Related Rates-Circle Inscribed a Square

Question

A circle is inside a square.

The radius of the circle is decreasing at a rate of 2 meters per day and the sides of the square are decreasing at a rate of 1 meter per day.

When the radius is 5 meters, and the sides are 18 meters, then how fast is the AREA outside the circle but inside the square changing?

The rate of change of the area enclosed between the circle and the square is square meters per day.

Answer 1

The first thing to do is to find an expression for the area between the circle and the square.

 

Let's call the side of the square s and the radius of the circle r.  The area of the square is s² and the area of the circle is πr².  Since the circle is inside the square, the area between them is the area of the square minus the area of the circle, or s² - πr².

 

We can calculate the rate of change of the area between by taking the derivative of that expression with respect to t.  Since both s and r are functions of t, the derivative will involve the chain rule:

 

 

D_{t (}s² - πr²) = 2s ds/dt - 2πr dr/dt.

 

They've given us all the values we need to figure this out:

 

s = 18

ds/dt = -1

r = 5

dr/dt = -2

 

So the area between the circle and the square is changing at a rate of:

 

2s ds/dt - 2πr dr/dt =

2 (18)(-1) - 2π (5)(-2) =

-36 + 20π

 

That's approximately equal to 26.83 square meters per day.