Related Rates (Two Formulas)

Question

The volume of a sphere is increasing at a constant rate of 198 cubic feet per second. At the instant when the volume of the sphere is 18 cubic feet, what is the rate of change of the surface area of the sphere? The volume of a sphere can be found with the equation V=4/3​πr^3 and the surface area can be found with S=4πr^2. Round your answer to three decimal places (if necessary).

Answer 1

V = (4/3)pi(r^3)= 18 ft^3

V' = 4pir^2r' = 198 ft^3/sec

SA = 4pir^2

SA' = 8pirr'

when V=18,

r^3 = 18(3/4pi) = 27/2pi,

r = 3(1/2pi)^(1/3)= about 1.6258

and r' = 198/4pir^2 = 99/2pi(3)(1/2pi)^(2/3)

=33/2pi(1/2pi)^(2/3)= about 5.9612

rr' = about 1.6258(5.9612) = about 9.6917

SA' = 8pirr' =8pi(9.6917)

= about 243.579 ft^2/sec