PROFESSOR: Hi.
Well, I hope you're ready
for second derivatives.
We don't go higher than that
in many problems, but the
second derivative is
an important--
the derivative of the derivative
is an important
thing to know, especially in
problems with maximum and
minimum, which is the big
application of derivatives, to
locate a maximum or a minimum,
and to decide which one it is.
And I can tell you right away,
locating a maximum, minimum,
is the first derivative's job.
The first derivative is 0.
If I have a maximum or a
minimum, and we'll have
pictures, somewhere in the
middle of my function I'll
recognize by derivative
equals 0.
Slope equals 0, that the
function is leveling off,
either bending down or bending
up, maximum or minimum.
OK, and it's the second
derivative that
tells me which it is.
The second derivative tells me
the bending of the graph.
OK, so we now will have
three generations.
The big picture of calculus
started with two functions:
the distance and the speed.
And we discussed in detail the
connection between them.
How to recover the speed if we
know the distance, take the
derivative.
Now comes the derivative of
the speed, which in that
language, in the
distance-speed-time language,
the second derivative is the
acceleration, the rate at
which your speed is changing,
the rate at which you're
speeding up or slowing down.
And this is the way I
would write that.
If the speed is the
first derivative--
df dt--
this is the way you write the
second derivative, and you say
d second f dt squared.
d second f dt squared.
OK, so that's you could say the
physics example: distance,
speed, acceleration.
And I say physics because, of
course, acceleration is the a
in Newton's Law f equals ma.
For a graph, like these graphs
here, I won't especially use
those physics words.
I'll use graph words.
So I would say function
one would be the
height of the graph.
And in this case, that height is
y equals x squared, so it's
a simple parabola.
Here would be the slope.
I would use the word "slope"
for the second function.
And the slope of y equals x
squared we know is 2x, so we
see the slope increasing.
And you see on this picture
the slope is increasing.
As x increases, I'm going
up more steeply.
Now, it's the second
derivative.
And what shall I call that?
Bending.
Bending is the natural
word for the second
derivative on a graph.
And what do I--
the derivative of 2x is 2, a
constant, a positive constant,
and that positive constant tells
me that the slope is
going upwards and that the
curve is bending upwards.
So in this simple case,
we connect these three
descriptions of our function.
It's positive.
It's slope is positive.
And its second derivative--
bending--
is positive.
And that gives us a function
that goes like that.
Now, let me go to a different
function.
Let me take a second example
now, an example where not
everything is positive.
But let's make it familiar.
Take sine x.
So sine x starts
out like that.
So this is a graph of sine x up
to 90 degrees, pi over 2,
so that's y equals sine x.
OK, what do you think
about its slope?
We know the derivative of sine
x, but before we write it
down, look at the graph.
The slope is positive, right?
But the slope actually
starts out at 1.
Better make it look a little
more realistic.
That's a slope of 1 there.
So the slope starts at 1 and the
slope drops to a slope of
0 up there.
So a slope of 1.
I see here is a 1.
Here I'm graphing y prime.
dy dx I sometimes write as y
prime, just because it's
shorter, and particularly, it'll
be shorter for a second
derivative.
So y prime, we know the
derivative of sine x is cos x,
which is pretty neat actually,
that we start with a familiar
function, and then we get its
twin, its other half.
And the cosine is the slope of
the sine curve, and it starts
at 1, a slope of 1, and it comes
down to 0, as we know
the cosine does.
So that's a graph
of the cosine.
And now, of course, we have
three generations.
I'm going to graph
y double prime.
Let me put it up here. y double
prime, the second
derivative, the derivative
of the cosine of x
is minus sine x.
OK, let's just--
from the picture, what
am I seeing here?
I'm seeing a slope of 0.
I'm taking now the slope
of the slope.
So here it starts at 0.
The slope is downwards, so the
second derivative is going to
be negative.
Oh, and it is negative,
minus sign x.
So the slope starts at 0 and
ends at minus 1 because that
now comes down at a
negative slope.
The slope is negative.
I'm going downhill, and that's
a graph of the second
derivative.
And which way is our
function bending?
It's bending down.
As I go along, the slope
is dropping.
And I see that in
the slope curve.
It's falling.
And I see it in the
bending curve
because I'm below 0 here.
This is bending down, where
that one was bending up.
I could introduce the word
convex for something that
bends upwards, and bending down,
I could introduce the
word concave. But those
are just words.
The graphs are telling us much
more than the words do.
OK, so do you see that picture
bending down, but going up?
So the slope is positive here,
but the second derivative, the
slope is dropping.
So the second derivative-- and
you have to pay attention to
keep them straight.
The second derivative is telling
us that the original
one is bending down.
OK, let me continue these graphs
just a little beyond 90
degrees, pi over 2, because
you'll see something
interesting.
So what happens in the next
part of the graph?
So this is going--
the sine curve, of course,
continues on its way downwards.
So the slope is going negative,
as I know the cosine
curve will do, as the cosine
curve will come like that.
The slope down to minus
1, the slope--
do you see here?
The slope is negative, so on
this slope graph, I'm below 0.
And the slope is 0.
Let me put a little mark at
these points here, at these
three points.
Those are important points.
In fact, that is a maximum,
of course.
The sine curve hits
its maximum at 1.
At that point when it hits its
maximum, what's its slope?
When you hit a maximum, you're
not going up anymore.
You haven't started down.
The slope is 0 right there.
What's the second derivative?
What's the bending
at a maximum?
The bending tells you that the
slope is going down, so the
bending is negative.
The bending is negative
at a maximum.
Good.
OK, now I'm going to continue
this sine curve for another 90
degrees, the cosine curve, and
I'll continue the bending
curve, so I have minus sine
x, which will go back up.
OK, now what?
Now what?
And then, of course, it
would continue along.
OK, there's something
interesting happening at 180
degrees, at pi.
Can I identify that point?
So there's 180 degrees.
Something's happening there.
I don't see--
I don't quite know how to say
what yet, but something's
happening there.
It's got to show up here, and
it has to show up here.
So whatever is happening is
showing up by a point where y
double prime, the second
derivative, is 0.
That's my new little
observation, not as big a deal
as maximum or minimum.
This was a max here.
And we identified it as a max
because the second derivative
was negative.
Now I'm interested
in this point.
Can you see what's happening
at this point as far as
bending goes?
This curve is bending down.
But when I continue, the
bending changes to up.
This is a point where
the bending changes.
The second derivative changes
sign, and we see it here.
Up to this square point,
the bending is below 0.
The bending is downwards
as I come to here.
But then there's something
rather special that--
you see, can I try to
blow that point up?
Here the bending is down, and
there it turns to up, and
right in there with the--
this is called--
so this is my final word
to introduce--
inflection point.
Don't ask me why.
An inflection point is a
point where the second
derivative is 0.
And what does that mean?
That means at that moment, it
stopped bending down, and it's
going to start bending up.
The second derivative is
passing through 0.
The sign of bending
is changing.
It's changing from concave
here to convex there.
That's a significant
point on the graph.
Not as big a thing as
the max or the min
that we had over there.
So let me draw one more example
and identify all these
different points.
OK, so here we go.
I drew it ahead of time because
it's got a few loops,
and I wanted to get
it in good form.
OK, here it is.
This is my function: x cubed
minus x squared.
Well, before I look at the
picture, what would be the
first calculus thing I do?
I take the derivative.
y prime is the derivative of
x cubed, is three x squared
minus the derivative of x
squared, which is 2x.
And now today, I take the
derivative of that.
I take the second derivative,
y double prime.
So the second derivative is
the derivative of this.
x squared is going to give me
2x, and I have a 3, so it's
all together 6x.
And minus 2x, the slope of
that is minus 2, right?
Cubic, quadratic, linear, and
if I cared about y triple
prime, which I don't,
constant.
And then the fourth derivatives
and all the rest
would be 0 for this case.
OK, now somehow, those
derivatives, those formulas
for y, y prime, y double prime
should tell me details about
this graph.
And the first thing I'm
interested in and the most
important thing is
max and min.
So let me set y prime to be--
which is 3x squared minus 2x.
I'll set it to be 0 because I
want to look for max, or min.
And I look for both at the same
time by setting y prime
equals 0, and then I find out
which I've got by looking at y
double prime.
So let me set y prime to be 0.
What are the solutions?
Where are the points on the
curve where it's stationary?
It's not climbing and
it's not dropping?
Well, I see them on
the curve here.
That is a point where
the slope is 0.
And I see one down here.
There is a point where the slope
is 0, but I can find
them with algebra.
I solve 3x squared equals
to 2x, and I see
it's a quadratic equation.
I expect to find two roots.
One of them is x equals 0, and
the other one is what?
If I cancel those x's to find
a non-zero, canceling those
x's leaves me with 3x equals
2 or x equals 2/3.
Yeah, and that's what
our graph shows.
OK, now we can see on the
graph which is a max
and which is a min.
And by the way, let
me just notice, of
course, this is the max.
But let me just notice that
it's what I would
call a local maximum.
It's not the absolute top of
the function because the
function later on is climbing
off to infinity.
This would be way a maximum
in its neighborhood, so a
maximum, and it's only
a local max.
And what do I expect to see at
a maximum at x equals 0?
I expect to see the slope 0 at
x equals 0, which it is.
Check.
And at a maximum, I need to know
the second derivative.
OK, here's my formula.
At x equals 0, I see y double
prime if x is 0 is minus 2.
Good.
Negative second derivative tells
me I'm bending down, as
the graph confirms, and the
place where the slope is 0 is
a maximum and not a minimum.
What about the other one?
What about at x equals 2/3?
At that point, y double prime,
looking at my formula here for
y double prime, is what?
6 times 2/3 is 4 minus
2 is plus 2.
4 minus 2 is 2.
So this will be--
this is positive, so I'm
expecting a min.
At x equals 2/3, I'm
expecting a min.
And, of course, it is.
And again, it's only
a local minimum.
The derivative can only tell
you what's happening very,
very close to that point.
The derivative doesn't know that
over here the function is
going further down.
So this is a min, and
again, a local min.
OK, those are maximum
and minimum
when we know the function.
Oh yeah, I better do the
inflection point.
Do you remember what the
inflection point is?
The inflection point is when
the bending changes from--
up to here I see that
bending down.
From here, I see
it bending up.
So I will not be surprised if
that's the point where the
bending is changing, and 1/3
is the inflection point.
And now how do we find
an inflection point?
How do we identify this point?
Well, y double prime
was negative.
y double prime was positive.
At that point, y double
prime is 0.
This is an inflection point.
And it is.
At x equals to 1/3, I
do have 6 times 1/3.
2 subtract 2, I have 0.
So that is truly an
inflection point.
And now I know all
the essential
points about the curve.
And these are the quantities--
oh!
Say you're an economist. You're
looking now at the
statistics for the US economy
or the world economy.
OK, I suppose we're in a--
we had a local maximum there,
a happy time a little while
ago, but it went downhill,
right?
If y is, say, the gross product
for the world or gross
national product,
it started down.
The slope of that curve
was negative.
The bending was even negative.
It was going down faster
all the time.
Now, at a certain moment, the
economy kept going down, but
you could see some
sign of hope.
And what was the sign of hope?
It was the fact that it
started bending up.
And probably that's where we are
as I'm making this video.
I suspect we're still going
down, but we're bending up.
And at some point, hopefully
tomorrow, we'll hit minimum
and start really up.
So I don't know.
I would guess we're somewhere
in there, and
I don't know where.
If I knew where, mathematics
would be even more useful than
it is, which would
be hard to do.
OK, so that's an example
of how the second
derivative comes in.
Now, I started by giving this
lecture the title Max and Min
and saying those are the biggest
applications of the
derivative.
Set the derivative
to 0 and solve.
Locate maximum points,
minimum points.
That's what calculus is most--
many of the word problems, most
of the ones I see in use,
involve derivative equals 0.
OK, so let me take a
particular example.
So these were graphs, simple
functions which I chose: sine
x, x squared, x cubed
minus x squared.
Now let me tell you the problem
because this is how
math really comes.
Let me tell you the problem, and
let's create the function.
OK, so much it's the problem
I faced this
morning and every morning.
I live here.
So OK, so here's home.
And there is a-- the Mass Pike
is the fast road to MIT.
So let me put in the Mass Pike
here, and let's say that's
MIT, and I'm trying to get there
as fast as possible.
OK, so for part of the time, I'm
going to have to drive on
city streets.
I do have to drive on city
streets, and then I get to go
on the Mass Pike, which is,
let's say, twice as fast. The
question is should I go directly
over to the fast road
and then take off?
Let's take off on
a good morning.
The Mass Pike could be twice
as slow, but let's assume
twice as fast. Should
I go straight over?
Probably not.
That's not the best way.
I should probably pick up the
Mass Pike on some road.
I could go directly to MIT on
the city streets at the slow
rate, say 30 miles an hour or 30
kilometers an hour and 60,
so speeds 30 and 60
as my speeds.
OK, so now I should have
put in some measure.
Let's call that distance
a, whatever it is.
Maybe it's about three miles.
And let me call--
so that's the direct distance.
If I just went direct to the
turnpike, I would go a
distance a at 30 miles
an hour, and then
I would go a distance--
shall I call that b?--
at 60.
So that's one possibility.
But I think it's not the best.
I think better to--
and you know better than me.
I think I should probably
angle over
here and pick up this--
my question is where should
I join the Mass Pike.
And let's--
so we get a calculus problem,
let's model it.
Suppose that I can join it
anywhere I like, not just at a
couple of entrances.
Anywhere.
And the question is where?
So calculus deals with the
continuous choice of x.
So that is the unknown.
I could take that as
the unknown x.
That was a key step, of course,
deciding what should
be the unknown.
I could also have taken this
angle as an unknown, and that
would be quite neat, too.
But let me take that x.
So this distance is
then b minus x.
So that's what I travel
on the Mass Pike,
so my time to minimize.
I'm trying to minimize
my time.
OK, so on this Mass Pike when
I travel at 60, I have
distance divided by 60
is the time, right?
Am I remembering correctly?
Let's just remember.
Distance is speed times time.
That's the one we know.
And then if I divide by the
speed, the time is the
distance divided by the speed,
the distance divided by the
speed on the pike.
And now I have the distance
on the city streets.
OK, so that speed is
going to be 30.
So the time is going to be a bit
longer for the distance,
and what is that distance?
OK, that was a.
This was x.
Pythagoras is the great leveler
of mathematics.
That's the distance on
the city streets.
And now what do I do?
I've got an expression
for the time.
This is the quantity I'm
trying to minimize.
I minimize it by taking its
derivative and set the
derivative to 0.
Take the derivative and set
the derivative to 0.
So now this is where I use
the formulas of calculus.
So the derivative, now I'm ready
to write the derivative,
and I'll set it to 0.
So the derivative of that, b is
a constant, so I have minus
1/60; is that OK?
Plus whatever the derivative
of this is.
Well, I have 1/30.
I always take the
constant first.
Now I have to deal with
that expression.
That is some quantity
square root.
The square root is the 1/2
power, so I have 1/2 times
this quantity to one
lower power.
That's the minus 1/2 power.
That means that I still have a
square root, but now it's a
minus 1/2 power.
It's down here.
And then the chain rule says
don't forget the derivative of
what's inside, which is 2x.
OK, depending on what order
you've seen these videos and
read text, you know the chain
rule, or you see it now.
It's a very, very valuable rule
to find derivatives as
the function gets complicated.
And the thing to remember,
there will be a proper
discussion of the chain rule.
It's so important.
But you're seeing it here that
the thing to remember is take
also the derivative of what's
inside the a squared plus x
squared, and the derivative of
the x squared is the 2x.
OK, and that I have
to set to 0.
And, of course, I'm going
to cancel the 2's, and
I'll set it to 0.
What does that mean
"set to zero"?
Here's something minus.
Here's something plus.
I guess what I really want
is to make them equal.
When the 1/60 equals this
messier expression, at that
point the minus term cancels
the plus term.
I get 0 for the derivative,
so I'm looking for
derivative equals 0.
That's my equation now.
OK, now I just have
to solve it.
All right, let's see.
If I wanted to solve that,
I would probably multiply
through by 60.
Can I do this?
I'll multiply both
sides by 60.
That will cancel the 30 and
leave an extra 2, so
I'll have a 2x here.
And let me multiply also by
this miserable square root
that's in the denominator
to get it up there.
I think that's what I've got.
That's the same equation as
this one, just simplified.
Multiply through by 60.
Multiply through by square
root of a squared plus x
squared, and it's
looking good.
All right, how am I going
to solve that?
Well, the only mess up
is the square root.
Get rid of that by squaring
both sides.
So now I square both sides,
and I get a squared plus x
squared, and the square
of 2x is 4x squared.
All right, now I have an
equation that's way better.
In fact, even better
if I subtract x
squared from both sides.
My equation is telling
me that a squared
should be 3x squared.
In other words, this
good x is--
now I'm ready to take the square
root and find x itself.
So put the 3 here.
Take the square root.
I'm getting a over the
square root of 3.
So there is a word problem, a
minimum problem, where we had
to create the function to
minimize, which was the time,
trying to get to work as
quickly as possible.
After naming the key quantity x,
then taking the derivative,
then simplifying, that's where
the little work of calculus
comes in, in the end getting
something nice, solving it,
and getting the answer a
over square root of 3.
So we now know what to do
driving in if there's an
entrance where we
want to get it.
And actually, it is a
beautiful answer.
If this is a over the square
root of 3, this will turn out
to be 30 degrees, pi over 6--
I think.
Yeah, I think that's right.
So that's the conclusion
from calculus.
Drive at a 30-degree angle.
Hope that there's a road
going that way--
sorry about that point--
and join the turnpike.
And probably the reason
for that nice
answer, 30 degrees, came--
I can't help but imagine that
because I chose 30 and 60
here, a ratio of 1:2, and then
somehow the fact that the sine
of 30 degrees is 1/2,
those two facts
have got to be connected.
So I change these 30 and 60
numbers, I'll change my
answer, but basically, the
picture won't change much.
And there's another little
point to make to really
complete this problem.
It could have happened that the
distance on the turnpike
was very small and that
this was a dumb move.
That 30-degree angle could
be overshooting
MIT if MIT was there.
So that's a case in which the
minimum time didn't happen
where the derivative
bottomed out.
If MIT was here, the good idea
would be go straight for it.
Yeah, the extra part on the
turn-- you wouldn't drive on
the turnpike at all.
And that's a signal that somehow
in the graph, which I
didn't graph this function, but
if I did, then this stuff
would be locating the minimum
of the graph.
But this extra example where you
go straight for MIT would
be a case in which the minimum
is at the end.
And, of course, that
could happen.
You could have a graph that just
goes down, and then it
ends, so the minimum is there.
Even though the graph looks like
it's still going down,
the graph ended.
What can you do?
That's the best point
there is.
OK, so that is a--
can I recap this lecture
coming first over here?
So the lecture is about maximum
and minimum, and we
learned which it is by the
second derivative.
So then we had examples.
There was an example of a
minimum when the second
derivative was positive.
Here was an example of a local
maximum when the second
derivative was negative.
Here with the sine and cosine,
those are nice examples.
And it takes some patience
to go through them.
I suggest you take another
simple function, like start
with cosine x.
Find its maximum.
Find its minimum.
Find its inflection points so
the inflection points are
where the bending is 0 because
it's changing from bending one
way to bending the other way.
We didn't need an inflection
test--
so actually, I didn't complete
the lecture, because I didn't
compute the second derivative
and show that
this was truly a minimum.
I could have done that.
I would have had to take the
derivative of this, which
would be one level messier,
and look at its sign.
I wouldn't have to
set it to 0.
I would be looking at the sign
of the second derivative.
And in this problem, it would
be safely come out positive
sign, meaning bending upwards,
meaning that this point I've
identified by all these steps
was truly the minimum time,
not a maximum.
OK, that's a big part of
important calculus
applications.
Thanks.
NARRATOR: This has been
a production of MIT
OpenCourseWare and
Gilbert Strang.
Funding for this video was
provided by the Lord
Foundation.
To help OCW continue to provide
free and open access
to MIT courses, please
make a donation at
ocw.mit.edu/donate.
