Question

A snowball is melting 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?

−16π in2/hour
−24π in2/hour
−32π in2/hour
−48π in2/hour (I think it's this one)

Answer 1

@paki

Answer 2

surface area=4*pi*r*r
given dr/dt=2 inch/h

Answer 3

so ds/dt=4*pi*(2*r)*(dr/dt)

Answer 4

okay so 8*pi*(3)(-2) right?

Answer 5

or 4*pi*(2*3)*2

Answer 6

hello?

Answer 7

yes ds/dt=-2 ans -48*pi

Answer 8

Thank you ;)