A circle is inside a square. The radius of the circle is increasing at a rate of 2 meters per hour and the sides of the square are decreasing at a rate of 4 meters per hour.

Question

A circle is inside a square. The radius of the circle is increasing at a rate of 2 meters per hour and the sides of the square are decreasing at a rate of 4 meters per hour. 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? The rate of change of the area enclosed between the circle and the square is __________ square meters per hour.

Please, I need help with this question.

Answer 1

Suppose the radius of the circle is r and each side of the square is s.

From the problem, we know that the radius is increasing at a rate of 2 m/hour. So dr/dt = 2.

The sides are decreasing so four sides are decreasing at the same time. The rate of change is ds/dt = -1

(why? because 4m /4 sides = 1).

Now, find the formula for the area. The area outside the circle and inside the square is

A(t) = s^2 - pi* r^2.

Since s(t) and r(t) are both functions of time, you can take derivative with respect to t for both sides of the

equation. Then you have

A'(t) = 2*s*s'(t) - 2*pi*r*r'(t)

We want to know the rate of change of area when r = 5 and s = 24/4 sides = 6. Therefore, you have

A'(t) = 2*6*(-1) - 2*pi*5*2 = -12 - 20*pi = -(12+20*pi)

Thus, the area is decreasing at a rate 12+20*pi