Connect a spherical balloon’s volume and radius rates

The Student leaves this method blank; the Teacher supplies $dV/dr=4\pi r^2$ and $dV/dt=(dV/dr)(dr/dt)$. Thus $dr/dt=\dfrac{dV/dt}{dV/dr}=\dfrac{10}{4\pi r^2}=\dfrac{5}{2\pi r^2}$. At $r=5$, this is $\dfrac{1}{10\pi}\text{ cm/s}$.

Both copies print this method: differentiating $V=\frac43\pi r^3$ with respect to time gives $dV/dt=4\pi r^2\,dr/dt$. Substitute $dV/dt=10$ and $r=5$ to obtain $10=4\pi(5)^2\,dr/dt$, hence $dr/dt=1/(10\pi)\text{ cm/s}$. The positive sign means the radius is increasing.