Q. 17 - Calculus

Question

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

Step-by-Step Solution

Answer

The derivative $\frac{dV}{dt}$ in terms of h and $\frac{dh}{dt}$ is $\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{3}{4}{\mathrm{\pi h}}^2\frac{\mathrm{dh}}{\mathrm{dt}}$.

Step 1. Given information.

Given that the radius r, height h, volume V of a cylinder are functions of time t, and the height of the cylinder is always twice its radius.

That is, the height is $h=2r$.

From $h=2r$, the radius is $r=\frac{h}{2}$.

Step 2. Formula used.

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

Step 3. Apply the value of r.

Apply the value $r=\frac{h}{2}$ in $V={\mathrm{\pi r}}^2\mathrm{h}$ as follows.

$$\begin{gathered} V={\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{V}=\mathrm{\pi }{\left(\frac{\mathrm{h}}{2}\right)}^2\mathrm{h} \\ \mathrm{V}=\frac{1}{4}{\mathrm{\pi h}}^3 \end{gathered}$$

Step 4. Apply the differentiation.

Apply the differentiation to $\mathrm{V}=\frac{1}{4}{\mathrm{\pi h}}^3$ with respect to t as follows.

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{V}\right)=\frac{d}{dt}\left(\frac{1}{4}{\mathrm{\pi h}}^3\right) \\ \frac{dV}{dt}=\frac{1}{4}\mathrm{\pi }\left(3{\mathrm{h}}^2\frac{\mathrm{dh}}{\mathrm{dt}}\right) \\ \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{3}{4}{\mathrm{\pi h}}^2\frac{\mathrm{dh}}{\mathrm{dt}} \end{gathered}$$

Step 5. Conclusion.

The derivative $\frac{dV}{dt}$ in terms of h and $\frac{dh}{dt}$ is $\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{3}{4}{\mathrm{\pi h}}^2\frac{\mathrm{dh}}{\mathrm{dt}}$.