Active Calculus

Subsection 3.5.1 Related Rates Problems

In problems where two or more quantities can be related to one another, and all of the variables involved are implicitly functions of time, \(t\text{,}\) we are often interested in 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.

Example 3.5.1.

Sand is being dumped by a conveyor belt onto a pile so that the sand forms a right circular cone, as pictured in Figure 3.5.2. How are the instantaneous rates of change of the sand's volume, height, and radius related to one another?

Solution.

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.

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

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

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.

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 grows² as its radius increases;

• how the location of the base of a ladder and its height along a wall change³ as the ladder slides;

• how the water level changes⁴ in a conical tank as it fills with water at a constant rate (compare the problem in Activity 3.5.2);

• how a skateboarder's shadow changes⁵ as he moves past a lamppost.

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.

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.

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

and hence

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.

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.

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.