Find the slope of the normal to the curve x=a | 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 $x=a \cos ^{3} 	heta$ and $y=a \sin ^{3} 	heta$. First,we find the derivatives with respect to $

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the parametric equations of the curve are $x=a \cos ^{3} 	heta$ and $y=a \sin ^{3} 	heta$. First,we find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = 3a \cos^{2} 	heta (-\sin 	heta) = -3a \cos^{2} 	heta \sin 	heta$ $\frac{dy}{d	heta} = 3a \sin^{2} 	heta (\cos 	heta) = 3a \sin^{2} 	heta \cos 	heta$ The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^{2} 	heta \cos 	heta}{-3a \cos^{2} 	heta \sin 	heta} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$. At $	heta = \frac{\pi}{4}$,the slope of the tangent is $m_{T} = -	an(\frac{\pi}{4}) = -1$. The slope of the normal $m_{N}$ is given by $-\frac{1}{m_{T}}$. Therefore,$m_{N} = -\frac{1}{-1} = 1$.

Find the slope of the normal to the curve $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ at $\theta=\frac{\pi}{4}$.

Similar Questions

If $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$,then at $t = \frac{\pi}{4}$,find $\frac{dy}{dx}$.

If $\tan x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$,then the value of $\frac{dy}{dx}$ is

The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ equals

The $2^{nd}$ derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is

If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$ and $\tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$,then the value of $\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module