If x=frac{1-t^2}{1+t^2} and y=frac{2at}{1+t^2 | vedclass

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

Question

(D) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$.Substitute $t=	an 	heta$,then $x=\cos 2	heta$ and $y=a \sin 2	heta$.Differentiating wi

Vedclass · Accepted Answer

$\frac{a(t^2-1)}{2t}$

Vedclass · Answer

(D) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$.Substitute $t=	an 	heta$,then $x=\cos 2	heta$ and $y=a \sin 2	heta$.Differentiating with respect to $	heta$:$\frac{dx}{d	heta} = -2 \sin 2	heta$ and $\frac{dy}{d	heta} = 2a \cos 2	heta$.Using the chain rule,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{2a \cos 2	heta}{-2 \sin 2	heta} = -a \cot 2	heta$.Since $\cot 2	heta = \frac{1}{	an 2	heta} = \frac{1-	an^2 	heta}{2 	an 	heta} = \frac{1-t^2}{2t}$,$\frac{dy}{dx} = -a \left( \frac{1-t^2}{2t} ight) = \frac{a(t^2-1)}{2t}$.

If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$,then $\frac{dy}{dx}=$

Similar Questions

The derivative of $f(\sec x)$ with respect to $g(\tan x)$ at $x=\frac{\pi}{4}$,where $f^{\prime}(\sqrt{2})=4$ and $g^{\prime}(1)=2$,is

If $x = \cos 2t + \log(\tan t)$ and $y = 2t + \cot 2t$, then $\frac{dy}{dx} = $

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then:

If $x = a \sec^{2} \theta$ and $y = a \tan^{2} \theta$,then find $\frac{d^{2} y}{d x^{2}}$.

If $x = \log p$ and $y = \frac{1}{p}$, then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module