A melting spherical block of ice has its surface area changing at the rate of 24 inch2 / sec at the time the radius is 3 inches. how fast is the volume of the ice changing at that moment

The surface area of the sphere is:

A = 4 * pi * r ^ 2

Deriving we have:

A ' = 8 * pi * r * r'

We clear the speed:

r ' = A' / (8 * pi * r)

Substituting values:

r ' = 24 / (8 * pi * 3)

r ' = 1 / pi inch / sec

Then, the volume of the sphere is:

V = (4/3) * (pi) * (r ^ 3)

Deriving we have:

V ' = (3) * (4/3) * (pi) * (r ^ 2) * (r')

Substituting values:

V ' = (3) * (4/3) * (pi) * (3 ^ 2) * (1 / pi)

Rewriting:

V ' = (4) * (9)

V ' = 36 inch3 / sec

Answer:

the volume of the ice is changing at 36 inch3 / sec at that moment