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.

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:

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

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

This is when . So

Applying implicit differentiation.

We have two variables, A and r, so:

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

answer: r there any options?

step-by-step explanation:

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6