
## Section3.5Related Rates

###### 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?

In most of our applications of the derivative so far, we have been interested in the instantaneous rate at which one variable, perhaps called $y\text{,}$ changes with respect to another, perhaps called $x\text{,}$ leading us to compute and interpret $\frac{dy}{dx}\text{.}$ We next consider situations where several variable quantities are related, but where each quantity is implicitly a function of time, which will be represented by the variable $t\text{.}$ Through knowing how the quantities are related, we will be interested in determining how their respective rates of change with respect to time are related.

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 Example3.47.

###### Example3.47

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 $d = 16\text{?}$ Why? These questions have been broken up into the following parts to guide you through the solution process.

1. Recall that the volume of a sphere of radius $r$ is $V = \frac{4}{3} \pi r^3\text{.}$ 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 $t\text{,}$ with respective derivatives $\frac{dV}{dt}$ and $\frac{dr}{dt}\text{.}$ Differentiate both sides of the equation $V = \frac{4}{3} \pi r^3$ with respect to $t$ to find a formula for $\frac{dV}{dt}$ that depends on both $r$ and $\frac{dr}{dt}\text{.}$

2. By differentiating the volume function with respect to time, we have related the rates of change of $V$ and $r\text{.}$ 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 $\frac{dr}{dt}$ or $\frac{dV}{dt}\text{?}$ Why?

3. 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 $\frac{dV}{dt}\text{,}$ 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 $d = 12\text{?}$

4. How is the situation different when $d = 16\text{?}$ When is the radius changing more rapidly, when $d = 12$ or when $d = 16\text{?}$

Hint
1. Remember that we need to apply the chain rule to the right side of the equation to differentiate $r$ with respect to $t\text{.}$

2. Look at the units.

3. Use your answer to part (b) to help fill in all missing information. Only one variable should be left.

4. Which radius results in a larger rate? Note that this question asks about the rate of change of the radius, not of the volume.

1. $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.$

2. $\frac{dV}{dt}=20$ cubic inches per second.

3. $20=4\pi (6)^2 \frac{dr}{dt},$ so $\frac{dr}{dt}=\frac{5}{36\pi}$ inches per second when the diameter is 12 inches.

4. $\frac{dr}{dt}=\frac{5}{64\pi}$ inches per second when the diameter is 16 inches. Thus, the radius is changing more rapidly when the diameter is 12 inches.

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

\begin{equation*} \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}. \end{equation*}
2. Since the units are cubic inches per second, this is a volume rate. So, $\frac{dV}{dt}=20$ cubic inches per second.

3. Since $\frac{dV}{dt}=20$ and $r=6$ when $d=12\text{,}$ we have $20=4\pi(6)^2\frac{dr}{dt}\text{.}$ So, $\frac{dr}{dt}=\frac{5}{36\pi}$ inches per second.

4. Doing the same thing as in part (c), we find that $\frac{dr}{dt}=\frac{5}{64\pi}$ inches per second when the diameter is $16$ inches (so $r=8$). 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.

### SubsectionRelated Rates Problems

In problems where two or more quantities can be related to one another and all of the variables involved are implicit functions of time, $t\text{,}$ we are often interested in examining how their rates are related; we call these related rates problems. Once we have an equation establishing the relationship among the variables, we differentiate implicitly with respect to time to find connections among the rates of change.

###### Example3.48

Sand is being dumped by a conveyor belt onto a pile so that the sand forms a right circular cone, as pictured below in Figure3.49. As sand falls from the conveyor belt, several features of the sand pile will change: the volume ($V$) of the pile will grow, the height ($h$) will increase, and the radius ($r$) will get bigger, too. All of these quantities are related to one another, and the rate at which each is changing is related to the rate at which sand falls from the conveyor.

Find an equation relating the rates at which the volume, radius, and height each change with respect to time.

Hint

Recall that the volume of a right circular cone is

\begin{equation*} V=\frac13\pi r^2h\text{.} \end{equation*}

Remember that all of $V\text{,}$ $r\text{,}$ and $h$ are functions of $t\text{.}$ You will need the chain rule (among others) to differentiate.

\begin{equation*} \frac{dV}{dt}=\frac23\pi rh\frac{dr}{dt}+\frac13\pi r^2\frac{dh}{dt}\text{.} \end{equation*}
Solution

We begin by identifying which variables are changing and how they are related. In this problem, we observe that the radius and height of the pile are related to its volume by the standard equation for the volume of a cone,

\begin{equation*} V = \frac{1}{3} \pi r^2 h\text{.} \end{equation*}

Viewing each of $V\text{,}$ $r\text{,}$ and $h$ as functions of $t\text{,}$ we differentiate implicitly to arrive at an equation that relates their respective rates of change. Taking the derivative of each side of the equation with respect to $t\text{,}$ we find

\begin{equation*} \frac{d}{dt}[V] = \frac{d}{dt}\left[\frac{1}{3} \pi r^2 h\right]\text{.} \end{equation*}

On the left, $\frac{d}{dt}[V]$ is simply $\frac{dV}{dt}\text{.}$ On the right, the situation is more complicated, as both $r$ and $h$ are implicit functions of $t\text{.}$ Hence we need the product and chain rules. We find that12Note particularly how we are using ideas from Section2.7 on implicit differentiation. There we found that when $y$ is an implicit function of $x\text{,}$ $\frac{d}{dx}[y^2] = 2y \frac{dy}{dx}\text{.}$ The same principles are applied here when we compute $\frac{d}{dt}[r^2] = 2r \frac{dr}{dt}\text{.}$

\begin{align*} \frac{dV}{dt} &= \frac{d}{dt}\left[\frac{1}{3} \pi r^2 h\right]\\ &= \frac{1}{3} \pi \frac{d}{dt}\big[r^2\big]h + \frac{1}{3} \pi r^2\frac{d}{dt}[h]\\ &= \frac{1}{3} \pi 2r\frac{dr}{dt}h + \frac{1}{3} \pi r^2 \frac{dh}{dt}\text{.} \end{align*}

Thus, the equation

\begin{equation*} \frac{dV}{dt} = \frac{2}{3} \pi rh \frac{dr}{dt}+ \frac{1}{3} \pi r^2 \frac{dh}{dt} \end{equation*}

relates the rates of change of $V\text{,}$ $h\text{,}$ and $r$ with respect to time.

If we are given sufficient additional information, we may then find the value of one or more of these rates of change at a specific point in time.

###### Example3.50

In the setting of Example3.48, suppose we also know the following: (a) sand falls from the conveyor in such a way that the height of the pile is always half the radius, and (b) sand falls from the conveyor belt at a constant rate of 10 cubic feet per minute. How fast is the height of the sandpile changing at the moment the radius is 4 feet?

Hint

Use your equation from Example3.48 that says

\begin{equation*} \frac{dV}{dt} = \frac{2}{3} \pi rh \frac{dr}{dt}+\frac{1}{3} \pi r^2 \frac{dh}{dt}\text{.} \end{equation*}

Alternatively, you may try substituting $h=\frac12r$ into the conical volume formula $V=\frac13\pi r^2h$ and retaking the derivative (you would not need the product rule this time, but would still need the chain rule).

When the radius is $4$ feet, the height is changing at a rate of $\frac{5}{8\pi}\approx0.199$ feet per minute.

Solution

First, we note that we want to know the value of $\frac{dh}{dt}$ when $r=4\text{.}$

The information that the height is always half the radius tells us that for all values of $t\text{,}$ $h = \frac{1}{2}r\text{.}$ Differentiating with respect to $t\text{,}$ it follows that $\frac{dh}{dt} = \frac{1}{2} \frac{dr}{dt}\text{.}$ These relationships enable us to relate $\frac{dV}{dt}$ to just one of $r$ or $h\text{.}$ Substituting the expressions involving $r$ and $\frac{dr}{dt}$ for $h$ and $\frac{dh}{dt}\text{,}$13Note that even though we are looking for the rate at which the height changes, the information we know involves the radius, so it is easier to first solve things in terms of $r$ and $\frac{dr}{dt}$ and then transition back to $\frac{dh}{dt}\text{.}$ Also, observe that rather than substituting expressions into the formula for $\frac{dV}{dt}$ we could instead substitute into the formula for $V$ and recalculate the derivative. Doing so would yield $V=\frac13\pi r^2\big(\frac12r\big)=\frac16\pi r^3$ and $\frac{dV}{dt}=\frac12\pi r^2\frac{dr}{dt}\text{,}$ which is the same relation that we have below. we now have that

\begin{align*} \frac{dV}{dt}\amp=\frac23\pi rh\frac{dr}{dt}+\frac13\pi r^2\frac{dh}{dt}\\ \amp = \frac{2}{3} \pi r \left(\frac{1}{2}r \right) \frac{dr}{dt}+\frac13\pi r^2\left(\frac12\frac{dr}{dt}\right)\\ \amp=\frac12\pi r^2\frac{dr}{dt}\text{.} \end{align*}

Since sand falls from the conveyor at the constant rate of 10 cubic feet per minute, the value of $\frac{dV}{dt}\text{,}$ the rate at which the volume of the sand pile changes, is $\frac{dV}{dt} = 10$ ft$^3$/min. We are interested in how fast the height of the pile is changing at the instant when $r = 4\text{,}$ so we substitute $r = 4\text{,}$ $\frac{dV}{dt} = 10\text{,}$ and $\frac{dr}{dt}=\left.\frac{dr}{dt}\right|_{r=4}$ into our equation above to find

\begin{equation*} 10 = \frac{1}{2} \pi (4)^2 \left. \frac{dr}{dt} \right|_{r=4}= 8\pi \left. \frac{dr}{dt} \right|_{r=4}\text{.} \end{equation*}

Only the value of $\left. \frac{dr}{dt} \right|_{r=4}$ remains unknown; solving, we find

\begin{equation*} \left. \frac{dr}{dt} \right|_{r=4} = \frac{10}{8\pi} \approx 0.39789 \ \text{ft/min}\text{.} \end{equation*}

Recall that we were interested in how fast the height of the pile was changing at this instant, so we want to know $\frac{dh}{dt}$ when $r = 4\text{.}$ Since $\frac{dh}{dt} = \frac{1}{2} \frac{dr}{dt}$ for all values of $t\text{,}$ it follows that

\begin{equation*} \left. \frac{dh}{dt} \right|_{r=4} =\frac12\left.\frac{dr}{dt}\right|_{r=4}= \frac{5}{8\pi} \approx 0.19894 \ \text{ft/min}\text{.} \end{equation*}

In Example3.50, it is important to note the difference between the notations $\frac{dr}{dt}$ and $\left. \frac{dr}{dt} \right|_{r=4}\text{.}$ The former represents the rate of change of $r$ with respect to $t$ at an arbitrary value of $t\text{,}$ while the latter is the rate of change of $r$ with respect to $t$ at a particular moment, the moment when $r = 4\text{.}$

Our work with both of the sandpile problems above is similar in many ways to our approach in Example3.47, and these steps are typical of most related rates problems. In certain ways, they also resemble work we do in applied optimization problems, and here we summarize the main approach for consideration in subsequent problems.

###### Note3.51
• 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, such as $\left. \frac{dr}{dt} \right|_{r=4}$ or $r'(4)\text{.}$

When identifying variables and drawing a picture, it is important to think about the dynamic ways in which the quantities change. Sometimes a sequence of pictures can be helpful; for some pictures that can be easily modified as applets built in Geogebra, see the following links,14We again refer to the work of Prof.Marc Renault of Shippensburg University, found at http://gvsu.edu/s/5p. which represent

• how a circular oil slick's area grows as its radius increases (http://gvsu.edu/s/9n);

• how the location of the base of a ladder and its height along a wall change as the ladder slides (http://gvsu.edu/s/9o);

• how the water level changes in a conical tank as it fills with water at a constant rate (http://gvsu.edu/s/9p; compare to the problem in Example3.52 below); and

• how a skateboarder's shadow changes as he moves past a lamppost (http://gvsu.edu/s/9q).

Drawing well-labeled diagrams and envisioning how different parts of the figure change is a key part of understanding related rates problems and being successful at solving them.

###### Example3.52

A water tank has the shape of an inverted circular cone (point down) with a base of radius 6 feet and a depth of 8 feet. Suppose that water is being pumped into the tank at a constant instantaneous rate of 4 cubic feet per minute.

1. Draw a picture of the conical tank, including a sketch of the water level at a point in time when the tank is not yet full. Say that $r$ is the radius of the water's surface and $h$ is the depth of the water at a given time, $t\text{,}$ and label these variables on your figure.

2. What equation relates the radius and height of the water, and why?

3. Determine an equation that relates the volume of water in the tank at time $t$ to the depth $h$ of the water at that time.

4. Through differentiation, find an equation that relates the instantaneous rate of change of water volume with respect to time to the instantaneous rate of change of water depth at time $t\text{.}$

5. Find the instantaneous rate at which the water level is rising when the water in the tank is 3 feet deep.

6. When is the water rising most rapidly: at $h = 3\text{,}$ $h = 4\text{,}$ or $h = 5\text{?}$

Hint

2. Recall that the volume of a cone is $V = \frac{1}{3} \pi r^2 h\text{.}$ Use (b) to eliminate $r$ from this equation.

3. Remember to differentiate implicitly with respect to $t\text{.}$

4. Use $h = 3$ and the fact that the value of $\frac{dV}{dt}$ is given.

5. Consider a similar situation at http://gvsu.edu/s/9p. Why does this phenomenon occur? How can you use calculus to prove your answer?

1. $r = \frac{3}{4}h\text{.}$

2. $V = \frac{3}{16} \pi h^3\text{.}$

3. $\frac{dV}{dt} = \frac{9}{16} \pi h^2 \frac{dh}{dt} \text{.}$

4. $\left. \frac{dh}{dt} \right|_{h=3} = \frac{64}{81\pi} \approx 0.2515$ feet per minute.

5. $h'(3)\gt h'(4)\gt h'(5)\text{.}$

Solution
1. Letting $r$ represent the water's radius at time $t$ and $h$ the water's depth, we see the following situation:

2. Observe that the right triangle with legs of length $h$ and $r$ is similar to the right triangle with legs of length $8$ and $6\text{,}$ respectively, based on how the water assumes the shape of the tank. Thus $\frac{r}{h} = \frac{6}{8}\text{,}$ so that $r = \frac{3}{4}h\text{.}$

3. Since the water in the tank always takes the shape of a circular cone, the volume of water in the tank at time $t$ is given by $V = \frac{1}{3}\pi r^2 h\text{.}$ Because we have established that $r = \frac{3}{4}h\text{,}$ it follows that

\begin{equation*} V = \frac{1}{3}\pi \left( \frac{3}{4}h \right)^2 h = \frac{3}{16} \pi h^3\text{.} \end{equation*}
4. Differentiating with respect to $t\text{,}$ we now find

$$\frac{dV}{dt} = \frac{9}{16} \pi h^2 \frac{dh}{dt}\text{,}\label{watervolume}\tag{3.1}$$

which relates the rates of change of $V$ and $h\text{.}$

5. It is given in the problem setting that water is entering the tank at a rate of 4 cubic feet per minute, hence $\frac{dV}{dt} = 4\text{,}$ and we are interested in the rate of change of the water's depth when $h = 3\text{.}$ Substituting these values into Equation(3.1), we find that

\begin{equation*} 4 = \frac{9}{16} \pi (3)^2 \left. \frac{dh}{dt} \right|_{h=3}\text{,} \end{equation*}

so that $\left. \frac{dh}{dt} \right|_{h=3} = \frac{64}{81\pi} \approx 0.2515$ feet per minute.

6. Note that we can solve Equation(3.1) for $\frac{dh}{dt}\text{.}$ Doing so with the substitution of the constant rate $\frac{dV}{dt}=4\text{,}$ we find that

\begin{equation*} \frac{dh}{dt}=\frac{64}{9\pi h^2}\text{,} \end{equation*}

so $\frac{dh}{dt}$ is inversely proportional to $h^2\text{.}$ In other words, $\frac{dh}{dt}$ will be decreasing as $h^2$ increases (which is when $h$ increases, since $h\gt0$). Thus, the water is rising faster at a depth of 3 feet than at a depth of 4 or 5 feet. We can also see this by calculating $h'(4)$ and $h'(5)$ and comparing these values to our answer from (e).

Recognizing which geometric relationships are relevant in a given problem is often the key to finding the function to optimize. For instance, although the problem in Example3.52 is about a conical tank, the most important fact is that there are two similar right triangles involved. In another setting, we might use the Pythagorean Theorem to relate the legs of the triangle. But in the conical tank, the fact that the water fills the tank so that that the ratio of radius to depth is constant turns out to be the important relationship. In other situations where a changing angle is involved, trigonometric functions may provide the means to find relationships among various parts of the triangle.

###### Example3.53

A television camera is positioned 4000 feet from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. In addition, the auto-focus of the camera has to take into account the increasing distance between the camera and the rocket. We assume that the rocket rises vertically.15A similar problem is discussed and pictured dynamically at http://gvsu.edu/s/9t. Exploring the applet at the link will be helpful to you in answering the questions that follow.

1. Draw a figure that summarizes the given situation. What parts of the picture are changing? What parts are constant? Introduce appropriate variables to represent the quantities that are changing.

2. Find an equation that relates the camera's angle of elevation to the height of the rocket, and then find an equation that relates the instantaneous rate of change of the camera's elevation angle to the instantaneous rate of change of the rocket's height (where all rates of change are with respect to time).

3. Find an equation that relates the distance from the camera to the rocket to the rocket's height, as well as an equation that relates the instantaneous rate of change of distance from the camera to the rocket to the instantaneous rate of change of the rocket's height (where all rates of change are with respect to time).

4. Suppose that the rocket's speed is 600 ft/sec at the instant it has risen 3000 feet. How fast is the distance from the television camera to the rocket changing at that moment? If the camera is following the rocket, how fast is the camera's angle of elevation changing at that same moment?

5. If from an elevation of 3000 feet onward the rocket continues to rise at 600 feet/sec, will the rate of change of distance with respect to time be greater when the elevation is 4000 feet than it was at 3000 feet, or less? Why?

Hint
1. Let $\theta$ represent the camera angle and note that one leg of the right triangle is constant. Which two are changing?

2. Think trigonometrically.

3. Think like Pythagoras.

4. Use the facts that $h = 3000$ and $\left.\frac{dh}{dt}\right|_{h=3000} = 600$ in your preceding work (assuming $h$ denotes the height of the rocket).

5. You can answer this question intuitively or by changing the value of $h$ in your work in (d).

1. $\frac{dh}{dt} = 4000 \sec^2 (\theta) \frac{d\theta}{dt}\text{.}$

2. $h \frac{dh}{dt} = z \frac{dz}{dt}\text{.}$

3. $\left. \frac{dz}{dt} \right|_{h=3000} = 360$ feet per second; $\left. \frac{d\theta}{dt} \right|_{h=3000} = \frac{12}{125}$ radians per second.

4. $\left.\frac{dz}{dt}\right|_{h=4000}\gt\left.\frac{dz}{dt}\right|_{h=3000}\text{.}$

Solution
1. Letting $\theta$ be the camera's elevation angle, $h$ the rocket's height, and $z$ the distance from the camera to the rocket, we have the following situation at a given point in time:

2. To relate $\theta$ and $h\text{,}$ observe that the tangent function is a good choice since $\tan(\theta) = \frac{h}{4000}\text{,}$ so that

\begin{equation*} h = 4000 \tan(\theta)\text{.} \end{equation*}

Differentiating implicitly with respect to $t\text{,}$ we find that

$$\frac{dh}{dt} = 4000 \sec^2 (\theta) \frac{d\theta}{dt}\text{.}\label{rocketanglerates}\tag{3.2}$$
3. To relate $z$ and $h\text{,}$ the Pythagorean Theorem is natural to consider. This tells us

\begin{equation*} h^2 + (4000)^2 = z^2\text{.} \end{equation*}

Differentiating both sides implicitly with respect to $t\text{,}$ it follows that

\begin{equation*} 2h \frac{dh}{dt} = 2z \frac{dz}{dt}\text{,} \end{equation*}

and thus

$$h \frac{dh}{dt} = z \frac{dz}{dt}\text{.}\label{rocketrates}\tag{3.3}$$
4. Using the given fact that the rocket's speed is 600 ft/sec at the instant it has risen 3000 feet, we know that $\left. \frac{dh}{dt} \right|_{h=3000} = 600\text{.}$ Note further in the triangle that when $h = 3000\text{,}$ use of the Pythagorean Theorem yields $z=5000\text{.}$ Substituting this information into Equation(3.3) from (c), we find that

\begin{equation*} 3000 \cdot 600 = 5000 \cdot \left. \frac{dz}{dt} \right|_{h=3000}\text{.} \end{equation*}

Solving for $\left. \frac{dz}{dt} \right|_{h=3000}$ gives us

\begin{equation*} \left. \frac{dz}{dt} \right|_{h=3000} = \frac{1800}{5} = 360 \ \text{feet/sec}\text{.} \end{equation*}

To answer the question about how fast the camera angle is changing, we use the same information but now in Equation(3.2), from (b). Observe that when $h = 3000\text{,}$ it follows that $\cos(\theta) = \frac{4}{5}\text{,}$ so $\sec(\theta) = \frac{5}{4}\text{.}$ Thus

\begin{equation*} 600 = 4000 \cdot \frac{25}{16} \left. \frac{d\theta}{dt} \right|_{h=3000}\text{,} \end{equation*}

and hence

\begin{equation*} \left. \frac{d\theta}{dt} \right|_{h=3000} = \frac{6 \cdot 16}{40 \cdot 25} = \frac{12}{125} \ \text{radians per second}\text{.} \end{equation*}
5. Recalling that $h \frac{dh}{dt} = z \frac{dz}{dt}\text{,}$ it follows that

\begin{equation*} \frac{dz}{dt} = \frac{h}{z} \frac{dh}{dt}\text{.} \end{equation*}

Using this equation when $h = 3000$ and $\frac{dh}{dt} = 600$ led us to conclude that $\left. \frac{dz}{dt} \right|_{h=3000} = \frac{3}{5} \cdot 600 = 360$ feet/sec. If we instead use $h = 4000\text{,}$ it follows that $z = 4000\sqrt{2}\text{,}$ so that

\begin{equation*} \left. \frac{dz}{dt} \right|_{h=4000} = \frac{4}{4\sqrt{2}} \cdot 600 \approx 424.26 \ \text{feet/sec}\text{.} \end{equation*}

Indeed, $\frac{dz}{dt}$ is an increasing function of $h$ provided that $\frac{dh}{dt}$ is constant because we can write $z = \sqrt{h^2 + 4000^2}\text{,}$ making

\begin{equation*} \frac{dz}{dt} = \frac{h}{\sqrt{h^2 + 4000^2}} \frac{dh}{dt} = \frac{h}{\sqrt{h^2 + 4000^2}} 600\text{.} \end{equation*}

It is straightforward to verify that $\frac{h}{\sqrt{h^2 + 4000^2}}$ is an increasing function of $h\text{;}$ hence any constant multiple of it is as well.

In addition to finding instantaneous rates of change at particular points in time, we can often make more general observations about how particular rates themselves will change over time. For instance, when a conical tank is filling with water at a constant rate, it seems obvious that the depth of the water should increase more slowly over time. Note how carefully we must phrase the relationship: we mean to say that while the depth, $h\text{,}$ of the water is increasing, its rate of change, $\frac{dh}{dt}\text{,}$ is decreasing (both as a function of $t$ and as a function of $h$). We make this observation by solving the equation that relates the various rates for one particular rate, without substituting any particular values for known variables or rates. For instance, in the conical tank problem in Example3.52, we established that

\begin{equation*} \frac{dV}{dt} = \frac{9}{16} \pi h^2 \frac{dh}{dt}\text{,} \end{equation*}

and hence

\begin{equation*} \frac{dh}{dt} = \frac{16}{9\pi h^2} \frac{dV}{dt}\text{.} \end{equation*}

Provided that $\frac{dV}{dt}$ is constant, it is immediately apparent that as $h$ gets larger, $\frac{dh}{dt}$ will get smaller but remain positive. Hence, the depth of the water is increasing at a decreasing rate.

###### Example3.54

As pictured in the applet athttp://gvsu.edu/s/9q, a skateboarder who is 6 feet tall (while on the skateboard) rides under a 15 foot tall lamppost at a constant rate of 3 feet per second. We are interested in understanding how fast the length of his shadow is changing at various points in time.

1. Draw an appropriate right triangle that represents a snapshot in time of the skateboarder, lamppost, and his shadow. Let $x$ denote the horizontal distance from the base of the lamppost to the skateboarder and $s$ represent the length of his shadow. Label these quantities, as well as the skateboarder's height and the lamppost's height on the diagram.

2. Observe that the skateboarder and the lamppost represent parallel line segments in the diagram, and thus similar triangles are present. Use similar triangles to establish an equation that relates $x$ and $s\text{.}$

3. Use your work in (b) to find an equation that relates $\frac{dx}{dt}$ and $\frac{ds}{dt}\text{.}$

4. At what rate is the length of the skateboarder's shadow increasing at the instant the skateboarder is 8 feet from the lamppost?

5. As the skateboarder's distance from the lamppost increases, is his shadow's length increasing at an increasing rate, increasing at a decreasing rate, or increasing at a constant rate?

6. Which is moving more rapidly: the skateboarder or the tip of his shadow? Explain, and justify your answer.

Hint
1. Note that the lengths of the legs of the right triangle will be $15$ for the vertical one and $x + s$ for the horizontal one.

2. The small triangle formed by the skateboarder and his shadow, with legs $6$ and $s\text{,}$ is similar to the large triangle that has the lamppost as one of its legs.

3. Simplify the equation in (b) as much as possible before differentiating implicitly with respect to $t\text{.}$

4. Find $\left. \frac{ds}{dt} \right|_{x=8}\text{.}$

5. Does the equation that relates $\frac{dx}{dt}$ and $\frac{ds}{dt}$ involve $x\text{?}$ Is $\frac{dx}{dt}$ changing or constant?

6. Let $y$ represent the location of the tip of the shadow, so that $y = x + s\text{.}$

1. $3s = 2x\text{.}$

2. $3 \frac{ds}{dt} = 2\frac{dx}{dt}\text{.}$

3. $\left. \frac{ds}{dt} \right|_{x=8} = 2$ feet per second.

4. The shadow length increases at a constant rate.

5. The tip of the shadow is moving at a rate of $5$ ft/sec.

Solution
1. The given information leads us to construct the following diagram:

2. The two right triangles shown in the diagram in (a) are similar: the small triangle is formed by the skateboarder and his shadow, and has legs of length $6$ and $s\text{;}$ the large triangle is formed by the lampost and the distance from the post to the shadow's tip, and has legs of length $15$ and $x+s\text{.}$ Similarity gives us that the ratios of leg lengths are equal, so we have

\begin{equation*} \frac{s}{6} = \frac{s+x}{15}\text{.} \end{equation*}

Simplifying yields $15s = 6s + 6x\text{,}$ or

\begin{equation*} 3s = 2x\text{.} \end{equation*}
3. Differentiating with respect to $t\text{,}$ it is immediate that

\begin{equation*} 3 \frac{ds}{dt} = 2\frac{dx}{dt}\text{.} \end{equation*}
4. Since $\frac{ds}{dt} = \frac{2}{3} \frac{dx}{dt}\text{,}$ and $\frac{dx}{dt} = 3\text{,}$ it follows that $\frac{ds}{dt} = 2$ for every value of $t$ (and $x$). Thus, $\left. \frac{ds}{dt} \right|_{x=8} = 2$ feet per second.

5. Because $\frac{ds}{dt}$ is constant, the shadow's length is increasing at a constant rate (irrespective of the distance from the skateboarder to the lamppost).

6. Let $y$ represent the location of the tip of the shadow, so that $y = x + s\text{.}$ Observe that we can now compute $\frac{dy}{dt}$ in terms of $\frac{dx}{dt}$ and $\frac{ds}{dt}\text{,}$ with

\begin{equation*} \frac{dy}{dt} = \frac{dx}{dt} + \frac{ds}{dt} = 3 + 2 = 5 \end{equation*}

feet per second, and hence the tip of the shadow is moving more rapidly than the skateboarder himself.

Even though the shadow's length is growing slower than the skateboarder is moving, the tip of the shadow is moving both due to growth and due to the skateboarder's movement, giving us the faster rate.

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.

###### Example3.55

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.

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.

\begin{equation*} \left. \frac{dz}{dt} \right|_{x = 45} = \frac{100}{\sqrt{5}} \approx 44.7214 \ \text{ft/sec} \text{.} \end{equation*}

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.

\begin{equation*} \left. \frac{ds}{dt} \right|_{x = 45} = \frac{430}{\sqrt{17}} \approx 104.2903 \ \text{ft/sec} \text{.} \end{equation*}
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 $x^2 + 90^2 = z^2\text{;}$ differentiating with respect to $t\text{,}$ we have

\begin{equation*} 2x\frac{dx}{dt} = 2z\frac{dz}{dt}\text{.} \end{equation*}

At the instant the ball is halfway to third base, we know $x = 45$ and $\left. \frac{dx}{dt} \right|_{x = 45} = 100\text{.}$ Moreover, by Pythagoras, $z^2 = 90^2 + 45^2\text{,}$ so $z = 45\sqrt{5}\text{.}$ Thus,

\begin{align*} \left.\frac{dz}{dt}\right|_{x=45}=\mathstrut\amp\left.\left(\frac xz\frac{dx}{dt}\right)\right|_{x=45}\\ =\mathstrut\amp\frac{45}{45\sqrt5}(100)\\ \approx\mathstrut\amp44.7214 \ \text{ft/sec}\text{.} \end{align*}

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 $t\text{,}$ 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, $x\text{,}$ $r\text{,}$ and $s$ form the sides of a right triangle, so that

$$x^2 + r^2 = s^2\text{,}\label{baseballPT}\tag{3.4}$$

by the Pythagorean Theorem.

Differentiating each side of Equation(3.4) with respect to $t\text{,}$ it follows that the three rates of change are related by the equation

\begin{equation*} 2x \frac{dx}{dt} + 2r \frac{dr}{dt} = 2s \frac{ds}{dt}\text{.} \end{equation*}

We are given that at the instant $x = 45\text{,}$ $r = \frac{90}{8}=\frac{45}4\text{,}$ so by Pythagoras, $s = \frac{45}{4}\sqrt{17}\text{.}$ In addition, at this same instant we know that $\left. \frac{dx}{dt} \right|_{x = 45} = 100$ and $\left. \frac{dr}{dt} \right|_{x = 45} = 30\text{.}$ Applying this information, we end up with

\begin{align*} \left.\frac{ds}{dt}\right|_{x=45}=\mathstrut\amp\left.\left(\frac xs\frac{dx}{dt}+\frac rs\frac{dr}{dt}\right)\right|_{x=45}\\ =\mathstrut\amp\frac{45}{\big(\frac{45}4\sqrt{17}\big)}(100)+\frac{\big(\frac{45}4\big)}{\big(\frac{45}4\sqrt{17}\big)}(30)\\ =\mathstrut\amp\frac{400}{\sqrt{17}}+\frac{30}{\sqrt{17}}\\ =\mathstrut\amp\frac{430}{\sqrt{17}}\\ \approx\mathstrut\amp104.2903 \ \text{ft/sec}\text{.} \end{align*}

Therefore, the distance between the ball and the runner is at this instant increasing at a rate of about $104.29$ feet per second.

### SubsectionSummary

• 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 $x\text{,}$ $y\text{,}$ and $z\text{,}$ then these quantities are related by the Pythagorean Theorem: $x^2 + y^2 = z^2\text{.}$ It follows by implicitly differentiating with respect to $t$ that their rates are related by the equation

\begin{equation*} 2x \frac{dx}{dt} + 2y\frac{dy}{dt} = 2z \frac{dz}{dt}\text{,} \end{equation*}

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

### SubsectionExercises

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?

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 Figure3.56.

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.

A baseball diamond is a square with sides $90$ feet long. Suppose a baseball player is advancing from second to third base at the rate of $24$ feet per second, and an umpire is standing on home plate. Let $\theta$ be the angle between the third baseline and the line of sight from the umpire to the runner. How fast is $\theta$ changing when the runner is $30$ feet from third base?

Sand is being dumped off a conveyor belt onto a pile in such a way that the pile forms in the shape of a cone whose radius is always equal to its height. Assuming that the sand is being dumped at a rate of $10$ cubic feet per minute, how fast is the height of the pile changing when there are $1000$ cubic feet on the pile?