Find the slope of the normal to the curve x=1 | vedclass

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

Question

(A) Given the parametric equations of the curve are $x = 1 - a \sin 	heta$ and $y = b \cos^{2} 	heta$. First,we find the derivatives with respect to

Vedclass · Accepted Answer

$-\frac{a}{2b}$

Vedclass · Answer

(A) Given the parametric equations of the curve are $x = 1 - a \sin 	heta$ and $y = b \cos^{2} 	heta$. First,we find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = -a \cos 	heta$ $\frac{dy}{d	heta} = b \cdot 2 \cos 	heta \cdot (-\sin 	heta) = -2b \sin 	heta \cos 	heta$ Now,the slope of the tangent $\frac{dy}{dx}$ is: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-2b \sin 	heta \cos 	heta}{-a \cos 	heta} = \frac{2b}{a} \sin 	heta$ At $	heta = \frac{\pi}{2}$,the slope of the tangent is: $m_{tangent} = \frac{2b}{a} \sin(\frac{\pi}{2}) = \frac{2b}{a} \cdot 1 = \frac{2b}{a}$ The slope of the normal is the negative reciprocal of the slope of the tangent: $m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{2b/a} = -\frac{a}{2b}$

Find the slope of the normal to the curve $x=1-a \sin \theta, y=b \cos^{2} \theta$ at $\theta=\frac{\pi}{2}$.

Similar Questions

If $x = a \cos \theta$ and $y = a \sin \theta$,then $\frac{d^2 y}{d x^2} =$ . . . . . . . (where $a \neq 0$ and $\theta \neq k \pi, k \in Z$)

If $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$,then $\left(\frac{d^2 y}{dx^2}\right)_{\theta=\pi / 2}=$

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

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

If $x = a(\cos \theta + \theta \sin \theta)$, $y = f(\theta)$, $f(2\pi) = 0$, $\frac{dy}{dx} = \frac{\tan \theta}{\theta}$, $\theta \neq 0$ and $\theta \neq (2n+1)\frac{\pi}{2}$, then $f\left(\frac{\pi}{3}\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module