If x=a(theta+sin theta) and y=a(1-cos theta), | vedclass

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

Question

(B) Given $x = a(	heta + \sin 	heta)$ and $y = a(1 - \cos 	heta)$.First,differentiate $x$ with respect to $	heta$:$\frac{dx}{d	heta} = a(1 + \cos

Vedclass · Accepted Answer

$	an \frac{	heta}{2}$

Vedclass · Answer

(B) Given $x = a(	heta + \sin 	heta)$ and $y = a(1 - \cos 	heta)$.First,differentiate $x$ with respect to $	heta$:$\frac{dx}{d	heta} = a(1 + \cos 	heta)$.Next,differentiate $y$ with respect to $	heta$:$\frac{dy}{d	heta} = a(0 - (-\sin 	heta)) = a \sin 	heta$.Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \sin 	heta}{a(1 + \cos 	heta)} = \frac{\sin 	heta}{1 + \cos 	heta}$.Using trigonometric identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$:$\frac{dy}{dx} = \frac{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \cos^2 \frac{	heta}{2}} = \frac{\sin \frac{	heta}{2}}{\cos \frac{	heta}{2}} = 	an \frac{	heta}{2}$.Thus,the correct option is $B$.

If $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$,then $\frac{dy}{dx} = $ . . . . . . .

Similar Questions

The angle made by the tangent of the curve $x = a(t + \sin t \cos t)$; $y = a(1 + \sin t)^2$ with the $x$-axis at any point on it is

Find the rate of change of $\sqrt{x^2 + 16}$ with respect to $\frac{x}{x - 1}$ at $x = 3$.

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

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{d y}{d x}$:
$x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$

If $x=3 \tan t$ and $y=3 \sec t$,then the value of $\frac{d^2 y}{d x^2}$ at $t=\frac{\pi}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module