Question

Let A(t) be the area of a circle with radius r(t), at time t in min. Suppose the radius is changing at the rate of drdt=6 ft/min. Find the rate of change of the area at the moment in time when r=9 ft

Answer 1

Answer:

The rate of change is $108\pi$ ft^(2)/min

Step-by-step explanation:

The area of a circle is given by the following equation:

$A(t) = \pi r^{2}$

To solve this question, we have to realize the implicit differentiation in function of t. We have two variables, A and r. So

$\frac{dA(t)}{dt} = 2\pi r \frac{dr}{dt}$

We have that:

$\frac{dr}{dt} = 6, r = 9$.

We want to find $\frac{dA}{dt}$

So

$\frac{dA(t)}{dt} = 2\pi*9*6$

$\frac{dA}{dt} = 108\pi$

Since the area is in square feet, the rate of change is in ft^(2)/min.

So the rate of change is $108\pi$ ft^(2)/min