UPSC NDA Math Model Paper 8

Given, at any instant t, for a sphere, r denotes the radius, S denotes the surface area. Surface area $S = 4\pi r^2$ Differentiating w.r.to t, we get $\Rightarrow \frac{ds}{dt}=4\pi \cdot 2 r \frac{dr}{dt}$ $\Rightarrow \frac{ds}{dt}=8\pi r \frac{dr}{dt}$ $\Rightarrow \frac{dr}{dt}=\frac{1}{8\pi r} \frac{ds}{dt}$ ....(1) Volume of sphere $V = \frac{4}{3} \pi r^{3}$ Differentiating w.r.to $t$, we get $\Rightarrow \frac{dV}{dt}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{dr}{dt}$ $\Rightarrow \frac{dV}{dt}=4\pi r^{2} \frac{dr}{dt}$ From equation (1), we have $\Rightarrow \frac{dV}{dt}=4\pi r^{2} \frac{1}{8\pi r} \frac{ds}{dt}$ $\Rightarrow \frac{dV}{dt}=\frac{r}{2} \frac{ds}{dt}$ Hence, if at any instant t, for a sphere, r denotes the radius, S denotes the surface area and V denotes the volume, then $\frac{dV}{dt}=\frac{1}{2} r \frac{dS}{dt}$