A is a point on the circle with radius 8 and centre 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 $O A$ and $\angle \mathrm{POM}= heta$. When $O M=4$ and $\frac{d heta}{d t}=6$ radians $/ \mathrm{sec}$, then…

Question

A is a point on the circle with radius 8 and centre 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 $O A$ and $\angle \mathrm{POM}=	heta$. When $O M=4$ and $\frac{d 	heta}{d t}=6$ radians $/ \mathrm{sec}$, then the rate of change of PM is (in units/sec)

Aicharya · Accepted Answer

When $O M=4 \Rightarrow \cos 	heta=\frac{O M}{O P}=\frac{1}{2}$ $\begin{aligned} & 	ext { Now, } P M=O P \sin 	heta \Rightarrow P M=8 \sin 	heta \ & \Rightarrow \frac{d(P M)}{d t}=8 \cos 	heta \frac{d 	heta}{d t} \ & \Rightarrow 8 	imes \frac{1}{2} 	imes 6=24\end{aligned}$

A is a point on the circle with radius 8 and centre at $O$. A particle P is moving on the circumference of…

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