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?
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
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.