CUET UG 2025 — Mathematics Calculus

The rate of change of volume with respect to surface area is $\dfrac{dV}{dS}$dSdV​ when the radius is 6 cm.

For a sphere with radius $r$r:

Volume: $V = \dfrac{4}{3}\pi r^3$V=34​πr3

Surface Area: $S = 4\pi r^2$S=4πr2

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Using the chain rule:

$$\dfrac{dV}{dS} = \dfrac{dV}{dr} \times \dfrac{dr}{dS}$$dSdV​=drdV​×dSdr​

$$\dfrac{dV}{dS} = \dfrac{dV}{dr} \div \dfrac{dS}{dr}$$dSdV​=drdV​÷drdS​

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Finding $\dfrac{dV}{dr}$drdV​:

$$V = \dfrac{4}{3}\pi r^3$$V=34​πr3

$$\dfrac{dV}{dr} = 4\pi r^2$$drdV​=4πr2

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Finding $\dfrac{dS}{dr}$drdS​:

$$S = 4\pi r^2$$S=4πr2

$$\dfrac{dS}{dr} = 8\pi r$$drdS​=8πr

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Calculating $\dfrac{dV}{dS}$dSdV​:

$$\dfrac{dV}{dS} = \dfrac{4\pi r^2}{8\pi r}$$dSdV​=8πr4πr2​

$$\dfrac{dV}{dS} = \dfrac{r}{2}$$dSdV​=2r​

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At $r = 6$r=6 cm:

$$\dfrac{dV}{dS} = \dfrac{6}{2}$$dSdV​=26​

$$\dfrac{dV}{dS} = 3$$dSdV​=3

Therefore, the rate of change of volume with respect to surface area is $3$3 cm³/cm² when the radius is 6 cm.