Question

The height of a cone increases by k%, its semi-vertical angle remaining the same. What is the approximate percentage increase (i) in total surface area, and (ii) in the volume, assuming that k is small?

Answer 1

Let h be the height, y be the surface area, V be the volume, l be the slant height and r be the radius of the cone.

$$\begin{gathered} \mathrm{Let}∆h\mathrm{be}\mathrm{the}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{height},∆r\mathrm{be}\mathrm{the}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{base}\mathrm{and}∆l\mathrm{be}\mathrm{the}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{slant}\mathrm{height}. \\ \mathrm{Semi}-\mathrm{vertical}\mathrm{angle}\mathrm{ramaining}\mathrm{the}\mathrm{same}. \\ \therefore \frac{∆h}{h}=\frac{∆r}{r}=\frac{∆l}{l} \\ \mathrm{Also},\frac{∆h}{h}\times 100=k \\ \mathrm{Then},\frac{∆h}{h}\times 100=\frac{∆r}{r}\times 100=\frac{∆l}{l}\times 100=k...\left(1\right) \\ \left(\mathrm{i}\right)\mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\text{the}\mathrm{cone},T=\pi rl+\pi r^2 \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}r,\mathrm{we}\mathrm{get} \\ \frac{dT}{dr}=\pi l+\pi r\frac{dl}{dr}+2\pi r \\ \Rightarrow \frac{dT}{dr}=\pi l+\pi r\frac{l}{r}+2\pi r\left[\mathrm{From}\left(1\right),\frac{dl}{dr}=\frac{∆l}{∆r}=\frac{l}{r}\right] \\ \Rightarrow \frac{dT}{dr}=\pi l+\pi l+2\pi r \\ \Rightarrow \frac{dT}{dr}=2\pi \left(l+r\right) \\ \therefore ∆T=\frac{dT}{dr}∆r=2\pi \left(l+r\right)\times \frac{kr}{100}=\frac{2kr\pi \left(l+r\right)}{100} \\ \therefore \frac{∆T}{T}\times 100=\frac{\left({\displaystyle \frac{2kr\pi \left(l+r\right)}{100}}\right)}{2\pi r\left(l+r\right)}\times 100=2k\% \\ \mathrm{Hence},\mathrm{the}\mathrm{percentage}\mathrm{increase}\mathrm{in}\mathrm{total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cone}\mathrm{is}2k. \\ \left(\mathrm{ii}\right)\mathrm{Volume}\mathrm{of}\mathrm{cone},V=\frac{1}{3}\pi r^{2}h \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.h,\mathrm{we}\mathrm{get} \\ \frac{dV}{dh}=\frac{1}{3}\pi r^2+\frac{1}{3}\pi h2r\frac{dr}{dh} \\ \Rightarrow \frac{dV}{dh}=\frac{1}{3}\pi r^2+\frac{1}{3}\pi h2r\frac{r}{h}\left[\mathrm{From}\left(1\right),\frac{dr}{dh}=\frac{∆r}{∆h}=\frac{r}{h}\right] \\ \Rightarrow \frac{dV}{dh}=\frac{1}{3}\pi r^2+\frac{2}{3}\pi r^2 \\ \Rightarrow \frac{dV}{dh}=\pi r^2 \\ \therefore ∆V=\frac{dV}{dh}dh=\pi r^2\times \frac{kh}{100}=\frac{k\pi r^{2}h}{100} \\ \therefore \frac{∆V}{V}\times 100=\frac{\left({\displaystyle \frac{k\pi r^{2}h}{100}}\right)}{{\displaystyle \frac{1}{3}\pi r^{2}h}}\times 100=3k\% \\ \mathrm{Hence},\mathrm{the}\mathrm{percentage}\mathrm{increase}\mathrm{in}\text{the}\mathrm{volume}\text{of the}\mathrm{cone}\mathrm{is}3k. \end{gathered}$$