calculus
posted by Anonymous .
The radius of a circle is increasing at a nonzero rate, and at a certain instant, the rate of increase in the area of the circle is numerically equal to the increase in its circumstance. At this instant, the radius of the cirle is...
Here is what I did: (unsure)
pi(r^2)=2pi(r)
r^2r=0
r(r2)=0
r=2

no, it is the rates of increase that are equal, so you have to set the derivatives equal.
dA/dr = 2pi(r)
dC/dr = 2pi
so 2pi(r) = 2 pi
r = 1
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