List-I માં આપેલ પ્રચલિત સ્વરૂપમાં વક્રોની સિસ | vedclass

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(C) આપણે $	heta = \frac{\pi}{3}$ પર $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$ શોધવાની જરૂર છે.$(i)$ $x = a(	heta - \sin 	heta), y = a(1 - \co

Vedclass · Accepted Answer

$(i)$ $\rightarrow$ $D$, (ii) $\rightarrow$ $C$, (iii) $\rightarrow$ $B$, (iv) $\rightarrow$ $A$

Vedclass · Answer

(C) આપણે $	heta = \frac{\pi}{3}$ પર $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$ શોધવાની જરૂર છે.$(i)$ $x = a(	heta - \sin 	heta), y = a(1 - \cos 	heta)$$\frac{dx}{d	heta} = a(1 - \cos 	heta) = 2a \sin^2(\frac{	heta}{2})$$\frac{dy}{d	heta} = a \sin 	heta = 2a \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$$\frac{dy}{dx} = \frac{2a \sin(	heta/2) \cos(	heta/2)}{2a \sin^2(	heta/2)} = \cot(\frac{	heta}{2})$. $	heta = \frac{\pi}{3}$ પર, $\frac{dy}{dx} = \cot(\frac{\pi}{6}) = \sqrt{3}$. $(i)$ $ightarrow$ $C$.(ii) $x = 3\cos 	heta - 2\cos^3 	heta, y = 3\sin 	heta - 2\sin^3 	heta$$\frac{dx}{d	heta} = -3\sin 	heta + 6\cos^2 	heta \sin 	heta = 3\sin 	heta(2\cos^2 	heta - 1) = 3\sin 	heta \cos(2	heta)$$\frac{dy}{d	heta} = 3\cos 	heta - 6\sin^2 	heta \cos 	heta = 3\cos 	heta(1 - 2\sin^2 	heta) = 3\cos 	heta \cos(2	heta)$$\frac{dy}{dx} = \frac{3\cos 	heta \cos(2	heta)}{3\sin 	heta \cos(2	heta)} = \cot 	heta$. $	heta = \frac{\pi}{3}$ પર, $\frac{dy}{dx} = \cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}$. (ii) $ightarrow$ $D$.(iii) $x = 3\cos 	heta - \cos^3 	heta, y = 3\sin 	heta - \sin^3 	heta$$\frac{dx}{d	heta} = -3\sin 	heta + 3\cos^2 	heta \sin 	heta = -3\sin 	heta(1 - \cos^2 	heta) = -3\sin^3 	heta$$\frac{dy}{d	heta} = 3\cos 	heta - 3\sin^2 	heta \cos 	heta = 3\cos 	heta(1 - \sin^2 	heta) = 3\cos^3 	heta$$\frac{dy}{dx} = \frac{3\cos^3 	heta}{-3\sin^3 	heta} = -\cot^3 	heta$. $	heta = \frac{\pi}{3}$ પર, $\frac{dy}{dx} = -(\cot(\frac{\pi}{3}))^3 = -(\frac{1}{\sqrt{3}})^3 = -\frac{1}{3\sqrt{3}}$. (iii) $ightarrow$ $B$.(iv) $x = a \log \sin 	heta, y = a 	an 	heta$$\frac{dx}{d	heta} = a \cot 	heta, \frac{dy}{d	heta} = a \sec^2 	heta$$\frac{dy}{dx} = \frac{a \sec^2 	heta}{a \cot 	heta} = \frac{1}{\cos^2 	heta} \cdot \frac{\sin 	heta}{\cos 	heta} = \frac{\sin 	heta}{\cos^3 	heta} = 	an 	heta \sec^2 	heta$. $	heta = \frac{\pi}{3}$ પર, $\frac{dy}{dx} = 	an(\frac{\pi}{3}) \sec^2(\frac{\pi}{3}) = \sqrt{3} \cdot (2)^2 = 4\sqrt{3}$. (iv) $ightarrow$ $A$.આમ, $(i)$ $ightarrow$ $C$, (ii) $ightarrow$ $D$, (iii) $ightarrow$ $B$, (iv) $ightarrow$ $A$.

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) $x = a \log \sin \theta, y = a \tan \theta$	$(D)$ $\frac{1}{\sqrt{3}}$
	$(E)$ $\frac{1}{3\sqrt{3}}$

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