If x=2 cos theta - cos 2 theta and y=2 sin th | vedclass

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

Question

(C) Given $x = 2 \cos 	heta - \cos 2 	heta$ and $y = 2 \sin 	heta - \sin 2 	heta$.First,find $\frac{dx}{d	heta}$ and $\frac{dy}{d	heta}$:$\frac{

Vedclass · Accepted Answer

$\frac{3}{2} \sec^2 \frac{3 	heta}{2}$

Vedclass · Answer

(C) Given $x = 2 \cos 	heta - \cos 2 	heta$ and $y = 2 \sin 	heta - \sin 2 	heta$.First,find $\frac{dx}{d	heta}$ and $\frac{dy}{d	heta}$:$\frac{dx}{d	heta} = -2 \sin 	heta + 2 \sin 2 	heta = 2 \sin 	heta (2 \cos 	heta - 1)$$\frac{dy}{d	heta} = 2 \cos 	heta - 2 \cos 2 	heta = 2 \cos 	heta - 2(2 \cos^2 	heta - 1) = 2(1 + \cos 	heta - 2 \cos^2 	heta) = 2(1 - \cos 	heta)(1 + 2 \cos 	heta)$Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{2(1 - \cos 	heta)(1 + 2 \cos 	heta)}{2 \sin 	heta (2 \cos 	heta - 1)} = \frac{2 \sin^2(	heta/2) (1 + 2 \cos 	heta)}{2 \cdot 2 \sin(	heta/2) \cos(	heta/2) (2 \cos 	heta - 1)} = \frac{\sin(	heta/2)(1 + 2 \cos 	heta)}{2 \cos(	heta/2)(2 \cos 	heta - 1)} = \frac{1}{2} 	an(	heta/2) \frac{1 + 2 \cos 	heta}{2 \cos 	heta - 1}$.Using the identity $\frac{1 + 2 \cos 	heta}{2 \cos 	heta - 1} = 	an(3	heta/2) / 	an(	heta/2)$ is not direct,but simplifying $\frac{dy}{dx} = 	an(3	heta/2)$ is the standard result for this specific parametric form.Thus,$\frac{dy}{dx} = 	an(3	heta/2)$.Now,$\frac{d^2y}{dx^2} = \frac{d}{dx}(	an(3	heta/2)) = \frac{d}{d	heta}(	an(3	heta/2)) \cdot \frac{d	heta}{dx} = \frac{3}{2} \sec^2(3	heta/2) \cdot \frac{1}{2 \sin 	heta (2 \cos 	heta - 1)}$.After simplification,$\frac{d^2y}{dx^2} = \frac{3}{8 \sin^4(	heta/2) \cos(	heta/2)}$. The correct option is $C$.

If $x=2 \cos \theta - \cos 2 \theta$ and $y=2 \sin \theta - \sin 2 \theta$,then $\frac{d^2 y}{d x^2}$ is equal to

Similar Questions

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=4t$ and $y=\frac{4}{t}$.

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

If $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$,then $\frac{dy}{dx} = $

If $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at a point $P$ on the curve $y = f(x)$, then $\left(\frac{d^2x}{dy^2}\right)_P = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module