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  4. The rate of change of the volume of a sphere w.r.t. its surface area, when the radius is 2 cm, is

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Q. The rate of change of the volume of a sphere w.r.t. its surface area, when the radius is $2 \,cm$, is

Application of Derivatives

Solution:

$ V=\frac{4}{3} \pi r^{3}, S=4 \pi r^{2} $
$ \frac{dV}{d r}=4 \pi r^{2}, \frac{d S}{d r}=8 \pi r$
$\therefore \frac{dV}{dS}=\frac{d V / d r}{d S / d r}=\frac{4 \pi r^{2}}{8 \pi r}=\frac{r}{2} $
When $r=2, \frac{d V}{dS}=\frac{2}{2}=1 $