ISI2015-MMA-86

Question

The coordinates of a moving point $P$ satisfy the equations $$\frac{dx}{dt} = \tan x, \:\:\:\: \frac{dy}{dt}=-\sin^2x, \:\:\:\:\: t \geq 0.$$ If the curve passes through the point $(\pi/2, 0)$ when $t=0$, then the equation of the curve in rectangular co-ordinates is

• A. $y=1/2 \cos ^2 x$
• B. $y=\sin 2x$
• C. $y=\cos 2x+1$
• D. $y=\sin ^2 x-1$

Answer 1

$\underline{\mathbf{Answer: A}}$

$\underline{\mathbf{Solution:}}$

$\begin{align} \mathrm{\frac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = -\frac{\sin^2x}{\tan x} =-\frac{1}{2}\sin2x }\end{align}$

$\therefore \mathrm{y = \frac{1}{4}\cos2x + c}$

It is given that the curve passes through the points $\left ( \frac{\pi}{2}, 0\right )$

$\therefore \mathrm {c = \frac{1}{4}}$

$\therefore$ The curve would be :

$\mathrm{y = \frac{1}{4} \cos2x + \frac{1}{4}}$

or,

$\mathrm{4y = 1 + \cos 2x = 1 + 2\cos^2x -1 = 2 \cos^2x }$

$\mathrm{\Rightarrow y = \frac{1}{2} \cos^2x}$

$\therefore \mathbf A$ is the correct option.

Comments

Excellent solution.

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Thanks.