Related Rates: Learn It 1

  • Show how quantities change using derivatives and explore how these changes are connected
  • Apply the chain rule to calculate how one changing quantity affects another

Related-Rates Problem-Solving

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.

Setting up Related-Rates Problems

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, [latex]V[/latex], is related to the rate of change in the radius, [latex]r[/latex]. In this case, we say that [latex]\frac{dV}{dt}[/latex] and [latex]\frac{dr}{dt}[/latex] are related rates because [latex]V[/latex] is related to [latex]r[/latex]. 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.

A spherical balloon is being filled with air at the constant rate of [latex]2 \, \frac{\text{cm}^3}{\text{sec}}[/latex] (Figure 1). How fast is the radius increasing when the radius is [latex]3\, \text{cm}[/latex]?

Three balloons are shown at Times 1, 2, and 3. These balloons increase in volume and radius as time increases.
Figure 1. As the balloon is being filled with air, both the radius and the volume are increasing with respect to time.

The volume of a sphere of radius [latex]r[/latex] centimeters is

[latex]V=\frac{4}{3}\pi r^3 \, \text{cm}^3[/latex]

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

[latex]V(t)=\frac{4}{3}\pi [r(t)]^3 \, \text{cm}^3[/latex]

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

[latex]V^{\prime}(t)=4\pi [r(t)]^2 \cdot r^{\prime}(t)[/latex]

The balloon is being filled with air at the constant rate of 2 cm3/sec, so [latex]V^{\prime}(t)=2 \, \text{cm}^3 / \sec[/latex]. Therefore,

[latex]2 \, \text{cm}^3 / \sec =(4\pi [r(t)]^2 \, \text{cm}^2) \cdot (r^{\prime}(t) \, \text{cm/sec})[/latex],

which implies

[latex]r^{\prime}(t)=\dfrac{1}{2\pi [r(t)]^2} \, \text{cm/sec}[/latex]

When the radius [latex]r=3 \, \text{cm}[/latex],

[latex]r^{\prime}(t)=\dfrac{1}{18\pi} \, \text{cm/sec}[/latex]

Watch the following video to see the worked solution to the example above.

What is the instantaneous rate of change of the radius when [latex]r=6 \, \text{cm}[/latex]?

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

How to: Solve a Related-Rates Problem

  1. Variable Assignment: Assign symbols to all variables involved in the problem. If it helps, draw a diagram to visualize the scenario.
  2. Information Setup: Clearly state the given information and what needs to be determined, using the assigned variables.
  3. Equation Development: Formulate an equation that connects the variables. This relationship should encapsulate the dynamics of the problem.
  4. Differentiation: Apply the chain rule to differentiate the equation with respect to time. This step transforms the static equation into one that describes rates of change.
  5. Substitution and Solution: Plug in all known values of rates and other variables into the differentiated equation and solve for the unknown rate.

We are able to solve related-rates problems using a similar approach to implicit differentiation. 

Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that quantity will behave as a constant and its derivative will not appear in the new equation found in step 4. We examine this potential error in the first example on the next page.