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

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

Question

(B) Given,$x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$.On differentiating $x$ and $y$ with respect to $t$,we get:$\frac{dx}{dt} = \frac{(

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given,$x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$.On differentiating $x$ and $y$ with respect to $t$,we get:$\frac{dx}{dt} = \frac{(1+t^{2})(-2t) - (1-t^{2})(2t)}{(1+t^{2})^{2}} = \frac{-2t - 2t^{3} - 2t + 2t^{3}}{(1+t^{2})^{2}} = \frac{-4t}{(1+t^{2})^{2}}$.$\frac{dy}{dt} = \frac{(1+t^{2})(2a) - (2at)(2t)}{(1+t^{2})^{2}} = \frac{2a + 2at^{2} - 4at^{2}}{(1+t^{2})^{2}} = \frac{2a(1-t^{2})}{(1+t^{2})^{2}}$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a(1-t^{2})}{(1+t^{2})^{2}} 	imes \frac{(1+t^{2})^{2}}{-4t}$.$\frac{dy}{dx} = \frac{2a(1-t^{2})}{-4t} = \frac{a(1-t^{2})}{-2t} = \frac{a(t^{2}-1)}{2t}$.

If $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$,then $\frac{d y}{d x}$ is equal to

Similar Questions

The slope of the tangent to the curve $x=t^{2}+3t-8$,$y=2t^{2}-2t-5$ at the point $(2,-1)$ is

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to

If $x = at^2$ and $y = 2at$,then find $\frac{d^2 x}{dy^2}$.

The $2^{nd}$ derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is

The derivative of $\cos^{3} x$ with respect to $\sin^{3} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module