The rate of change of volume of a sphere with respect to its surface area, when radius is 4 cm, is equal to Option 1: 64π cm^3/cm^2 Option 2: 32π cm^3/cm^2 Option 3: 1/2 cm^3/cm^2 Option 4: 2 cm^3/cm^2

Answer: 2 cm^3/cm^2. Solution: The rate of change of volume with respect to surface area is dV/dS when radius is 4 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)/(dS/dr) V = 4/3 r^3 dV/dr = 4/3 x 3r^2 dV/dr = 4 r^2 S = 4 r^2 dS/dr = 4 x 2r dS/dr = 8 r dV/dS = (dV/dr)/(dS/dr) dV/dS = (4 r^2)/(8 r) dV/dS = r/2 At r = 4 cm: dV/dS = 4/2 dV/dS = 2 cm Therefore, the rate of change of volume with respect to surface area when radius is 4 cm equals 2 cm.

Calculus

Application of Derivatives

Medium

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The rate of change of volume of a sphere with respect to its surface area, when radius is 4 cm, is equal to

64π $cm^3/cm^2$

32π $cm^3/cm^2$

$\frac{1}{2} cm^3/cm^2$

2 $cm^3/cm^2$

✅ Correct Option: 4

The rate of change of volume with respect to surface area is $\frac{dV}{dS}$ when radius is 4 cm.

For a sphere with radius $r$ :

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

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

Using the chain rule:

$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$

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

$\frac{dV}{dr} = \frac{4}{3}\pi \times 3r^2$

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

$S = 4\pi r^2$

$\frac{dS}{dr} = 4\pi \times 2r$

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

$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$

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

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

At $r = 4$ cm:

$\frac{dV}{dS} = \frac{4}{2}$

$\frac{dV}{dS} = 2$ cm

Therefore, the rate of change of volume with respect to surface area when radius is 4 cm equals 2 cm.

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