Question

Suppose the radius r, height h, and volume V of a cylinder are functions of time t. How is $\frac{dV}{dt}$ related to$\frac{dh}{dt}$ if the radius of the cylinder is constant?

Short answer

The derivatives $\frac{dV}{dt}$and $\frac{dh}{dt}$are related by$\frac{dV}{dt}={\mathrm{\pi r}}^2\frac{\mathrm{dh}}{\mathrm{dt}}$.

Step-by-step solution

Step 1. Formula used.

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

Step 2. Apply the differentiation.

Apply the differentiation to V with respect to t and keeping ras constant as follows.

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

Step 3. Conclusion.

The derivatives$\frac{dV}{dt}$and$\frac{dh}{dt}$are related by$\frac{dV}{dt}={\mathrm{\pi r}}^2\frac{\mathrm{dh}}{\mathrm{dt}}$.