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

(A) Given the parametric equations of the curve: $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$. First,find the derivative $\frac{dy}{dx}$ using the

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the parametric equations of the curve: $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$. First,find the derivative $\frac{dy}{dx}$ using the chain rule: $\frac{dx}{d	heta} = 3a \cos^2 	heta (-\sin 	heta) = -3a \cos^2 	heta \sin 	heta$ $\frac{dy}{d	heta} = 3a \sin^2 	heta (\cos 	heta) = 3a \sin^2 	heta \cos 	heta$ $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^2 	heta \cos 	heta}{-3a \cos^2 	heta \sin 	heta} = -	an 	heta$ At $	heta = \pi / 4$,the slope of the tangent is $m_t = -	an(\pi / 4) = -1$. The slope of the normal $m_n$ is given by $-\frac{1}{m_t}$. $m_n = -\frac{1}{-1} = 1$.

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

Similar Questions

If $x = a \cos \theta$ and $y = b \sin \theta$,then find the value of $\left[\frac{d^2 y}{d x^2}\right]_{\theta = \frac{\pi}{4}}$.

If the parametric equations of a curve are given by $x = \cos \theta + \log \tan \frac{\theta}{2}$ and $y = \sin \theta$,then the points for which $\frac{dy}{dx} = 0$ are given by

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

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

Find $\frac{dy}{dx}$,if $y=12(1-\cos t)$ and $x=10(t-\sin t)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module