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.

Answer 2

The altitude \(h\) of the cone is given as \(20 \text{ cm}\). The semi-vertical angle is \(\theta = 30^\circ\) and is increasing at a rate of \(\frac{d\theta}{dt} = 2^\circ/\text{sec}\). We need to find the rate at which the radius \(r\) of the base is increasing.
From the geometry of the cone, \(\tan\theta = \frac{r}{h}\). Therefore, we have:
\[ r = h \cdot \tan\theta \] Differentiating with respect to time \(t\), we get:
\[\frac{dr}{dt} = h \cdot \sec^2\theta \cdot \frac{d\theta}{dt}\] Plugging in the values, \(h = 20\), \(\theta = 30^\circ\), and \(\frac{d\theta}{dt} = 2^\circ/\text{sec}\), we have to convert degrees to radians:
\(\frac{2\pi}{180}\text{ radians/sec} = \frac{\pi}{90}\text{ radians/sec}\)
Now \[ \sec^2\theta = \sec^2 30^\circ = \frac{4}{3} \]as \(\sec 30^\circ = \frac{2}{\sqrt{3}}\). Plug these into the differentiated equation:
\[\frac{dr}{dt} = 20 \cdot \frac{4}{3} \cdot \frac{\pi}{90}\] Simplifying: \[\frac{dr}{dt} = \frac{80 \pi}{270}\] \[\frac{dr}{dt} = \frac{80 \times 3.14}{270}\] \[\frac{dr}{dt} = \frac{251.2}{270}\approx \frac{160}{3}\text{ cm/sec}\] Therefore, the radius of the base is increasing at a rate of \(\frac{160}{3}\text{ cm/sec}\).