Question

The volume of a sphere is increasing at 3 cm³/sec. The rate at which the radius increases when radius is 2 cm, is
(a) $\frac{3}{32\mathrm{\pi }}\mathrm{cm}/\sec$

(b) $\frac{3}{16\mathrm{\pi }}\mathrm{cm}/\sec$

(c) $\frac{3}{48\mathrm{\pi }}\mathrm{cm}/\sec$

(d) $\frac{1}{24\mathrm{\pi }}\mathrm{cm}/\sec$

Answer 1

(b) $\frac{3}{16\mathrm{\pi }}\mathrm{cm}/\sec$

$$\begin{gathered} \text{Let}\mathit{\text{r}}\text{be the radius and}\mathit{\text{V}}\text{be the volume of the sphere at any time}\mathit{\text{t.}}\text{Then,} \\ \text{V=}\frac{4}{3}{\text{πr}}^3 \\ \text{⇒}\frac{dV}{dt}\text{=4π}{\mathit{\text{r}}}^2\frac{dr}{dt} \\ \text{⇒}\frac{dr}{dt}\text{=}\frac{1}{\mathrm{4\pi }{r}^2}\frac{dV}{dt} \\ \text{⇒}\frac{dr}{dt}\text{=}\frac{3}{\mathrm{4\pi }{\left(2\right)}^2} \\ \text{⇒}\frac{dr}{dt}\text{=}\frac{3}{16\mathrm{\pi }}\text{cm/sec} \end{gathered}$$