Question

The altitude of a cone is 20 cm and its semi-vertical angle is \(30^\circ\). If the semi-vertical angle is increasing at the rate of \(2^\circ\) per second, then the radius of the base is increasing at the rate of:

• A. 30 cm/sec
• B. \(\frac{160}{3}\) cm/sec
• C. 10 cm/sec
• D. 160 cm/sec

Hint:

When dealing with rates of change of geometrical quantities, ensure that you use the correct trigonometric relations and convert angular velocity to radians if necessary.

Answer 1

The Correct Option is B

Step 1: Let \(\theta\) be the semi-vertical angle and \(r\) be the radius of the cone at time \(t\). The relationship between \(r\) and \(\theta\) is: \[ r = 20 \tan \theta \] Step 2: Differentiate with respect to time \(t\): \[ \frac{dr}{dt} = 20 \sec^2 \theta \cdot \frac{d\theta}{dt} \] Given that \(\frac{d\theta}{dt} = 2^\circ = \frac{\pi}{90} \, {radians/sec}\) and \(\theta = 30^\circ\), we find: \[ \frac{dr}{dt} = 20 \sec^2 30^\circ \times \frac{\pi}{90} = 20 \times \left(\frac{4}{3}\right) \times \frac{\pi}{90} = \frac{160}{3} \, {cm/sec} \] Therefore, the radius is increasing at the rate of \(\frac{160}{3}\) cm/sec.