Question

If $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1+\cos 2t)$, then find $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$.

• A. $-\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $-\frac{b}{a}$
• D. None of these

Answer 1

The correct option is B $\frac{b}{a}$

Given, $x=a\sin 2t(1+\cos 2t)$
and $y=b\cos 2t(1+\cos 2t)$
Differentiate $x$ w.r.t. $t$,

$\frac{dx}{dt}=2a\cos 2t(1+\cos 2t)+a\sin 2t(-2\sin 2t)$
$\frac{dx}{dt}=2a\cos 2t+2a{\cos }^{2}2t-2a{\sin }^{2}2t$
$\frac{dx}{dt}=2a\cos 2t+2a\cos 4t$
Now, $\frac{dy}{dt}=-2b\sin 2t(1+\cos 2t)+b\cos 2t(-2\sin 2t)$
$\frac{dy}{dt}=-2b\sin 2t-2b\sin 2t\cos 2t-2b\sin 2t\cos 2t$
$\frac{dy}{dt}=-2b\sin 2t-2b\sin 4t$
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-2b[\sin 2t+\sin 4t]}{2a[\cos 2t+\cos 4t]}$
${\left[\frac{dy}{dx}\right]}_{t=\pi /4}=\frac{-b}{a}\left[\frac{\sin \pi /2+\sin \pi }{\cos \pi /2+\cos \pi }\right]$
$=\frac{-b}{a}\left[\frac{1+0}{0-1}\right]$
$=\frac{b}{a}$