The volume of a sphere is decreasing at a constant rate of 560 cubic meters per minute. At the instant when the radius of the sphere is 6 meters, what is the rate of change of the radius?

Answer 1

Let

r(t) be the radius of the sphere at time t,

r'(t) be the rate of change of the radius at time t,

V(t) = 4πr³(t)/3 be the volume of the sphere at time t,

V'(t) = 4πr²(t)r'(t) = 560 m³/min be the rate of change of the volume at time t.

From that expressing

r'(t) = V'(t)/4πr²(t)

Substituting actual numbers for the particular time t0 when r(t0) = 6m

r'(t0) = Let

r(t) be the radius of the sprere at time t,

r'(t) be the rate of change of the radius at time t,

V(t) = 4πr³(t)/3 be the volume of the sphere at time t,

V'(t) = 4πr²(t)r'(t) = 560 m³/min be the rate of change of the volume at time t.

From that expressing

r'(t) = V'(t)/4πr²(t)

Substituting actual numbers for the particular time t0 when r(t0) = 6m

r'(t0) = Let

r(t) be the radius of the sprere at time t,

r'(t) be the rate of change of the radius at time t,

V(t) = 4πr³(t)/3 be the volume of the sphere at time t,

V'(t) = 4πr²(t)r'(t) = 560 m³/min be the rate of change of the volume at time t.

From that expressing

r'(t) = V'(t)/4πr²(t)

Substituting actual numbers for the particular time t0 when r(t0) = 6m

r'(t0) = 560/(4×π×6²) = 1.2 m/min