Question

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

Answer 1

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