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

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(B) Given the parametric equations of the curve: $x=	heta+\sin 	heta$ and $y=1+\cos 	heta$. First,find the derivative $\frac{dy}{dx}$ with respect

Vedclass · Accepted Answer

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

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(B) Given the parametric equations of the curve: $x=	heta+\sin 	heta$ and $y=1+\cos 	heta$. First,find the derivative $\frac{dy}{dx}$ with respect to $	heta$: $\frac{dx}{d	heta} = 1+\cos 	heta$ and $\frac{dy}{d	heta} = -\sin 	heta$. Thus,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-\sin 	heta}{1+\cos 	heta}$. At $	heta = \frac{\pi}{2}$,the slope of the tangent is $m_t = \frac{-\sin(\pi/2)}{1+\cos(\pi/2)} = \frac{-1}{1+0} = -1$. The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$. Now,find the point $(x, y)$ at $	heta = \frac{\pi}{2}$: $x = \frac{\pi}{2} + \sin(\frac{\pi}{2}) = \frac{\pi}{2} + 1$ and $y = 1 + \cos(\frac{\pi}{2}) = 1 + 0 = 1$. The equation of the normal is $(y - y_1) = m_n(x - x_1)$: $(y - 1) = 1(x - (1 + \frac{\pi}{2}))$. $y - 1 = x - 1 - \frac{\pi}{2}$. $x - y - \frac{\pi}{2} = 0$,which simplifies to $2x - 2y - \pi = 0$.

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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