Calculus Math Problem

Question

Assume that the radius r of a sphere is expanding at a rate of

60 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 at t = 2 min, assuming that r = 10 at t = 0.

Answer 1

A = 4 π r ²

d A / d t = 8 π r d r / d t

d r / d t = 60 cm / min

r = 60 t + C ; r = 10 ; t = 0

10 = 60 ( 0 ) + C

C = 10

r = 60 t + 10. ; t = 2 min

r = 60 (2) + 10

r = 130

d A / d t = 8 π r d r / d t

= 8 π ( 130 ) (60)

d A / d t = 62400 π rate of change