Question

If the vertical angle of a cone is \(60^\circ\) and the rate of change of its total surface area is \(2\sqrt{3}\,\text{cm}^2/\text{sec}\), then the rate of change of its volume (in \(\text{cm}^3/\text{sec}\)) when its radius is \(5\) cm is:

• A. \(15\)
• B. \(10\)
• C. \(5\)
• D. \(9\)

Hint:

In cone problems involving the vertical angle, first convert it into the semi-vertical angle and immediately establish relations among \(r\), \(h\), and \(l\) using trigonometry.

Answer 1

The Correct Option is C

Concept: For a cone: \[ V=\frac13\pi r^2 h \] and total surface area is: \[ S=\pi r(r+l) \] where:

• \(r\) = radius

• \(h\) = height

• \(l\) = slant height
The vertical angle is \(60^\circ\), so the semi-vertical angle is: \[ 30^\circ \] Using trigonometric relations, we can establish relationships between \(h\), \(r\), and \(l\).

Step 1: Use the geometry of the cone. The semi-vertical angle is: \[ 30^\circ \] From the right triangle formed inside the cone: \[ \tan30^\circ=\frac{r}{h} \] Since: \[ \tan30^\circ=\frac1{\sqrt3} \] we get: \[ \frac1{\sqrt3}=\frac{r}{h} \] Thus, \[ h=\sqrt3\,r \] Now compute slant height: \[ l=\sqrt{r^2+h^2} \] \[ =\sqrt{r^2+3r^2} \] \[ =\sqrt{4r^2} \] \[ =2r \]

Step 2: Find the total surface area in terms of \(r\). \[ S=\pi r(r+l) \] Substituting \(l=2r\), \[ S=\pi r(r+2r) \] \[ S=3\pi r^2 \] Differentiate with respect to time: \[ \frac{dS}{dt}=6\pi r\frac{dr}{dt} \] Given: \[ \frac{dS}{dt}=2\sqrt3 \] and \(r=5\), \[ 2\sqrt3=6\pi(5)\frac{dr}{dt} \] \[ 2\sqrt3=30\pi\frac{dr}{dt} \] Hence, \[ \frac{dr}{dt}=\frac{\sqrt3}{15\pi} \]

Step 3: Find the volume in terms of \(r\). \[ V=\frac13\pi r^2h \] Since \(h=\sqrt3 r\), \[ V=\frac13\pi r^2(\sqrt3 r) \] \[ V=\frac{\sqrt3}{3}\pi r^3 \] Differentiate: \[ \frac{dV}{dt} = \sqrt3\pi r^2\frac{dr}{dt} \] Substitute \(r=5\): \[ \frac{dV}{dt} = \sqrt3\pi(25)\left(\frac{\sqrt3}{15\pi}\right) \] \[ = \frac{75}{15} \] \[ =5 \] Therefore, \[ \boxed{5} \]