Question

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min. The volume of a sphere is V = 4 3 πr3 and its surface area is 4πr2. Determine the rate of change in surface area when r = 20 cm.

Answer 1

Answer:

The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.

Step-by-step explanation:

The area of a sphere is given by the following formula:

$A = 4\pi r^{2}$

In which A is the area, measured in cm², and r is the radius, measured in cm.

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min.

This means that $\frac{dr}{dt} = 40$

Determine the rate of change in surface area when r = 20 cm.

This is $\frac{dA}{dt}$ when $r = 20$. So

$A = 4\pi r^{2}$

Applying implicit differentiation.

We have two variables, A and r, so:

$\frac{dA}{dt} = 8r\pi \frac{dr}{dt}$

$\frac{dA}{dt} = 8*20\pi*40$

$\frac{dA}{dt} = 20106.19$

The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.