For the curve y = 3 sin theta cos theta,x = e | vedclass

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(C) Given,$y = 3 \sin 	heta \cos 	heta = \frac{3}{2} \sin 2	heta$. $\frac{dy}{d	heta} = \frac{3}{2} \cdot 2 \cos 2	heta = 3 \cos 2	heta$. Given

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$\frac{\pi}{4}$

Vedclass · Answer

(C) Given,$y = 3 \sin 	heta \cos 	heta = \frac{3}{2} \sin 2	heta$. $\frac{dy}{d	heta} = \frac{3}{2} \cdot 2 \cos 2	heta = 3 \cos 2	heta$. Given $x = e^{	heta} \sin 	heta$. $\frac{dx}{d	heta} = e^{	heta} \sin 	heta + e^{	heta} \cos 	heta = e^{	heta} (\sin 	heta + \cos 	heta)$. Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3 \cos 2	heta}{e^{	heta} (\sin 	heta + \cos 	heta)}$. The tangent is parallel to the $x-$axis when $\frac{dy}{dx} = 0$,which implies $3 \cos 2	heta = 0$. $\cos 2	heta = 0 \Rightarrow 2	heta = \frac{\pi}{2}$ (since $0 \leq 	heta \leq \pi$,$0 \leq 2	heta \leq 2\pi$). $2	heta = \frac{\pi}{2}$ or $2	heta = \frac{3\pi}{2}$. $	heta = \frac{\pi}{4}$ or $	heta = \frac{3\pi}{4}$. Checking the denominator $e^{	heta} (\sin 	heta + \cos 	heta)$ at $	heta = \frac{3\pi}{4}$: $\sin \frac{3\pi}{4} + \cos \frac{3\pi}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$. Since the denominator becomes zero at $	heta = \frac{3\pi}{4}$,the derivative is undefined. Thus,the only valid solution is $	heta = \frac{\pi}{4}$.

For the curve $y = 3 \sin \theta \cos \theta$,$x = e^{\theta} \sin \theta$,$0 \leq \theta \leq \pi$,the tangent is parallel to the $x-$axis when $\theta$ is

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