If a snowball rolls down a hill and its surface area increaces at a rate of 18cm^2/min, find the rate at which the radius increases when the diameter is 12cm. be shure to specify the units of measure for your answer. the formula for the surface area of a sphere is s=4pir^2

dr/dt = 9/(24pi)  cm per minute

9/(24pi) is approximately equal to 0.119366

Work Shown:

Given info

dS/dt = 18 cm^2/min is the rate of change of the surface area

r = 6 cm is the radius, from the fact that the diameter is 12 cm

Use the surface area equation given, apply the derivative, plug in the given values and then isolate dr/dt which represents the rate of change for the radius

S = 4*pi*r^2

dS/dt = 2*4*pi*r*dr/dt

dS/dt = 8*pi*r*dr/dt

18 = 8*pi*6*dr/dt

18 = 48*pi*dr/dt

48pi*dr/dt = 18

dr/dt = 18/(48pi)

dr/dt = (9*2)/(24*2pi)

dr/dt = 9/(24pi)

The units are cm per minute, which can be written as cm/min.

answer: 43/157

this is the answer and if you try to prove me wrong your a meanie