Question

If a spherical balloon has a variable diameter \(3x+\frac{9}{2}\) units, then the rate of change of its volume with respect to \(x\) is

• A. \(\frac{27π}{4}(2x+3)^2\)
• B. \(\frac{2π}{3}(2x+3)^2\)
• C. \(\frac{27π}{8}(2x+3)^2\)
• D. \(\frac{27π}{2}(2x+3)^2\)

Hint:

Write the radius from the diameter, find the volume in terms of x, then differentiate.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The volume of a sphere is \(V = \dfrac{4}{3}\pi r^3\). We need \(\dfrac{dV}{dx}\), so we express \(V\) in terms of \(x\) and differentiate.

Step 2: Find the radius.
Diameter \(= 3x + \dfrac{9}{2}\), so \(r = \dfrac{3x}{2} + \dfrac{9}{4} = \dfrac{3}{4}(2x + 3)\).

Step 3: Write the volume.
\[ V = \frac{4}{3}\pi\cdot\frac{27}{64}(2x + 3)^3 = \frac{9\pi}{16}(2x + 3)^3 \]

Step 4: Differentiate.
\[ \frac{dV}{dx} = \frac{9\pi}{16}\cdot 3(2x+3)^2\cdot 2 = \frac{27\pi}{8}(2x+3)^2 \]

Step 5: Check the options.
Option (C) matches. Option (A) has a coefficient \(\dfrac{27\pi}{4}\), which is double the correct one, and option (D) has \(\dfrac{27\pi}{2}\), which is four times it.

Final Answer:
\(\dfrac{dV}{dx} = \dfrac{27\pi}{8}(2x+3)^2\), option (C). \[ \boxed{\frac{27\pi}{8}(2x+3)^2} \]