If x = a(t + sin t) and y = a(1 - cos t),then | vedclass

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(A) Given the parametric equations: $x = a(t + \sin t)$ $y = a(1 - \cos t)$ We need to find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First,differentiate

Vedclass · Accepted Answer

$	an (t/2)$

Vedclass · Answer

(A) Given the parametric equations: $x = a(t + \sin t)$ $y = a(1 - \cos t)$ We need to find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First,differentiate $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}[a(1 - \cos t)] = a(0 - (-\sin t)) = a \sin t$ Next,differentiate $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}[a(t + \sin t)] = a(1 + \cos t)$ Now,calculate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t}$ Using trigonometric identities $\sin t = 2 \sin(t/2) \cos(t/2)$ and $1 + \cos t = 2 \cos^2(t/2)$: $\frac{dy}{dx} = \frac{2 \sin(t/2) \cos(t/2)}{2 \cos^2(t/2)} = \frac{\sin(t/2)}{\cos(t/2)} = 	an(t/2)$ Therefore,the correct option is $A$.

List-$I$	List-$II$
$(i)$ $x = a(\theta - \sin \theta), y = a(1 - \cos \theta)$	$(A)$ $4\sqrt{3}$
(ii) $x = 3\cos \theta - 2\cos^3 \theta, y = 3\sin \theta - 2\sin^3 \theta$	$(B)$ $-\frac{1}{3\sqrt{3}}$
(iii) $x = 3\cos \theta - \cos^3 \theta, y = 3\sin \theta - \sin^3 \theta$	$(C)$ $\sqrt{3}$
(iv) $x = a \log \sin \theta, y = a \tan \theta$	$(D)$ $\frac{1}{\sqrt{3}}$
	$(E)$ $\frac{1}{3\sqrt{3}}$

If $x = a(t + \sin t)$ and $y = a(1 - \cos t)$,then $\frac{dy}{dx}$ equals

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