The equation of the normal to the curve x = t | vedclass

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Question

(C) Given the parametric equations of the curve: $x = 	heta + \sin 	heta$ and $y = 1 + \cos 	heta$.First,find the derivatives with respect to $	he

Vedclass · Accepted Answer

$2x - 2y - \pi = 0$

Vedclass · Answer

(C) Given the parametric equations of the curve: $x = 	heta + \sin 	heta$ and $y = 1 + \cos 	heta$.First,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = 1 + \cos 	heta$ and $\frac{dy}{d	heta} = -\sin 	heta$.The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-\sin 	heta}{1 + \cos 	heta}$.At $	heta = \frac{\pi}{2}$,the point $(x, y)$ is:$x = \frac{\pi}{2} + \sin \frac{\pi}{2} = \frac{\pi}{2} + 1$ and $y = 1 + \cos \frac{\pi}{2} = 1$.The slope of the tangent at $	heta = \frac{\pi}{2}$ is $m_t = \frac{-\sin(\pi/2)}{1 + \cos(\pi/2)} = \frac{-1}{1 + 0} = -1$.The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.The equation of the normal at $(\frac{\pi}{2} + 1, 1)$ is:$y - 1 = 1(x - (\frac{\pi}{2} + 1))$$y - 1 = x - \frac{\pi}{2} - 1$$x - y - \frac{\pi}{2} = 0$,which can be written as $2x - 2y - \pi = 0$.

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

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