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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
∙ 16y ago
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 footby 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.
Wiki User
∙ 16y ago
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