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

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

Question

(D) Given the parametric equations: $x = 1 - a \sin 	heta$ and $y = b \cos^2 	heta$. First,find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx

Vedclass · Accepted Answer

$-a/2b$

Vedclass · Answer

(D) Given the parametric equations: $x = 1 - a \sin 	heta$ and $y = b \cos^2 	heta$. First,find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$. $\frac{dx}{d	heta} = \frac{d}{d	heta}(1 - a \sin 	heta) = -a \cos 	heta$. $\frac{dy}{d	heta} = \frac{d}{d	heta}(b \cos^2 	heta) = b(2 \cos 	heta)(-\sin 	heta) = -2b \cos 	heta \sin 	heta$. Therefore,$\frac{dy}{dx} = \frac{-2b \cos 	heta \sin 	heta}{-a \cos 	heta} = \frac{2b}{a} \sin 	heta$. At $	heta = \pi / 2$,the slope of the tangent is $m_t = \frac{2b}{a} \sin(\pi / 2) = \frac{2b}{a}$. The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{2b/a} = -\frac{a}{2b}$.

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

Similar Questions

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

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

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

If $x = a \sin \theta$ and $y = b \cos \theta$, then find $\frac{d^2y}{dx^2}$.

The derivative of $\cot ^{-1} x$ with respect to $\log (1+x^{2})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module