A = πr^2 . We could find C in terms of r and then plug it into the equation for A, or we could
work with the equations separately and then relate them. Let’s do both and compare.
Method 1: First, we find C in terms of r: r = . Now, we plug this in for r in the equation for
A: A = π = . Next, we take the derivative of the equation with respect to t:
= . Next, we plug in C = 100π and = 40, and solve:
(40) = 2,000. Therefore, the answer is 2,000 ft^2 /s.
Method 2: First, we take the derivative of C with respect to t: = 2π . Next, we plug
in = 40 and solve for : 40 = 2π , so = . Next, we take the derivative of A
with respect to t: = 2πr . Now, we can plug in for and r and solve for . Note
that when the circumference is 100π, r = 50: = 2π(50) = 2,000. Therefore, the
answer is 2,000 ft^2 /s.
Which method is better? In this case they are about the same. Method 1 is going to be more
efficient if it is easy to solve for one variable in terms of the other, and it is also easy to take
the derivative of the resulting expression. Otherwise, we will prefer to use Method 2 (See
Example 3 on this page).
- in./s
We are given the rate at which the volume is increasing, = 27π and are looking for the rate
at which the radius is increasing, . Thus, we need to find a way to relate the volume of a
sphere to its radius. Recall that the volume of a sphere is V = πr^3 . All we have to do is take
the derivative of the equation with respect to t: = 4πr^2 . Now we
substitute = 27π and r = 3: 27π = 4π(3)^2 . If we solve for , we get = in./s.
