Question

If $x=2\cos t+\cos 2t$, $y=2\sin t-\sin 2t$ then $\frac{dy}{dx}$ at $t=\frac{\pi }{4}$ is

• A. $1-\sqrt{2}$
• B. $-(1+\sqrt{2})$
• C. $\sqrt{2}$
• D. $1/\sqrt{2}$

Answer 1

Answer: (A)

$1-\sqrt{2}$

Given $x=2\cos t+\cos 2t$

derivative will be $\frac{d}{dt}\left(x\right)=\frac{d}{dt}(2\cos t+\cos 2t)=-2\sin t-2\sin 2t$

$y=2\sin t-\sin 2t$

$\frac{d}{dt}\left(y\right)=\frac{d}{dt}(2\sin t-\sin 2t)=2\cos t-2\cos 2t$

$\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

$=\frac{2\cos t-2\cos 2t}{-2\sin t-2\sin 2t}$

${\left(\frac{dy}{dx}\right)}_{t=\frac{\pi }{4}}=\frac{2\times \frac{1}{\sqrt{2}}-2\times 0}{-2\times \frac{1}{\sqrt{2}}-2\times 1}$

$=\frac{\sqrt{2}}{-2-\sqrt{2}}$

$=\frac{-1}{1+\sqrt{2}}$
$=1-\sqrt{2}$ ....... [By rationalization]