Related Rate (Calculus)

Question

A circle is inside a square.

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

When the radius is 5 meters, and the sides are 24 meters, then how fast is the AREA outside the circle but inside the square changing?

Answer 1

Let the area of the section between the square and the circle be as follows: A= x² -πr². Take the derivative wrt time to get dA/dt = 2xdx/dt - π2rdr/dt and now plug in the given data to get dA/dt=2•24•(-4)- 2π•5•5 m/min and then do the calculation to get -192 m/min - 50π m/min this gives dA/dt= -349.08 m/min for how fast the inner area is changing.