Question

The altitude of a cone is 20 cm and its semi vertical angle is 30°. If the semi vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of

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

Hint:

When dealing with related rates, always start by expressing the quantities in terms of each other, then differentiate with respect to time using the chain rule.

Answer 1

The Correct Option is C

Let \( r \) be the radius of the base and \( h \) be the altitude of the cone. We are given: - \( h = 20 \) cm (the altitude), - The semi vertical angle \( \theta = 30^\circ \), - The rate of change of the semi vertical angle is \( \frac{d\theta}{dt} = 2^\circ \) per secon(D) We need to find the rate at which the radius \( r \) is increasing, i.e., \( \frac{dr}{dt} \). Step 1: Relationship between radius and height From the right triangle formed by the radius \( r \), the height \( h \), and the slant height \( l \), we can use the tangent of the semi vertical angle: \[ \tan \theta = \frac{r}{h} \] Since the height \( h \) is constant, we can rewrite this as: \[ r = h \tan \theta \] Substitute \( h = 20 \) cm: \[ r = 20 \tan \theta \] Step 2: Differentiate with respect to time Now, differentiate both sides of the equation with respect to time \( t \): \[ \frac{dr}{dt} = 20 \frac{d}{dt} (\tan \theta) \] Using the chain rule: \[ \frac{dr}{dt} = 20 \sec^2 \theta \frac{d\theta}{dt} \] Step 3: Substitute known values We are given \( \frac{d\theta}{dt} = 2^\circ \), but we must convert it to radians: \[ 2^\circ = \frac{2\pi}{180} \text{ radians} = \frac{\pi}{90} \text{ radians per second} \] Now, substitute the values \( \theta = 30^\circ \) and \( \frac{d\theta}{dt} = \frac{\pi}{90} \) into the equation. First, calculate \( \sec^2 30^\circ \): \[ \sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}, \quad \sec^2 30^\circ = \frac{4}{3} \] Thus: \[ \frac{dr}{dt} = 20 \cdot \frac{4}{3} \cdot \frac{\pi}{90} \] \[ \frac{dr}{dt} = \frac{160}{3} \text{ cm/sec} \] Therefore, the radius is increasing at the rate of \( \frac{160}{3} \) cm/se(C) Thus, the correct answer is \( \boxed{\frac{160}{3}} \) cm/se(C)