#### To determine

**To find:** The derivative dAdt in terms of drdt.

#### Answer

The derivatives dAdt=2πrdrdt.

#### Explanation

**Given:**

The area of the circle is *A* and its radius is *r*.

If the area of the circle expands as the time passes.

**Derivative rules:**

*Chain rule*: dydx=dydu⋅dudx.

**Calculation:**

Let *A* be an area of the circle with radius *r* and time variable is *t*.

Here, the area of the circle expands as time passes. That is, the area of the circle is changed as time *t* is changed.

The area of the circle is changed means the radius of the circle should be changed.

Thus,the area of the circle is a function of the time variable *t* and the radius of the circle also a function of the time variable *t*.

The area of the circle is A=πr2.

Differentiate with respect to time variable *t*.

dAdt=ddt(πr2)=πddt(r2)

Apply the chain rule and simplify the terms,

dAdt=πddt(r2)=πddr(r2)drdt=2πrdrdt

Therefore, the derivatives dAdt=2πrdrdt.

#### To determine

**To find:** The area of spill increasing when the radius is 30 m.

#### Answer

The area of spill increasing at the rate dAdt=60π m2/s when the radius is 30 m.

#### Explanation

**Given:**

The radius of the oil spill increases at a constant rate of 1 m/s.

**Calculation:**

Note that, the rates of change are derivatives.

The radius of the oil spill increases at a constant rate of 1 m/s. That is, drdt=1 m/s.

Obtain the area of spill increasing when the radius is 30 m.

That is, the area is change when the radius is change.

Here, the area and radius are both depends on time variable *t*.

From part (a), dAdt=2πrdrdt.

Substitute the radius r=30 m and drdt=1 m/s in dAdt=2πrdrdt.

dAdt=2π(30 m)(1 m/s)=60π m2/s

Therefore, the area of spill increasing at the rate dAdt=60π m2/s when the radius is 30 m.