#### To determine

**To find:** The rate of change of the volume of the sphere where the radius of the sphere is increasing at a rate of 4 mm/s and the diameter of the sphere is 80 mm.

#### Answer

The rate of change of the volume of the sphere is dVdt=25600 mm3/s.

#### Explanation

**Given:**

The diameter of the sphere is 80 mm.

The rate of change of the radius of the sphere is 4 mm/s.

**Formula used:**

(1) Chain rule dydx=dydu⋅dudx

(2) Volume of the sphere of radius r, V=43πr3.

**Calculation:**

Let V be the volume and *r* be the radius of the sphere respectively.

Therefore, the volume is, V=43πr3.

The volume of the sphere increases as the time t increases.

Therefore, the radius of the sphere also increases as the time t increases.

Thus, the volume V and the radius r are the functions of the time variable t.

Obtain the derivative dVdt.

Differentiate the volume V with respect to the time variable t.

dVdt=ddt(43πr3)=(43π)ddt(r3)=(43π)(3r2)drdt=(4πr2)drdt

Since the diameter of the sphere is 80 mm, the radius is, 40 mm.

Substitute 40 for *r* and drdt=4 in dVdt.

dVdt=(4π)(402)(4)=(4π)(1600)(4)=(4π)(6400)=25600π mm3/s

Thus, the rate of change of the volume of the sphere is dVdt=25600 mm3/s.