Related Rates Problem (Analytical Geometry & Calculus I)- Goal: Find the rate of change of the area enclosed between the circle and the square

Question

A circle is inside a square.

The radius of the circle is increasing at a rate of 3 meters per hour and the sides of the square are increasing at a rate of 1 meter per hour.

When the radius is 6 meters, and the sides are 21 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

This area A = a² - πr². dA/dt = 2ada/dt - 2πrdr/dt = 2·21m·1m/h - 2π·6m·3m/h = - 71.1 m²/h

Answer 2

Let x be the side dimension of the square and r be the radius of the circle

A = SquareArea - CircleArea = x²-πr²

dA/dt = 2x dx/dt - 2πr dr/dt

Just substitute x=21, r=6, dx/dt=1 and dr/dt =3 to evaluate dA/dt