If the vertical angle of a cone is \(60^\circ\) and the rate of change of its total surface area is \(2\sqrt{3}\,\text{cm}^2/\text{sec}\), then the rate of change of its volume (in \(\text{cm}^3/\text{sec}\)) when its radius is \(5\) cm is:
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In cone problems involving the vertical angle, first convert it into the semi-vertical angle and immediately establish relations among \(r\), \(h\), and \(l\) using trigonometry.
Step 1: Use the cone geometry. The vertical angle is $60^\circ$, so the semi-vertical angle is $30^\circ$. Then $\tan30^\circ=\dfrac{r}{h}=\dfrac{1}{\sqrt3}$, giving $h=\sqrt3\,r$. Step 2: Find the slant height. \[ l=\sqrt{r^2+h^2}=\sqrt{r^2+3r^2}=2r. \] Step 3: Write the surface area in terms of $r$. Total surface area $S=\pi r(r+l)=\pi r(r+2r)=3\pi r^2$. Step 4: Use the given rate to find $\dfrac{dr}{dt}$. $\dfrac{dS}{dt}=6\pi r\dfrac{dr}{dt}$. With $\dfrac{dS}{dt}=2\sqrt3$ and $r=5$, \[ 2\sqrt3=30\pi\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{\sqrt3}{15\pi}. \] Step 5: Write the volume in terms of $r$. $V=\dfrac13\pi r^2 h=\dfrac13\pi r^2(\sqrt3 r)=\dfrac{\sqrt3}{3}\pi r^3$, so $\dfrac{dV}{dt}=\sqrt3\pi r^2\dfrac{dr}{dt}$. Step 6: Put in $r=5$. \[ \frac{dV}{dt}=\sqrt3\pi(25)\cdot\frac{\sqrt3}{15\pi}=\frac{3\cdot25}{15}=5. \] \[ \boxed{5} \]