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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?

###### Wiki User

###### 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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