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?
To find how fast the radius is increasing, we can use the formula for the area of a circle:
A = πr^2
where A is the area and r is the radius.
Differentiating both sides of the equation with respect to time t, we get:
dA/dt = 2πr(dr/dt)
The left side of the equation is given as 96π square meters per second, which is the rate at which the area is increasing.
Substituting the given values, we have:
96π = 2π(64π)(dr/dt)
Simplifying and canceling out the π terms, we get:
96 = 128(dr/dt)
Dividing both sides by 128, we have:
(dr/dt) = 96/128
Simplifying further, we get:
(dr/dt) = 0.75 meters per second
Therefore, when the area of the region is 64π square meters, the radius is increasing at a rate of 0.75 meters per second.
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.