If x = 2cos t - cos 2t and y = 2sin t - sin 2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$.Differentiating with respect to $t$:$\frac{dx}{dt} = -2\sin t + 2\sin 2t$$\frac{dy}{dt}

Vedclass · Accepted Answer

$\sqrt{2} + 1$

Vedclass · Answer

(A) Given $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$.Differentiating with respect to $t$:$\frac{dx}{dt} = -2\sin t + 2\sin 2t$$\frac{dy}{dt} = 2\cos t - 2\cos 2t$Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos t - 2\cos 2t}{-2\sin t + 2\sin 2t} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$.At $t = \frac{\pi}{4}$:$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\cos \frac{\pi}{2} = 0$,$\sin \frac{\pi}{2} = 1$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.Substituting these values:$\frac{dy}{dx} = \frac{\frac{1}{\sqrt{2}} - 0}{1 - \frac{1}{\sqrt{2}}} = \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{2}-1}{\sqrt{2}}} = \frac{1}{\sqrt{2}-1}$.Rationalizing the denominator:$\frac{1}{\sqrt{2}-1} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2} + 1$.

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

Similar Questions

Find $\frac{dy}{dx}$,if $x = 2 \cos \theta - \cos 2 \theta$ and $y = 2 \sin \theta - \sin 2 \theta$.

If $x = \sin(t)$, $y = \cos(pt)$, then $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = $

If $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at a point $P$ on the curve $y = f(x)$, then $\left(\frac{d^2x}{dy^2}\right)_P = $

If $x=a\left\{\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right\}$ and $y=a \sin \theta$, then $\frac{dy}{dx}$ is equal to

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module