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The rate of change volume of a sphere with respect to its surface area when the radius is 5 meters is
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$\frac{5}{2}$
$\begin{aligned} & V=\frac{4}{3} \pi r^3 \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=4 \pi r^2 \frac{\mathrm{d} r}{\mathrm{~d} t} \\ & S=4 \pi r^2 \\ & \Rightarrow \frac{\mathrm{d} s}{\mathrm{~d} t}=8 \pi r \frac{\mathrm{d} r}{\mathrm{~d} t} \\ & \text { Now } \frac{\mathrm{d} v}{\mathrm{~d} s}=\frac{\frac{\mathrm{d} v}{\mathrm{~d} t}}{\frac{\mathrm{ds}}{\mathrm{d} t}}=\frac{r}{2}=\frac{5}{2}\end{aligned}$
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