Question

In Exercises 19–26, write down an equation that relates the two quantities described. Then use implicit differentiation to obtain a relationship between the rates at which the quantities change over time.

The volume V and radius r of a cylinder with a fixed height of 10 units.

Short answer

The equation that relates the volume Vand the radius r of the cylinder is $V=10{\mathrm{\pi r}}^2$.

The derivative$\frac{dV}{dt}$and$\frac{dr}{dt}$are related by $\frac{dV}{dt}=20\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}$.

Step-by-step solution

Step 1. Given information.

The height of a cylinder is 10 units.

Step 2. Formula used.

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

Step 3. Apply the value of h.

Apply the value of $h=10$in $V={\mathrm{\pi r}}^2\mathrm{h}$as follows.

$$\begin{gathered} V={\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{V}={\mathrm{\pi r}}^2\left(10\right) \\ \mathrm{V}=10{\mathrm{\pi r}}^2 \end{gathered}$$

The equation that relates the volumeV and radiusr of the cylinder is$V=10{\mathrm{\pi r}}^2$.

Step 4. Apply the differentiation.

Apply the differentiation to $V=10{\mathrm{\pi r}}^2$ with respect to t as follows.

$$\begin{gathered} \frac{d}{dt}\left(V\right)=\frac{d}{dt}\left(10{\mathrm{\pi r}}^2\right) \\ \frac{dV}{dt}=20\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}} \end{gathered}$$

The derivative$\frac{dV}{dt}$and$\frac{dr}{dt}$are related by.

Step 5. Conclusion.

The equation that relates the volume V and the radius r of the cylinder is $V=10{\mathrm{\pi r}}^2$.

The derivative$\frac{dV}{dt}$and$\frac{dr}{dt}$are related by$\frac{dV}{dt}=20\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}$.