Calculus
posted by Rina on .
A certain calculus student hit Mrs. Evans in the head with a snowball. If the snowball is melting at the rate of 10 cubic feet per minute, at what rate is the radius of the snowball changing when the snowball is 1 foot in radius?

That is a pretty big snowball !
V = (4/3) pi r^3
dV/dt = (4/3) pi 3 r^2 dr/dt
10 = 4 pi r^2 dr/dt
since r = 1
dr/dt = 10/(4 pi)
note : using common sense we knew that. The rate of change of the volume is the surface area 4 pi r^2 times the rate of radius change