A right circular cone has a fixed height of 9 cm. The rate of change of its volume with respect to the radius of its base, when the radius is 4 cm, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(24\pi\) cm2
Let r be the radius of the base. With height h = 9 cm, the volume is
\( \displaystyle V = \frac{1}{3}\pi r^2 h = 3\pi r^2 \)
Differentiating with respect to r by the power rule:
\( \displaystyle \frac{dV}{dr} = 6\pi r \)
\( = 6\pi \times 4 = 24\pi \) cm2
(cm3 of volume per cm of radius.)
Why the other options are incorrect
Option (2) \(48\pi\) cm2 — Incorrect. This is the volume \(3\pi \times 16\) itself, not its rate of change.
Option (3) \(12\pi\) cm2 — Incorrect. This is half the rate; the factor 2 from the derivative of \(r^2\) is left out.
Option (4) \(16\pi\) cm2 — Incorrect. This is the base area \(\pi r^2\) at r = 4, not dV/dr.