A snowball is melting so that the radius is changing at the rate of 2 inches per hour. How fast is the surface area of the snowball changing at the instant the snowball has a radius of 3 inches?

Question

Question options:

1)	-16π in2/hour
2)	-24π in2/hour
3)	-32π in2/hour
4)	-48π in2/hour

Answer 1

Surface Area = 4 * pi * R^2 = SA(R) and R is function of time t

dR/dt = -2

dSA/dt = 8 * pi * R * dR/dt

when R=3

dSA/dt = 8 * pi * 3 * -2 =- 48*pi

looks like option #4