A beach ball is deflating at a constant rate of 10 cubic centimeters per second. When the volume of the ball is 256/3(pi) cubic centimeters, what is the rate of change of the surface area? (S=4(pi)r^2 and V=4/3(pi)r^3)

Answer 1

V = (4/3) pi r^3
so when V = 256 * pi/(3)
256 pi/(3 ) = 4 pi r^3 /(3)
r^3 = 64
r = 4

d v = surface area of sphere * dr

dv = 4 pi r^2 dr = 64 pi dr

dv/dt = 64 pi dr/dt = 10 cm^3/s
dr/dt = 10/(64 pi)

so
dV/dt = 4 pi r^2 dr/dt = 10
so
dr/dt = 10/(4 pi )
dS/dt = 8 pi r dr/dt
= 32 pi (10/(4pi) )
= 80 cm^2/s

Answer 2

v = 256 pi/3 = (4/3) pi r^3
so r = 4
S = v(3/r) = 4 pi r^2 = 64 pi

v = (4/3) pi r^3 = S (r/3) = (1/3) S r

dv/dt = (1/3) [ S dr/dt + r dS/dt ]
but S dr/dt = dv/dt = 10
30 = 10 + 4 dS/dt
20 = 4 dS/dt
dS/dt = 5 cm^3/s