Can you please help me with this question

Question

A circle is inside a square.

The radius of the circle is increasing at a rate of 5 meters per hour and the sides of the square are decreasing at a rate of 4 meters per hour.

When the radius is 4 meters, and the sides are 20 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 hour.

Answer 1

Suppose the radius of the circle is r, and the side of the square is x

Then dr/dt = 5, and dx/dt = -4 from first sentence

The area outside the circle but inside the square A = x² - πr²

We need dA/dt when r=4 and x = 20

dA/dt = 2x dx/dt - 2πr dr/dt

Substitute in for x, dx/dt, r, dr/dt