Question:

The rate of change of the volume of a sphere with respect to its surface area $ S $ is

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In problems involving rates of change, remember to differentiate both the surface area and volume formulas with respect to the given variable and simplify to obtain the final result.
Updated On: Apr 16, 2025
  • \( \frac{1}{2} \sqrt{\frac{S}{\pi}} \)
  • \( \sqrt{\frac{S}{\pi}} \)
  • \( \frac{2}{3} \sqrt{\frac{S}{\pi}} \)
  • \( \frac{1}{4} \sqrt{\frac{S}{\pi}} \)
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The Correct Option is D

Solution and Explanation

Let the radius of the sphere be \( r \). The surface area \( S \) and the volume \( V \) of a sphere are given by the formulas: \[ S = 4\pi r^2 \quad \text{and} \quad V = \frac{4}{3}\pi r^3 \] Now, we want to find the rate of change of the volume with respect to the surface area. First, we differentiate both expressions with respect to \( r \): \[ \frac{dV}{dr} = 4\pi r^2 \quad \text{and} \quad \frac{dS}{dr} = 8\pi r \] The rate of change of the volume with respect to the surface area is: \[ \frac{dV}{dS} = \frac{\frac{dV}{dr}}{\frac{dS}{dr}} = \frac{4\pi r^2}{8\pi r} = \frac{1}{2} r \] Now, using the formula for the surface area \( S = 4\pi r^2 \), we solve for \( r \): \[ r = \sqrt{\frac{S}{4\pi}} \] Substitute this value of \( r \) into the expression for \( \frac{dV}{dS} \): \[ \frac{dV}{dS} = \frac{1}{2} \sqrt{\frac{S}{4\pi}} = \frac{1}{4} \sqrt{\frac{S}{\pi}} \] Thus, the correct answer is (D).
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