If x=e^{theta} sin theta and y=e^{theta} cos | vedclass

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

Question

(A) Given,$x=e^{	heta} \sin 	heta$ and $y=e^{	heta} \cos 	heta$. We differentiate both with respect to $	heta$: $\frac{dx}{d	heta} = e^{	heta}

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given,$x=e^{	heta} \sin 	heta$ and $y=e^{	heta} \cos 	heta$. We differentiate both with respect to $	heta$: $\frac{dx}{d	heta} = e^{	heta} \sin 	heta + e^{	heta} \cos 	heta = e^{	heta}(\sin 	heta + \cos 	heta)$ $\frac{dy}{d	heta} = e^{	heta} \cos 	heta - e^{	heta} \sin 	heta = e^{	heta}(\cos 	heta - \sin 	heta)$ Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{e^{	heta}(\cos 	heta - \sin 	heta)}{e^{	heta}(\cos 	heta + \sin 	heta)} = \frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta}$. Dividing numerator and denominator by $\cos 	heta$,we get $\frac{dy}{dx} = \frac{1 - 	an 	heta}{1 + 	an 	heta}$. At $(x, y) = (1, 1)$,we have $x = e^{	heta} \sin 	heta = 1$ and $y = e^{	heta} \cos 	heta = 1$. Dividing these,$\frac{y}{x} = \frac{e^{	heta} \cos 	heta}{e^{	heta} \sin 	heta} = \cot 	heta = 1$,so $	an 	heta = 1$. Substituting $	an 	heta = 1$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$.

If $x=e^{\theta} \sin \theta$ and $y=e^{\theta} \cos \theta$,where $\theta$ is a parameter,then $\frac{dy}{dx}$ at $(1,1)$ is equal to

Similar Questions

If $x = \sin t$ and $y = \cos pt$,then which of the following is true?

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

Find $\frac{dy}{dx}$,if $x = 2 \cos \theta - \cos 2 \theta$ and $y = 2 \sin \theta - \sin 2 \theta$.

If $x = at^2$ and $y = 2at$,then $\frac{dy}{dx} =$ . . . . . . ,where $t \neq 0$.

If $x=f(t)$ and $y=g(t)$,then the value of $\frac{d^{2} y}{d x^{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module