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.

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.

anonymous · Answer

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

anonymous · Answer

i got the same thing it wasnt right

anonymous · Answer

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