A spherical balloon is inflated at a rate of 10 cubic centimeters per second. How fast is the radius of the balloon increasing when the radius is 5 centimeters?
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We know that the volume of a sphere is given by the formula $V = \frac{4}{3}\pi r^3$, where $r$ is the radius.
We can take the derivative of both sides with respect to $t$ (time) to find the rate of change of the volume with respect to time: $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$.
Using the chain rule, we have: $\frac{dV}{dt} = \frac{d}{dr}(r^3)\cdot\frac{dr}{dt}\cdot\frac{4}{3}\pi$.
We are given that $\frac{dV}{dt} = 10 \text{ cm}^3/\text{s}$.
Plugging in $\frac{dV}{dt} = 10$, we have: $10 = 3r^2\cdot\frac{dr}{dt}\cdot\frac{4}{3}\pi$.
Simplifying, we have: $10 = 4\pi r^2\cdot\frac{dr}{dt}$.
Dividing by $4\pi r^2$, we get $\frac{dr}{dt} = \frac{10}{4\pi r^2}$.
When the radius is $5$ centimeters, we have $\frac{dr}{dt} = \frac{10}{4\pi (5)^2}$.
Calculating, we find $\frac{dr}{dt} \approx \boxed{0.08}$ centimeters per second.