The rate of change of volume of a sphere with respect to its surface area, when the radius is 6cm is: Option 1: 1 cm³/cm² Option 2: 6 cm³/cm² Option 3: 2 cm³/cm² Option 4: 3 cm³/cm²

Answer: 3 cm³/cm². Solution: The rate of change of volume with respect to surface area is dV/dS when the radius is 6 cm. For a sphere with radius r: Volume: V = 4/3 r^3 Surface Area: S = 4 r^2 Using the chain rule: dV/dS = dV/dr x dr/dS dV/dS = dV/dr dS/dr Finding dV/dr: V = 4/3 r^3 dV/dr = 4 r^2 Finding dS/dr: S = 4 r^2 dS/dr = 8 r Calculating dV/dS: dV/dS = (4 r^2)/(8 r) dV/dS = r/2 At r = 6 cm: dV/dS = 6/2 dV/dS = 3 Therefore, the rate of change of volume with respect to surface area is 3 cm³/cm² when the radius is 6 cm.

Calculus

Application of Derivatives

Medium

core

The rate of change of volume of a sphere with respect to its surface area, when the radius is 6cm is:

1 cm³/cm²

6 cm³/cm²

2 cm³/cm²

3 cm³/cm²

✅ Correct Option: 4

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

For a sphere with radius $r$ :

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

Surface Area: $S = 4\pi r^2$

Using the chain rule:

$\dfrac{dV}{dS} = \dfrac{dV}{dr} \times \dfrac{dr}{dS}$

$\dfrac{dV}{dS} = \dfrac{dV}{dr} \div \dfrac{dS}{dr}$

Finding $\dfrac{dV}{dr}$ :

$V = \dfrac{4}{3}\pi r^3$

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

Finding $\dfrac{dS}{dr}$ :

$S = 4\pi r^2$

$\dfrac{dS}{dr} = 8\pi r$

Calculating $\dfrac{dV}{dS}$ :

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

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

At $r = 6$ cm:

$\dfrac{dV}{dS} = \dfrac{6}{2}$

$\dfrac{dV}{dS} = 3$

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

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