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The radius of a circular ripple on water increases at a rate of four feet per second - so how fast is the area of disturbed water increasing when the radius is ten feet?

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2007-10-28 14:31:19

This question can be addressed by simple calculus. As time (t)

passes, the radius (r) of the circle increases at 4 feet per

second. dr/dt = 4 The area of the circle is equal to pi

multiplied by the square of the radius: A = pi x r2 From

which we can infer that dA/dr = 2pi x r The radius, then,

changes by four feet for every passing second. In turn, the area is

increasing at a given moment by (2pi x r) square feet for

every added radial foot. If we multiply square feet per foot

by feet per second we obtain square feet per second, as

required by the question. This is the chain rule of

differentiation.

dA/dt = dA/dr x dr/dt = 4 x 2pi x r = 8pi x

r If r = 10 feet, then the area is momentarily increasing at

8pi x 10 = 80pi = 251 square feet per second.


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