In our earlier discussions of rates of change, we only dealt with the rate of change of a single quantity at a time.

In many physical situations, examining the relationship between two quantities, each changing with respect to time, can aid in problem solving.

In a related rates problem, we attempt to find the rate of change of one quantity in terms of the rate of change of a second, related quantity that is, in some cases, easier to calculate.

In This Module

• We will explore different related rates problems.

Consider a circle whose radius is changing with respect to time.

As the radius changes, so does the area of the circle.

Let \(r(t)\) be the radius of the circle at time \(t\) and let \(A(t)\) be the area of the circle at time \(t\).

Of course, we know that these two quantities are not independent.

More precisely, they are connected by the formula \(A = \pi r^{2}\).

If the radius of a circle is increasing at a constant rate of \(2\) cm/s, at what rate is the area of the circle changing when the radius is \(3\) cm?

Solution

Since we know the relationship between the quantities \(A\) and \(r\), we can find the relationship between the rates of change of these quantities with respect to time.

We do so by differentiating both sides of the equation \(A = \pi r^{2}\), implicitly, with respect to time.

Remember that, although we do not explicitly include the variable \(t\) in the equation, \(A\) and \(r\) are functions of time and \(\pi\) is a constant.

If the radius of a circle is increasing at a constant rate of \(2\) cm/s, at what rate is the area of the circle changing when the radius is \(3\) cm?

Solution

We are given that \(\dfrac{dr}{dt}=2\) cm/s and we are asked to determine the value of \(\dfrac{dA}{dt}\).

Right away we see that even though the radius is changing at a constant rate, the area is not.

The rate of change of the area depends on both \(\frac{dr}{dt}\) (which is constant) and \(r\) (which is changing with respect to time).

We want the rate of change of the area of the circle at the instant when the radius is \(3\) cm.

Therefore, we want to calculate \(\dfrac{dA}{dt}\) when \(r = 3\).

If the radius of a circle is increasing at a constant rate of \(2\) cm/s, at what rate is the area of the circle changing when the radius is \(3\) cm?

Solution

Recall that \(\dfrac{dr}{dt} = 2\) cm/s. Substituting \(\dfrac{dr}{dt} = 2\) cm/s and \(r = 3\) cm into the above equation, we get

Checking our units for \(\dfrac{dA}{dt}\), we get (cm)\(\times\)(cm/s)\(=\)cm\(^2\)/s as we would expect.

Therefore, the area of the circle is increasing at a rate of \(12 \pi\) cm\(^2\)/s when the radius is \(3\) cm.