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 radius r of a cone with a fixed height of 5 units

Short answer

Equation relating the quantities: $\mathrm{S}=\mathrm{\pi r}\sqrt{{\mathrm{r}}^2+25}+{\mathrm{\pi r}}^2$

Implicit differentiation: $\frac{dS}{dt}=\left(\pi \sqrt{{\mathrm{r}}^2+25}+\frac{{\mathrm{\pi r}}^2}{\sqrt{{\mathrm{r}}^2+25}}+2\mathrm{\pi }r\right)\frac{dr}{dt}$

Step-by-step solution

Step 1. Given information

Surface area of the cone$=S$

Radius of the cone$=r$

Height of the cone$=5$

Step 2. Equation relating the quantities

Surface area of the cone is given by,

$$\begin{gathered} S=\mathrm{\pi r}\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2}+{\mathrm{\pi r}}^2 \\ \Rightarrow \mathrm{S}=\mathrm{\pi r}\sqrt{{\mathrm{r}}^2+5^2}+{\mathrm{\pi r}}^2 \\ \Rightarrow \mathrm{S}=\mathrm{\pi r}\sqrt{{\mathrm{r}}^2+25}+{\mathrm{\pi r}}^2 \end{gathered}$$

Step 3. Implicit differentiation with respect to time

From the above step,

$$\begin{gathered} S=\mathrm{\pi r}\sqrt{{\mathrm{r}}^2+25}+{\mathrm{\pi r}}^2 \\ \mathrm{Differentiating}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}, \\ \frac{dS}{dt}=\pi \sqrt{{\mathrm{r}}^2+25}\frac{dr}{dt}+\frac{\mathrm{\pi r}\left(2\mathrm{r}\right)}{2\sqrt{{\mathrm{r}}^2+25}}\frac{dr}{dt}+\left(2r\right)\pi \frac{dr}{dt} \\ \Rightarrow \frac{dS}{dt}=\left(\pi \sqrt{{\mathrm{r}}^2+25}+\frac{{\mathrm{\pi r}}^2}{\sqrt{{\mathrm{r}}^2+25}}+2\mathrm{\pi }r\right)\frac{dr}{dt} \end{gathered}$$