Question

The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. The rate of change of lateral surface when the radius = 7 cm and altitude 24 cm is
(a) 54π cm²/min
(b) 7π cm²/min
(c) 27 cm²/min
(d) none of these

Answer 1

$\mathrm{Let}r\mathrm{be}\mathrm{the}\mathrm{radius},h\mathrm{be}\mathrm{the}\mathrm{height}\mathrm{and}S\mathrm{be}\mathrm{the}\mathrm{lateral}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{at}\mathrm{any}\mathrm{time}t.$

$$\begin{gathered} \mathrm{Given}:\frac{dr}{dt}=3\mathrm{cm}/\min \mathrm{and}\frac{dh}{dt}=-4\mathrm{cm}/\min \\ \mathrm{Here}, \\ {l}^2=h^2+r^2 \\ \Rightarrow l=\sqrt{{\left(24\right)}^2+{\left(7\right)}^2} \\ \Rightarrow l=\sqrt{625} \\ \Rightarrow l=25 \\ \mathit{\text{S}}\text{=πrl} \\ \Rightarrow S^2={\left(\mathrm{\pi rl}\right)}^2 \\ \Rightarrow S^2={\mathrm{\pi }}^2{r}^2\left(h^2+r^2\right) \\ \Rightarrow S^2={\mathrm{\pi }}^2{r}^4+{\mathrm{\pi }}^2{h}^2{r}^2 \\ \Rightarrow 2S\frac{dS}{dt}=4{\mathrm{\pi }}^2{r}^3\frac{dr}{dt}+2{\mathrm{\pi }}^2{r}^{2}h\frac{dh}{dt}+2{\mathrm{\pi }}^2{h}^{2}r\frac{dr}{dt} \\ \Rightarrow 2\mathrm{\pi }rl\frac{dS}{dt}=2{\mathrm{\pi }}^{2}rh\left[\frac{2r^2}{h}\frac{dr}{dt}+r\frac{dh}{dt}+h\frac{dr}{dt}\right] \\ \Rightarrow 25\frac{dS}{dt}=24\mathrm{\pi }\left[\frac{2{\left(7\right)}^2}{24}\times 3-7\times 4+24\times 3\right]\left[\mathrm{Given}:r=7,h=24\right] \\ \Rightarrow 25\frac{dS}{dt}=24\mathrm{\pi }\left[\frac{49}{4}-28+72\right] \\ \Rightarrow 25\frac{dS}{dt}=24\mathrm{\pi }\left[\frac{49+288-112}{4}\right] \\ \Rightarrow \frac{dS}{dt}=24\mathrm{\pi }\left[\frac{225}{100}\right] \\ \Rightarrow \frac{dS}{dt}=24\mathrm{\pi }\left(2.25\right) \\ \Rightarrow \frac{dS}{dt}=54\mathrm{\pi }{\mathrm{cm}}^2/\sec \end{gathered}$$