suppose that a tumor on a person's body is spherical in shape. if the radius of the tumor is 0.5cm the radius is increasing at the rate of 0.001cm per day. what is the rate of increasing volume of that tumor at that time?

Answer 1

V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
so when r = .5, dr/dt = .01
just plug it in...

dV/dt = 4π(.25)(.01) = ..... cm^3/day

Answer 2

V = (4/3)πr^3
dV/dt = (4/3)π(3)r^2(dr/dt)

r=0.5 cm => r^2 = (0.5)^2 = 0.25 squared cm

dr/dt = 0.001 cm

dV/dt = 4π(0.25)(0.001)
= 0.003142 cubic cm per day