We know that the formula for the area of a circle is $A = \pi r^2$ where $A$ is the area and $r$ is the radius. Since the area is increasing at a rate of $96\pi$ square meters per second, we have $\frac{dA}{dt} = 96\pi$. We are asked to find $\frac{dr}{dt}$ when $A = 64\pi$.
Differentiating the area equation with respect to time $t$ using the chain rule, we have $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}.$ Since $\frac{dA}{dt} = 96\pi$, we have $96\pi = \frac{dA}{dr} \cdot \frac{dr}{dt}$.
When $A = 64\pi$, we can substitute this value into the equation to get $96\pi = \frac{dA}{dr} \cdot \frac{dr}{dt}$ $\implies 96\pi = 2r \cdot \frac{dr}{dt}.$ Since $A = \pi r^2$, we have $64\pi = \pi r^2 \implies 64 = r^2 \implies r = 8$ (the radius cannot be negative).
Finally, we substitute $r = 8$ into the equation to get $ 96\pi = 2r \cdot \frac{dr}{dt} \implies 96\pi = 2 \cdot 8 \cdot \frac{dr}{dt} \implies 96\pi = 16 \cdot \frac{dr}{dt} \implies \frac{dr}{dt} = \boxed{6\pi}$ meters per second.
The area of a circular region is increasing at a rate of
96pi square meters per second. When bthe area of the region is 64pi square meters, how fast, in meters per second, is the radius 64pi of the region increasing?
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