### Video Transcript

In this video, weβll learn how to
use derivatives to find the relation between the rates of two or more quantities in
related rates problems.

In related rates problems, the idea
is to compute the rate of change of one quantity in terms of the rate of change of
another quantity, which can sometimes be more easily measured. This works because if two related
quantities are changing over time, then the rates at which they change and, of
course, their derivatives are also related.

Consider a spherical balloon being
inflated with air for instance. Both the radius π and the volume
π£ of the balloon are increasing over time. And since the volume of a sphere is
in direct proportion to the cube of its radius, the rate of change of the volume and
the rate of change of the radius are also related. Note also that the sign of the
derivative can indicate to us whether the quantity is increasing or decreasing over
time, depending on whether itβs positive or negative, respectively. These problems can be solved using
implicit differentiation, which makes use of the chain rule, so make sure you have a
sound understanding of how this works before watching this video.

Now for related rates problems,
itβs usually best to jump straight into a problem and see how it all works.

The radius of a circle is
increasing at a rate of three millimeters per second. Find the rate of change of the area
of the circle when the radius of the circle is 15 millimeters.

Remember, when we think about the
rate of change of a quantity, weβre thinking about its derivative with respect to
time. So in this case, letting the area
of the circle be π΄, the rate of change of the area will be dπ΄ by dπ‘, in other
words, the derivative of the area with respect to time. The problem is we know the formula
for the area of a circle in terms of its radius. Its area is equal to π times the
radius squared. This means we can differentiate π΄
with respect to π, but weβre going to struggle differentiating π΄ with respect to
π‘.

Luckily, implicit differentiation,
which is a special version of the chain rule, can help. This says that since π΄ is a
function in π, the derivative of π΄ with respect to π‘ is equal to dπ΄ by dπ times
dπ by dπ‘. And so we really need to find
expressions or values for dπ΄ by dπ and dπ by dπ‘. Of course, we already identified
that π΄ is equal to ππ squared, so dπ΄ by dπ is fairly straightforward. π is, of course, a constant. And so we can use the general power
rule for differentiation to find dπ΄ by dπ. Itβs two ππ.

But what about dπ by dπ‘? Well, in fact, weβre told that the
radius of the circle increases at a rate of three millimeters per second. This means that dπ by dπ‘, the
rate of change of the radius, is positive, since itβs increasing, three. And so dπ΄ by dπ‘ is the product of
these two expressions. Itβs two ππ times three, which is
equal to six ππ. The rate of change of area of the
circle with respect to time is now given as a function of its radius.

Our final job is to substitute π
equals 15 into this expression. And this will give us the
instantaneous rate of change of the area when the radius is this value. dπ΄ by dπ‘ at π equals 15 then is
six times π times 15, which is equal to 90π. Since this is the rate of change of
the area, which will be in square millimeters, with respect to time, the rate of
change is 90π square millimeters per second.

So weβve seen how the formula for
the area of a circle can help us find the rate of change given information about the
rate of change of the radius. In our next example, weβll look at
how we can link volume and surface area of a spherical balloon.

A spherical balloon leaks
helium at a rate of 48 cubic centimeters per second. What is the rate of change of
its surface area when its radius is 41 centimeters?

Ultimately, in this question,
weβre looking to find the rate of change of a quantity. And we know that the rate of
change is to do with its derivative with respect to time. Now, in fact, in this case, we
want the rate of change of the surface area. So if we let the surface area
be equal to π΄, its rate of change is dπ΄ by dπ‘.

We do have a slight problem,
though. The surface area of a sphere
whose radius is π units is four ππ squared. And so to find an expression
for the rate of change of π΄, dπ΄ by dπ‘, weβre going to use implicit
differentiation. This says that the derivative
of π΄ with respect to π‘ is equal to the derivative of π΄ with respect to π
times dπ by dπ‘. Now dπ΄ by dπ can be
calculated using the power rule for differentiation. Since π is a constant, dπ΄ by
dπ is two times four ππ, which is eight ππ. And so dπ΄ by dπ‘ is the
product of eight ππ and dπ by dπ‘.

Usually weβd look to find some
value or expression for the rate of change of radius with respect to time. Thatβs dπ by dπ‘. In this question, though, weβre
given information about the rate at which helium leaks from the balloon. This is given in cubic
centimeters per second. So actually, this is the rate
of change of its volume. Specifically, if we let the
volume of our sphere be π£, weβre told information about dπ£ by dπ‘. In fact, the balloon is leaking
helium, so its derivative, the rate of change of π£ with respect to time, is
going to be negative. Itβs negative 48. This doesnβt help us
particularly at this point with finding dπ by dπ‘, but we can perform a similar
process using implicit differentiation but this time for dπ£ by dπ‘.

We know that dπ£ by dπ‘ could
be calculated by finding the product of dπ£ by dπ and dπ by dπ‘. We know the value of dπ£ by
dπ‘. Itβs negative 48. So, in fact, if we can find an
expression or value for dπ£ by dπ, we can rearrange to find a formula or value
for dπ by dπ‘. And we know, in fact, that the
volume of a sphere with radius π is four-thirds ππ cubed. And so its derivative with
respect to π, dπ£ by dπ, is three times four-thirds ππ squared, which is
four ππ squared. And at this stage, we can also
replace dπ£ by dπ‘ with negative 48. And so now we see that we can
find a formula for dπ by dπ‘ by dividing both sides of this equation by four
ππ squared. Thatβs negative 48 over four
ππ squared, which becomes negative 12 over ππ squared.

And now we see we can replace
dπ by dπ‘ in our earlier formula for dπ΄ by dπ‘. When we do, we get that dπ΄ by
dπ‘ is eight ππ times negative 12 over ππ squared. We can simplify by cross
canceling one value of π and one π, since we know π is not gonna be equal to
zero. And so we have an expression
for dπ΄ by dπ‘ in terms of π. Itβs negative 96 over π. The rate of change of the
surface area when π is equal to 41 then is found by substituting π equals 41
in. And itβs negative 96 over
41. The units for the rate of
change will be square centimeters per second.

And actually this answer makes
a lot of sense. A negative rate of change means
that the surface area is decreasing over time. And we know that the balloon is
leaking helium, so it must be getting smaller. dπ΄ by dπ‘ is negative 96 over
41 square centimeters per second.

In our next example, weβre going to
demonstrate how to link two different rate of change values with a third rate of
change value. This is within the context of
finding the area of a rectangle.

The length of a rectangle is
increasing at a rate of 15 centimeters per second and its width at a rate of 13
centimeters per second. Determine the rate at which the
area of the rectangle increases when the length of the rectangle is 25
centimeters and its width is 12 centimeters.

Weβre given quite a lot of
information about the area and information about the length and width of a
rectangle. Letβs begin by defining the
length of our rectangle to be equal to π centimeters and its width to be equal
to π€. And so itβs area π΄ will be
equal to its length times its width ππ€. Thatβs not quite enough,
though. Weβre looking to find
information about the rate of change of the area, in other words, the derivative
of area π΄ with respect to π‘. And of course, the information
we have about the area of the rectangle is in terms of its length and its
width.

Luckily, we can use a
combination of implicit differentiation and the product rule to find an
expression for dπ΄ by dπ‘ in terms of π, π€, and their respective
derivatives. Letβs begin with the product
rule. π΄ itself is the product of
what we assume are two differentiable functions π and π€. Now we can make that assumption
because in the question weβre given the rate of change of π and the rate of
change of π€. So the derivative with π΄ with
respect to π‘ according to the product rule is π times dπ€ dπ‘ plus π€ times
dπ by dπ‘.

And this is really useful
because weβre told the length of the rectangle increases at a rate of 15
centimeters per second. And so the derivative of π
with respect to π‘ must be positive since itβs increasing and it must be equal
to 15. Similarly, we can say that the
derivative of π€ with respect to π‘, the rate of change of its width, is 13. Substituting these values into
our earlier expression for dπ΄ by dπ‘, and we get 13π plus 15π€. And actually thatβs everything
we need because weβre told that the length of the rectangle at a given point is
25 centimeters and its width is 12.

And so weβre going to
substitute these values into our expression for dπ΄ by dπ‘. That gives us 13 times 25 plus
15 times 12, which is 505. This is the rate of change of
its area. And area will be in square
centimeters. And weβre also told that the
time is given in seconds. And so we see the rate at which
the area of the rectangle is increasing is 505 square centimeters per
second. Note that even if we hadnβt
been told that it was increasing, we could infer that from the sign of the
derivative; itβs positive.

In our previous examples, weβve
looked at geometric applications of related rates problems. Itβs important to realize that we
can also apply similar processes when working with equations of curves. Letβs see what that might look
like.

A particle is moving along the
curve six π¦ squared plus two π₯ squared minus two π₯ plus five π¦ minus 13
equals zero. If the rate of change of its
π₯-coordinate with respect to time as it passes through the point negative one,
three is two, find the rate of change of its π¦-coordinate with respect to time
at the same point.

So letβs begin by looking at
what the question actually wants us to find. It wants us to find the rate of
change of its π¦-coordinate. Now we recall that the rate of
change of a quantity is its derivative with respect to time. So in this case, weβre looking
to find dπ¦ by dπ‘. And in fact, weβre given
information about the rate of change of the π₯-coordinate with respect to
time. Weβre given information about
dπ₯ by dπ‘. At a given point, at the point
negative one, three, the value of dπ₯ by dπ‘ is equal to two. And so what weβre going to do
first is begin by differentiating the entire equation of our curve. Weβll need to use implicit
differentiation to do so.

Remember, weβre differentiating
with respect to π‘. And the derivative of zero is,
of course, zero. We can then go through and
differentiate term by term. And of course, weβre going to
use implicit differentiation to do so. Letβs begin with six π¦
squared. We can differentiate six π¦
squared with respect to π¦. It will be two times six
π¦. That simplifies to 12π¦. And so that means the
derivative of this expression with respect to π‘ will be 12π¦ times dπ¦ by
dπ‘. We perform a similar process
for two π₯ squared. We differentiate it with
respect to π₯. And thatβs two times two π₯ or
four π₯. And then we multiply that by
dπ₯ by dπ‘.

Next, we differentiate negative
two π₯ with respect to π‘. And to do so, we differentiate
with respect to π₯ and then times that by dπ₯ by dπ‘. In a similar way, the
derivative of five π¦ with respect to π‘ is five dπ¦ by dπ‘. Then the derivative of our
constant, negative 13, is simply equal to zero. Weβre now going to factor for
dπ¦ by dπ‘ and dπ₯ by dπ‘ just to neat some things up a little bit. When we do, we find that dπ¦ by
dπ‘ times 12π¦ plus five plus dπ₯ by dπ‘ times four π₯ minus two is equal to
zero.

Since weβre trying to find the
value of dπ¦ by dπ‘ at a given point, weβre going to rearrange to make dπ¦ by
dπ‘ the subject. We begin by subtracting the
term containing dπ₯ by dπ‘ from both sides; then we divide through by 12π¦ plus
five. And so our expression for dπ¦
by dπ‘ is as shown. All thatβs left is to perform a
few substitutions. Firstly, we know the value of
dπ₯ by dπ‘ is equal to two. And weβre interested in finding
the rate of change of the π¦-coordinate with respect to time at the same point,
when π₯ is equal to negative one and π¦ is equal to three.

And so we substitute all of
these values into our expression for dπ¦ by dπ‘. We get negative two times four
times negative one minus two over 12 times three plus five. And that gives us a value of 12
over 41. We arenβt given any units for
the rate of change of the π₯-coordinate. Although, we might assume that
itβs units per time. And so the rate of change of
the π¦-coordinate with respect to time at the same point is 12 over 41 or 12
over 41 units per time.

Now that weβve demonstrated a
number of examples on how to answer related rates problems, letβs recap the key
points from this lesson.

In this lesson, we learned that if
two related quantities change over time, the rates at which they change, and such
their derivatives, are also related. We saw that implicit
differentiation, that special version of the chain rule, is hugely important in
related rates problems. But we can also use other rules for
differentiations such as the product rule. Finally, we saw how considering the
sign of the derivative can give us information about whether that quantity itself is
increasing or decreasing. If the final derivative, the rate
of change, is positive, then the quantity is increasing. And if itβs negative, the quantity
is decreasing.