Video Transcript
In this video, we’ll 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 𝑟 and the volume
𝑣 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 it’s 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,
it’s 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, we’re thinking about its derivative with respect to
time. So in this case, letting the area
of the circle be 𝐴, the rate of change of the area will be d𝐴 by d𝑡, 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 𝐴
with respect to 𝑟, but we’re going to struggle differentiating 𝐴 with respect to
𝑡.
Luckily, implicit differentiation,
which is a special version of the chain rule, can help. This says that since 𝐴 is a
function in 𝑟, the derivative of 𝐴 with respect to 𝑡 is equal to d𝐴 by d𝑟 times
d𝑟 by d𝑡. And so we really need to find
expressions or values for d𝐴 by d𝑟 and d𝑟 by d𝑡. Of course, we already identified
that 𝐴 is equal to 𝜋𝑟 squared, so d𝐴 by d𝑟 is fairly straightforward. 𝜋 is, of course, a constant. And so we can use the general power
rule for differentiation to find d𝐴 by d𝑟. It’s two 𝜋𝑟.
But what about d𝑟 by d𝑡? Well, in fact, we’re told that the
radius of the circle increases at a rate of three millimeters per second. This means that d𝑟 by d𝑡, the
rate of change of the radius, is positive, since it’s increasing, three. And so d𝐴 by d𝑡 is the product of
these two expressions. It’s two 𝜋𝑟 times three, which is
equal to six 𝜋𝑟. 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 𝑟
equals 15 into this expression. And this will give us the
instantaneous rate of change of the area when the radius is this value. d𝐴 by d𝑡 at 𝑟 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 we’ve 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, we’ll 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,
we’re 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 𝐴, its rate of change is d𝐴 by d𝑡.
We do have a slight problem,
though. The surface area of a sphere
whose radius is 𝑟 units is four 𝜋𝑟 squared. And so to find an expression
for the rate of change of 𝐴, d𝐴 by d𝑡, we’re going to use implicit
differentiation. This says that the derivative
of 𝐴 with respect to 𝑡 is equal to the derivative of 𝐴 with respect to 𝑟
times d𝑟 by d𝑡. Now d𝐴 by d𝑟 can be
calculated using the power rule for differentiation. Since 𝜋 is a constant, d𝐴 by
d𝑟 is two times four 𝜋𝑟, which is eight 𝜋𝑟. And so d𝐴 by d𝑡 is the
product of eight 𝜋𝑟 and d𝑟 by d𝑡.
Usually we’d look to find some
value or expression for the rate of change of radius with respect to time. That’s d𝑟 by d𝑡. In this question, though, we’re
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 𝑣, we’re told information about d𝑣 by d𝑡. In fact, the balloon is leaking
helium, so its derivative, the rate of change of 𝑣 with respect to time, is
going to be negative. It’s negative 48. This doesn’t help us
particularly at this point with finding d𝑟 by d𝑡, but we can perform a similar
process using implicit differentiation but this time for d𝑣 by d𝑡.
We know that d𝑣 by d𝑡 could
be calculated by finding the product of d𝑣 by d𝑟 and d𝑟 by d𝑡. We know the value of d𝑣 by
d𝑡. It’s negative 48. So, in fact, if we can find an
expression or value for d𝑣 by d𝑟, we can rearrange to find a formula or value
for d𝑟 by d𝑡. And we know, in fact, that the
volume of a sphere with radius 𝑟 is four-thirds 𝜋𝑟 cubed. And so its derivative with
respect to 𝑟, d𝑣 by d𝑟, is three times four-thirds 𝜋𝑟 squared, which is
four 𝜋𝑟 squared. And at this stage, we can also
replace d𝑣 by d𝑡 with negative 48. And so now we see that we can
find a formula for d𝑟 by d𝑡 by dividing both sides of this equation by four
𝜋𝑟 squared. That’s negative 48 over four
𝜋𝑟 squared, which becomes negative 12 over 𝜋𝑟 squared.
And now we see we can replace
d𝑟 by d𝑡 in our earlier formula for d𝐴 by d𝑡. When we do, we get that d𝐴 by
d𝑡 is eight 𝜋𝑟 times negative 12 over 𝜋𝑟 squared. We can simplify by cross
canceling one value of 𝜋 and one 𝑟, since we know 𝑟 is not gonna be equal to
zero. And so we have an expression
for d𝐴 by d𝑡 in terms of 𝑟. It’s negative 96 over 𝑟. The rate of change of the
surface area when 𝑟 is equal to 41 then is found by substituting 𝑟 equals 41
in. And it’s 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. d𝐴 by d𝑡 is negative 96 over
41 square centimeters per second.
In our next example, we’re 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.
We’re given quite a lot of
information about the area and information about the length and width of a
rectangle. Let’s begin by defining the
length of our rectangle to be equal to 𝑙 centimeters and its width to be equal
to 𝑤. And so it’s area 𝐴 will be
equal to its length times its width 𝑙𝑤. That’s not quite enough,
though. We’re looking to find
information about the rate of change of the area, in other words, the derivative
of area 𝐴 with respect to 𝑡. 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 d𝐴 by d𝑡 in terms of 𝑙, 𝑤, and their respective
derivatives. Let’s begin with the product
rule. 𝐴 itself is the product of
what we assume are two differentiable functions 𝑙 and 𝑤. Now we can make that assumption
because in the question we’re given the rate of change of 𝑙 and the rate of
change of 𝑤. So the derivative with 𝐴 with
respect to 𝑡 according to the product rule is 𝑙 times d𝑤 d𝑡 plus 𝑤 times
d𝑙 by d𝑡.
And this is really useful
because we’re told the length of the rectangle increases at a rate of 15
centimeters per second. And so the derivative of 𝑙
with respect to 𝑡 must be positive since it’s increasing and it must be equal
to 15. Similarly, we can say that the
derivative of 𝑤 with respect to 𝑡, the rate of change of its width, is 13. Substituting these values into
our earlier expression for d𝐴 by d𝑡, and we get 13𝑙 plus 15𝑤. And actually that’s everything
we need because we’re told that the length of the rectangle at a given point is
25 centimeters and its width is 12.
And so we’re going to
substitute these values into our expression for d𝐴 by d𝑡. 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 we’re 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 hadn’t
been told that it was increasing, we could infer that from the sign of the
derivative; it’s positive.
In our previous examples, we’ve
looked at geometric applications of related rates problems. It’s important to realize that we
can also apply similar processes when working with equations of curves. Let’s see what that might look
like.
A particle is moving along the
curve six 𝑦 squared plus two 𝑥 squared minus two 𝑥 plus five 𝑦 minus 13
equals zero. If the rate of change of its
𝑥-coordinate with respect to time as it passes through the point negative one,
three is two, find the rate of change of its 𝑦-coordinate with respect to time
at the same point.
So let’s begin by looking at
what the question actually wants us to find. It wants us to find the rate of
change of its 𝑦-coordinate. Now we recall that the rate of
change of a quantity is its derivative with respect to time. So in this case, we’re looking
to find d𝑦 by d𝑡. And in fact, we’re given
information about the rate of change of the 𝑥-coordinate with respect to
time. We’re given information about
d𝑥 by d𝑡. At a given point, at the point
negative one, three, the value of d𝑥 by d𝑡 is equal to two. And so what we’re going to do
first is begin by differentiating the entire equation of our curve. We’ll need to use implicit
differentiation to do so.
Remember, we’re differentiating
with respect to 𝑡. And the derivative of zero is,
of course, zero. We can then go through and
differentiate term by term. And of course, we’re going to
use implicit differentiation to do so. Let’s begin with six 𝑦
squared. We can differentiate six 𝑦
squared with respect to 𝑦. It will be two times six
𝑦. That simplifies to 12𝑦. And so that means the
derivative of this expression with respect to 𝑡 will be 12𝑦 times d𝑦 by
d𝑡. We perform a similar process
for two 𝑥 squared. We differentiate it with
respect to 𝑥. And that’s two times two 𝑥 or
four 𝑥. And then we multiply that by
d𝑥 by d𝑡.
Next, we differentiate negative
two 𝑥 with respect to 𝑡. And to do so, we differentiate
with respect to 𝑥 and then times that by d𝑥 by d𝑡. In a similar way, the
derivative of five 𝑦 with respect to 𝑡 is five d𝑦 by d𝑡. Then the derivative of our
constant, negative 13, is simply equal to zero. We’re now going to factor for
d𝑦 by d𝑡 and d𝑥 by d𝑡 just to neat some things up a little bit. When we do, we find that d𝑦 by
d𝑡 times 12𝑦 plus five plus d𝑥 by d𝑡 times four 𝑥 minus two is equal to
zero.
Since we’re trying to find the
value of d𝑦 by d𝑡 at a given point, we’re going to rearrange to make d𝑦 by
d𝑡 the subject. We begin by subtracting the
term containing d𝑥 by d𝑡 from both sides; then we divide through by 12𝑦 plus
five. And so our expression for d𝑦
by d𝑡 is as shown. All that’s left is to perform a
few substitutions. Firstly, we know the value of
d𝑥 by d𝑡 is equal to two. And we’re interested in finding
the rate of change of the 𝑦-coordinate with respect to time at the same point,
when 𝑥 is equal to negative one and 𝑦 is equal to three.
And so we substitute all of
these values into our expression for d𝑦 by d𝑡. 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 aren’t given any units for
the rate of change of the 𝑥-coordinate. Although, we might assume that
it’s units per time. And so the rate of change of
the 𝑦-coordinate with respect to time at the same point is 12 over 41 or 12
over 41 units per time.
Now that we’ve demonstrated a
number of examples on how to answer related rates problems, let’s 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 it’s negative, the quantity
is decreasing.
