# 4.1 Related rates

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• Express changing quantities in terms of derivatives.
• Find relationships among the derivatives in a given problem.
• Use the chain rule to find the rate of change of one quantity that depends on the rate of change of other quantities.

We have seen that for quantities that are changing over time, the rates at which these quantities change are given by derivatives. If two related quantities are changing over time, the rates at which the quantities change are related. For example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing. In this section, we consider several problems in which two or more related quantities are changing and we study how to determine the relationship between the rates of change of these quantities.

In many real-world applications, related quantities are changing with respect to time. For example, if we consider the balloon example again, we can say that the rate of change in the volume, $V,$ is related to the rate of change in the radius, $r.$ In this case, we say that $\frac{dV}{dt}$ and $\frac{dr}{dt}$ are related rates    because V is related to r . Here we study several examples of related quantities that are changing with respect to time and we look at how to calculate one rate of change given another rate of change.

## Inflating a balloon

A spherical balloon is being filled with air at the constant rate of $2{\phantom{\rule{0.2em}{0ex}}\text{cm}}^{3}\text{/}\text{sec}$ ( [link] ). How fast is the radius increasing when the radius is $3\phantom{\rule{0.2em}{0ex}}\text{cm}?$

The volume of a sphere of radius $r$ centimeters is

$V=\frac{4}{3}\pi {r}^{3}{\text{cm}}^{3}.$

Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, $t$ seconds after beginning to fill the balloon with air, the volume of air in the balloon is

$V\left(t\right)=\frac{4}{3}\pi {\left[r\left(t\right)\right]}^{3}{\text{cm}}^{3}.$

Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation

$V\prime \left(t\right)=4\pi {\left[r\left(t\right)\right]}^{2}{r}^{\prime }\left(t\right).$

The balloon is being filled with air at the constant rate of 2 cm 3 /sec, so $V\prime \left(t\right)=2{\phantom{\rule{0.2em}{0ex}}\text{cm}}^{3}\text{/}\text{sec}.$ Therefore,

$2{\text{cm}}^{3}\text{/}\text{sec}=\left(4\pi {\left[r\left(t\right)\right]}^{2}{\text{cm}}^{2}\right)·\left(r\prime \left(t\right)\text{cm/s}\right)\text{,}$

which implies

$r\prime \left(t\right)=\frac{1}{2\pi {\left[r\left(t\right)\right]}^{2}}\phantom{\rule{0.1em}{0ex}}\text{cm/sec}.$

When the radius $r=3\phantom{\rule{0.2em}{0ex}}\text{cm,}$

$r\prime \left(t\right)=\frac{1}{18\pi }\phantom{\rule{0.1em}{0ex}}\text{cm/sec}.$

What is the instantaneous rate of change of the radius when $r=6\phantom{\rule{0.2em}{0ex}}\text{cm}?$

$\frac{1}{72\pi }\phantom{\rule{0.2em}{0ex}}\text{cm/sec},$ or approximately 0.0044 cm/sec

Before looking at other examples, let’s outline the problem-solving strategy we will be using to solve related-rates problems.

## Problem-solving strategy: solving a related-rates problem

1. Assign symbols to all variables involved in the problem. Draw a figure if applicable.
2. State, in terms of the variables, the information that is given and the rate to be determined.
3. Find an equation relating the variables introduced in step 1.
4. Using the chain rule, differentiate both sides of the equation found in step 3 with respect to the independent variable. This new equation will relate the derivatives.
5. Substitute all known values into the equation from step 4, then solve for the unknown rate of change.

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