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 = 4/3πr^3 and its surface area is 4πr^2. Determine the rate of change in surface area when r = 10 cm.

Answer 1

SA = 4π r²
dA/dt = 8π r dr/dt
= 8 * 10 * 40 * 3.14 = 10,053 cm²/min

Answer 2

i got the same thing it wasnt right

Answer 3

The only variable to calculate here is r, I don't see any wrong with r.
Check the round off!