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

Question

A circle is inside a square.

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

When the radius is 5 meters, and the sides are 18 meters then how fast is the area between the circle and the square changing?

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

(Enter your answer as a decimal number rounded to 2 places.)

Answer 1

Area of the square is s².

Area of the containing circle is πr².

So the area between the circle and the square is the difference in their respective areas.

A = s² - πr².

At this point rates of change are with respect to t.

dA/dt = 2s ds/dt - 2πr dr/dt

We are given that ds/dt = -1 m/hr and dr/dt = 1 m/hr.

To find the rate of change of A (with respect to t) at the time when s = 18 and r = 5, substitute the indicated values.

dA/dt = 2(18)(-1) - 2π(5)(+1)

dA/dt = -36 - 10π

Note that if units are included:

dA/dt = 2(18 m)(-1 m/hr) - 2π(5 m)(1 m/hr)

the result is in m²/hr which is in fact the rate at which the area is decreasing.