The equation of the normal to the curve x=the | vedclass

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

Question

(B) Given the parametric equations of the curve: $x=	heta+\sin 	heta$ and $y=1+\cos 	heta$. First,find the derivative $\frac{dy}{dx}$ with respect

Vedclass · Accepted Answer

$2x-2y-\pi=0$

Vedclass · Answer

(B) Given the parametric equations of the curve: $x=	heta+\sin 	heta$ and $y=1+\cos 	heta$. First,find the derivative $\frac{dy}{dx}$ with respect to $	heta$: $\frac{dx}{d	heta} = 1+\cos 	heta$ and $\frac{dy}{d	heta} = -\sin 	heta$. Thus,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-\sin 	heta}{1+\cos 	heta}$. At $	heta = \frac{\pi}{2}$,the slope of the tangent is $m_t = \frac{-\sin(\pi/2)}{1+\cos(\pi/2)} = \frac{-1}{1+0} = -1$. The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$. Now,find the point $(x, y)$ at $	heta = \frac{\pi}{2}$: $x = \frac{\pi}{2} + \sin(\frac{\pi}{2}) = \frac{\pi}{2} + 1$ and $y = 1 + \cos(\frac{\pi}{2}) = 1 + 0 = 1$. The equation of the normal is $(y - y_1) = m_n(x - x_1)$: $(y - 1) = 1(x - (1 + \frac{\pi}{2}))$. $y - 1 = x - 1 - \frac{\pi}{2}$. $x - y - \frac{\pi}{2} = 0$,which simplifies to $2x - 2y - \pi = 0$.

The equation of the normal to the curve $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\frac{\pi}{2}$ is

Similar Questions

For $x \neq -1, y \neq -1$, if $x = \frac{1 - \sqrt[3]{y}}{1 + \sqrt[3]{y}}$, then $\frac{dx}{dy} =$

If $x = \sin^{-1}(3t - 4t^3)$ and $y = \cos^{-1}(\sqrt{1 - t^2})$,then $\frac{dy}{dx}$ is equal to

If $x=e^{\theta} \sin \theta$ and $y=e^{\theta} \cos \theta$,where $\theta$ is a parameter,then $\frac{dy}{dx}$ at $(1,1)$ is equal to

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

If,for $a \neq 0$,$x = a(1 - \sin t)$ and $y = a(t + \cos t)$,then $\frac{d^2 y}{d x^2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module