Question 13 of 82intermediate🔧 ApplyNumerical4 marks

A balloon, which always remains spherical, has a variable diameter 3/2 (2x + 1). Find the rate of change of its volume with respect to x.

Correct Answer

The rate of change of volume with respect to x is 27π8(2x+1)2\frac{27\pi}{8}(2x+1)^2.

Exercise: EXERCISE 6.1 | Q: 13 | (Page 5)
For More Understanding

Explanation

The problem asks for the rate of change of volume of a spherical balloon with respect to x, given its diameter as a function of x. Using the formula for volume of a sphere and differentiating with respect to x (either directly or via chain rule) yields the result. The steps follow the pattern of Example 1 in the context, where the rate of change of area with respect to radius is found by differentiation.

Solution Steps

  1. Step 1: Express radius in terms of x. Diameter = 32(2x+1)\frac{3}{2}(2x+1). Since the balloon is spherical, radius r=diameter2=34(2x+1)r = \frac{\text{diameter}}{2} = \frac{3}{4}(2x+1).

  2. Step 2: Write the volume formula. Volume of a sphere: V=43πr3V = \frac{4}{3}\pi r^3.

  3. Step 3: Substitute r into V. V=43π[34(2x+1)]3=43π⋅2764(2x+1)3=9π16(2x+1)3V = \frac{4}{3}\pi \left[\frac{3}{4}(2x+1)\right]^3 = \frac{4}{3}\pi \cdot \frac{27}{64}(2x+1)^3 = \frac{9\pi}{16}(2x+1)^3.

  4. Step 4: Differentiate V with respect to x. dVdx=9π16⋅3(2x+1)2⋅2=27π8(2x+1)2\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x+1)^2 \cdot 2 = \frac{27\pi}{8}(2x+1)^2.

  5. Step 5: State the final answer. The rate of change of volume with respect to x is 27π8(2x+1)2\frac{27\pi}{8}(2x+1)^2.

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