Q. 21 - Calculus

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 surface area S and height h of a cylinder with a fixed radius of 2 units. 

Step-by-Step Solution

Answer

The equation that relates the surface area S and the height of the cylinder h is$\mathrm{S}=4\mathrm{\pi h}+8\mathrm{\pi }$.

The derivative $\frac{dS}{dt}$ and $\frac{dh}{dt}$ are related by  $\frac{dS}{dt}=4\mathrm{\pi }\frac{\mathrm{dh}}{\mathrm{dt}}$.

Step 1. Given information.

The radius of the cylinder is 2 units.

Step 2. Formula used.

The surface area of the cylinder is $S=2\mathrm{\pi r}\left(\mathrm{h}+\mathrm{r}\right)$ sq. units.

Step 3. Apply the value of r .

Apply the value of $r=2$ in $S=2\mathrm{\pi r}\left(\mathrm{h}+\mathrm{r}\right)$ as follows.

$$\begin{gathered} S=2\mathrm{\pi r}\left(\mathrm{h}+\mathrm{r}\right) \\ S=2\mathrm{\pi }\left(2\right)(\mathrm{h}+2) \\ \mathrm{S}=4\mathrm{\pi }\left(\mathrm{h}+2)\right) \\ \mathrm{S}=4\mathrm{\pi h}+8\mathrm{\pi } \end{gathered}$$

The equation that relates the surface area S and the height of the cylinder h is $\mathrm{S}=4\mathrm{\pi h}+8\mathrm{\pi }$.

Step 4. Apply the differentiation.

Apply the differentiation to $S=4\mathrm{\pi h}+8\mathrm{\pi }$ as follows.

$$\begin{gathered} \frac{d}{dt}\left(S\right)=\frac{d}{dt}\left(4\mathrm{\pi h}+8\mathrm{\pi }\right) \\ \frac{dS}{dt}=4\mathrm{\pi }\frac{\mathrm{dh}}{\mathrm{dt}}+0 \\ \frac{\mathrm{dS}}{\mathrm{dt}}=4\mathrm{\pi }\frac{\mathrm{dh}}{\mathrm{dt}} \end{gathered}$$

The derivative $\frac{dS}{dt}$ and $\frac{dh}{dt}$ are related by $\frac{dS}{dt}=4\mathrm{\pi }\frac{\mathrm{dh}}{\mathrm{dt}}$.

Step 5. Conclusion.

The equation that relates the surface area S and the height of the cylinder h is$\mathrm{S}=4\mathrm{\pi h}+8\mathrm{\pi }$.

The derivative $\frac{dS}{dt}$ and $\frac{dh}{dt}$ are related by $\frac{dS}{dt}=4\mathrm{\pi }\frac{\mathrm{dh}}{\mathrm{dt}}$.