If x and y are connected parametrically by th | vedclass

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

Question

(A) Given equations are $x = a(\cos 	heta + 	heta \sin 	heta)$ and $y = a(\sin 	heta - 	heta \cos 	heta)$.First,differentiate $x$ with respect t

Vedclass · Accepted Answer

$	an 	heta$

Vedclass · Answer

(A) Given equations are $x = a(\cos 	heta + 	heta \sin 	heta)$ and $y = a(\sin 	heta - 	heta \cos 	heta)$.First,differentiate $x$ with respect to $	heta$:$\frac{dx}{d	heta} = a \left[ \frac{d}{d	heta}(\cos 	heta) + \frac{d}{d	heta}(	heta \sin 	heta) ight]$$= a [-\sin 	heta + (	heta \cos 	heta + \sin 	heta \cdot 1)]$$= a [-\sin 	heta + 	heta \cos 	heta + \sin 	heta] = a 	heta \cos 	heta$.Next,differentiate $y$ with respect to $	heta$:$\frac{dy}{d	heta} = a \left[ \frac{d}{d	heta}(\sin 	heta) - \frac{d}{d	heta}(	heta \cos 	heta) ight]$$= a [\cos 	heta - (	heta(-\sin 	heta) + \cos 	heta \cdot 1)]$$= a [\cos 	heta + 	heta \sin 	heta - \cos 	heta] = a 	heta \sin 	heta$.Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a 	heta \sin 	heta}{a 	heta \cos 	heta} = 	an 	heta$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$.

Similar Questions

If $x=\cos ^3 \theta-\sin ^3 \theta$ and $y=\sqrt[3]{\cos \theta}-\sqrt[3]{\sin \theta}$,then the value of $\frac{d y}{d x}$ at $\theta=\frac{\pi}{4}$ is

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{dy}{dx} = $

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

At any two points of the curve represented parametrically by $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$,the tangents are parallel to the $x$-axis. The values of the parameter $t$ corresponding to these points differ from each other by:

If $x=3 \tan t$ and $y=3 \sec t$,then the value of $\frac{d^2 y}{d x^2}$ at $t=\frac{\pi}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module