## Calculus: Early Transcendentals 8th Edition

$\frac{dh}{dt} = \frac{8}{9\pi} \approx 0.283 cm/s$
Volume of a cone: $V = \frac{1}{3} \pi r^{2}h$ From proportions: $\frac{r}{h} = \frac{3}{10}$ Solve for $r$: $r = \frac{3}{10}h$ $V = \frac{1}{3} \pi (\frac{3}{10}h)^{2}h$ $V = \frac{3}{100} \pi h^{3}$ $\frac{dV}{dt} = \frac{3}{100} \pi 3h^{2} \times \frac{dh}{dt}$ $\frac{dV}{dt} = \frac{3}{100} \pi 3(5)^{2} \times \frac{dh}{dt}$ $\frac{dV}{dt} = \frac{9}{4} \pi \times \frac{dh}{dt}$ Given: $\frac{dV}{dt} = 2$ $2 = \frac{9}{4} \pi \times \frac{dh}{dt}$ Solve for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{2}{\frac{9\pi}{4}}$ $\frac{dh}{dt} = \frac{8}{9\pi} \approx 0.283 cm/s$