Question

if V denotes the volume and S is the surface area of a sphere , then find the rate of change of V w.r.t S when the radius of the sphere is 2 cm

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{r}\mathrm{be}\mathrm{the}\mathrm{radius},\mathrm{S}\mathrm{be}\mathrm{the}\mathrm{Surface}\mathrm{area}\mathrm{and}\mathrm{V}\mathrm{be}\mathrm{the}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{sphere}\mathrm{at}\mathrm{any}\mathrm{instant}\mathrm{of}\mathrm{time}\mathrm{t}. \\ \mathrm{Now},\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{Differentiating}\mathrm{V}\mathrm{w}.\mathrm{r}.\mathrm{t}.\text{'}\mathrm{r}\text{'},\mathrm{we}\mathrm{get} \\ \frac{\mathrm{dV}}{\mathrm{dr}}=4{\mathrm{\pi r}}^2..........\left(1\right) \\ \mathrm{Now},\mathrm{S}=4{\mathrm{\pi r}}^2 \\ \mathrm{Differentiating}\mathrm{S}\mathrm{w}.\mathrm{r}.\mathrm{t}.\text{'}\mathrm{r}\text{'},\mathrm{we}\mathrm{get} \\ \frac{\mathrm{dS}}{\mathrm{dr}}=8\mathrm{\pi r}..........\left(2\right) \\ \mathrm{Dividing}\left(1\right)\mathrm{by}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{\mathrm{dV}}{\mathrm{dr}}÷\frac{\mathrm{dS}}{\mathrm{dr}}=\frac{4{\mathrm{\pi r}}^2}{8\mathrm{\pi r}} \\ \Rightarrow \frac{\mathrm{dV}}{\mathrm{dS}}=\frac{\mathrm{r}}{2} \\ \Rightarrow {\left[\frac{\mathrm{dV}}{\mathrm{dS}}\right]}_{\mathrm{r}=2}=\frac{2}{2}=1 \\ \mathrm{So},\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{volume}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{surface}\mathrm{area}\mathrm{is}\mathbf{1}\mathbf{}{\mathbf{cm}}^{\mathbf{3}}\mathbf{/}{\mathbf{cm}}^{\mathbf{2}}\mathbf{}\mathbf{.} \end{gathered}$$