If the rate of change of volume of sphere is equal to the rate of change of its radius, find the radius of the sphere.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{sphere} \mathrm{V}= \frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{Given}, \\ \mathrm{dV}=\mathrm{dr} \\ \mathrm{or} \frac{\mathrm{dV}}{\mathrm{dr}}=1 ---\left(1\right) \\ \mathrm{V}= \frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{differentiating} \mathrm{both} \mathrm{sides} \mathrm{w}.\mathrm{r}.\mathrm{t} \mathrm{r} \\ \frac{\mathrm{dV}}{\mathrm{dr}}=\frac{4}{3}\mathrm{\pi }\times 3{\mathrm{r}}^2 \\ \mathrm{from} 1, \mathrm{we} \mathrm{get} \\ 1=\frac{4}{3}\mathrm{\pi }\times 3{\mathrm{r}}^2 \\ 1=4{\mathrm{\pi r}}^2 \\ {\mathrm{r}}^2=\frac{1}{4\mathrm{\pi }} \\ \mathrm{r}=\frac{1}{2\sqrt{\mathrm{\pi }}} \\ \mathbf{Hence}\mathbf{ }\mathbf{radius}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{sphere}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{r}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\sqrt{\mathbf{\pi }}} \end{gathered}$$

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