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**HallsofIvy** Was $\displaystyle V= \pi h^3$ given in the problem?

The volume of a cone, of height h and base radius r, is given by $\displaystyle V= \frac{1}{3}\pi r^2 h$. Here, you are told that the "semi-vertical angle" is 60 degrees so using $\displaystyle \frac{r}{h}= tan(60)= \sqrt{3}$. $\displaystyle r= \sqrt{3}h$, $\displaystyle r^2= 3h^2 and $V= \pi h^3[/tex]. That is where that formula came from.

To change to a "rate of change" formula, differentiate both sides with respect to t, using the chain rule.

$\displaystyle \frac{dV}{dt}= \frac{d\left(\pi h^3\right)}{dt}= 3\pi h^2 \frac{dh}{dt}$

Using the fact that $\displaystyle \frac{dV}{dt}= .3$, you can solve that for $\displaystyle \frac{dh}{dt}$, as a function of h.

The area of the base is, of course, $\displaystyle A= \pi r^2= 3\pi h^2$ so $\displaystyle \frac{dA}{dt}= 6\pi h\frac{dh}{dt}$.