think that more rectangles would give more errors and the approximation would get worse and worse and ..

P: How big do you think each error is?

S: I haven't the foggiest idea.

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P: Let's estimate. If f(x) and g(x) were constants, then there'd be no error, but if f(x) is increasing at a maximum rate A (meaning y = f(x) is increasing A times more rapidly than x), then y changes by about Ah when x changes by an amount h. Similarly, if g(x) is changing at the rate B (that's g'(x), by the way) then y = g(x) changes by an amount Bh, roughly. The error in area, for each rectangle, is no bigger than the sum of the shaded areas, and each is roughly a triangle so we can estimate using (1/2) (base)(height) and get (1/2)(h)(Ah) = (A/2)h2 and (1/2)(h)(Bh) = (B/2)h2 which adds up to (A/2+B/2)h2 and putting h=(b-a)/n we get (A/2+B/2)(b-a)2/n2. That means that the error in each triangle is proportional to 1/n2 and although we have n such triangles we'd get a total error proportional to 1/n which does go to zero as n∞. Nice, eh?

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S: Triangles? Did you say the shaded areas were triangles? But ...

P: No, no ... they're approximately triangles. It's like assuming that f(x) and g(x) were replaced (for purposes of this estimation) by their linear, tangent line approximation ... surely you remember the linear approximation? Besides, triangles sometimes come in handy for estimating. Here, let me give you a problem you can't solve but which you can estimate, using triangles. Ready? There is this train track and it's one long piece of steel 1,000 metres long and it's pinned down at each end so the ends can't move. Then it gets hot and the increased temperature expands the metal by 2 metres and the track buckles in the centre. Got the picture?

S: A picture is worth .. | P: Okay, here's the picture. My question is: how much does it buckle in the centre? If we call it "h" then we can estimate this using a triangle. See the triangle? We have h2 \+ 5002 = 5012 (using Pythagorus) so h2 = 5012 \- 5002 = 1001 so h = ≈ ... uh, did you bring your calculator?

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S: Nope.

P: Okay, we'll do a linear approximation on this one. Let's see ... 1001 is pretty close to 900 which is 302, but 312 is 961 so I'll use it and get the tangent line to at x = 961. That's f(961) + f'(961) (x - 961) and for f(x) = x1/2, we have f'(x) = 1/2so f(961) + f'(961) (x - 961) = + = 31 + so ≈ 31 + when x is close to 961. I plug in x = 1001 and get ≈ 31 + ≈ 31.6 which is good enough for me.

S: For me too. (I hope he doesn't mean 31.6 metres!?) Say, do I have to know this for the final exam?

P: Train tracks? No. Linear approximation? Yes. Now pay attention, because if you're satisfied that this "elemental rectangle" gimmick gives the correct definite integral, then I'll show you something really neat.

S: I can hardly wait.

We consider the same problem as above: the area enclosed by y = f(x) and y = g(x). Rather than subdividing the area into vertical "elemental rectangles" (located at x, between x = a and x = b), we subdivide into horizontal "elemental rectangles" ... located at y between y = c and y = d.

 | To find the area of this horizontal rectangle we use (width) (height) where the height is just dy and the width is the "right x-value" _minus_ the "left x-value". That means we need to solve y = f(x) for x in terms of y ... and y = g(x) for x in terms of y. Suppose this gives x = F(y) and x = G(y) respectively. Then, for the rectangle located at "y", the width is G(y) - F(y) so the area of this "elemental rectangle" is dy and all such rectangles, from y = c to y = d gives the definite integral:

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**Example:** Calculate the area enclosed by y = x and y = using horizontal rectangles.

**Solution:** We first sketch the curves and identify the area. Then we find the points of intersection by solving y = x and y = . Then we imagine a single, typical horizontal elemental rectangle located

at y, with height dy, and compute its area. For a given y we need to know the _right x-value_ and the _left x-value_ which means we need to solve y = x and y = for x in terms of y; that's easy. **x = y** and **x = y** 2. Hence our horizontal rectangle has area (y - y2) dy and we all such areas from y = 0 to y = 1 and get

= [- ]= - = and (surprise!) it's the same area as we computed using vertical rectangles. |

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S: So why use horizontal rectangles at all?

P: I knew you'd ask that. Sometimes it's easier. Let's go back to the area enclosed by y = x, y = 1 and y = sin x. We sketch the curves, find the points of intersection (we've already done that, earlier) and, knowing that we're going to use

horizontal rectangles we immediately solve for x in terms of y. It's fortunate that we've already considered the arcsine function because

y = sin x, solved for x, gives x = arcsin y (the inverse function, remember?). Okay, now we sketch a typical horizontal rectangle located at "y", between y = 0 and y = 1. Its area is:

(xright \- xleft) dy = (arcsin y - y) dy and all such rectangles gives the definite integral: . Nice, eh?

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If you look back at this problem when we used vertical rectangles, we had to divide the x-interval into two pieces: 0 ≤ x ≤ 1 and 1 ≤ x ≤ π/2 because the expression for elemental area changed. But horizontal rectangles always go from the line x = y to the curve x = arcsin y, so there's one integral to evaluate instead of two!

S: But how would you find the antiderivative of arcsin y? I mean, what's ?

P: Uh ... good question. You'll have to wait till later.

S: Thanks.

**Example:** Find the area enclosed by y = arctan x, y = and the tangent line to y = arctan x at x = 0.

**Solution:** The tangent line has slope = arctan x = = 1, at x = 0 and y = arctan 0 = 0 (when x = 0) so it's = 1 or simply y = x. Now we sketch the curves y = arctan x, y = and y = x, identify the area, and find

the points of intersection: y = x intersects y = π/4 at (π/4,π/4) and y = arctan x intersects y = π/4 at (1,π/4) and, of course, y = x and y = arctan x intersect at (0,0). Now we draw a typical horizontal rectangle located at "y", between y = 0 and y = π/4. Its height is dy and its width is (xright \- xleft) so its area is: |

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(tan y - y) dy and gives:

= [ _ln_ |sec y| - ]= - = _ln_ \- .

S: Hah! You didn't check for reasonableness!

P: Okay. Checking for reasonableness, we compare to the area of the triangle (which looks much like the shaded region in the diagram): (1/2) (base) (height) = (1/2) (1 - π/4) (π/4) which is roughly (1/2) (1 - 3/4)(3/4) = 3/32 ≈ .1 whereas ln \- is roughly ... uh ... well ...

S: Gotcha!

P: I'll use the linear, tangent line approximation to estimate ln . Let's see ... is near 1, so I'll find the tangent line to y = ln x at x = 1 and that has a slope of ln x = = 1 (at x = 1) so the line is = 1 and that's y = x - 1 (since I know that ln 1 = 0) and putting x = , I estimate ln ≈ - 1 and I know that is roughly 1.4 so I get ln ≈ .4 and subtracting ≈ ≈ .3, I get .1 which checks out and now you're really impressed, right?

S: zzzzz ... uh, say ... you seem to be fond of approximations. Everything is an approximation. Isn't anything exact in math?

P: Sure, but what would you do with the exact value of for example. As a decimal it has an infinite number of digits. Do you really want to see them all? We can use Newton's method to get, say, a million digits and that should be good enough. Besides, sometimes approximate answers can give exact results when you use limits. We've already seen that haven't we? From the Riemann SUM, which is an approximation, we take a limit and get the area under the curve exactly.

S: Can't we just go on?

LECTURE 16

AREAS IN POLAR COORDINATES

In order to compute the area enclosed by curves in rectangular coordinates (y = , y = arcsin x, x = a, x = π/4 and curves like that) we subdivide into elemental rectangles and using the definite integral. In order to compute the area enclosed by curves in polar coordinates (r = cos , r = , r = 3 + 4 sin  and curves like that) we do a similar thing ... except that rectangular "elemental areas" are now inappropriate. To see what _is_ appropriate, let's repeat the prescription for finding the area "under" y = f(x) (i.e. from y = 0 to y = f(x)) from x = a to x = b ... except we'll word it slightly differently so we can repeat it for polar coordinates:

(1) Subdivide a ≤ x ≤ b into n subintervals of length h = .

(2) On each subinterval pick a convenient value of x, say xk on the kth subinterval, and assume y = f(x) is constant at the value f(xk) over the kth subinterval.

(3) Determine the area from y = 0 to y = f(xk), on the kth subinterval.

(4) Sum all such areas to get a Riemann SUM.

(5) Take the limit at n∞ and hence h0.

(6) This limit (assuming there _is_ a limit) is the required AREA. |

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Now, in polar coordinates:

(1) Subdivide a ≤  ≤ b into n subintervals of length h = .

(2) On each subinterval pick a convenient value of , say k on the kth subinterval, and assume r = f() is constant at the value f(k) over the kth subinterval.

(3) Determine the area from r = 0 to r = f(k), on the kth subinterval.

(4) Sum all such areas to get a Riemann SUM.

(5) Take the limit at n∞ and hence h0.

(6) This limit (assuming there _is_ a limit) is the required AREA.

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The difference is in step (3) where the "elemental area" is a sector of a circle rather than a rectangle (and this seems appropriate for polar coordinates, right?). To deduce this area (in case you haven't memorized such) we recall the area of a circle of radius r: π r2. If we take just a fraction of this area, say the fraction , then we get the area of a sector where the central angle is "h" (instead of "2π"). Hence the sector has area π r2 = r2 h and that's what we use for the elemental area. The Riemann SUM would be and its limit would become the definite integral: rather than .

The area enclosed by r = f(),  =  and  =  is given by:

**S:** So how come you switched from a, b to , ?

**P:** Everybody uses Greek letters for angles ... didn't you know that? Pay attention:

It's important to regard as the AREA SWEPT OUT BY THE RADIUS r = f() as  goes from  to  = . To emphasize this point we'll sketch some polar curves and shade the area which gives:

Here, the polar curve r = cos  is traversed once when

 goes from 0 to ... or from - to |

Here the integral

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Here the limaçon reaches r = 0 when  = so

gives the area shown. |

Here r traces out the inner loop as  goes from to so gives the area of this inner loop.

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It's important to note that, for example, would give twice the area of the circle and

will give the area within the limaçon with the inner loop counted twice!

Remember the diagram ===>>>

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**Example:** Express, as a definite integral, the area enclosed by the curve r = cos2sin .

**Solution:** We sketch the curve (a lemniscate ... we've done this one before ...) and observe that the curve is traversed in its entirety when  goes from 0 to π (and no more!). The area swept out is given in the diagram.

S: I'm not sure I get this "swept out" business. Sounds like a broom or ...

P: Pay attention: you imagine the radius r as a straight line from the origin to a point on r = f(). Then, as  changes, this straight line sweeps out an area. That can't be hard to understand. In fact, one of Kepler's three laws of planetary motion says that the area swept out is the same for any given time period. Here the origin is the sun and a planet is revolving about the sun along some path r = f() and each month (or day or year) the area swept out is the same and that means that when the planet is close to the sun and r is small it must be travelling quickly whereas when r is large it travels slowly so as to keep the area swept out the same for ... are you listening?

S: zzzz

**Example:** Find the area _outside_ r = 2 + 2 cos  but _inside_ r = 1.

**Solution:** We sketch both polar curves and find the points of intersection by solving r = 2 + 2 cos  = 1. This gives cos  = - so  = and  = .

Now the integral will give the area swept out as "r" traces the limaçon while

gives the area swept out when r traces out the circle. We want the difference:

\- = - . Put cos2 = and get

\- = - [5+8sin+sin2]= - -= -

S: Reasonableness! Reasonableness!

P: You check it for reasonableness.

S: Huh? Well ... let's see ... are there any triangles around? How about this one =====>>> ?

The base is 1 and the height is ... y = r sin  and that's ... uh, (1) sin (2π/3) = sin 120˚ = /2 so the area is = /4, right? And is maybe 1.5 or something so I get .4 or something. Good?

P: How does that compare with the exact answer?

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S: The exact answer is 3.5- 5π/3 and that's about (3.5)(1.5) - (5)(3)/3 = 5.25 - 5 = .25 which is way off ... so your answer is wrong!

P: You said is maybe 1.5, but it's about 1.73 so that'd change your /4 estimate to about .43 and the exact answer to about .86 and ...

S: See? You are wrong!

P: ... and you only estimated the upper half of the area so if I double your estimate I'd get ...

S: Hey, it's .86, right on the button! See?

P: Hmmm. Now if you only had me around when you write your final exam.

LECTURE 17

TECHNIQUES OF INTEGRATION

We collect those "indefinite" integrals which we already know (and which you should remember!):

= + C (n ≠ -1) = + C (a ≠ 0) = _ln_ | x | + C

= - cos x + C = sin x +C = _ln_ | sec x | +C

= _ln_ | sec x + tan x | + C = arctan + C = arcsin + C

We could add to this list by differentiating a great number of functions then we'd know the antiderivatives of the answers. i.e. since = x + _ln_ | cos x | = - x tan x + _ln_ | cos x |

then = x _ln_ | cos x | + C. However, we need a more systematic way of evaluating integrals (and we'll say "integrals" and mean "indefinite integrals" unless otherwise indicated).

Note, however, that every formula for differentiation will yield a formula for integration (and, again, we mean "indefinite" integration unless otherwise indicated). For example we have = f'(x) + g'(x) so we have the integration formula: = f(x) + g(x) + C ... not too exciting.

**THE METHOD OF SUBSTITUTION** **:**

As a "Differentiation Rule" we also have the Chain Rule: f(u) = f'(u) where u is some (differentiable!) function of x. This gives the "Integration Rule": = f(u) + C, which, if we could understand what it's saying, looks pretty interesting.

Note: if we replace dx by du (as though we were "cancelling the dx", the above result reads:

= f(u) + C which is pretty obvious (i.e. f(u) is clearly the antiderivative of f'(u) !) ... so maybe the Chain Rule is of little value, but then again, this rule works for any differentiable functions f & u so maybe there's something to it ... so we should try a few f's and u's.

• If f(u) = sin u then = sin u, for any (differentiable) function "u". If u = x2, we get:

.

• If f(u) = eu and u = sin x we get the result .

• How to identify the function "u", given an integral to evaluate: ?

Look for a function of x whose derivative is "next to dx" (i.e. is a factor of integrand), and let _u = this function_!!

That's the method of substitution we'll talk about:

• I = = ? Let u = x2 (since its derivative, 2x, is "next to dx"), so = 2x hence du = dx =

2x dx. Now write I = = = sin u +C = sin x2 \+ C (as obtained above). Note: normally the indefinite integral would most likely arise as , so you could rewrite it as , or simply ignore constant multipliers and set u = x2, du = dx = 2x dx so du = x dx giving = etc.

• I = = ? Note that u = has derivative which (except for a constant factor "2") can be placed "next to dx" (i.e. is a factor of the integrand). Then put u = and du = dx = dx

so put dx = 2 du, giving I = = - 2 cos u + C = - 2 cos + C.

• = ? Let u = ln x (its derivative, = , is a factor of the integrand) and du = dx = dx and reduce this tough integral to simply = + C = + C.

• = ? Let u = 1+x2 (its derivative, 2x, is a factor of the integrand ... except for the "2" which we ignore ... it'll appear, soon) then du = dx = 2x dx hence du = x dx and get: = _ln_ |u| \+ C

= _ln_ (1+x2) + C and we conclude that = = _ln_ (1+x2) + C ... which we can verify by noting that = .

• = ? Let u = x2+1 (its derivative, = 2x , is a factor of the integrand ... except for the "2" which we ignore ... it'll appear, soon) then du = dx = 2x dx hence du = x dx and get: = = + C = u1/2 \+ C = + C.

• = ? Let u = ex (its derivative, = ex , is a factor of the integrand) and get

= so now let v = cos u (its derivative, = - sin u , is a factor of the integrand) and get = - _ln_ | v | + C = - _ln_ | cos u | + C = - _ln_ | cos ex | + C (which, of course, we should verify by differentiation).

• = ? Let u = sin x so du = dx = cos x dx and we get = tan u + C

= tan (sin x) + C.

• = ? Let u = _ln_ x then du = dx = dx and we get = _ln_ | u | + C = _ln_ | _ln_ x | + C.

• = ? Let u = sin x2 then du = dx = 2x sin x2 dx and we get = eu \+ C

= esin x2 \+ C.

If we inspect the integrals that we've been able to evaluate with this "rule", then we see that there is always some function appearing in the integrand and the derivative of this function also appears there, right beside the "dx". If you can find a function in the integrand (call it g(x)) whose derivative is also in the integrand "next to dx" (that is, dx is also there, under the ∫-sign ) i.e. the integral has the form

then let u = g(x) and replace g(x) wherever it occurs in the integrand by u (which will certainly make for a simpler integrand!) AND replace dx = dx by du (and that simplifies things even further!). The resultant integral will have the form and be easier to integrate ... usually. Just remember:

• = ? Let u = x + 1 so du = dx = (1) dx = du and get but we haven't got rid of all the x's yet, so we note that x = u - 1 and get, finally, = = - 2u + _ln_ | u | + C

= 2 \- 2(x+1) + _ln_ | x+1 | + C.

•

S: Hold on a minute. I don't get this at all. You can do a million examples and I still wouldn't get it. How can you let

u = something then let du = dx ? Are you cancelling the dx's, is that what you're doing? That isn't legal ... is it? I mean ...

P: Okay, that's a fair question. But remember, you can always check the answer by differentiating. If I say that

= 2 \- 2(x+1) + ln | x+1 | + C, then just differentiate the right-side and see if you get . See?

S: Aren't I supposed to understand what I'm doing or am I just supposed to turn some crank and out pops ...

P: Look here. I said that this arises from the Chain Rule: f(u) = f'(u) which is equivalent to

= f(u) + C. See the dx in the integrand? Good. Now watch this. We'll change it into just du. It's clear that = f(u) + C, right? (After all, if you differentiate f(u), then integrate it, you get back to f(u) ... except for a constant C.) Okay, this means that = and now see what's happened to the dx ? It's now just du, so we bypass this rigmarole and simply replace du by dx ... which is precisely what we've been doing.

S: And if I don't like it, I can just differentiate the right-side to prove it, right?

P: Sure. Remember, differentiation is a SCIENCE but integration is an ART ... perhaps a black art. Besides, checking your answer is what we like to do anyway, right? Besides, you can't argue with success ... and we've been pretty successful in evaluating the above integrals. Did I ever tell you the story of Oliver Heaviside?

S: Oh no, please don't.

P: This is interesting ... and won't be on the final exam, I promise. Heaviside was an electrical engineer who lived about the turn of the century (the 20th century). He invented a scheme for solving the darndest problems; it's called the Heaviside calculus. Mathematicians scoffed: "You can't do this!" and "How do you justify that?" and so on. His response? "Check my answer. It works."

S: I'm afraid to ask ... but give me an example.

P: Well ... let's see. Suppose we use the notation Dy in place of , and D2y in place of , and so on. Then we see that Deax = a eax and D2eax = a2eax and, continuing, D100eax = a100eax so we can just replace D by "a" and get the umpteenth derivative of eax: . "Just replace D by a". That's a Heaviside Rule. Also, if we use the notation (D+2) eax we'd mean D eax \+ 2 eax = (a + 2) eax and (D+2)2eax would mean (D+2)(D+2)eax = (D+2)((a + 2) eax) = (performing the differentiation) = (a2 \+ 2a + 4) eax which is precisely (a+2)2eax. In other words, (D+2)2eax = (a+2)2eax and again we just replace D by "a". You can check it out for yourself. (D+2)3eax = (a3 +6a2+24a +8)eax = (a+2)3eax and so on.

S: That's Heaviside calculus?

P: Part of it. Pay attention. Now we do something different. Suppose we differentiate a product. That is, we perform D(eax u(x)) where u(x) is some function of x. According to the PRODUCT rule we'd get D(eax u) = eax u' \+ a eax u or, using our slick D-notation, we'd get: D(eax u) = eax (u' + a u) which we'd write as D(eax u) = eax (D+a) u. Similarly D2(eax u) = eax (u'' + 2a u' + a2u) = eax (D+a)2u. See the rule now? No? One more time then. We differentiate eax u three times and we'd get a factor eax in every term which we'd factor out and get: eax(u''' + 3a u'' + 3a2 u' \+ a2 u) which is precisely eax(D3u +3aD2u +3a2Du +a2u) which is precisely eax(D+a)3u. Now you see the rule:

which is called the Heaviside Shift Theorem. The eax gets shifted to the left, past the D, and "a" gets added to D leaving the operation (D+a)n to be performed on the function u(x).

S: That's Heaviside calculus?

P: Listen! There's more! Now if Dy means then y should mean ... what? Well, whatever y means, when we

differentiate it via D we should get y, right? After all, we just "cancel the D's". That means that y must be nobody else but . Now, if we perform the operation eax we get = then if we perform that operation again we'd get eax = eax = . See? Just replace D by "a". Now comes the good part.

S: Can I go now?

P: The best is yet to come! We want to evaluate which Heaviside would write as ex x2 and then he'd use his Shift Theorem to get ex x2 (where we "shifted the ex" and "added 1 to D"). What to do with x2? Well, we can divide D + 1 into 1 by long division (either that, or recognize as the sum of a geometric series with first term "1" and common ratio -D ). In any case we'd get:= ex x2 = ex x2 = ex x2

= ex where we performed the indicated differentiations on x2 and we'd conclude that

= ex \+ C and we finally added the "C".

S: You gotta be kidding. Can you really do all that?

P: As Oliver would say: "Check my answer. It works."

S: Can we get back to "let u = this and du = that"?

P: First, I have one more technique to show you ... for evaluating indefinite integrals. This is perhaps the most important technique available and it's really very nice.

**INTEGRATION BY PARTS** **:**

We consider one last "Differentiation Rule", hoping it will result in an "Integration Rule" ... and the rule we consider is a familiar one: uv = u + v . Integrating both sides with respect to x, we get an "Integration Rule": = + . Now stand back. By we mean a function which, when differentiated, yields uv and that's clearly uv itself. Our "Integration Rule" then becomes: uv = + which we can rearrange to read:

. Before we say anything else, let's substitute some functions for u and v and see if this "Rule" is getting us anywhere.

Examples:

• Let u = x and v = sin x. Then we substitute = 1 and = cos x and = uv - becomes:

which is very nice because is easy, so we get

= x sin x + cos x + C, and we can verify that we do indeed have the antiderivative of x sin x because = x sin x.

• Let u = x2 and v = sin x. Then we substitute = 2x and = cos x and = uv - becomes:

= x2 sin x - or . In other words, we've replaced the problem of evaluating by the problem of evaluating the simpler integral

but, of course, we've already evaluated this one by the same technique so we can write, finally:

= x2 sin x - 2+ C (adding the C after the last integral sign has gone).

• Let u = arctan x and v = x, so = and = 1 so the "Rule" = uv - becomes:

= x arctan x - _ln_ (1+x2) + C (where we put =

_ln_ (1+x2) because we've already done this one, earlier, using our method of substitution).

S: Please, no more! Are you really thinking about starting with this ... this "Rule",= uv - , then just plugging in functions for u and v and seeing what we get? I mean, is that what I'm supposed to ...

P: I just want to check to see if this Integration Rule which derives from the Product Rule for differentiation is going to be useful ... and we can see that it is, because how else would you integrate arctan x? In general, you'll run across and you'll have to figure out what is u and what is v and ...

S: You're kidding, right? I mean, you already knew the answer ... who's u and who's v ... then you invented the integral, yet you expect me to start with the integral and ...

P: Okay, that's a fair comment, so let's look at our new Integration Rule ... carefully:

INTEGRATION BY PARTS FORMULA

= uv -

Comments:

• The original integral, having the form , is replaced by uv - which involves another integral ... hopefully simpler than the original! But we have done a partial integration, because "uv" has come out from under an sign ... so this technique is called INTEGRATION by PARTS.

• The original integral involves the integration of a certain _product_ : ... which, in our examples above, were x cos x or x2 cos x or (1) arctan x. After applying the Integration by Parts formula given above, we are still left with an integral, but now the two factors (u) and reappear transformed: the first factor, (u), reappears differentiated, , whereas the second factor, , reappears integrated, (v), giving the remaining integral . This transformation gives us the clue as to who's u and who's v. In fact, we want to choose them so the remaining integral is simpler.

• Since "u" reappears differentiated, we let u = _the factor we want to reappear differentiated_. What's left in the original integrand must then be .

**Examples:** (a) = **?** (b) = **?**

Solutions:

(a) Since _ln_ x is a frightening (transcendental) function with a very nice derivative, we let u = _ln_ x (because we'd rather have the derivative = ). That means that = x so v = . Having identified all four players in this drama: u, , v and , we can substitute into the Integration by Parts formula = uv - to get: = -

= _ln_ x - and you must admit that the remaining integral is pretty easy. Finally, then,

= _ln_ x - + C (which we can verify by differentiating the right-side, to get x _ln_ x).

(b) In , the function arctan x has a "nice" derivative so we let u = arctan x, hence must be x (since the original integrand must be u so if we know u we also know ). We need all four of u, v, and so we calculate = and v = . Substituting into the "Formula" we get = = uv - = arctan x - . We pause long enough to evaluate the remaining integral: = = = x - arctan x so we finally get:

= arctan x - + C.

S: Whoa! What happened there? I mean, = and all that jazz?

P: Cute, eh? I just added and subtracted "1" from the numerator, then split it up into two pieces, each being easy to integrate, namely and whose integral I've memorized. Do you like that?

S: No! Besides, you said you let u = arctan x because it has a "nice" derivative. There's another factor in and it's x and it has an even "nicer "derivative so I'd let u = x.

P: And what does that leave for ?

S: Huh?

P: The original integrand is x arctan x which is u . If u = x, then must be arctan x. Hence, to find v itself, you'd have to integrate arctan x. See? My way, letting u = arctan x, eliminates the need for integrating arctan x.

S: But we know from an earlier example ... it's ... uh, x arctan x - ln (1+x2), so doing it my way, I'd let u = x and = arctan x so = 1 and v = = x arctan x - ln (1+x2) and I'd use the magic integration-by-parts formula and get: = x- ... mamma mia, that's not so nice ... uh, I leave it as an exercise for the prof.

It is convenient, for purposes of "memorizing the formula", to rewrite = uv - in the form where we've replaced dx by dv and dx by du. (We've done this _replacement_ before, in our "method of substitution".) Although easier to remember, I don't think it's easier to use!

**Example** : = ?

**Solution:** Let u = _ln_ x, then = . What's left under the integral sign is x2 dx which must be dv, that is, dv = x2 dx so v = = . Now we use = uv - and get = _ln_ x - = _ln_ x - + C (which, as usual, can be verified by differentiating the right-side to get x2 _ln_ x).

S: I like = uv - much better than = uv - .

P: Then use it! In fact, when I was your age my calculus prof gave us the formula as:

which scared me, I admit, but now that I'm much older and a little wiser I see what it's saying: if the original integral has a product, f(x) g(x), you use Integration by Parts and get another integral where the first factor reappears differentiated, that's f'(x), and the second reappears integrated, that's If you remember that, then you can determine "who's u and who's v" in your head. For example, suppose I wanted to integrate . I could use integration by parts and get two possible "second integrals", either

where the 2nd factor was differentiated, or where the 1st factor was differentiated. I'd choose the first type, letting u = the factor I want to reappear differentiated, namely u = arcsin x.

Similarly, for I could use Integration by Parts and replace this integral with ... the integrand is (the integral of x) (the derivative of ex) ... OR with where the integrand is

(the derivative of x) (the integral of ex))). I'd choose the latter, letting u = the factor I want to reappear differentiated, namely u = x.

Assorted Examples in Integration:

• For = , let u = x+1, du = dx = dx and get =

= _ln_ |u| + + C = _ln_ |x+1| + + C

• For , let u = 1 + x2/3 so du = x-1/3 dx so dx = x1/3 du and get

= = = + C = + C

S: Wait! I thought we look for a function whose derivative is "next to dx" and let u = this function and so on. But I don't see that that's what you're doing.

P: Okay, let me give you another technique: look for the hairiest function around and u = that. Then, by hook or crook, get rid of every "x" and "dx" and hope the resultant integration in terms of "u" and "du" is easier. In the first example above, after recognizing x2+2x+1 = (x+1)2, I didn't like (x+1)2 in the denominator ... it'd be much easier if it were simply x2 living there ... so I let u = x+1 and get a u2 in the denominator. And replacing the numerator and the "dx" by the appropriate expressions in "u" didn't make things any worse ... and I got an easy integral in terms of "u". See? In the next example I didn't like the x2/3 so I could have let u = x2/3 and du = (2/3)x-1/3dx and get but then I'd get 1+u in the denominator which I'd like to replace by a single variable so I'd continue with a second substitution, letting v = 1+u and so on ... hence I thought it'd be easier to simply let u = 1 + x2/3 right away and ..

S: That's confusing. How would I know to do that?

P: Experience. Do dozens of integrals and you eventually get the hang of it. Remember, integration is an art. You have to have a feel for it and you get this "feel" by doing lots of examples ... yourself ... not just watching me or reading the examples in some book. Remember, too, that math is not a spectator sport. Get involved. Anyway, let me say something about the last example above. I eventually eliminated all references to "x" and "dx" and got an easy integral in "u" and "du":

= + C = + C which I would most likely write

as + C absorbing the constant into the arbitrary constant C.

S: And you forgot the absolute value signs in the ln !

P: Well, actually, I didn't because | 1 + x2/3 | = 1 + x2/3 since this function is always positive.

S: How would I know that?

P: x2/3 is a square. It's the square of x1/3. That makes it positive, or at least non-negative. Everybody knows that ... even you.

S: Can you do an example which doesn't work out? I mean, sometimes you make a mistake, right? Sometimes, you let u = this and you should have let u = that.

**Example:** Evaluate ... if you can.

• Try u = ex \+ 1 so ex = u - 1 hence x = _ln_ (u - 1) (taking the _ln_ of each side) and dx = du = du then we'd get which actually seems worse that the original integral.

• Now let's try u = _ln_ x, then x = eu and dx = du = eu du and we'd get which is awful!

• Let's try ... let's try ... uh ...

S: Give up?

P: Yeah, I can't do this one.

S: Welcome to the club.

P: That's not surprising, you know. If you just invent some function, f(x), chances are I couldn't evaluate in terms of known functions. But I could always invent a brand new function. In fact, let me invent the following function: P(x) = then what do you think P(x) is?

S: I haven't the foggiest.

P: Remember the Fundamental theorem, and how we managed to evaluate definite integrals without resorting to Riemann SUMS? In trying to evaluate we invented the function A(t) = and discovered that A(t) = f(t). We should make a fuss about this so you'll remember it:

= f(u)

In fact, I even changed the labels on the variables so you'd recognize it even if the variables aren't called "x" and "t". For example:

• If H(w) = then H'(w) = esin w.

• If G(k) = then G(k) = tan k2.

• = _ln ln_ x.

• If P(x) = then P(x) = . Hence, I now have a function whose derivative is just as I promised. Hence, = P(x) + C.

S: You can't do that! Just because you gave it a name ... called it P(x) ... does that mean you've actually evaluated this integral? Can I do that too? I mean, if you give me some integral I can't evaluate, can I just call it P(x) ... and will I get full marks on an exam?

P: No. I'll only give you integrals which can be evaluated in terms of known functions, and that's how I want you to evaluate them. I'm only telling you all this ... inventing new functions ... so you'll see that it can be done and you won't ever say "this integral can't be evaluated" when you really mean "this integral can't be evaluated in terms of known functions". Let me see you evaluate .

S: Easy. It's ln | x | + C.

P: And before the natural log function was invented, what would you say about this integral?

S: Uh ... it can't be evaluated in terms of known functions, right?

P: Right. Now suppose I define P(x) = ... which is, after all, the area associated with the curve y = from t = 1 to t = x ... and if I determined this area very accurately for each x (using, for example, Riemann sums with plenty of rectangles) and if I made a table of values of P(x) and gave this function a name, say the Ponzo function, and I found where it was increasing and decreasing (because, you see, I know the derivative of P(x)), and I managed to get it into a bunch of calculus books ... then, in future, when students ran across the above integral they'd just write the answer as P(x) + C because P(x) would be a "known function". See? It's just like writing answers ln | x | + C or sin x + C or ex \+ C. See? You're happy only because somebody has already invented the ln and sine functions.

S: I guess so.

Examples:

(a) = ?

(b) = ?

(c) = ?

Solutions:

(a) =- (using a property of definite integrals)

= - where u = t2. Then gives

\- (where we've used the Chain Rule)

= _ln_ (1+u2) (2t) - _ln_ (1+t2) = 2t _ln_ (1+t4) - _ln_ (1+t2)

Note: In the above, we want to use = f(u) but this requires:

(i) the lower limit must be constant ... and that's the reason for writing = -

(ii) the upper limit is a variable,

(iii) we're differentiating with respect to this variable upper limit (that's why we put u = t2)

(b) In , we'll use = uv - and we'll let u = _ln_ t so dv = dt. Then du = dt = dt and v = = t hence we get uv - = t _ln_ t - t + C (which we verify by differentiating).

(c) = [ _ln_ x _ln_ ( _ln_ x) - _ln_ x ] (we've already evaluated the indefinite integral, earlier)

=( _ln_ e2 _ln_ ( _ln_ e2) - _ln_ e2) - ( _ln_ e _ln_ ( _ln_ e) - _ln_ e) = (2 _ln_ 2 \- 2) - (0 - 1) = 2 _ln_ 2 - 1

When we evaluate a definite integral (as opposed to an indefinite integral), we must first find an antiderivative of the integrand (i.e. the indefinite integral), then evaluate at it the upper limit, then at the lower limit, then subtract. If we make a substitution, u = g(x) say, then integrate, we get a function of u so we have to go back to x again to evaluate at the upper and lower limits. This is sometimes inconvenient, especially if you make more than one substitution to determine the antiderivative. It's nice to know, then, that you can carry your limits along with you. Just ...

S: I haven't the faintest idea what you're talking about. Can't you give me an example?

Consider . We make the substitution u = 1+x2 so that du = dx = 2x dx, hence x dx = du and we get = = _ln_ | u | (and we omit the constant of integration because we want to evaluate a definite integral). Now we return to the original variable, x, and get:= _ln_ (1+x2)]= _ln_ 2 - _ln_ 1. HOWEVER (and here's the nice part) we could also have stuck with the "u" variable, changed the limits and written _ln_ |u | ]= _ln_ 2 - _ln_ 1. See how it works? When x = 0, then u = 1+02 = 1 and when x = 1 then u = 1+12 = 2 so the u-limits are 1 and 2. In other words, when we changed variables (from x to u) we can also _change limits_ , writing:

etc. etc.

**Examples:** Evaluate: (a) (b)

(c) (d) (e)

(f) (g) (h)

(i)

Solutions:

(a) = . Let u = 1-3x so du = -3 dx and get = + C = + C

(b) = . Let u = 1-3x so du = -3 dx. Also, when x goes from 1 to 2, u goes from -2 to -5 so we get = = = = .

(c) = 0 because sin t3 is an ODD function.

(d) = = = x - arctan x + C

S: You've already done that one.

P: I wanted you to see "add and subtract" again. Pay attention.

(e) . Let x = tan t so dx = dx = sec2t dt (we make this substitution because we see not only tan t but also its derivative, sec2t). When t goes from 0 to π/4, x goes from tan 0 = 0 to tan π/4 = 1 so we get which we recognize as the area under a quarter-circle of radius 1 (since y = is the upper half of the circle x2+y2=1, and for x between 0 and 1 we get the area of a 1/4-circle) namely π12 = .

(f) . Let u = _ln_ t so du = and when t goes from 1 to e, u goes from _ln_ 1 = 0 to _ln_ e = 1. We get

= = = .

(g) = . Let u = x-1 and du = dx. For x =, u = - and for x = 1, u = 0 so we get

== = - = 0 - (- ) =

S: That's pretty tricky. Should I know all this? I'll never remember it.

P: Completing the squares is useful;. Remember it! See how many examples I'm doing? You'll get it by osmosis.

(h) . Let u = so du = dx = and we get 2 = 2 _ln_ | sec u | + C

(i) . Let u = arcsin z so (using the Chain rule) we get = = esin u = eu .

LECTURE 18

APPLICATIONS OF THE DEFINITE INTEGRAL

**VOLUMES** **:**

Now that we're experts in integration, we should evaluate some integrals which represent something other than areas.

Consider the problem of determining the volume of a solid. If it were a right-circular cylinder we'd have a formula: πr2h (where r = the radius, h = the length). If it were a sphere we'd also have a formula: 4πr3/3. In fact, there are many solids for which we have formulas ... BUT, we need a method in cases where the solid is irregular. Just as we did when we considered the problem of the area enclosed by curves, we subdivide the volume into simpler volumes for which we do have a formula. (Remember the rectangles, both vertical and horizontal ... and the circular sectors for areas enclosed by polar curves?)

  | Before we go on, let's consider the volume generated by a plane area, A, which moves a distance L in a direction perpendicular to the plane. The volume generated is AL. If the plane area is a circle, then A = πr2 and the volume of the "right-circular cylinder" is, as we've seen, πr2h. That's just a special case. For any plane area A, the solid is _still_ called a cylinder and the volume is AL.

---|---

Now consider a more general solid.

  | Perhaps the simplest method is to cut the solid into many very thin slices. Each slice is very nearly cylindrical with some cross-sectional area, say A, and a thickness, say h. The volume of such a slice is (very nearly) Ah. To make things more precise, we'll assume that the volume is cut by a plane (or a knife) at a distance x from some fixed plane. The cross-sectional area will, of course, depend upon x. We'll call it A(x) and the thickness of the slice, at the place x, we'll call dx (which is a better name, for now, than "h"). Then A(x) dx is a good approximation to the volume of the thin slice. If we SUM the volumes of all such slices

---|---

we'll get the total volume of the solid, and the definite integral does the for us.

We get . In other words, the volume of a solid is the integral of the cross-sectional area.

S: You're going too fast. I don't see how you'll get the exact volume of the solid. Maybe you'll get an approximation, but the exact volume? Not likely!

P: Okay, we'll do this from scratch. (Aren't you tired of constantly going back to Riemann SUMs?) The volume lies between x = a and x = b (where "x" measures the distance of the slices from some fixed plane) so now we subdivide the interval [a,b] into n subintervals of length h = and in each subinterval we pick a convenient value of x, say x1 in the first subinterval, x2 in the second and so on (does this sound familiar?) and now we consider the volume of the first slice of width h and cross-sectional area A(x1), the second slice of width h and cross-sectional area A(x2) and so on and the SUM of volumes of all such slices is A(x1) h + A(x2) h + ... A(xn) h = which we recognize as a Riemann SUM so now we let n∞ so h0 (does this sound familiar?) and get, in the limit, , the total volume of the solid. See?

S: Well ... not really.

P: Okay, we'll do it differently. Suppose we let A(x) be the cross-sectional area at the place x and let V(x) be the volume lying to the left of x. (i.e. from "a" to "x".) Now we'll differentiate V(x) via . Note that V(x+h) - V(x) is just the volume from x to x+h and we can write:

Aminimum h < V(x+h) - V(x) < Amaximum h

where Aminimum is the minimum and Amaximum the maximum cross-sectional area between x and x+h. What we've done is found a cylindrical volume

|

---|---

bigger and smaller than V(x+h) - V(x). Now ...

S: Hey ... let me do it! This does seem familiar! We SQUEEZE, right? Okay ... uh, let's see ... we write

Aminimum h < V(x+h) - V(x) < Amaximum h

then Aminimum < < Amaximum

then Aminimum < < Amaximum

then ... uh ...

P: Great! Keep going! The guy in the middle is just and now you've got to guarantee that the limits on each side are the same. That's how we SQUEEZE.

S: Are they the same? I mean, what is the limit of Aminimum?

P: Remember when we did this before, for areas? Go back and look at your notes. The method is exactly the same. The limits of Aminimum and Amaximum are just A(x) because they're the cross-sectional areas at some points between x and x+h and h is going to zero so ...

S: Yeah, I got it, but we'd get A(x)< < A(x), or maybe A(x) < < A(x) and how does that give us ?

P: When taking limits we'd get A(x) ≤ ≤ A(x) and that makes = A(x).

S: How'd the < change all of a sudden to ≤ ?

P: Hmmm ... good question. Let's do this for something simpler. If x is a number less than 1, say 0 < x < 1, then it's greater than 0 and smaller than x2. (If you don't believe me, try a few squares and convince yourself that x < x2.) Okay, we then have 0 < x < x2. Now let x0. Do you think that 0 < x < x2. No. In fact all three are precisely 0, so when taking limits across inequalities we can't guarantee that < doesn't change to ≤, so we always write ≤ instead of <, just in case ... sort of insurance. In our case we'd get A(x) ≤ ≤ A(x) hence V(x) is an antiderivative of A(x). In fact, it's the special antiderivative that has the value 0 when x = a ... and that makes V(x) = . See?

= = A(x) (Remember? It's the integrand evaluated at the upper limit.) In particular, the total volume of our solid is V(b) or which, of course, is the same as .

S:S: I meant to ask you about that. You seem to rather cavalier with what you call the variable under the integral sign. You call it "t", then you call it "x" and sometimes you even ...

P: = = F(b) - F(a). Remember? Now watch this: = = F(b) - F(a) and =

= F(b) - F(a). See? They're all the same number. It doesn't matter what you call the variable of integration. When the smoke clears, it's gone anyway. The variable of integration is sometimes called a "dummy variable". See?

S: Then why don't you always use the same name, like x ... which I like best.

P: Well ... uh, that's a good question. You see, it's like this. Nobody likes to write because the x in the upper limit gets confused with the x under the integral. The upper limit is a particular value of the other x. You see ... uh, it's much better to distinguish between them and write so we're considering the area from x = a to x = t (and it'd be bad manners to write "from x = a to x = x").

S: I've heard lots of people say that: "from x = a to x = x".

P: Well, you won't hear me say it!

S: But didn't you say ... when we were slicing that solid ... that we measure distance and call it x and slice the solid at the place x, and that'd mean at x = x, wouldn't it?

P: Uh ... I think we should go on. We've still got a ways to go before the end of this lecture. Pay attention.

Examples:

**Volume of a cylinder** **:** Assume the cylinder lies between x = 0 and x = L and we slice it by planes perpendicular to the axis of the cylinder. The cross-sectional area is constant at some value A. Then V = = = AL, as expected. For a right circular cylinder, A = πr2 and the volume is V = πr2L. |

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**Volume of a cone** **:**

The area A0 lies in a plane. A point P lies a distance h above the plane. Every point on the perimeter of the area is joined to P, forming a cone. We slice the cone by planes parallel to the plane of the area, a distance x from the vertex of the cone. The cross-sectional area is a scaled-down version of the base area and has the value A(x) = 2A0. The volume is V = = = = A0 h. If it's a right circular cone, A0 = πr2 and the volume is V = πr2h, as expected.

|

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S: Wait. The cross-section has area A(x) = A0, not 2A0. How do you get ...

P: Suppose the area is a circle, so A0 = πr2. If we slice half-way from vertex to base the cross-section is again a circle with what radius?

S: I'd say r.

P: Right! And what's the area of a circle of radius r? It's π 2 = πr2. See? It's of the base area, not . In fact, if we scale every length by a factor then an area will be as large. If we scale down a solid so every length is reduced to then the volume is reduced to 3 = . See how it works? If every metre of length is reduced by a certain factor then every quantity that's measured in metres2 (like SURFACE AREA or CROSS-SECTIONAL AREA) gets reduced by that factor squared. Quantities measured in metres3 (like VOLUME) get reduced by that factor cubed. Anyway, we see that the well-known formula for the volume of a cone, namely holds for any cone, not just right circular cones. Nice, eh?

S: If you say so.

**VOLUMES OF SOLIDS OF REVOLUTION** **:**

  | The simplest solids to deal with are clearly those which have particularly simple cross-sectional areas (since we have to integrate this to get the volume). One class of solids which has this nice feature is _Solids of Revolution_. We begin with an area in the x-y plane, enclosed by given curves ... for now we assume that the area is bounded by y = f(x), x = a, x = b and the x-axis (y = 0) as shown. Then we revolve this area about the x-axis, sweeping out a solid of revolution. Now, when we slice this volume by planes perpendicular to the x-axis the cross-sections are just circles! Indeed, if we locate the plane by its distance from the y-axis (x = 0), the radius of the circular cross-section is just y = f(x)

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so A(x) = πy2 and if the thickness of each slice is dx then volume of the slice at the place "x" is just

A(x) dx = π y2 dx. The total volume of the solid of revolution is then: V = = π

A nicer way to think of this is to avoid revolving the _entire_ area at one time ... just subdivide the area into vertical rectangles and revolve these one at a time. Each such rectangle sweeps out a disc of radius f(x) (since it's located at the place x and the y-coordinate _is_ the radius) and thickness dx. The volume is then πy2 dx and all such disc-volumes (for every elemental rectangle, from x = a to x = b) gives

V = π as before. The advantage of this |

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interpretation of the formula π is that we can also consider subdividing the area into _other_ elemental areas (such as horizontal rectangles!) and revolving each.

**Example:** Determine the volume of a cone of height H and base radius R.

**Solution:** We recognize the cone as a solid of revolution: the area enclosed by y = x , x = 0, x = H, y = 0 is revolved about the x-axis. The volume is then:

V = π= π 2 = π 2 = πR2H

|

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S: You just did a cone. Do you like cones?

P: Yes ... now pay attention.

**Example:** Determine the volume of a sphere.

**Solution:** We first recognize that a sphere is a solid of revolution: just revolve the semi-circle y = about the x-axis, from x = -a to x = a. The volume is V = π= π= π= πa3

**Example:** Calculate the volume of a torus (or donut).

**Solution:** To generate a torus we revolve the area within a circle, x2+(y-b)2 = a2, about the x-axis. We can solve this for y = b ± , giving an upper half-circle and a lower half-circle. Now we subdivide the area into vertical rectangles (located at the place x, with thickness dx) and revolve each, generating a disc-with-a-hole (or a "washer"). If the outer radius is R and the inner radius r, the washer has volume dx (since the cross-sectional area is the area between two circles). But R is the y-value of the upper half-circle: R = b + and r is the y-value of the lower half-circle: r = b - , hence the volume of the washer is:

π(R2 \- r2) dx = π(R - r)(R + r)dx = 4πb dx and the total volume of the torus is the from x = -a to x = a, giving: V = 4πb.

Although this integral seems difficult, we recognize it as the area beneath a semi-circle of radius "a", namely π a2. Finally, then, V = 2π2ba2 which certainly has the dimensions of volume: if "a" and "b" are measured in metres, then ba2 is in metres3 as it should be. Also, if a = 0 we get V = 0 as expected (and that gives us more confidence in our answer). Finally (and this is really interesting), we can write this volume as (πa2) (2πb) which is just the area of the circle (πa2) multiplied by the distance moved by its centre (around a circle of radius b, hence a distance 2πb) ... and _that_ gives us even more confidence in our answer.

S: Why do you need all this confidence? Don't you have any faith in your answer? |

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P: I have faith only if the answer seems reasonable ... and this answer does seem reasonable.

S: I like that last thing you said: the volume is just the area multiplied by the distance ... uh, how did it go?

P: Revolve an area about a line (like the circular area about the x-axis) and the volume of the solid of revolution is just (area)(distance travelled by the centre of the area). In our case, area = πa2 and the distance travelled is 2πb. Nice, eh?

S: Hey, can I use that instead of all this integration? I'd never be able to do the donut your way.

P: Sure, do this one: find the volume when the area enclosed by y = x and y = x2 is revolved about the x-axis.

S: I first draw a picture ... uh, y = x, I know that one, but y = x2 ... that one starts at (0,0) and gets bigger as x gets bigger so I'll just sketch a curve which gets bigger. Then I find the points of intersection; I solve y = x2 and y = x and I get x2 = x so x = 1. I conclude that ...

P: What! x = 1? Is that all? What about the intersection at x = 0? You can't just go from x2 = x to x = 1.

S: Sure I can. I divided by x, don't you see? Pay attention. I get ...

P: You can divide by x only if x ≠ 0 and that means you've just thrown out a solution, namely x = 0. Look. Write x2 = x as x2 \- x = 0 so x (x - 1) = 0 so either x = 0 or x = 1. See? I get both solutions.

S: Do you want me to do this, or what? Okay, I get intersections at (0,0) and (1,1). Now I find the area between the curves by using vertical rectangles ... the width is dx and the height ... uh, how do I know which curve is bigger? I mean, which is the upper y-value? I won't worry about that now, I'll just assume y = x2 is bigger so that gives me rectangles with an area of (x2 \- x) dx and I add them all up ... them (as you're so fond of saying) ... and I get the area

= = - = = . I conclude that the area is , so now I ...

P: What! You got a negative answer. So where did you go wrong? I'll tell you where you went wrong, you ...

S: I know, I know ... y = x is bigger, so who cares? I know the area is positive, so I just forget the (-) sign. Pay attention. Now I'm going to find the volume when I revolve this area about the x-axis. I just use (area)(distance travelled by the centre of the area) and the area = 1/6 and distance travelled is ... is ... how am I supposed to know where the centre of the area is? I guess it's half-way ... uh, at (1/2,1/2) ... so that'd make a distance travelled of about 2π(1/2) or just π, so the volume is (1/6)π. How's that?

P: Terrible! You don't even know if x2 is larger or smaller than x and you're just guessing the location of the centre of the area and ...

S: Okay, you do it.

  | P: First I make a reasonable diagram (and I know that x2 is smaller than x when x lies in 0 < x < 1 since I just have to try one value, say x=.5 then x2=.25 is smaller). Then I subdivide into vertical rectangles of area (x - x2) dx and from 0 to 1 and get as the area. Next I find the volume by revolving each rectangle about the x-axis. Each gives a "washer" of inner radius y = x2 and outer radius y = x so the volume is πdx and I these from x = 0 to x = 1 and get the total volume

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V = π= π=

S: I thought you were going to use (area)(distance travelled by the centre of the area) to find the volume. But I know why you didn't use it ... you don't know where the centre of the area is, right?

P: Sure I do. It's located at ... uh, the y-coordinate of the centre of area is ... give me a second ... it's .

S: How'd you get that? I mean, how do you know the centre of area is at y = ?

P: Pay attention: I use (area)(distance travelled by the centre of the area) and I 've computed the area as and the centre of area I take to be at y = so it travels around a circle of radius so I get a volume of (area)(distance travelled ) which is = and, as we've seen, it's the correct volume. See? y = works!

S: But that's cheating! I bet you just figured out where the centre of area had to be to give the right volume. Am I right?

P: Yes. And that's as good a way as any to define what we mean by the "centre of area". It's located at a point such that (area)(distance travelled by the centre of the area) gives the volume. For example, the y-coordinate of the centre of area ... let's call it y- ... satisfies A 2π y- = V (where A = area and V = volume) so we have a formula for finding the centre of area: y- = . Nice, eh? You see, it makes sense, this definition of "centre of area", and it works when the area is a circle so that gives us some faith in using this definition and besides, if I define the centre of area in this manner who can argue with me? It's a definition, after all, and we can't argue with a definition! It just has to be a reasonable definition which agrees with what we'd expect the centre of area to be in simple cases. See? Just use y- = .

S: Can we patent that formula ... or has somebody already discovered it?

P: It's called:

  |

THE THEOREM OF PAPPUS

A planar region of area A is revolved about a line

(in the same plane as the region).

The volume of the solid of revolution is given by:

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Given any two of V, y- and A we can compute the third from A = 2π y- A.

Examples:

  | Since the area of a semi-circle is A = πr2 and the volume (when revolved about the diameter) is V = πr3 , then the centre of area (or CENTROID) is located at y- = = r (the distance from the diameter).

We revolve a right-triangle about a side, generating a right-circular cone. The area of the triangle is A = = a b and the volume of the cone is V = = πa2 b, so the CENTROID of the triangle is located a distance = a from the line about which we revolved.

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S: Hey! I knew that! The centroid of a triangle is 1/3 of ... of something. Say, can I use this theorem is you ask me to compute a volume?

P: Only if I say "Use the Theorem of Pappus to compute ..."

S: But why? If it's easier, then I should be allowed to use it!

P: If I want to test you on your ability to set up a definite integral and evaluate it, then I won't allow you to use Pappus. However, by all means use Pappus to test for "reasonableness".

S: But if I get the right answer, surely ...

P: No! Getting the right answer isn't as important as knowing a method of solution. You can write

= ∞ - ∞ = 0 and get the right answer, but it's not worth any marks because the method is all wrong. On the other hand, even if you get the wrong answer you may still get full marks. For example, you solve a complicated problem ... demonstrating your ability to solve this type of problem ... then you end up with (2)(3) which you inadvertently write as "5". I'd give you full marks because I know you can multiply "2" times "3" correctly ... I think.

S: Thanks.

Example:

Compute the volume generated when the region enclosed by y = x, y = x2 and y = is revolved about the x-axis.

**Solution:** The points of intersection are (0,0), (.5,.5) and (,.5) as shown in the diagram. For 0 ≤ x ≤ .5, the vertical rectangles rise from y = x2 to y = x, and in .5 ≤ x ≤ they rise from y = x2 to y = .5 so we'll need two definite integrals. When we revolve the rectangles in [0,.5] we get "washers" with volume π((x)2 \- (x2)2) dx and when we revolve the rectangles in [.5,] the volumes are π((1)2-(x2)2) dx. |

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The total volume is then:

V = π+ π= π+ π

= - which is roughly πwhich is roughly 3(.7 - .5) ≈ .6, and we should check this for reasonableness. However, since we'll be doing this problem again, differently, we'll wait. |

LECTURE 19

MORE APPLICATIONS OF THE DEFINITE INTEGRAL

Volumes of solids of revolution using horizontal rectangles

In the previous problem we were faced with a region which had to be subdivided into two parts: 0 ≤ x ≤ and ≤ x ≤ . There were different expression for the height of the vertical rectangles depending upon where the rectangle was located. Had we taken horizontal rectangles (as we often do when computing AREA rather than VOLUME) and revolved each about the x-axis, then we'd get a single definite integral for the volume. That's because a horizontal rectangle, located at the place "y" (between y = 0 and y = .5), _always_ runs from y = x (on the left) to y = x2 (on the right). |

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Okay, at the place y the right-most x-value is x =and the left-most is x = y so the width of the rectangle is xright \- xleft = - y (compare this to

yupper \- ylower when we took vertical rectangles!). The height of the horizontal rectangle is dy (it's an increment in "y" rather than an increment in "x", so we call it dy). If we wanted the AREA of the region we'd just |

via . However, we want to revolve each horizontal rectangle about the x-axis, giving a VOLUME. The solid generated is a cylindrical shell. (A _very thin_ cylindrical shell). We need to know the volume of such a shell (as we needed to know the volume of a "disc" or a "washer" when we took vertical rectangles).

  | Imagine such a shell as a rectangular piece of paper rolled into a cylinder. To find its volume we cut it, unroll it, and recognize that its volume is just

( _width_ )( _length_ )( _thickness_ ) = (W)(2πR)(t) where "t" is the thickness and the _length_ is just the circumference of a circle of radius R. It's much like painting the sides of a cylindrical soup can, then asking for the volume of paint: the thickness is _very_ small.

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  | For our horizontal rectangle we need only identify the _width_ , _height_ and _thickness_. They're - y , 2πy and dy respectively. Then the volume of each cylindrical shell is just 2πy (- y) dy and the total volume of all such shells, obtained by revolving all horizontal rectangles from y = 0 to y = 1, is the , namely:

V = 2π = 2π

= 2π = - .

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S: Hold on! Last time you got - !

P: Well, I guess they're the same number ... wouldn't you think?

S: But they're not the same! I mean, they sure don't look the same. I mean ...

P: Well look carefully through both solutions and see if you can find any error. That's left as an exercise for the student ... and guess who's the student?

When we use horizontal rectangles we need to solve for x, from the equations for the curves which enclose the region. That's because we need the width of the rectangle located at some y-value and that's xright \- xleft. In general, if the region we're revolving is bounded by, say y = f(x) and y = g(x) as shown below, then, for vertical rectangles we'd get a thin washer with volume π(g2(x) - f2(x)) dx (assuming that y = g(x) is the upper curve) and for horizontal rectangles we'd get a thin cylindrical shell with volume 2πy (F(y) - G(y)) dy where y = f(x), when solved for x gives x = F(y) and y = g(x) gives x = G(y).

  |

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**NOTE:** F and f are inverse functions ... and g and G are also inverse functions ... and it's not always clear which will give the simpler integrals to evaluate: horizontal or vertical rectangles.

Example:

Compute the volume when the region enclosed by y = sin x, y = x and y = 1 is revolved about the x-axis.

**Solution:** If we take vertical rectangles we'd get the volume expressed as

V = π+ πboth of which are pretty easy (the only tough integral is but if we write sin2x = it's easy) ... BUT there are two integrals. If we try horizontal rectangles we solve y = x for x = y (pretty easy) and y = sin x for x = arcsin y (and right away we expect a problem!). The volume (expressed in terms of horizontal rectangles, revolved) is V = 2πwhich, although it's only a single integral, is more difficult (because of the arcsine function).

S: Aren't there any real problems we can ... uh, you can solve? I mean, problems which aren't just math problems?

P: Using definite integrals? Sure. Lots of them. Remember, the definite integral can represent things other than AREA or VOLUME. This area/volume stuff is just so we can picture what's happening.

**Example:** The speed of an object depends upon the time according to v(t) = t2 \+ t metres/second, where t is measured in seconds. Determine the distance travelled during the time interval from t = 1 to t = 3.

**Solution:** If x(t) is the distance travelled in a time t, then = v = t2 \+ t so x(t) = = + + C. Further, x = 0 when t = 0 so (substituting) we find 0 = 0 + C so C = 0 and x(t) = + . When t = 1,

x(1) = + and when t = 3, x(3) = + . The distance travelled during the time from t = 1 to t = 3 is

x(3) - x(1) = - = and this is just . We conclude that the distance can be found directly by integrating the speed from t = 1 to t = 3 ... a definite integral: .

Note that v(t) is the rate of change of distance measured in, say, _km/hour_ , and x(t) is the distance travelled (in _km_ ) after a time t. Hence v(t) dt is the number of _kilometres_ during dt _hours_ and gives the total number of _kilometres_. This understanding goes with a number of similar problems.

• If A(z) is the change in the number of _acres per foot_ , at z _feet_ , then A(z) dz is the number of _acres_ for a change of dz _feet_ and is the total change in the number of _acres_ from z = a to z = b _feet_.

• If U(T) is the change in _pressure per degree_ Celsius, when the temperature is T˚, then U(T) dT is the pressure change when the temperature changes from T to T+dT and gives as the total change in pressure when the temperature changes from T = a to T = b.

• If the growth is M(t) _cm per month_ , after t _months_ , then M(t) dt is the number of _centimetres_ of growth during the time from t to t+dt and is the total growth from t = a to t = b _months_.

In general, if W is a rate of change, measured in _whatzits_ per _doodle_ , and if D, the amount of _doodles_ changes from D = a to D = b, then the total change is W x (b - a) _doodles_ = W (b - a) _whatzits_. Unfortunately, if the rate of change, W, depends upon the number of _doodles_ , say W = f(D), then we assume that D changes in microscopic steps during which W = f(D) is essentially constant ... during each microscopic step. When the amount of _doodles_ changes from D to D+dD, the change in _whatzits_ is f(D) dD. The total change is then obtained by all these microscopic changes: . Note that if f(D) is constant at the value W, the integration yields W (b - a) as expected.

**Example:** In drilling a well, the cost per metre depends upon the type of sand, gravel or rock which must be excavated. Suppose the cost is C(x) _dollars/metre_ at a depth x _metres_. (As x changes, the type of material changes, hence the cost changes.) Express, as a definite integral, the cost in digging a well of H _metres_.

**Solution:** In digging from x to x+dx, the cost per metre is C(x) so the cost for just this short piece is

C(x) dx dollars. The total cost, for H metres, is dollars.

S: I don't understand this at all.

P: Patience.

**Example:** The force required to extend a spring a distance of x cm is F(x) Newtons. (The more the extension, the greater the force required ... and we'll measure the force in "Newtons".) If WORK is measured as ( _force_ )( _distance_ ), how much work is done in extending the spring a distance h metres.

**Solution:** In extending from x to x+dx, the force is F(x) and the work is F(x) dx. The total work in extending from x = 0 to x = h is

S: I still don't ...

P: Okay, here's how it goes. We're digging a well and we're at a depth of x metres. At this depth the cost is $C for every metre, so digging a small, incremental distance dx will cost $C dx. The next small step of length dx will also cost $C dx , but C changes, so we just the costs via . See? The definite integral does it all for us. It's like finding the area under a curve y = f(x). We consider only the part from x to x+dx and assume that f(x) is constant there, at the value f(x). The area under the curve is then approximated by f(x) dx (i.e. the area of a vertical rectangle). When we from x = a to x = b we get the exact area, namely . See how it works? Consider extending the spring. We do it in small steps of length dx (like digging the well in small steps of length dx). When we're at a length x the force is F(x) and even though it changes with x we assume it's constant over the small interval dx and approximate the work by (force)(distance) = f(x) dx. Now all such incremental work, from x = 0 to h and get WORK =and it's exact.

S: Seems fishy that you can write down an approximation then, poof, you get the exact answer.

P: Go back and review Riemann SUMS and taking the limit as h0 and n∞ and all that good stuff. It's this limiting procedure which gives us the exact answer. Remember? The errors actually go to zero. That's why I never say "we sum this or that" I always say "we this or that". Buried in that expression is a world of ingenuity. Let's do the well-digging or the work problem by constructing a Riemann SUM then taking the limit. Are you interested?

S: Will it be on the final exam?

P: Forget it. We'll go on with another application.

**Example:** The cost of manufacturing an item depends upon the number of items manufactured. For the first few items the cost is high and the profit low, but as we produce more items the cost decreases hence the profits (when we sell the items) increase. Suppose the profit for the nth item is p(n) dollars per item. Express, as a definite integral, the profit in producing (and selling) K items.

**Solution:** If we've produced n items and produce just a few more items, say dn items, then the profit is approximately p(n) dn. Now we the profits for all K items and get the total profit as .

S: Come on! "We produce just a few more items, like maybe dn more"? I thought dx and dy and dn were supposed to be really small. Are you talking about 1/1000 or an item or what?

P: That's a good question. In fact, a very good question. How come you never asked when we maximized the yield of apples from our orchard? Remember? We asked "How many more trees should be added to maximize the yield". I invented the problem so there would be an integer number of trees to add. You should have asked "What if we had to add 1/1000 of a tree, what then?" You see, if we are to apply the methods of calculus we have to assume that the independent variable can take on any value (in some domain). Then we can talk about "dx" and so on. If the independent variable is the number of trees or the number of items then it's clearly an integer so we fudge things a bit and consider it a "continuous variable". When we get an answer like "add 57.8 trees" we just add 58 more trees. See? Actually, if our manufacturing plant produced tens of thousands of items the graph of p(n) versus n might look like the the diagram below ... and if we expand a piece we'd see that p(n) is a bunch of points, one point for each integer "n".

In fact, the actual total profit is really "the profit for the first item" + "the profit for the second" and so on. In other words, it's p(1) + p(2) + ... + p(K). However, it's often easier to use and recognize that it's a reasonable approximation if n is large.

S: Aha! Approximation! I said your just gave approximations, didn't I?

P: Okay, in this case we do get an approximation, but not because of the definite integral. If we have an equation for p(n), such as p(n) = 5 + , then = = 5K +ln K and this would be exactly the area under the curve and exactly the total profit if n were a continuous variable. The mathematics can't tell that "n" is an integer; it thinks it's any real number. It's we who have introduced the error (fortunately, a small error in many cases of interest) by resorting to the methods of calculus.

S: No, it's not we. I didn't introduce anything. It's you who introduced the error ... and here's another one for you. Your graph of p(n) is decreasing and you said the profit would increase. Ha!

P: Pay attention. Here's another application.

S: Wait. You said it's easier to use than to use p(1) + p(2) + ... + p(K). You're kidding, right? I mean, integrating is easier than adding?

P: Okay, let's see you add p(1) + p(2) + ... when p(n) = 5 + . You'd get (5+) + (5+) + (5+) + ... + (5+) and what's that add up to? There are 5's which add up to 5K, by what about + + + ... + ? That's not easy. In fact (and this may surprise you), if K is large, it adds up to ln K very nearly.

S: Nothing surprises me ... and what's "very nearly".

P: Remember when we talked about one thing being "close to" another? We agreed that the way to compare them is to take their ratio and see if it's very close to 1. In this case, the ratio is very close to 1, when K is large. In fact, you may be interested to know that = 1.

S: I'm not.

P: Pay attention.

Consider the graph of y = from x = 1 to x = n. We divide it up into rectangles as shown.

First we consider n rectangles whose total area is larger than the area under the curve, then rectangles whose total area is smaller than the area under the curve. In each case the width of each rectangle is "1", but the heights differ. In the first case, the heights (given by y = ) are and and etc. so the total area of these rectangles is just: 1 + + + ... + . The second set of rectangles have heights and and etc. so the total area of rectangles is + + + ... + and this is smaller than the area under the curve. But we know exactly the area under the curve; it's = _ln_ n so (using the smaller rectangles):

\+ + + ... + < _ln_ n hence 1 + + + ... + < 1 + _ln_ n hence

< \+ 1 hence ≤ 1.

Now use the larger rectangles and get:

_ln_ n < 1 + + + ... + hence + _ln_ n < 1 + + + ... + hence

\+ 1 < hence 1 ≤ .

P: Well? What's our conclusion? Did you notice that we used the ol' SQUEEZE again?

S: zzzzzz

P: Well, at least you see that the definite integral can be used for lots of things, not just AREAs. We'll do some more:

**AVERAGE VALUE OF A FUNCTION** **:**

Suppose we wish to determine the average value of a function. To be specific, suppose the temperature during a 24 hour period varies according to T = 20 + 10 sin so it starts out (at t = 0) at 20˚C and varies from a minimum of 10˚C to a maximum of 30˚C, then starts all over again, 24 hours later. What's the average temperature? Clearly we'd want to measure the temperature many times throughout the day and average them. Suppose we measure the temperature n times and get T1, T2, ....,Tn. Then the average would be . To be more accurate we should measure T every second or perhaps every 1/10 second, etc. What would we get if we measured T an infinite number of times? If we could do that, we'd have the average temperature exactly.

We subdivide the 24 hour time interval into n subintervals of length h = and measure the temperature at times t = h, 2h, 3h, .... , nh (which, of course, is the same as t = 24). The average is then and we'd like to compute the limit of this as n∞. To this end we rewrite it as:

= where we've multiplied numerator and denominator by h and put nh = 24. Now we let n∞ and recognize the limit of a Riemann SUM which yields: . For our particular T = 20 + 10 sin we'd get an average temperature of = 20˚C which is not too surprising, I guess.

We generalize this:

The average value of f(x) over the interval x = a to x = b is

One interesting thing about this "average" value is the following:

Suppose y = f(x) is positive on [a,b] and we graph it. The "area under f(x)" would be ( _width_ )( _height_ ) provided f(x) were a constant (i.e. the graph was a horizontal line, so f(x) = constant). If f(x) weren't a constant, what _height_ should we use in order to get the correct value for the area? In other words, using _width_ = b - a, what _height_ would make the area under the curve equal to the rectangular area: ( _width_ )( _height_ ) = (b - a) ( _height_ ). Well, since the area is

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, we'd want (b - a) ( _height_ ) = so that the correct _height_ is .

It's just the "average height" as given by our formula for "average"... so we think it's a good definition of "average".

**Example:** The velocity of a vehicle is given by v(t) = 50 + 20e-t km/hour. Calculate its average velocity over the time interval from t = 0 to t = 3 hours.

Solution:

vaverage = = [50t - 20e-t]= = km/hour.

Note that is the total distance travelled by the vehicle during the time interval from t = t1 to t = t2, so

= is just what we'd expect the average velocity to be!

**A PARADOX** :

Consider the volume when the area under y = is revolved about the x-axis, from x = 1 to x = L. We have (using vertical rectangles) V = π= π= π . Now consider the area under the curve, namely A = = = _ln_ L. |

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Notice that revolving the curve y = about the x-axis generates a kind of container which, when filled with paint, would require π metres3 of paint (assuming both x and y are measured in metres). As L∞ we'd never need more than π m3 of paint ... even though the container becomes infinite in length. Now look at the area:

_ln_ L∞ as L∞. That means that even though the infinite container is filled with paint (π m3 worth of paint), it's not enough paint to cover this infinite area (which, after all, lies entirely within the container!).

S: Do you expect me to believe that?

P: Do you see anything wrong with my reasoning?

S: I guess not ... but I still don't believe it. I mean, is the math is so stupid that ...

P: It's never the math that's stupid, it's us.

S: Speak for yourself.

P: I bet you think it always takes more and more paint to cover bigger and bigger areas. Pick a big area.

S: I pick 1010 metres2.

P: Okay, I cover it with paint 10-30 metres thick and that'll take (area)(thickness) = 1010 10-30 = 10-20 metres3 of paint. Not much, eh? And if you picked a larger area I'd just use a much thinner coat of paint and could paint your larger area with even less paint!

S: So?

P: So let's talk about real paint with molecules having a radius of ... what's a good symbol for a small radius?

S: How about ? I always liked . He's small but he gets around and ...

P: Pay attention. We'll fill our "infinite" container with paint having molecules which are  metres in radius. That means the paint will only fill the container until the radius of the neck of the container is y = = . That means L = in our formula for volume, so the volume is V = π (1 - ) metres3. Now our area is ln L = ln and covered with paint of molecular radius  will give a thickness 2 and hence a volume of: (area)(thickness) = 2 ln metres3. Did you know that 2 ln is always less than π (1 - )? That means there is always enough paint in the container to cover the area. What do you think of that? You see, comparing an area with a volume is wrong, wrong wrong! They're don't even have the same dimensions. It's like comparing apples to oranges. There really is no paradox.

S: Do I have to know this for the final ...

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LECTURE 20

IMPROPER INTEGRALS

At the end of the last lecture we considered , the area under y = from x = 1 to x = L ... then we let L∞. It would be tempting to write the infinite area as , but then we'd recall the definition of the "definite integral"and realize that this collection of symbols means the limit, as n∞, of a Riemann SUM of n rectangular areas each of the form f(xi) h, where xi is some point in the ith subinterval (so f(xi) gives the height of the rectangle) and h is the width of each subinterval (hence the width of each rectangular area), and, for , we have h = . Now what happens if b = ∞, as in ? Is h = ? Certainly not. How then do we subdivide the infinite interval 1 ≤ x < ∞ so that we have n rectangles each with a finite (not infinite!) area??!%$#*

S: What's wrong with an infinite area? That's what it is, right? I mean, when you did this problem you said the area is infinite, so what's wrong with ...

P: Okay, you have a point there. So let me consider the volume of revolution when the above area is revolved about the x-axis, namely π. Now, as L∞, we actually get a finite volume, yet if we look back at the definition of the definite integral ... as a limit of a Riemann SUM ... we'd run into the same problem if we let L∞ and wrote the volume as π. The interval is infinite in extent hence if we subdivide into n subintervals (say 100 or 1000 subintervals), we'd have rectangles of infinite width.

S: But you already got an answer for that problem. You said ... if I remember correctly ...

V = π= π= π then you let L∞ and got the answer V = π. What's wrong with that? No Riemann SUM, no subintervals, no infinite widths, no infinite anything!

P: You know, you're becoming very clever. That's exactly what we'll do with an integral where the upper limit is ∞.

The collection of symbols means the limit of a Riemann SUM ... that's the definition ... we have no choice ... and that's where we have a problem because each term in the Riemann SUM is infinite! However, if we do as suggested, we avoid this problem. Hence, we will DEFINE (and this is important!) this type of so-called IMPROPER INTEGRAL as follows:

DEFINITION of an IMPROPER INTEGRAL

=

provided the limit exists

In other words, if somebody gives you an integral such as π, you just evaluate πwhich is a perfectly _proper_ integral (nothing infinite anywhere) ... and you have a warm feeling all over because πhas a perfectly acceptable definition in terms the limit of a Riemann SUM, with subintervals of perfectly acceptable width h = ... then, having evaluated π, you let L∞! If you get a limit (as is the case here), you can say " πconverges to π ", or simply π= π.

S: What's all this "converges to π" stuff? I mean, ...

P: I admit it seems a little unusual to introduce this phrase, but it'll become clearer in the next calculus course. In the meantime, to impress your friends, you can say diverges but converges, and I might ask you to see if a certain "improper integral" converges or diverges.

**Example:** Which of the following improper integrals converge?

(a) (b) (c) (d)

Solutions:

(a) By definition, = = == ∞, hence diverges.

(b) = = = = 1, so converges to 1.

(c) = = but cos L has no limit, so doesn't converge (hence, it diverges).

(d) Here we write = + and consider each integral separately. We know that converges to 1 (as seen above, in part (b)), so we investigate = =

= = ∞ , so , and consequently , diverges.

S: Hold on! First, why do you split as + ? Why not = + ? Second, ...

P: First things first. I could certainly have split the infinite interval, -∞ < x < ∞, at x = 1 rather than at x = 0. I'd still find that diverges. It really doesn't matter where I split it. Not only that, I could have left the integral in one piece and considered then let A  -∞ (that'd give me ) then let B∞ (to get ). See?

S: No.

P: Okay, since we already have a definition for an integral where one limit is infinite, let's agree to always split at zero when we have an integral from -∞ to ∞, then use the previous definition on each improper integral:

= +

S: I can't really believe that such animals ever, ever show up in a real problem. I mean, integrals which go from -∞ to ∞. I mean ...

P: Oh, they do. If f(x) is a probability density, meaning that f(x) dx is the probability that a certain random variable lies between x and x + dx, then is the probability that the variable lies between A and B and if all values of the variable in -∞ to ∞ are possible (with perhaps different probabilities), then = 1, meaning that there is a 100% chance of the variable being in -∞ to ∞ ... and all probability density functions, f(x), must satisfy this constraint, namely that = 1. For example, the famous "normal" or Gaussian probability density,

f(x) = e-x2/2, describes (among other things) the distribution of random errors in a measurement and satisfies

= 1 (as it must). In fact, would be the probability that the random variable lies between -1 and 1. In fact ...

S: Wait! I haven't the foggiest idea of what you're talking about.

P: Sorry ... got carried away ... but it doesn't matter as long as you understand that improper integrals do arise in real problems. Do you?

S: I guess so.

**NOTE:** is defined as the limit of as L∞, and if f(x) ≥ 0 we can interpret as the area "under y = f(x) from x = a to x = L". Clearly, if is to converge, the area must remain finite as L∞ and this allows us ... sometimes ... to see clearly whether converges. For example, consider

= . As L increases, so does . To show that converges, we need only show that the area doesn't become infinite as L∞. But

e-x2 ≤ e-x when x ≥ 1 (since x2 ≥ x when x ≥ 1) so the area under y = e-x2 is less than the area under y = e-x (from x = 1 to x = L) and this latter area is = = e-1\- e-L = - which is less than . We conclude that is always less than , hence cannot become infinite as L∞ ... so converges ... and we showed this without having to evaluate the integral .

In fact, if we stare at the "area under y = f(x) from x = a to x = L", it seems clear that, in order for to converge, the curve y = f(x) must approach zero very rapidly.

We've already seen in example (a) above, that diverges and that's because y = , although it _does_ approach zero, doesn't do it quickly enough. On the other hand, because of the explosive growth of ex, y = e-x =  0 very rapidly ... and we saw in example (b) that , in fact, converges.

On the other hand, in example (c), y = sin x oscillates continually between -1 and +1 and the "area"

= cos(-1) - cos L oscillates between a minimum of cos (-1) -1 and a maximum of cos(-1) + 1 and never approaches any limiting value. On the other hand ...

S: How many hands have we got here? Can't we just do more problems?

Another Kind of Improper Integral

If the limits of integration (in ) are infinite (a = -∞ or b = ∞ ... or both) we have an IMPROPER INTEGRAL and we now have a mechanism for dealing with these integrals. If a = -∞ and/or b = ∞ it means the extent of variation of "x" is infinite. It means the graph extends an infinite distance in the "x"-direction. A picture is

worth a thousand words, so look at the picture ===>>

Here we consider an area of infinite extent in the "x"-direction, from say x = 0 to x = ∞. (I keep putting "x" in quotes for a reason!).

Now we rotate the area shown (actually, we interchange x and y), so that it's infinite in extent in the "y"-direction!

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Here's the picture ===>>

Remember, it's the same area! If the first is finite, then so is the second. If the first is infinite, so is the second. However, we now have a different kind of problem. For this second area, it will have a form such as where y = f(x) has a vertical asymptote at x = 0, that is f(x)  ∞ as x  0+. Such an integral as this cannot be defined in terms of a limit of a Riemann |

SUM. Why? Because now the elemental rectangles can have infinite _height_! (Previously, they had infinite _width_.) However, we avoid the problem (of not being able to define these integrals in terms of a limit of Riemann SUMs) by defining them as follows:

If f(x) is discontinuous at x = a, then=

In other words, we avoid integrating from x = a and instead begin at x = L (where L > a). This avoids the place where f(x) is discontinuous and the integral is well-defined and we're happy. Then we let L sneak up on x = a (from values greater than a, of course, hence L a+) and pray there is a limiting value. If so, that's the value of . If not, then this improper integral doesn't have a value! (i.e it diverges).

**Example:** Evaluate , if possible.

**Solution:** This is an improper integral (since f(x) =  ∞ as x  0+) so we consider = = 2 - 2and let L0+ to get = 2. The "area under y = from x = 0 to x = 1" is infinite in extent (in the "y"-direction), nevertheless it has a finite value.

We illustrate, _graphically_ , what we do with these two types of improper integrals. Do you know why?

S: Yeah. A picture is worth a thousand words.

Remember, for to have a value (i.e to "exist" or to "converge"), f(x) must get small enough fast enough. We've already seen that y = doesn't get small enough fast enough, because diverges (to infinity). (We saw that when we considered the area under y = , remember?) On the other hand, the volume of revolution gave rise to an integral which did converge. What does that tell you?

S: Huh?

P: Pay attention. It means that f(x) = doesn't get small enough fast enough, but f(x) = does! Now, a problem for you. If f(x) = , how big do we have to make the number p in order that converge?

S: Well, that's pretty easy. If p = 2, it converges and if p = 1 it doesn't.

P: And if p = 1.5 ?

S: Oh, yeah, well ... we write = =

P: No, no! You haven't been listening! You can't just substitute x = ∞! You have to consider and let L∞.

S: Fussy, fussy. Okay, I get == - L1-p and now I let L∞ and get since L1-p drops dead as L∞, right?

P: Even if p = 1?

S: No, I'm talking about p>1. That'd make 1-p < 0 so L1-p is really L raised to a negative power and that'd make it go to zero as L∞. Right?

P: You've been eating smart pills again. That's very good. So what's your conclusion?

S: I conclude that ... uh ... I don't understand the question.

P: You'd conclude that, if p>1, then y = decrease rapidly enough for the area under the curve to be finite. That is,

converges if p > 1, to . If p = 1, the area is infinite.

S: I knew that.

P: Guess what?

S: What.

P: We're finished the course!

S: Yeah! Yeah! Yeah!

SOLUTIONS TO "ASSORTED PROBLEMS"

1. (a) = = = = 3+x (for x ≠ 2)  5 as x 

(b) = = sin  sin = 1 as x  2

(c) f(t) = |sin t| = -sin t near x = - hence f'(x) = - cos x = - cos(-) (at x = - ) = -

2. y = = when 2x2\- 3x - 2 = 0, or (2x+1)(x-2) = 0 hence curves intersect when x = and x = 2. For a quick sketch of y = , note that it has a vertical asymptote at x = -1/2 and that y = 1 when x = 0 and y ≈ x2/x = x when x is large (neglecting the constants). The area is then = - . |

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In this remaining integral we set u = x+1 so du = dx = (1) dx = dx and change the limits: when x = -1/2, u = 1/2 and when x = 2, u = 3. We get for this integral:

= = .

  | After evaluating we get an area:

\- 2 _ln_ 3 + 2 _ln_ or - 2 _ln_ 6 ≈ 2

and we check for reasonableness by comparing with the area of the triangle shown: area of triangle = (1/2)(base)(height) = (1/2)(5/2)(5/2 - 1) = 15/8 ≈ 2 which seems okay.

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3. =. Let u=x+2, du=dx and get = _ln_ | u | - +C = _ln_ |x+2| - + C.

4. (a) Using xn+1 = xn \- with f(x) = x3-x2+x+22 we get:

xn+1 = xn \- . The only root is slightly

larger than 3, so we use x1=3 and get

x2=3.045789, x3=3.044724, x4=3.044723, x5=3.044723

and we conclude that the root is 3.04472 to five dec. places.

(b) With f(x) = x _ln_ x - 6 we get: xn+1=xn \- .

There is one root between 4 and 5. If we use x1=4.3 we

get x2=4.189351, x3=4.188760, x4=4.188760 and we

conclude that the root is 4.18876 to five dec. places.

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5. (a) = = = = = 1 (which is not surprising since the ratio is just 2 which has a limit of 1).

(b) = = = = -

ex ≈ 1 + x + which is the quadratic approximation at x = 0.

6. (a)

Using vertical rectangles: Using horizontal rectangles:

Area = + Area =

(b)

Using horizontal rectangles: Using vertical rectangles:

Volume = π Volume = 2π

= π = 2π

7. (a) = = does not exist (since ±∞ depending upon whether x0+ or 0-).

(b) = = where we've divided numerator and denominator by x4 (the highest power of x). Now x4 = hence the denominator becomes

\+  +0 as x  -∞ whereas the numerator  1. Hence the ratio  ∞.

8. (a) Let A = arctan(-) so that tan A = - and A must lie in -π/2 < A < π/2 (the range of the arctan function). Clearly, A lies in the fourth quadrant. From the diagram, cos= cos A = .

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(b) = + = + = = - .

9. (a) = = = = .

(b) = = .

10. (a) = = arcsin L = arcsin 1 = .

(b) = = u sin u - v sin v = 2x3 sin x2 \- 2x3 sin(-x3) = 4x3 sin x2 where we've set u = x2 and v = -x2. (Note: we could have written = 2 as well, since the integrand is an EVEN function of x.)

11. Write y = f(x) = , then x = f(y) = . Were we able to solve this latter equation for y we'd have y = g(x), the inverse of f(x).

(a) f'(x) = = ≥ 0 when x ≥ 0 so f has an inverse there (... and everywhere else for that matter!).

(b) To find y = g(x) when x = 2 we must solve 2 = and the solution is clearly y = g(2) = 1. (Note: the value x = 2 was chosen so you could find y by inspection!)

(c) To find = g'(x) when x = 2 (and y = 1) we differentiate x = implicitly, giving

1 = then substitute y = 1 to get 1 = 4 so = g'(x) = when x = 2.

(Note: Don't solve for dy/dx, then substitute; substitute first then find dy/dx ... it's much easier!)

12. We want to determine whether eπ > πe or, to put it differently, whether e1/e > π1/π (raising both sides of the inequality to the power ). Hence we consider the function y = x1/x to see if it's increasing or decreasing when x goes from x = e to x = π. We use logarithmic differentiation: _ln_ y = _ln_ x so =

= which is positive (hence _ln_ y is increasing) when _ln_ x < 1, i.e. when x < e, and decreases thereafter. In particular, _ln_ y, hence y, is smaller at x = π than at x = e, so π1/π < e1/e or, to put it differently, πe < eπ. (In fact, x1/x has its maximum value at x = e).

13. Let (x,y) be a point on y = (3 - x2)/2. Then the distance from (0,0) to (x,y) is L = which we want to minimize. In fact, it's simpler to minimize L2 = x2 \+ y2. First substitute x2 = 3 - 2y (the equation of the curve, rewritten) and get L2 = 3 - 2y \+ y2 so L2 = -2 + 2y < 0 when y < 1 (so L2 is decreasing) then L2 > 0 when y > 1 (so L2 is |

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increasing). Hence the minimum occurs at y = 1, hence x = ±= ±1. Note that as you travel along the curve (from the third quadrant where y is negative), L2 is first negative so L2 is decreasing ... to a minimum at (-1,1), then L2<0 so L2 increases (because y > 1), then it decreases again until y = 1 again (at (1,1).

(A picture is worth a thousand words!)

14. (a) = = .

(b) = x + cos ( _ln_ x) = cos ( _ln_ x) - sin ( _ln_ x) .

15. Area = ... using vertical rectangles

= = - 4 = 8.

Area = = = - , and of course we get the negative of the area because is negative on [-4,-2] .. so the area is actually |

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16. = 0 since the integrand is an ODD function. (Remember the test for an ODD function: f(-x) = f(x) which, for our f(x), becomes: f(-x) = sin (-x)3 = sin(-x3) = - sin x3 = - f(x)).

17. (a) If y = ax then _ln_ y = x _ln_ a and = _ln_ a so = y _ln_ a = ax _ln_ a.

(b) If y = xx then _ln_ y = x _ln_ x and = 1 + _ln_ x so = y (1 + _ln_ x) = xx ( 1 + _ln_ x).

18. Let f(x) = x1/3. Then f(64) = 641/3 = 4 and f'(x) = x-2/3 so f'(64) = = = . The linear approximation is: f(x) = x1/3  f(64) + f'(64) (x - 64) = 4 + . At x = 62 get the approximation: 621/3 ≈ 4 -

19. . Let u = ex so x = _ln_ u and dx = du = du. Then get = = arctan ex \+ C.

20. f(x) = = provided x ≠ -1. Hence we need only sketch y = 1 - and put an "open circle" at x = -1. Since y = 1, there is a horizontal asymptote: y = 1. Since y = ±∞, the y-axis is a vertical asymptote. There are no critical points. |

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21. We sketch, and note the points of intersection: (0,0) and (π/4,1).

The area (using vertical rectangles) is

= = + _ln_ . |

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22.

(a) The area inside r = sin  is: = = = . This agrees (of course!) with π(1/2)2.

(b) The area inside r =1- sin  and inside the circle r = 1 is: 2... taking double the area to the right of the y-axis and noting that the area inside the circle is swept out when  runs from -π/2 to 0 and the limaçon area when  runs from 0 to π/2. This gives +and putting sin2 = - gives +

= += - 2 (about - 2 = 1.75 which is somewhat larger than half the circular area: π 12 ≈ 1.5, so we're happy with our answer!).

(c) The area inside the smaller loop of the limaçon r = 1- 2 sin  is: 2taking twice the area from  = π/6 (which is the first time r goes to zero, after which r goes negative, sweeping out the inner loop) to  = π/2 (when r is its most negative). This gives = using, once again, sin2= 1/2 - (1/2)cos2. We get = π - .

(d) The area inside the lemniscate r2 = cos 2 is: 4= 4= 4.

23. If r is the radius and h the height, the surface area is 2πr2 \+ 2πrh (the two ends plus the curved side). But r and h are related: volume = πr2h = 300 (given) so we substitute h = 300/πr2 and get the surface area as A(r) = 2πr2+600/r. To minimize we consider A'(r) = 4πr - 600/r2 = (4πr3 \- 600)/r2 which is first negative when r is small, (so the area is decreasing), then positive for large r (the area is increasing), so the minimum occurs for A'(r) = 0, or r3=600/4π, or r = (150/π)1/3 and h = 300/πr2 = 2 (150/π)1/3. (Note: h = 2r gives the minimum amount of aluminum no matter what the volume of the can!)

24.

  | If the dimensions are x and y, then the area is xy. However, there's a relation between x and y, namely 2x + y = 1000 (given length of fence), and if we substitute y = 1000 - 2x we get the area A(x) = x (1000 - 2x) = 1000x - 2x2. To maximize we consider A'(x) = 1000 - 4x which is first positive when x is

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small (meaning the area is increasing), then negative when x is large (so the area is decreasing), hence the maximum area occurs when A'(x) = 0, or x = 250 meters ... hence y = 1000 - 2x = 500 metres.

25. We're given the rate of change where x gives the location of the man, and we want to know where y is the length of shadow. To find the relation between x and y we note the similar triangles: = so that (A-B) y = B x. Differentiating gives (A-B) = B and, substituting, we |

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get = = = 1 m/s. (Note: most "related rate" problems ask for when x has some value. Here, = 1 m/s for _every_ value of x.)

26. Let the minute and hour hands have lengths "a" and "b" respectively, and let their angular displacements from the 12 o'clock position be  and  respectively. (I like symbols better than numbers!) The distance between them is given by the "cosine law" for triangles: l2 = a2 \+ b2 \- 2ab cos ( \- ). We know that = = radians/hour and = = 2π radians/hour and we wish to know when  = 0 and  = . Differentiating the relation between l,  and  gives: |

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2l = 2ab sin ( \- ) and substituting what we're given (including l = = 5, at 3 o'clock) gives = sin = - cm/hour. (A negative sign, so the tips are getting closer together.)

27. For f(x) = cos x, f'(x) = - sin x, f''(x) = - cos x and f'''(x) = sin x. The "familiar" angle nearest to 57˚ is 60˚, or x = π/3, so we get the cubic approximation: f() + f'() (x-) + f''() + f'''()

= - - + and, for x - = - 3 = - , we get cos 57˚ ≈ .54463888 (compared to the "exact" value of .54463903).

28.

  | Let the top of the picture be "b" metres above your eye, the bottom "a" metres. If you stand a distance "x" metres from the wall, the angle subtended by the picture at your eye is  \-  (see diagram) where tan  = and tan  = . We maximize  \-  = f(x) = arctan - arctan for x > 0. First, we compute f'(x) = - = which is first positive (when x is small) so that the angle is

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increasing, then f'(x) < 0 when x is large ... so the maximum occurs when f'(x) = 0, at x = . For b = 3 and a = 1 we get x = metres. (Note: although we're given the dimensions "a" and "b", I choose to give them labels so I can check the dimensions ... and am less likely to make errors in arithmetic: I can always multiply "a" by "b" and get "ab" ... I'm not so sure about 3 and 1.)

29. To find a point of intersection of xy = and x2 \- y2 = 1, put y = into the second equation, giving: x2 \- = 1, or x4 \- x2 \- 2 = 0, or (x2 \- 2)(x2+1) = 0, hence x2 \- 2 = 0 and x = ±, so y = = ±1. There are two points of intersection: (,1) and (-,-1). At these points we expect the tangent lines to be perpendicular: gives = - = - for the first curve (at _both_ points of intersection) and gives

= = . The product of these slopes is -1, hence they are perpendicular.

30. (a) gives + = 0 so = - . At the point (a,b), the tangent has slope - and the tangent line is: = - or y - b = - or y = b - .

(b) Put x = 0 to get the y-intercept, namely y = b + a = b + .

Put y = 0 to get the x-intercept, namely x = a + b= a + .

(c) The sum of the intercepts is a + 2+ b = 2 = 12 = 1.

(Note: + = 1, since (a,b) lies on the curve.)

31. Curves r=, r=cos  intersect where f()=-cos =0. A plot of f() indicates a single root near =1, so we start Newton's method with =1.

The Newton iterations satisfy: |

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n+1 = n \- = n \- . Starting with 1=1 we get: 2 = 0.75036387, 3 = 0.73911289,

4 = 0.73908513, 5 = 0.73908513 and conclude that the point of intersection occurs at  = 0.73908513 radians (about 42 degrees) so r =  = 0.73908513 as well.

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32. Consider f(x) = π(1 - x) - 2x _ln_ = π(1 - x) + 2x _ln_ x (noting that _ln_ = - _ln_ x). To show that f(x) > 0, we'll find the minimum value of f(x) and hope that it's positive. First we investigate where f(x) is increasing or decreasing using f'(x) = -π + 2(1 + _ln_ x) which is negative at first (when x is near 0 since _ln_ x  -∞ when x0+) then positive when x is large. The minimum then occurs when -π + 2(1 + _ln_ x) = 0 or _ln_ x = - 1 or x = eπ/2 - 1. Substituting into π(1 - x) + 2x _ln_ x gives π - 2 eπ/2 - 1 which is positive.

33. (a) = = arctan L - arctan 0 = arctan L  as L∞.

(b) In , let u = so du = dx = and as x goes from 1 to L, u goes from to . We get = = 2 cos (1) - 2 cos which has no limit as L∞, hence doesn't exist (i.e this improper integral diverges).

(c) = + . We have that = = = = 1 and, in a similar manner,

= = = = 1. Hence = 1 + 1 = 2.

(d) = + and = = = = ∞ so diverges hence so does

(e) = = π3/2 \- L3/2 π3/2 as L0, so converges to π3/2.

(f) = = _ln_ | L - 1| - _ln_ 2 -∞ as L1- (recall that _ln_ | L - 1|  -∞ as | L - 1| 0+). Hence, has no value ... or "diverges" ... or "doesn't exist".

(g) For let u = _ln_ x and dv = dx so du = dx = dx and v = x and we get (integrating by parts)

= uv - = x _ln_ x - = x _ln_ x - x \+ C. Hence:

= = - = -1 + L - L _ln_ L (noting that _ln_ 1 = 0)

and now we must evaluate L _ln_ L. If we rewrite L _ln_ L as it then has the form and we can use l'Hopital's rule: = = = 0. Hence = = -1. (The fact that's it's negative should be no surprise: _ln_ x < 0 when 0 < x < 1).

* Leonard Euler (1707-1783) is responsible for the modern treatment of logarithms and exponential functions and introduced the notations sin(x), cos(x), etc. and f(x). A Swiss mathematician, he studied math under Johann Bernoulli but soon outstripped his teacher.
* This notation is due to Gottfried Wilhelm Leibniz (1646-1716), who created calculus ... along with Isaac Newton (1642-1727), who discovered the ideas independently. Newton called derivatives "fluxions" and used a notation more like x' (which is still used today). In 1924 a letter written by Newton was discovered in which Newton acknowledged that some of his early ideas came from Fermat (1601-1665) who found a method for constructing tangents to curves and maxima and minima.

* Isaac Newton (1642-1727) was one of the greatest geniuses of all time and, along with Archimedes and Gauss, one of the top three mathematicians. He discovered, among other things, the law of universal gravitation, the binomial theorem, the breaking of white light into its constituent colours and, of course, the differential and integral calculus ... and he did most of this in his twenties! (How does that make you feel?)
* Perhaps the third most important number is g, Euler's constant. It's defined by

g = ≈ .5772156649015

* The wealthy Marquis de l'Ho^pital had one of the famous Bernoullis as his math teacher. Apparently, in 1955, a letter was discovered which indicated that it was Bernoulli and not l'Ho^pital who was the author of this "rule" ... l'Ho^pital simply paid Bernoulli so that the Marquis could claim it as his own! Nevertheless, l'Ho^pital did publish the very first calculus book.

* This is often called the lemniscate of Bernoulli. The Bernoulli family included Jakob (1654-1705) who taught himself the Newton/Leibniz calculus and invented polar coordinates, and his younger brother Johann (1667-1748) who studied medicine as well as calculus, and Johann's son, Daniel (1700-1782) who became an outstanding mathematical physicist. There was great rivalry among the Bernoullis. As the story goes, Johann posed a problem to the mathematicians of the world, the Brachistochrone Problem: "A wire is bent into a curve joining two given points. A bead slides down the wire without friction. What curve will give the minimum time of descent?" The problem was solved by Newton, Leibniz (the curve is a CYCLOID) ... and Jakob. Johann was not pleased.

* Bernhard Riemann (1826-1866) was educated at Gottingen University, Germany, the home of the most gifted mathematician of all time: Carl Friedrich Gauss (1777-1855). Riemann (among many other things!) developed the concept of SUMs (Riemann SUMS ... what else?) as the basis for the definite integral.
