#### To determine

**To find:** The quantities in the given problem.

#### Answer

The quantities in the given problem is dSdt=−1 cm2/min.

#### Explanation

**Given:**

Given: the rate of decrease of surface area of the snow ball is 1cm2/min

If we let t be time (in min) and S be the surface area of the snow ball (in cm2) then given that dSdt=−1 cm2/min.

#### To determine

**To find:** The unknown.

#### Explanation

**Formula used:**

Surface area of the sphere with radius r:S=4πr2

**Calculation:**

Unknown: the rate of decrease of diameter of the snow ball when diameter of the snow ball is 10 cm.

If we let x is the diameter of the snow ball then, we want to find dxdt when x=10 cm.

#### To determine

**To draw:** The picture of the plane position at time *t*.

#### Explanation

Draw the picture of the plane position at time *t* is given below.

From Figure 1, it is observed that the circle.

#### To determine

**To write:** The equation that relates the quantities.

#### Answer

The Equation is dxdt=12πx⋅dSdt.

#### Explanation

Since r is the radius of the snow ball, the diameter is x=2r, then r=12x.

Therefore, the surface area of the snow ball in terms of diameter x is as follows.

S=4π(12x)2=4π(14x2)=πx2

Obtain dSdt when x=10 cm.

Differentiate S=πx2 with respect to the time t.

ddt(S)=ddt(πx2)dSdt =πddt(x2) =π(2x)dxdt =2πxdxdt

Therefore, on further simplification the rate of change of diameter is as follows.

dxdt=12πx⋅dSdt.

#### To determine

**To find:** The rate at which the diameter of the snow ball decreases when diameter of the snow ball is 20 cm.

#### Answer

The required rate at which the diameter of the snow ball decreases is dxdt=120π cm/min.

#### Explanation

Substitute −1 for dSdt and 10 for x in dxdt

dxdt=12π×10(−1)=−120π

Therefore, the required rate at which the diameter of the snow ball decreases is dxdt=120π cm/min.