A circle is inside a square. The radius of the circle is decreasing at a rate of 3 meters per hour and the sides of the square are decreasing at a rate of 5 meters per hour. When the radius is 4 meters, and the sides are 23 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.

Step-by-step explanation:

Let r = radius of circle

     s = length of a side of the square

dr/dt = -3m/hour

  ds/dt = 5m/hour

we know that

area of square= s^2

area of circle=πr^2

Let A be the area of outside

A = (area of square) - (area of circle)

          = s^2 - πr^2

dA/dt=?

when r = 4 and s = 23

 substituting we have

dA/dt = 2s(ds/dt) - 2πr(dr/dt)

             =2(23)(5) - 2π(4)(-3)

            = (230 + 24π) m^2/hour

            = (230 + 75.408) m^2/hour

            = 305.408m^2/hour

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