Active Calculus

As sand falls from the conveyor belt, several features of the sand pile will change: the volume of the pile will grow, the height will increase, and the radius 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.

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 that

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

(Note particularly how we are using ideas from Section 2.7 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 \(\frac{d}{dt}[r^2] = 2r \frac{dr}{dt}\text{.}\))

The equation

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

relates the rates of change of \(V\text{,}\) \(h\text{,}\) and \(r\text{.}\)

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.

Example 3.5.3.

In the setting of Example 3.5.1, 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?

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 \(\frac{dh}{dt}\text{,}\) we now have that

\begin{equation} \frac{dV}{dt} = \frac{1}{3} \pi r^2 \cdot \frac{1}{2} \frac{dr}{dt} + \frac{2}{3} \pi r \cdot \frac{1}{2}r \cdot \frac{dr}{dt}\text{.}\label{vCG}\tag{3.5.1} \end{equation}

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 \(r = 4\) and \(\frac{dV}{dt} = 10\) into Equation (3.5.1), to find

\begin{equation*} 10 = \frac{1}{3} \pi 4^2 \cdot \frac{1}{2} \left. \frac{dr}{dt} \right|_{r=4} + \frac{2}{3} \pi 4 \cdot \frac{1}{2}4 \cdot \left. \frac{dr}{dt} \right|_{r=4} = \frac{8}{3}\pi \left. \frac{dr}{dt} \right|_{r=4} + \frac{16}{3} \pi \left. \frac{dr}{dt} \right|_{r=4}\text{.} \end{equation*}

Only the value of \(\left. \frac{dr}{dt} \right|_{r=4}\) remains unknown. We combine like terms on the right side of the equation above to get \(10 = 8 \pi \left. \frac{dr}{dt} \right|_{r=4}\text{,}\) and solve for \(\left. \frac{dr}{dt} \right|_{r=4}\) to find

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

feet per second. Because we were interested in how fast the height of the pile was changing at this instant, 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

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

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 \(r = 4\text{.}\)

Had we known that \(h = \frac{1}{2}r\) at the beginning of Example 3.5.1, we could have immediately simplified our work by writing \(V\) solely in terms of \(r\) to have

\begin{equation*} V = \frac{1}{3} \pi r^2 \left(\frac{1}{2}h\right) = \frac{1}{6} \pi r^3\text{.} \end{equation*}

From this last equation, differentiating with respect to \(t\) implies

\begin{equation*} \frac{dV}{dt} = \frac{1}{2} \pi r^2 \frac{dr}{dt}\text{,} \end{equation*}

from which the same conclusions can be made.

Our work with the sandpile problem above is similar in many ways to our approach in Preview Activity 3.5.1, 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.

Note 3.5.4.

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 \(\left. \frac{dr}{dt} \right|_{r=4}\text{.}\)

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,¹ 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 the problem in Activity 3.5.2);
how a skateboarder's shadow changes as he moves past a lamppost http://gvsu.edu/s/9q.

We again refer to the work of Prof. Marc Renault of Shippensburg University, found at http://gvsu.edu/s/5p.

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.

Activity 3.5.2.

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.

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. Introduce variables that measure the radius of the water's surface and the water's depth in the tank, and label them on your figure.
Say that \(r\) is the radius and \(h\) the depth of the water at a given time, \(t\text{.}\) What equation relates the radius and height of the water, and why?
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.
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 \(t\text{.}\)
Find the instantaneous rate at which the water level is rising when the water in the tank is 3 feet deep.
When is the water rising most rapidly: at \(h = 3\text{,}\) \(h = 4\text{,}\) or \(h = 5\text{?}\)

Think about similar triangles.
Recall that the volume of a cone is \(V = \frac{1}{3} \pi r^2 h\text{.}\)
Remember to differentiate implicitly with respect to \(t\text{.}\)
Use \(h = 3\) and the fact that the value of \(\frac{dV}{dt}\) is given.
Consider http://gvsu.edu/s/9p. Why does this phenomenon occur?

\(r = \frac{3}{4}h\text{.}\)
\(V = \frac{3}{16} \pi h^3\text{.}\)
\(\frac{dV}{dt} = \frac{9}{16} \pi h^2 \frac{dh}{dt} \text{.}\)
\(\left. \frac{dh}{dt} \right|_{h=3} = \frac{64}{81\pi} \approx 0.2515\) feet per minute.

Letting \(r\) represent the water's radius at time \(t\) and \(h\) the water's depth, we see the following situation:
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, and thus \(\frac{r}{h} = \frac{6}{8}\text{,}\) so that \(r = \frac{3}{4}h\text{.}\)
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*}
Differentiating with respect to \(t\text{,}\) we now find

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

which relates the rates of change of \(V\) and \(h\text{.}\)
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 the equation that relates \(\frac{dV}{dt}\) and \(\frac{dh}{dt}\text{,}\) 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.

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 Activity 3.5.2 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.

Activity 3.5.3.

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. (A 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.)

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.
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).
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).
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?
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?

Let \(\theta\) represent the camera angle and note that one leg of the right triangle is constant. Which two are changing?
Think trigonometrically.
Think like Pythagoras.
Use the facts that \(h = 3000\) and \(\frac{dh}{dt} = 600\) in your preceding work.
You can answer this question intuitively or by changing the value of \(h\) in your work in (d).

\(\frac{dh}{dt} = 4000 \sec^2 (\theta) \frac{d\theta}{dt}\text{.}\)
\(h \frac{dh}{dt} = z \frac{dz}{dt}\text{.}\)
\(\left. \frac{dz}{dt} \right|_{h=3000} = 360 \ \text{feet/sec}; \) \(\left. \frac{d\theta}{dt} \right|_{h=3000} = \frac{12}{125} \) radians per second.
greater.

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:
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

\begin{equation*} \frac{dh}{dt} = 4000 \sec^2 (\theta) \frac{d\theta}{dt}\text{.} \end{equation*}
To relate \(z\) and \(h\text{,}\) the Pythagorean Theorem is natural to consider. By this famous result,

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

Differentiating both sides implicitly with respect to \(t\text{,}\) it follows

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

and thus \(h \frac{dh}{dt} = z \frac{dz}{dt}\text{.}\)
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 \(h = 3000\text{,}\) it follows \(z = 5000\text{,}\) since the base leg of the triangle is constant at 4000, by using the Pythagorean Theorem. Substituting this information from the instant \(h = 3000\) into the equation that relates the rates of change of \(z\) and \(h\) found in (c), we find that

\begin{equation*} 2 \cdot 3000 \cdot 600 = 2 \cdot 5000 \cdot \left. \frac{dz}{dt} \right|_{h=3000}\text{.} \end{equation*}

Solving for \(\left. \frac{dz}{dt} \right|_{h=3000}\) we have

\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 the equation we found in (b) that relates the rates of change of \(\theta\) and \(h\text{:}\)

\begin{equation*} \frac{dh}{dt} = 4000 \sec^2 (\theta) \frac{d\theta}{dt}\text{.} \end{equation*}

Observe that when \(h = 3000\text{,}\) in the 3000-4000-5000 right triangle, it follows that \(\cos(\theta) = \frac{4}{5}\text{,}\) so \(\sec(\theta) = \frac{5}{4}\text{.}\) Thus, using the instantaneous information,

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

and thus

\begin{equation*} \left. \frac{d\theta}{dt} \right|_{h=3000} = \frac{6 \cdot 16}{40 \cdot 25} = \frac{12}{125}\text{,} \end{equation*}

which is measured in radians per second.
Recalling that \(2h \frac{dh}{dt} = 2z \frac{dz}{dt}\text{,}\) 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 \(\left. \frac{dz}{dt} \right|_{h=3000} = \frac{3}{5} \cdot 600 = 360 \ \text{feet/sec}\text{.}\) 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 \(h\text{,}\) provided that \(\frac{dh}{dt}\) is constant, because we can write \(h^2 + 4000^2 = z^2\text{,}\) so \(z = \sqrt{h^2 + 4000^2}\text{,}\) 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 \(h\text{.}\)

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 Activity 3.5.2, we established that

\begin{equation*} \frac{dV}{dt} = \frac{1}{16} \pi h^2 \frac{dh}{dt}\text{,} \end{equation*}

and hence

\begin{equation*} \frac{dh}{dt} = \frac{16}{\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.

Activity 3.5.4.

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

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.
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 \(s\text{.}\)
Use your work in (b) to find an equation that relates \(\frac{dx}{dt}\) and \(\frac{ds}{dt}\text{.}\)
At what rate is the length of the skateboarder's shadow increasing at the instant the skateboarder is 8 feet from the lamppost?
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?
Which is moving more rapidly: the skateboarder or the tip of his shadow? Explain, and justify your answer.

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.
The small triangle formed by the skateboarder and his shadow, with legs \(6\) and \(s\) is similar to the large triangle that has the lamppost as one of its legs.
Simplify the equation in (b) as much as possible before differentiating implicitly with respect to \(t\text{.}\)
Find \(\left. \frac{ds}{dt} \right|_{x=8}\text{.}\)
Does the equation that relates \(\frac{dx}{dt}\) and \(\frac{ds}{dt}\) involve \(x\text{?}\) Is \(\frac{dx}{dt}\) changing or constant?
Let \(y\) represent the location of the tip of the shadow, so that \(y = x + s\text{.}\)

\(3s = 2x\text{.}\)
\(3 \frac{ds}{dt} = 2\frac{dx}{dt}\text{.}\)
\(\left. \frac{ds}{dt} \right|_{x=8} = 2\) feet per second.
at a constant rate.
Let \(y\) represent the location of the tip of the shadow; \(\frac{dy}{dt} = 5\) feet/sec.

The given information leads us to construct the following diagram:
The small triangle formed by the skateboarder and his shadow, with legs of length \(6\) and \(s\) is similar to the large triangle that has the lamppost as one of its legs (length 15) and horizontal leg of length \(x + s\text{.}\) Because the ratios of the lengths of the legs of these two triangles is equal, we have

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

Simplifying, we have \(15s = 6s + 6x\text{,}\) so that \(9s = 2x\text{,}\) or most simply, \(3s = 2x\text{.}\)
Differentiating with respect to \(t\text{,}\) it is immediate that \(3 \frac{ds}{dt} = 2\frac{dx}{dt}\text{.}\)
Since \(\frac{ds}{dt} = \frac{2}{3} \frac{dx}{dt}\text{,}\) and \(\frac{dx}{dt} = 3\text{,}\) it follows \(\frac{ds}{dt} = 2\) for every value of \(t\) (and \(x\)). Thus, \(\left. \frac{ds}{dt} \right|_{x=8} = 2\) feet per second.
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).
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 \(\frac{dy}{dt} = \frac{dx}{dt} + \frac{ds}{dt} = 3 + 2 = 5\) feet/sec, and hence the tip of the shadow is moving more rapidly than the skateboarder himself.

In the first three activities of this section, we provided guided instruction to build a solution in a step by step way. For the closing activity and the following exercises, most of the detailed work is left to the reader.

Activity 3.5.5.

A baseball diamond is \(90'\) square. 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 at the same instant the runner is \(1/8\) of the way to first base running at \(30\) feet/sec?

Let \(x\) denote the position of the ball along the third base line at time \(t\text{,}\) and \(z\) the distance from the ball to first base. Note that the basepaths meet at 90 degree angles.

Let \(x\) denote the position of the ball at time \(t\) and \(z\) the distance from the ball to first base, as pictured below.

\(\left. \frac{dz}{dt} \right|_{x = 45} = \frac{100}{\sqrt{5}} \approx 44.7214 \ \text{feet/sec} \text{.}\)

Let \(r\) be the runner's position at time \(t\) and let \(s\) be the distance between the runner and the ball, as pictured.

\(\left. \frac{ds}{dt} \right|_{x = 45} = \frac{430}{\sqrt{17}} \approx 104.2903 \ \text{feet/sec} \text{.}\)

We let \(x\) denote the position of the ball 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{equation*} 2 \cdot 45 \cdot 100 = 2 \cdot 45 \sqrt{5} \cdot \left. \frac{dz}{dt} \right|_{x = 45}\text{,} \end{equation*}

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

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 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

\begin{equation*} x^2 + r^2 = s^2\text{,} \end{equation*}

by the Pythagorean Theorem. Differentiating each side 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 \(x = 45\text{,}\) \(r = \frac{90}{8}\text{,}\) 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,

\begin{equation*} 2 \cdot 45 \cdot 100 + 2 \cdot \frac{45}{4} \cdot 30 = 2 \cdot \frac{45}{4}\sqrt{17} \cdot \left. \frac{ds}{dt} \right|_{x = 45} \end{equation*}

and therefore

\begin{equation*} \left. \frac{ds}{dt} \right|_{x = 45} = \frac{430}{\sqrt{17}} \approx 104.2903 \ \text{feet/sec}\text{.} \end{equation*}

Subsection 3.5.2 Summary

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.

Exercises 3.5.3 Exercises

1. Height of a conical pile of gravel.

2. Movement of a shadow.

3. A leaking conical tank.

4.

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?

The boat is approaching the dock at a rate of \(\frac{13}{6} \approx 2.167\) feet per second.

Using the given information, we construct the figure shown below.

As pictured, we know that \(5\) is the vertical height from the pulley to the level of the bow of the boat, and we let \(z\) be the length of the rope from the pulley to the bow of the boat, and \(x\) the horizontal distance from the dock to the bow of the boat.

We are given that the rope is being pulled in at \(2\) feet per second, and thus \(\frac{dz}{dt} = -2\) feet per second. Because we want to know how fast the boat is approaching the dock when the length of rope from bow to pulley is 13 feet, we want to know \(\left. \frac{dx}{dt} \right|_{z=13}\text{.}\) Thus, we need to relate the changing quantities \(z\) and \(x\text{.}\)

Because the rope, the post (vertically extended), and the horizontal distance from the bow of the boat to the post on the dock form a right triangle at all times, it follows that

\begin{equation*} x^2 + 5^2 = z^2\text{.} \end{equation*}

Having now related \(z\) and \(x\text{,}\) we differentiate this equation with respect to \(t\text{.}\) By the chain rule, we now see that

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

At the instant \(z = 13\text{,}\) \(x^2 + 5^2 = 13^2\text{,}\) and thus \(x = 12\text{.}\) Using all of the given information at the instant \(z = 13\) (including that \(\frac{dz}{dt} = -2\)),

\begin{equation*} 2(12) \left. \frac{dx}{dt} \right|_{z = 13} = 2(13)(-2)\text{.} \end{equation*}

Solving for \(\left. \frac{dx}{dt} \right|_{z = 13}\text{,}\)

\begin{equation*} \left. \frac{dx}{dt} \right|_{z = 13} = -\frac{13}{6} \end{equation*}

feet per second. Thus the boat is approaching the dock at a rate of \(\frac{13}{6} \approx 2.167\) feet per second.

5.

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 3.5.5.

Figure 3.5.5. The swimming pool.

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.

The depth of the water is increasing at

\begin{equation*} \left. \frac{dh}{dt}\right|_{h = 5} = 1.28 \end{equation*}

feet per minute. The depth of the water is increasing at a decreasing rate.

The variables in this problem are the volume \(V\) of water in the pool and the depth \(y\) of water in the pool (measured at the deepest point of the pool); each is implicitly a function of time, \(t\text{.}\) We are given that \(\frac{dV}{dt}\) is a constant \(800\) cubic feet per minute and want to find \(\frac{dh}{dt}\bigm|_{h=5}\text{.}\) To do this, we first need to relate \(V\) and \(h\text{.}\) The volume of the water in the pool at height \(h \lt 12\) is the volume of the triangular cross sectional area of height \(h\) (as shown in the figure below) times the width (25 feet) of the pool. The height of the triangular cross section is \(h\) and the length is distance of the dotted line indicated in the figure.

Placing the cross section on a coordinate system as in the figure, the hypotenuse of the triangle is the line connecting the points \((0,-15)\) and \((60, -3)\text{.}\) This line has slope \(\frac{12}{60} = 0.2\) and \(y\)-intercept \((0,-15)\text{.}\) So the equation of this line is \(y = 0.2x-15\text{.}\) It is the \(x\) coordinate of the point on this line corresponding to the \(y\)-coordinate \(h-15\) that is the length of the triangle whose height is \(h\text{.}\) So the length \(l\) is \(l = \frac{(h-15)+15}{0.2} = 5h\text{.}\) Thus, the volume of water in the pool at height \(h\) is

\begin{equation*} V = \left(\frac{1}{2}(h)(5h)\right)(25) = 62.5h^2\text{.} \end{equation*}

Both \(V\) and \(h\) are functions of time. Differentiating both sides of the equation with respect to \(t\) gives

\begin{equation*} \frac{dV}{dt} = 62.5(2h) \frac{dh}{dt}\text{,} \end{equation*}

\begin{equation*} \frac{dh}{dt} = \frac{1}{125h} \frac{dV}{dt}\text{.} \end{equation*}

When the depth of water in the pool is \(5\) feet, then the depth is increasing at the rate

\begin{equation*} \left. \frac{dh}{dt}\right|_{h = 5} = \frac{1}{125(5)} (800) = 1.28 \end{equation*}

feet per minute. Since \(\frac{dh}{dt}\) is inversely proportional to \(h\text{,}\) the rate at which the depth of the water increases slows as \(h\) increases, so the depth of the water is increasing at a decreasing rate.

6.

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?

\(\left. \frac{d\theta}{dt} \right|_{x = 30} = -0.24\) radians per second.

Let \(x\) represent the distance from the runner to third base along the path from second to third, let \(z\) be the distance from the umpire at home plate to the runner, and call \(\theta\) the angle between the third baseline and the line of sight from the umpire to the runner. See the figure below.

Each of \(x\text{,}\) \(z\text{,}\) and \(\theta\) is implicitly a function of \(t\text{.}\) We are given that the runner is advancing at a rate of \(24\) feet per second, so \(\frac{dx}{dt} = -24\text{.}\) Since we want to know \(\left. \frac{d\theta}{dt} \right|_{x = 30}\text{,}\) we want to relate \(x\) and \(\theta\text{.}\)

We first observe that by definition,

\begin{equation*} \tan(\theta) = \frac{x}{90} = \frac{1}{90}x\text{.} \end{equation*}

Having related \(x\) and \(\theta\text{,}\) we can differentiate both sides of the preceding equation with respect to \(t\text{.}\) By the chain rule,

\begin{equation*} \sec^2 (\theta) \frac{d\theta}{dt} = \frac{1}{90} \frac{dx}{dt}\text{,} \end{equation*}

and thus

\begin{equation*} \frac{d\theta}{dt} = \frac{1}{90 \sec^2 (\theta)} \frac{dx}{dt} \end{equation*}

At the instant \(x = 30\text{,}\) since \(x^2 + 90^2 = z^2\text{,}\) \(z = \sqrt{90^2 + 30^2} = \sqrt{9000} = 30\sqrt{10}\text{.}\) Hence, at this same instance, \(\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{30\sqrt{10}}{90} = \frac{\sqrt{10}}{3}\text{.}\) Recall we also know that \(\frac{dx}{dt} = -24\text{.}\) Applying this information to the equation that relates the rates of change of \(x\) and \(\theta\text{,}\)

\begin{equation*} \left. \frac{d\theta}{dt} \right|_{x = 30} = \frac{1}{90 \left( \frac{\sqrt{10}}{3} \right)^2} (-24) = -\frac{1}{100}(24) = -0.24 \end{equation*}

radians per second.

7.

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?

\(\left. \frac{dh}{dt}\right|_{V=1000} = \frac{10}{\pi \left( \sqrt[3]{\frac{3000}{\pi}} \right)^2} \approx 0.0328\) feet per minute.