The rate of change of the volume of a sphere | vedclass

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(C) Let the radius of the sphere be $r$. Volume of the sphere $V = \frac{4}{3} \pi r^3$. Surface area of the sphere $A = 4 \pi r^2$. We need to find t

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$1 	ext{ cm}^3 / 	ext{ cm}^2$

Vedclass · Answer

(C) Let the radius of the sphere be $r$. Volume of the sphere $V = \frac{4}{3} \pi r^3$. Surface area of the sphere $A = 4 \pi r^2$. We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dA}$. Using the chain rule,$\frac{dV}{dA} = \frac{dV/dr}{dA/dr}$. First,differentiate $V$ with respect to $r$: $\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3} \pi r^3) = 4 \pi r^2$. Next,differentiate $A$ with respect to $r$: $\frac{dA}{dr} = \frac{d}{dr}(4 \pi r^2) = 8 \pi r$. Now,calculate $\frac{dV}{dA} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$. Given the radius $r = 2 	ext{ cm}$,substitute this value into the expression: $\frac{dV}{dA} = \frac{2}{2} = 1 	ext{ cm}^3 / 	ext{ cm}^2$.

The rate of change of the volume of a sphere with respect to its surface area,when its radius is $2 \text{ cm}$,is

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