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)

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Answer:

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

Step-by-step explanation:

We are given the following in the question:

[tex]\dfrac{dr}{dt} = 4\text{ feet per minute}[/tex]

Instant radius = 11 feet

Area of circle =

[tex]A = \pi r^2[/tex]

where r is the radius of the circle.

Rate of change of area of circle =

[tex]\dfrac{dA}{dt} = \dfrac{d}{dt}(\pi r^2) = 2\pi r\dfrac{dr}{dt}[/tex]

Putting all the values, we get,

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

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