#### To determine

**To find:** The rate of change of the surface area of the spherical ball whose radius is 8 cm and the rate of change of the radius is 2 cm/min.

#### Answer

The rate of change of the surface area of the spherical ball is dSdt=128π cm2/min.

#### Explanation

**Given:**

The radius of the spherical ball is 8 cm.

The rate of change of the radius of the spherical ball is 2 cm/min.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

(2) Surface area of the spherical ball with radius r is, S=4πr2.

**Calculation:**

The surface area of the spherical ball is, S=4πr2.

The surface area and the radius of the spherical ball increases when the time t increases.

Thus, the surface area and the radius of the spherical ball is a function of the time variable t.

Differentiate surface area S with respect to the time variable t.

dSdt=ddt(4πr2)=(4π)ddt(r2)=4π(2rdrdt)=8πrdrdt

Substitutes r=8 and drdt=2 in dSdt.

dSdt=(8π)(8)(2)=(8π)(16)=(128)(π)=128π cm2/min

Therefore, the rate of change of the surface area of the spherical ball is dSdt=128π cm2/min.