Question

The vertical angle of a right circular cone is \( 60^\circ \). If water is being poured into the cone at the rate of \( \frac{1}{\sqrt{3}} \) m\(^3\)/min, then the rate (m/min) at which the radius of the water level is increasing when the height of the water level is 3m is:

• A. \( \frac{1}{3\sqrt{3}\pi} \)
• B. \( \frac{1}{9\sqrt{3}\pi} \)
• C. \( \frac{1}{9\pi} \)
• D. \( \frac{1}{3\pi} \)

Hint:

Use geometry to relate radius and height in a cone. - Differentiate the volume equation to find rates of change.

Answer 1

The Correct Option is C

Step 1: Establish relation between \( r \) and \( h \)
From the given vertical angle \( 60^\circ \), the half-angle is \( 30^\circ \), so: \[ \tan 30^\circ = \frac{r}{h} \Rightarrow r = \frac{h}{\sqrt{3}}. \] Step 2: Express volume in terms of \( h \)
The volume of a cone is: \[ V = \frac{1}{3} \pi r^2 h. \] Substituting \( r = \frac{h}{\sqrt{3}} \): \[ V = \frac{1}{3} \pi \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{1}{3} \pi \frac{h^3}{3} = \frac{\pi}{9} h^3. \] Step 3: Differentiate with respect to time
\[ \frac{dV}{dt} = \frac{\pi}{9} (3h^2) \frac{dh}{dt}. \] Given \( \frac{dV}{dt} = \frac{1}{\sqrt{3}} \): \[ \frac{1}{\sqrt{3}} = \frac{\pi}{9} \times 3 \times 9 \frac{dh}{dt}. \] \[ \frac{dh}{dt} = \frac{1}{9\pi}. \] Thus, the correct answer is \( \boxed{\frac{1}{9\pi}} \).