Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 78

* Refer to the problem in the book.

Illustration:







Solution:

Recall that the volume of the cone is $\displaystyle V = \frac{1}{3} \pi r^2 h$ and by using similar triangles,


$\begin{equation} \begin{aligned} \frac{r}{h} =& \frac{3}{10} \\ \\ r =& 0.3 h \end{aligned} \end{equation}$


Hence,


$\begin{equation} \begin{aligned} V =& \frac{1}{3} \pi (0.3h0^2)h \\ \\ V =& \frac{9}{300} \pi h^3 \end{aligned} \end{equation}$


Taking the derivative with respect to time,


$\begin{equation} \begin{aligned} \frac{dV}{dt} =& \frac{9 \pi}{300} (3h^2) \frac{dh}{dt} \\ \\ \frac{dV}{dt} =& \frac{9 \pi}{100} h^2 \frac{dh}{dt} \\ \\ \frac{dh}{dt} =& \frac{\displaystyle 100 \frac{dV}{dt}}{9 \pi h^2} ; \qquad \qquad \text{where $\large \frac{dV}{dt} = 2 cm^3 /s$ and $h = 5 cm$} \\ \\ \frac{dh}{dt} =& \frac{100(2)}{9 \pi (5)^2} \\ \\ \frac{dh}{dt} =& 0.2829 cm/s \end{aligned} \end{equation}$