Question

A is a point on the circle with radius 8 and center at O. A particle P is moving on the circumference of the circle starting from A. M is the foot of the perpendicular from P on OA and \( \angle POM = \theta \). When \( OM = 4 \) and \( \frac{d\theta}{dt} = 6 \) radians/sec, then the rate of change of PM is (in units/sec):

• A. \( 24\sqrt{3} \)
• B. \( 24 \)
• C. \( 15\sqrt{3} \)
• D. \( 48\sqrt{3} \)

Hint:

For problems involving circular motion, use differentiation techniques with trigonometric functions and apply chain rule properly.

Answer 1

The Correct Option is B

Step 1: Define the given quantities \[ OM = r \cos \theta, \quad PM = r \sin \theta. \] Given: \[ r = 8, \quad OM = 4 \Rightarrow \cos \theta = \frac{4}{8} = \frac{1}{2}. \] Thus, \( \theta = \frac{\pi}{3} \) and \( \sin \theta = \frac{\sqrt{3}}{2} \). Step 2: Differentiate \( PM = r \sin \theta \) \[ \frac{d}{dt} (PM) = r \cos \theta \cdot \frac{d\theta}{dt}. \] Substituting values: \[ \frac{d}{dt} (PM) = 8 \cdot \frac{6}{2} = 24. \]