Question

a circle is inside a square.
the radius of the circle is increasing at a rate of 1 meter per hour and the sides of the square are decreasing at a rate of 4 meters per hour.
when the radius is 3 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
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 outside the circle but inside the square, \( A \), is given by the area of the square minus the area of the circle: \( A = s^2 - \pi r^2 \).

Step2: Differentiate with respect to time \( t \)

Using the chain rule, we differentiate both sides with respect to \( t \). The derivative of \( s^2 \) with respect to \( t \) is \( 2s\frac{ds}{dt} \), and the derivative of \( \pi r^2 \) with respect to \( t \) is \( 2\pi r\frac{dr}{dt} \). So, \( \frac{dA}{dt} = 2s\frac{ds}{dt} - 2\pi r\frac{dr}{dt} \).

Step3: Substitute the given values

We know that \( r = 3 \) m, \( s = 15 \) m, \( \frac{dr}{dt} = 1 \) m/h, and \( \frac{ds}{dt} = - 4 \) m/h (negative because the side length is decreasing). Substituting these values into the derivative:

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Answer:

\( -120 - 6\pi \) (or approximately \( -138.85 \)) square meters per hour.