A_{circle} = πr^{2}

dA_{cir} / dt = 2πr dr/dt = 36π m^{2}/min

A_{square} = s^{2}

dA_{sq} / dt = 2s ds/dt = - 84 m^{2}/min

Enclosed area = A_{sq} - A_{cir}

The enclosed area is decreasing at a rate of - 84 - 36π m^{2}/min.

James H.

asked • 12/04/21A circle is inside a square.

The radius of the circle is increasing at a rate of 3 meters per minute and the sides of the square are decreasing at a rate of 2 meters per minute.

When the radius is 6 meters, and the sides are 21 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 minute.

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A_{circle} = πr^{2}

dA_{cir} / dt = 2πr dr/dt = 36π m^{2}/min

A_{square} = s^{2}

dA_{sq} / dt = 2s ds/dt = - 84 m^{2}/min

Enclosed area = A_{sq} - A_{cir}

The enclosed area is decreasing at a rate of - 84 - 36π m^{2}/min.

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