Question

The radius of a cone of height 9 units is changed from 2 units to 2.12 units. The exact change and approximate change in the volume of the cone are respectively

• A. \((1.4437)\pi, (1.44)\pi\)
• B. \((1.4832)\pi, (1.479)\pi\)
• C. \((1.4842)\pi, (1.48)\pi\)
• D. \((1.4832)\pi, (1.44)\pi\)

Hint:

Differentials (\(dy\)) provide a linear approximation of the change (\(\Delta y\)). For small \(\Delta x\), \(dy \approx \Delta y\). The difference represents the error in approximation.

Answer 1

The Correct Option is D

Step 1: Formula Setup: Volume of cone \(V(r) = \frac{1}{3}\pi r^2 h\). Given \(h = 9\). \(V(r) = \frac{1}{3}\pi r^2 (9) = 3\pi r^2\). Given \(r = 2\) and \(\Delta r = 2.12 - 2 = 0.12\).
Step 2: Approximate Change (\(dV\)): Using differentials: \(dV = \frac{dV}{dr} \Delta r\). \(\frac{dV}{dr} = 6\pi r\). At \(r=2\): \(dV = 6\pi(2) \cdot (0.12) = 12\pi \cdot 0.12 = 1.44\pi\).
Step 3: Exact Change (\(\Delta V\)): \(\Delta V = V(r + \Delta r) - V(r)\). \(\Delta V = 3\pi(2.12)^2 - 3\pi(2)^2 = 3\pi (2.12^2 - 4)\). Using \(a^2 - b^2 = (a-b)(a+b)\): \(\Delta V = 3\pi (2.12 - 2)(2.12 + 2) = 3\pi (0.12)(4.12)\). \(\Delta V = 0.36\pi (4.12) = 1.4832\pi\).
Required Answer: Exact: \(1.4832\pi\), Approximate: \(1.44\pi\).