Question

A is an end of diameter of a circle with centre $O$ and the radius $2$ units. If a particle ${}^{\prime }{p}^{\prime }$ starting from $A$, moves on a circle with a regular velocity $4radians/sec$ and $M$ is the foot of the perpendicular of ${}^{\prime }{p}^{\prime }$ on the diameter, then the rate at which $M$ is moving on the diamter, when it is at a distance of $1units$ from $O$ is

• A. $4\sqrt{3}units/sec$
• B. $-4\sqrt{3}units/sec$
• C. $2\sqrt{3}units/sec$
• D. $-4units/sec$

Answer 1

Answer: (A), (B)

The correct options are
A $4\sqrt{3}units/sec$
B $-4\sqrt{3}units/sec$

$p$ starts from point $A$ on the circumference such that perpendicular on the diameter such that $M$ is foot of perpendicular.

Given that speed of $p$, $\frac{d\theta }{dt}=4rad/s$

$M$ travels a distance of $x$ in time $t$ then

$OM=x$

Now, in $△OpM$

$\cos \theta =\frac{OM}{Op}$

$\cos \theta =\frac{x}{2}$

$\theta ={\cos }^{-1}\frac{x}{2}$

On differentiating with respect to $x$, we get

$\frac{d\theta }{dx}=\frac{-1}{\sqrt{1-{\left(\frac{x}{2}\right)}^2}}\times \left(\frac{1}{2}\right)$

$\frac{d\theta }{dx}=\frac{-1}{2\sqrt{\frac{4-x^2}{4}}}$

$\frac{d\theta }{dx}=\frac{\mp 1}{\sqrt{(4-x^2)}}$

$\frac{d\theta }{dt}.\frac{dt}{dx}=\frac{\mp 1}{\sqrt{(4-x^2)}}$

$4.\frac{dt}{dx}=\frac{\mp 1}{\sqrt{(4-x^2)}}$

$\frac{dx}{dt}=\mp 4\sqrt{(4-x^2)}$

${\frac{dx}{dt}|}_{x=1}=\mp 4\sqrt{3}unit/s$

Hence, this is the required solution.