Professor Dave again, let’s talk about related
rates.
There are plenty of real-world scenarios that
calculus can apply to, and now that we’ve
learned about implicit differentiation, we
are ready to look at one such application,
the concept of related rates.
Let’s say we are inflating a balloon, and
let’s pretend that the balloon is a perfect
sphere, so we essentially have a sphere that
is expanding at a constant rate.
Now say we want to describe the rate at which
the volume of the sphere is increasing, as
well as the rate at which the radius of the
sphere is increasing.
Both of these values are increasing due to
the expansion of the balloon, and their rates
of increase are related, hence related rates.
But one of these is probably much easier to
physically measure than the other, so rather
than trying to measure both, let’s measure
the one that we can, and compute the rate
of change in the other using calculus.
It’s probably easier to measure the change
in volume, so let’s say that we do measure
the volume over some period of time, and we
find that it is increasing by one hundred
cubic centimeters per second.
How can we use this to find out how fast the
radius is increasing at a particular instant,
like the moment when the diameter of the balloon
is fifty centimeters?
Well let’s say that volume is V and radius
is r.
Then dV over dt, or the change in volume over
time, equals one hundred cubic centimeters
per second, as we measured.
But what we want to know is dr over dt, or
the change in radius over time, at the moment
that the radius is twenty five centimeters,
since the radius is half the diameter.
To see how these values will relate, we need
an equation that relates V and r, and that
will be the equation for the volume of the
sphere.
When we learned geometry, we saw that this
formula is V equals four thirds pi r cubed.
Now as we said, what we are looking for is
dr over dt, so let’s differentiate this
with respect to t.
For V, that simply becomes dV over dt, but
for the right side we will need to use the chain rule.
This is because t is not present in this equation,
the radius is the variable, but the radius
is itself a function of t, which is why we
use the chain rule.
So we will have to differentiate this expression
with respect to r, and then multiply that
by the derivative of r with respect to t,
or dr over dt.
The derivative with respect to r is easy,
we just bring the three down here, and change
the three to a two, giving us four pi r squared,
and then that must be followed by dr over dt.
Well dr over dt is what we are trying to solve
for, because we are interested in how the
radius changes over time, so let’s solve
for that.
This means simply dividing both sides by four
pi r squared.
That means that dr over dt equals dV over
dt, times one over four pi r squared.
We plug in our value for dV over dt, which
is one hundred, and the value for r that we
are interested in, which is twenty five.
We evaluate, and we get one over twenty five
pi centimeters per second.
Let’s try another example.
Say we have a ladder resting against a wall.
The ladder is ten feet long, and it is sliding
away from the wall at a rate of one foot per second.
How fast is the top of the ladder sliding
down the wall when the bottom of the ladder
is six feet away from the wall?
Just as with any other example we could look
at, we will quickly see that we need to be
able to construct equations that represent
the scenario.
Looking at the diagram, we have a ladder with
a constant length of ten, and then these values
are the variables x and y, as they both change over time.
So that makes a right triangle with a known
hypotenuse and the two legs unknown.
We can relate the variables by the Pythagorean
theorem.
That would be x squared plus y squared equals
ten squared, or a hundred.
Now as we said, we want to know how these
parameters change over time, so let’s differentiate
both sides of this equation with respect to time.
The derivative of x squared with respect to
time will be two x times dx over dt.
For y, we get the same thing, two y times
dy over dt.
On the right, the derivative of any constant is zero.
Now we want to know about how fast the ladder
is sliding down the wall, which is described
by dy over dt, so let’s solve for dy over dt.
We take the term with all the x’s to the
other side, and then we divide by two y, and
we get negative x over y times dx over dt.
So all we have to do is plug in what we know,
and solve.
We are asking about dy over dt when the ladder
is six feet away from the wall.
That means x must be six.
Then, by the pythagorean theorem, we can solve
for y at that instant, and that must be eight.
We can also plug in dx over dt, because we
said that the ladder was sliding away from
the wall at one foot per second.
That means x is increasing by one foot every
second, so dx over dt is positive one.
We simplify, and we get negative three fourths
feet per second.
This is a negative value, because y is decreasing
over time, as the ladder slides down the wall.
So as we can see, related rates problems can
be rather straightforward.
The real challenge lies in understanding what
the problem is asking, and being able to first
draw a diagram.
Then we must write equations from that diagram
that relate the two variables in question.
Because we are describing rates, these will
always be functions of time, and we will be
taking derivatives of functions with respect to time.
This may require the chain rule, as we may
be taking the derivative of a function first
with respect to some variable, so we must
then multiply by the derivative of that variable
with respect to t, since we are ultimately
concerned with terms involving dt.
Once we have done that, we just plug in the
information we know, and solve for what we
are looking for, which will always be d something over dt.
Let’s check comprehension.
