Find the slope of the normal to the curve x=1 | vedclass

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(A) Given the parametric equations of the curve are $x = 1 - a \sin 	heta$ and $y = b \cos^{2} 	heta$. First,we find the derivatives with respect to

Vedclass · Accepted Answer

$-\frac{a}{2b}$

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(A) Given the parametric equations of the curve are $x = 1 - a \sin 	heta$ and $y = b \cos^{2} 	heta$. First,we find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = -a \cos 	heta$ $\frac{dy}{d	heta} = b \cdot 2 \cos 	heta \cdot (-\sin 	heta) = -2b \sin 	heta \cos 	heta$ Now,the slope of the tangent $\frac{dy}{dx}$ is: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-2b \sin 	heta \cos 	heta}{-a \cos 	heta} = \frac{2b}{a} \sin 	heta$ At $	heta = \frac{\pi}{2}$,the slope of the tangent is: $m_{tangent} = \frac{2b}{a} \sin(\frac{\pi}{2}) = \frac{2b}{a} \cdot 1 = \frac{2b}{a}$ The slope of the normal is the negative reciprocal of the slope of the tangent: $m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{2b/a} = -\frac{a}{2b}$

Find the slope of the normal to the curve $x=1-a \sin \theta, y=b \cos^{2} \theta$ at $\theta=\frac{\pi}{2}$.

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