Suppose the volume, V , of a spherical tumour with a radius of r = 2 cm uniformly grows at a rate of dV/dt= 0.3 cm^3/day where t is the time in days. At what rate is the surface area of the tumour increasing? The volume of a sphere is given by V =4 3πr^3and the surface area is given by A = 4πr^2.

Answer 1

v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt
so, dr/dt = (dv/dt) / 4pi r^2

a = 4pi r^2
da/dt = 8pi r dr/dt
= 8pi r (dv/dt)/4pir^2
= 2/r dv/dt

Now plug in your numbers.