A curve is given by the equations x = a cos t | vedclass

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(C) Given $x = a \cos 	heta + \frac{1}{2}b \cos 2	heta$ and $y = a \sin 	heta + \frac{1}{2}b \sin 2	heta$.First,find the derivatives with respect

Vedclass · Accepted Answer

$\cos 	heta = - \frac{a^2 + 2b^2}{3ab}$

Vedclass · Answer

(C) Given $x = a \cos 	heta + \frac{1}{2}b \cos 2	heta$ and $y = a \sin 	heta + \frac{1}{2}b \sin 2	heta$.First,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = -a \sin 	heta - b \sin 2	heta$$\frac{dy}{d	heta} = a \cos 	heta + b \cos 2	heta$Then,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \cos 	heta + b \cos 2	heta}{-a \sin 	heta - b \sin 2	heta}$.To find $\frac{d^2y}{dx^2}$,we use $\frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{d	heta}(\frac{dy}{dx}) \cdot \frac{d	heta}{dx}$.Setting $\frac{d^2y}{dx^2} = 0$ implies the numerator of $\frac{d}{d	heta}(\frac{dy}{dx})$ must be zero.After simplification,this leads to the condition:$a^2 + 2b^2 + 3ab(\cos 2	heta \cos 	heta + \sin 2	heta \sin 	heta) = 0$Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$:$a^2 + 2b^2 + 3ab \cos(2	heta - 	heta) = 0$$a^2 + 2b^2 + 3ab \cos 	heta = 0$Therefore,$\cos 	heta = - \frac{a^2 + 2b^2}{3ab}$.

$A$ curve is given by the equations $x = a \cos \theta + \frac{1}{2}b \cos 2\theta$ and $y = a \sin \theta + \frac{1}{2}b \sin 2\theta$. The points for which $\frac{d^2y}{dx^2} = 0$ are given by:

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