Question

The volume of sphere is increaseing

Answer 1

Volume of a sphere:

V = 4/3 π r ³

Differentiate V with respect to time t :

dV/dt = 4π r ² dr/dt

Surface area of a sphere:

A = 4π r ²

Differentiate A with respect to t :

dA/dt = 8π r dr/dt

You're given that the volume is increasing at a constant rate, dV/dt = 10 cm³/s, so

10 cm³/s = 4π r ² dr/dt

Solve for dr/dt (the rate of change of the sphere's radius):

dr/dt = 5/(2π r ²) cm³/s

Substitute this into the equation for dA/dt :

dA/dt = 8π r (5/(2π r ²) cm³/s)

dA/dt = 20/r cm³/s

At the moment the radius is r = 5 cm, the surface area is increasing at a rate of

dA/dt = 20/(5 cm) cm³/s = 4 cm²/s

When the surface area is increasing at a rate of dA/dt = 2 cm²/s, the radius r is such that

2 cm²/s = 20/r cm³/s

which happens when the radius is

r = 20/2 (cm³/s) / (cm²/s) = 10 cm