A is a point on the circle with radius 8 and | vedclass

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(B) Given that the radius of the circle is $OP = 8$.$M$ is the foot of the perpendicular from $P$ on $OA$,so $	riangle OMP$ is a right-angled triangl

Vedclass · Accepted Answer

$24$

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(B) Given that the radius of the circle is $OP = 8$.$M$ is the foot of the perpendicular from $P$ on $OA$,so $	riangle OMP$ is a right-angled triangle at $M$.In $	riangle OMP$,we have $\cos 	heta = \frac{OM}{OP}$.Given $OM = 4$ and $OP = 8$,so $\cos 	heta = \frac{4}{8} = \frac{1}{2}$.Also,$PM = OP \sin 	heta = 8 \sin 	heta$.To find the rate of change of $PM$ with respect to time $t$,we differentiate $PM$ with respect to $t$:$\frac{d(PM)}{dt} = \frac{d}{dt}(8 \sin 	heta) = 8 \cos 	heta \frac{d	heta}{dt}$.Given $\frac{d	heta}{dt} = 6 	ext{ rad/sec}$ and $\cos 	heta = \frac{1}{2}$,we substitute these values:$\frac{d(PM)}{dt} = 8 	imes \frac{1}{2} 	imes 6 = 24 	ext{ units/sec}$.

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