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A rock is thrown into a still pond. The circular ripples move outward from the point of impact of the rock so that the radius of the circle formed by a ripple increases at the rate of 4 feet per minute. Find the rate at which the area is changing at the instant the radius is 11 feet.When the radius is 11feet, the area is changing at approximately ____ square feet per minute. (round to the nearest thousandth)

User MattUebel
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1 Answer

3 votes

Answer:

When the radius is 11 feet, the area is changing at approximately 276.571 square feet per minute.

Explanation:

We are given the following in the question:


(dr)/(dt) = 4\text{ feet per minute}

Instant radius = 11 feet

Area of circle =


A = \pi r^2

where r is the radius of the circle.

Rate of change of area of circle =


(dA)/(dt) = (d)/(dt)(\pi r^2) = 2\pi r(dr)/(dt)

Putting all the values, we get,


(dA)/(dt) = = 2\pi (11)(4) = 88\pi = 276.571\text{ square feet per minute}

Thus, when the radius is 11 feet, the area is changing at approximately 276.571 square feet per minute.

User TinaW
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