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 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 $x = a \sin^3 t$ and $y = a \cos^3 t$, then the value of $\frac{d^2y}{dx^2}$ at $t = \frac{\pi}{3}$ is equal to

Differentiate $\sin ^{2} x$ with respect to $e^{\cos x}$.

The slope of the tangent to the curve $x = 3t^2 + 1, y = t^3 - 1$ at $x = 1$ is

What is the slope of the normal to the curve $x = a(\theta - \sin \theta)$,$y = a(1 - \cos \theta)$ at the point $\theta = \pi / 2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module