CC Related Rates

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Section 4.8 Related Rates

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Motivating Questions

If two quantities that are related, such as the radius and volume of a spherical balloon, are both changing as implicit functions of time, how are their rates of change related? That is, how does the relationship between the values of the quantities affect the relationship between their respective derivatives with respect to time?

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This section corresponds to 3.13 Related Rates in the workbook.

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In Section 4.7 we saw that

y

can be defined implicitly as a function of

x

and that it was possible to find the rate of change of

y

with respect to

x,

that is

\frac{d y}{d x},

without explicitly solving for

y .

It is also possible for both

x

and

y

to be defined as implicit functions of time

(t);

that is, both variables

x

and

y

implicitly change with time. When two variables change in time, it is possible to find the rate of change of one variable given the rate of change of the second variable; these are known as related rates.

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We will start by examining an example without context to understand the process, then move on to applications in which we use related rates.

Example 4.8.1.

Suppose that

x

and

y

are both implicitly functions of time

t

and are related to each other by the equation

x y^{2} + y = x^{2} + 17.

We can find the rate of change of y with respect to time

\frac{d y}{d t}

given

x = 2,

y = 3

and the rate of change of x with respect to time when

\frac{d x}{d t} = 13 .

Take the implicit time derivative of the equation:

\frac{d}{d t} [x y^{2} + y = x^{2} + 17] = \overset{using the product rule}{\overset{⏞}{\frac{d x}{d t} y^{2} + x (2 y \frac{d y}{d t})}} + \frac{d y}{d t} = 2 x \frac{d x}{d t}

Substitute the known values and solve for

\frac{d y}{d t} :

13 (3)^{2} + (2) (2 (3) \frac{d y}{d t}) + \frac{d y}{d t} = 4 (13)

117 + 12 \frac{d y}{d t} + \frac{d y}{d t} = 52

13 \frac{d y}{d t} = - 65

\frac{d y}{d t} = - 5

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Subsection 4.8.1 Related Rates Problems

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In looking at applications of related rates we will start with an application to Business. We have seen that we can determine Profit, Revenue, and Cost as functions of the number of items sold

(x);

however, in the real world the number of items that are produced and sold is always changing in time. We can apply the idea of related rates to understand how Profit, Revenue, or Cost change in time given how the number of items changes in time.

Example 4.8.2.

Suppose that the revenue, in dollars, from the sale of

x

items is

R (x) = 480 x - 0.4 x^{2},

and the cost, in dollars, of producing

x

items is

C (x) = 5000 + 0.6 x^{2} .

Find the rate of change of profit

\frac{d P}{d t}

when

40

items are sold (i.e.

x = 40)

and the rate of change of units is

\frac{d x}{d t} = 30

units per day.

Hint.

Start by finding the Profit function, then find the implicit time derivative.

Answer.

\frac{d P}{d t} = 12000

dollars per day.

Solution.

We begin by creating the profit function

P (x) = R (x) - C (x) = (480 x - 0.4 x^{2}) - (5000 + 0.6 x^{2}) = 480 x - x^{2} - 5000.

Then take the implicit time derivative of the profit function as an implicit equation

\frac{d}{d t} [P (x) = 480 x - x^{2} - 5000]

\frac{d P}{d t} = 480 \frac{d x}{d t} - 2 x \frac{d x}{d t} .

Then plug in the values that are given,

x = 40

and

\frac{d x}{d t} = 30 :

\frac{d P}{d t} = 480 (30) - 2 (40) (30) = 12000.

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Now we want to consider other applications of related rates. Before we start we will outline a strategy for tackling these type of problems.

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Strategy for Setting up and Solving Related Rates Problems.

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Identify the quantities in the problem that are changing and choose clearly defined variable names for them. Draw one or more figures that clearly represent the situation.
Determine all rates of change that are known or given and identify the rate(s) of change to be found.
Find an equation that relates the variables whose rates of change are known to those variables whose rates of change are to be found.
Differentiate implicitly with respect to $t$ to relate the rates of change of the involved quantities.
Evaluate the derivatives and variables at the information relevant to the instant at which a certain rate of change is sought. Use proper notation to identify when a derivative is being evaluated at a particular instant.

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For example, suppose that air is being pumped into a spherical balloon so that its volume increases at a constant rate. Since the balloon’s volume and radius are related, we ought to be able to discover how fast the radius is changing by knowing how fast the volume is changing. Can we determine how fast the radius of the balloon is increasing when the balloon is a certain size? We explore a scenario of this type in Example 4.8.3.

Example 4.8.3.

A spherical balloon with diameter

d

is being inflated at a constant rate of 20 cubic inches per second. How fast is the radius of the balloon changing at the instant the balloon’s diameter is 12 inches? Is the radius changing more rapidly when

d = 12

or when

d = 16 ?

Why? These questions have been broken up into the following parts to guide you through the solution process.

Recall that the volume of a sphere of radius $r$ is $V = \frac{4}{3} π r^{3} .$ Note that in the setting of this problem, both $V$ and $r$ are changing as $t$ (time) changes, and thus both $V$ and $r$ may be viewed as implicit functions of $t,$ with respective derivatives $\frac{d V}{d t}$ and $\frac{d r}{d t} .$ Differentiate both sides of the equation $V = \frac{4}{3} π r^{3}$ with respect to $t$ to find a formula for $\frac{d V}{d t}$ that depends on both $r$ and $\frac{d r}{d t} .$
By differentiating the volume function with respect to time, we have “related the rates” of change of $V$ and $r .$ Recall that the problem statement tells us that the balloon is being inflated at a constant rate of 20 cubic inches per second. Is this rate the value of $\frac{d r}{d t}$ or $\frac{d V}{d t} ?$ Why?
Note that when the diameter of the balloon is 12 inches, we know the value of the radius. In the formula you found in (a) for $\frac{d V}{d t},$ substitute the values we know for the relevant quantities and solve for the remaining unknown quantity. How fast is the radius changing at the instant $d = 12 ?$
How is the situation different when $d = 16 ?$ When is the radius changing more rapidly, when $d = 12$ or when $d = 16 ?$

Hint.

Remember that we need to apply the chain rule to the right side of the equation to differentiate $r$ with respect to $t .$
Look at the units.
Use your answer to part (b) to help fill in all missing information. Only one variable should be left.
Which radius results in a larger rate? Note that this question asks about the rate of change of the radius, not of the volume.

Answer.

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} .$
$\frac{d V}{d t} = 20$ cubic inches per second.
$20 = 4 π (6)^{2} \frac{d r}{d t},$ so $\frac{d r}{d t} = \frac{5}{36 π}$ inches per second when the diameter is 12 inches.
$\frac{d r}{d t} = \frac{5}{64 π}$ inches per second when the diameter is 16 inches. Thus, the radius is changing more rapidly when the diameter is 12 inches.

Solution.

Starting with the equation $V = \frac{4}{3} π r^{3},$ we differentiate both sides with respect to $t .$ We need to use the chain rule to differentiate the right side of the equation since $r$ is really $r (t),$ a function of $t .$ So, the equation we have is really $V (t) = \frac{4}{3} π [r (t)]^{3} .$ Differentiating with respect to $t$ we have

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} .$
Since the units are cubic inches per second, this is a volume rate. So, $\frac{d V}{d t} = 20$ cubic inches per second.
Since $\frac{d V}{d t} = 20$ and $r = 6$ when $d = 12,$ we have $20 = 4 π (6)^{2} \frac{d r}{d t} .$ Solving this equation for $\frac{d r}{d t},$ we have $\frac{d r}{d t} = \frac{5}{36 π}$ inches per second.
Doing the same thing as in part (c), we find that $\frac{d r}{d t} = \frac{5}{64 π}$ inches per second when the diameter is $16$ inches (the only difference in our equation here is that $r = 8$ ). This number is smaller than the one calculated in part (c), so the radius is changing more rapidly when the diameter is $12$ inches. Intuitively, this makes sense since the balloon’s volume is growing at a constant rate: as the balloon grows, a small change in the radius will have a larger impact on the change in volume; equivalently, the same change in volume corresponds to a smaller change in the radius when the balloon is large.

Example 4.8.4.

16

foot ladder is placed against a wall. To visualize this use follow the link: http://gvsu.edu/s/9o

gvsu.edu/s/9o

;

If the bottom of the ladder moves away from the wall at a constant rate, how would you describe the movement of the top of the ladder?
Let $x$ be the distance from the bottom of the wall to the bottom of the ladder, and $h$ be the distance from the bottom of the wall to the top of the ladder. Both of these quantities change as a function of time, $t .$ If the base $(x)$ is increasing, is the rate of change of the height of the ladder on the wall $\frac{d h}{d t}$ positive or negative?
Suppose the top of the ladder slides down at a constant rate of $4$ feet per second. How fast is the base of the ladder moving when the height is 12 feet?

Answer.

As the base of the ladder moves away from the wall at a constant rate, the top of the ladder slides down the wall with a constant rate.
$\frac{d h}{d t}$ is negative.
The base is moving away from the wall at a rate of $\frac{d x}{d t} = \frac{24}{\sqrt{112}}$ feet per second.

Solution.

As the base of the ladder moves away from the wall at a constant rate, the top of the ladder slides down the wall with a constant rate.
The rate of change of the height of the ladder on the wall would be negative, that is, $\frac{d h}{d t} < 0 .$ This is negative since the distance from the bottom of the wall to the top of the ladder, $h,$ is decreasing.
Here we want $\frac{d x}{d t},$ how fast the base of the ladder is moving away from the wall. We know $h = 12$ and $\frac{d h}{d t} = - 4 .$

We will use the Pythagorean Theorem, which says $x^{2} + h^{2} = 16^{2},$ to compute the length of the missing side $x$ by plugging in $h = 12 :$

$x^{2} + 12^{2} = 16^{2} ⟹ x^{2} = 16^{2} - 12^{2} ⟹ x = \sqrt{16^{2} - 12^{2}} = \sqrt{112}$

Then we take the implicit time derivative

$\frac{d}{d t} [x^{2} + h^{2} = 16^{2}]$

$2 x \frac{d x}{d t} + 2 h \frac{d h}{d t} = 0.$

Finally, substitute in what is known and solve for $\frac{d x}{d t}$

$2 (\sqrt{112}) \frac{d x}{d t} + 2 (12) (- 4) = 0 ⟹ \frac{d x}{d t} = \frac{24}{\sqrt{112}} .$

Thus the base is moving away from the wall at a rate of $\frac{d x}{d t} = \frac{24}{\sqrt{112}}$ feet per second.

Example 4.8.5.

A person is watching a boat sailing parallel to a straight shoreline, shown below. The boat stays at a constant

50

feet from the coastline, moving at a rate of

10

feet per minute. Let

y

be the distance from the boat to the person.

Figure 4.8.6. Boat sailing parallel to a straight shoreline

Is the rate of change of the distance from the person to the boat, $\frac{d y}{d t},$ positive or negative?
How fast is the distance $y,$ the distance from the person to the boat, changing the when $y = 130$ feet?

Answer.

The rate of change of the distance distance from the person to the boat is positive, i.e. $\frac{d y}{d t} > 0 .$
The distance from the person to the boat is increasing at a rate of $\frac{d y}{d t} = \frac{12}{\sqrt{13}}$ feet per minute.

Solution.

The rate of change of the distance from the person to the boat is positive, that is, $\frac{d y}{d t} > 0,$ since the distance is increasing as the boat sails away.
To find how fast the distance $y$ is changing, we need to compute $\frac{d y}{d t} .$ We know $y = 130$ and $\frac{d x}{d t} = 10$ feet per minute.

We will use the Pythagorean Theorem, which tells us that $x^{2} + 50^{2} = y^{2} .$ First we find length of the missing side $x$ by plugging in $y = 12 :$

$x^{2} + 50^{2} = 130^{2} ⟹ x^{2} = 130^{2} - 50^{2} ⟹ x = \sqrt{130^{2} - 50^{2}} = 120.$

Then we take the implicit time derivative, noting that both $x$ and $y$ are changing with time:

$\frac{d}{d t} [x^{2} + 50^{2} = y^{2}]$

$2 x \frac{d x}{d t} + 0 = 2 y \frac{d y}{d t} .$

Finally, we substitute in what is known and solve for $\frac{d y}{d t} :$

$2 (120) (10) = 2 (130) \frac{d y}{d t} ⟹ \frac{d y}{d t} = \frac{120}{\sqrt{130}} .$

Therefore, the distance from the person to the boat is increasing at a rate of $\frac{d y}{d t} = \frac{12}{\sqrt{13}}$ feet per minute.

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In most of the examples thus far, we provided guided instruction to build a solution in a step-by-step way. For the closing example and the following exercises, most of the detailed work is left to the reader.

Example 4.8.7.

A baseball diamond is square and measures

90

feet on each side. A batter hits a ball along the third base line and runs to first base. At what rate is the distance between the ball and first base changing when the ball is halfway to third base, if at that instant the ball is traveling

100

feet/sec? At what rate is the distance between the ball and the runner changing at the same instant, if the runner is at that point one-eigth of the way to first base and running at

30

feet/sec?

Hint.

Draw some pictures! Label everything you know with a constant, and everything that is changing with a variable and a corresponding rate of change (with respect to time,

t

). For the instant in question, you will also want to write down the values of the variables at that instant and identify what quantities you need to find. Note that the basepaths meet at right angles.

Answer.

Let

x

denote the position of the ball (as a distance from home plate) at time

t

and

z

the distance from the ball to first base, as pictured below.

{\frac{d z}{d t} |}_{x = 45} = \frac{100}{\sqrt{5}} \approx 44.7214 ft/sec .

Let

r

be the runner’s position (as a distance from home plate) at time

t

and let

s

be the distance between the runner and the ball, as pictured.

{\frac{d s}{d t} |}_{x = 45} = \frac{430}{\sqrt{17}} \approx 104.2903 ft/sec .

Solution.

We let

x

denote the position of the ball (as a distance from home plate) at time

t

and

z

the distance from the ball to first base, as pictured below.

By the Pythagorean Theorem, we know that

x^{2} + 90^{2} = z^{2};

differentiating with respect to

t,

we have

2 x \frac{d x}{d t} = 2 z \frac{d z}{d t} .

At the instant the ball is halfway to third base, we know

x = 45

and

{\frac{d x}{d t} |}_{x = 45} = 100 .

Moreover, by the Pythagorean Theorem,

z^{2} = 90^{2} + 45^{2},

z = 45 \sqrt{5} .

We can solve for

\frac{d z}{d t}

using the equation above and then plug in all of the known variables:

\begin{aligned} {\frac{d z}{d t} |}_{x = 45} = & {(\frac{x}{z} \frac{d x}{d t}) |}_{x = 45} \\ = & \frac{45}{45 \sqrt{5}} (100) \\ \approx & 44.7214 ft/sec . \end{aligned}

Therefore, the distance between the ball and first base is at this instant growing at a rate of about

44.72

feet per second.

For the second question, we still let

x

represent the ball’s position at time

t,

but now we introduce

r

as the runner’s position (also as a distance from home plate) at time

t

and let

s

be the distance between the runner and the ball. In this setting, as seen in the diagram below,

x,

r,

and

s

form the sides of a right triangle, so that

\begin{matrix} (4.8.1) & x^{2} + r^{2} = s^{2}, \end{matrix}

by the Pythagorean Theorem.

Differentiating each side of Equation (4.8.1) with respect to

t,

it follows that the three rates of change are related by the equation

2 x \frac{d x}{d t} + 2 r \frac{d r}{d t} = 2 s \frac{d s}{d t} .

We are given that at the instant

x = 45,

r = \frac{90}{8} = \frac{45}{4},

so solving for

s

in the equation above, we have

s = \frac{45}{4} \sqrt{17} .

In addition, at this same instant we know that

{\frac{d x}{d t} |}_{x = 45} = 100

and

{\frac{d r}{d t} |}_{x = 45} = 30 .

Solving for

\frac{d s}{d t}

and applying this information, we end up with

\begin{aligned} {\frac{d s}{d t} |}_{x = 45} = & {(\frac{x}{s} \frac{d x}{d t} + \frac{r}{s} \frac{d r}{d t}) |}_{x = 45} \\ = & \frac{45}{(\frac{45}{4} \sqrt{17})} (100) + \frac{(\frac{45}{4})}{(\frac{45}{4} \sqrt{17})} (30) \\ = & \frac{400}{\sqrt{17}} + \frac{30}{\sqrt{17}} \\ = & \frac{430}{\sqrt{17}} \\ \approx & 104.2903 ft/sec . \end{aligned}

Therefore, the distance between the ball and the runner is at this instant increasing at a rate of about

104.29

feet per second.

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Subsection 4.8.2 Summary

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When two or more related quantities are changing as implicit functions of time, their rates of change can be related by implicitly differentiating the equation that relates the quantities themselves. For instance, if the sides of a right triangle are all changing as functions of time, say having lengths $x,$ $y,$ and $z,$ then these quantities are related by the Pythagorean Theorem: $x^{2} + y^{2} = z^{2} .$ It follows by implicitly differentiating with respect to $t$ that their rates are related by the equation

$2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 2 z \frac{d z}{d t},$

so if we know the values of $x,$ $y,$ and $z$ at a particular time, as well as two of the three rates, we can deduce the value of the third.

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Exercises 4.8.3 Exercises

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1. Related rates with polynomial equations.

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Consider

x^{4} + 2 x y^{4} = - 185,

find

\frac{d y}{d t}

when

x = 5,

y = - 3,

and

\frac{d x}{d t} = 2 .

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\frac{d y}{d t}

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Round to three decimal places.

Hint.

Both variables

x

and

y

are now implicitly defined. Take the time derivative of

x^{4} + 2 x y^{4}

implicitly.

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2. Height of a conical pile of gravel.

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Gravel is being dumped from a conveyor belt at a rate of

30

cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is

17

feet high? Recall that the volume of a right circular cone with height

h

and radius of the base

r

is given by

V = \frac{1}{3} π r^{2} h .

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When the pile is

17

feet high, its height is increasing at feet per minute.

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3. Movement of a shadow.

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A street light is at the top of a 16 foot tall pole. A 6 foot tall woman walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 40 feet from the base of the pole?

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The tip of the shadow is moving at ft/sec.

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4. Docking a boat.

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A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow. If the rope is being pulled in at a rate of 2 feet per second, how fast is the boat approaching the dock when the length of rope from bow to pulley is 13 feet?

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5. A leaking conical tank.

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Water is leaking out of an inverted conical tank at a rate of

12500.0

{cm}^{3} / min

at the same time that water is being pumped into the tank at a constant rate. The tank has height

13.0 m

and the the diameter at the top is

7.0 m .

If the water level is rising at a rate of

15.0 cm / min

when the height of the water is

4.5 m,

find the rate at which water is being pumped into the tank in cubic centimeters per minute.

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Answer:

{cm}^{3} / min

Hint.

R

be the unknown rate at which water is being pumped in. Then you know that if

V

is volume of water,

\frac{d V}{d t} = R - 12500.0 .

Use geometry (similar triangles) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius

r

and height

h

is given by

\frac{1}{3} π r^{2} h

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6. Filling a swimming pool.

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A swimming pool is

60

feet long and

25

feet wide. Its depth varies uniformly from

3

feet at the shallow end to

15

feet at the deep end, as shown in the Figure 4.8.8.

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Suppose the pool has been emptied and is now being filled with water at a rate of

800

cubic feet per minute. At what rate is the depth of water (measured at the deepest point of the pool) increasing when it is

5

feet deep at that end? Over time, describe how the depth of the water will increase: at an increasing rate, at a decreasing rate, or at a constant rate. Explain.

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