Question

Suppose the radius r, height h, and volume V of a cylinder are functions of time t, and further suppose that the volume of the cylinder is always constant. Write $\frac{dr}{dt}$ in terms of r, h, and $\frac{dh}{dt}$.

Short answer

The derivative $\frac{dr}{dt}$ in terms of r, h, and$\frac{dh}{dt}$is$\frac{\mathrm{dr}}{\mathrm{dt}}=-\frac{\mathrm{r}}{2\mathrm{h}}\frac{\mathrm{dh}}{\mathrm{dt}}$.

Step-by-step solution

Step 1. Given information.

Given that the radius r, the height h,and the volumeV of the cylinder are functions of a time t.

Also given, the volume of the cylinder is always constant.

That is,$\frac{dV}{dt}=0$.

Step 2. Formula used.

The volume of the cylinder is given by the formula$V={\mathrm{\pi r}}^2\mathrm{h}$cu. units.

Step 3. Apply the differentiation.

Apply the differentiation to $V={\mathrm{\pi r}}^2\mathrm{h}$ with respect to t and keeping V as constant as follows.

role="math" localid="1648732789305" $$\begin{gathered} \frac{d}{dt}\left(V\right)=\frac{d}{dt}\left({\mathrm{\pi r}}^2\mathrm{h}\right) \\ 0=2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}\mathrm{h}+{\mathrm{\pi r}}^2\frac{\mathrm{dh}}{\mathrm{dt}} \\ 2\mathrm{\pi rh}\frac{\mathrm{dr}}{\mathrm{dt}}=-{\mathrm{\pi r}}^2\frac{\mathrm{dh}}{\mathrm{dt}} \\ \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{-{\mathrm{\pi r}}^2}{2\mathrm{\pi rh}}\frac{\mathrm{dh}}{\mathrm{dt}} \\ \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{-\mathrm{r}}{2\mathrm{h}}\frac{\mathrm{dh}}{\mathrm{dt}} \end{gathered}$$

Step 4. Conclusion.

The derivative $\frac{dr}{dt}$in terms ofr, hand$\frac{dh}{dt}$is$\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{-\mathrm{r}}{2\mathrm{h}}\frac{\mathrm{dh}}{\mathrm{dt}}$.