Question

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 increasing at a rate of 1 meter per hour.
when the radius is 6 meters, and the sides are 15 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.
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Explanation:

Step1: Define the area function

Let \(r\) be the radius of the circle and \(s\) be the side - length of the square. The area \(A\) outside the circle but inside the square is \(A=s^{2}-\pi r^{2}\).

Step2: Differentiate the area function with respect to time \(t\)

Using the chain rule, \(\frac{dA}{dt}=2s\frac{ds}{dt}-2\pi r\frac{dr}{dt}\).

Step3: Substitute the given values

We are given that \(\frac{dr}{dt}=- 3\) (negative because the radius is decreasing), \(\frac{ds}{dt}=1\), \(r = 6\), and \(s = 15\).
Substitute these values into the derivative formula:

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Answer:

\(30 + 36\pi\approx30+36\times3.14=30 + 113.04=143.04\)