Match the values of frac{dy}{dx} at theta = f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We need to find $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$ at $	heta = \frac{\pi}{3}$.$(i)$ $x = a(	heta - \sin 	heta), y = a(1 - \cos 	h

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We need to find $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$ at $	heta = \frac{\pi}{3}$.$(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})$$\frac{dy}{dx} = \frac{2a \sin(	heta/2) \cos(	heta/2)}{2a \sin^2(	heta/2)} = \cot(\frac{	heta}{2})$. At $	heta = \frac{\pi}{3}$, $\frac{dy}{dx} = \cot(\frac{\pi}{6}) = \sqrt{3}$. $(i)$ $ightarrow$ $C$.(ii) $x = 3\cos 	heta - 2\cos^3 	heta = \cos(3	heta), y = 3\sin 	heta - 2\sin^3 	heta = \sin(3	heta)$ (Wait, the given equations are $x = 3\cos 	heta - 2\cos^3 	heta$ and $y = 3\sin 	heta - 2\sin^3 	heta$. Let's differentiate directly).$\frac{dx}{d	heta} = -3\sin 	heta + 6\cos^2 	heta \sin 	heta = 3\sin 	heta(2\cos^2 	heta - 1) = 3\sin 	heta \cos(2	heta)$$\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)$$\frac{dy}{dx} = \frac{3\cos 	heta \cos(2	heta)}{3\sin 	heta \cos(2	heta)} = \cot 	heta$. At $	heta = \frac{\pi}{3}$, $\frac{dy}{dx} = \cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}$. (ii) $ightarrow$ $D$.(iii) $x = 3\cos 	heta - \cos^3 	heta, y = 3\sin 	heta - \sin^3 	heta$$\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$. At $	heta = \frac{\pi}{3}$, $\frac{dy}{dx} = -(\cot(\frac{\pi}{3}))^3 = -(\frac{1}{\sqrt{3}})^3 = -\frac{1}{3\sqrt{3}}$. (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$. At $	heta = \frac{\pi}{3}$, $\frac{dy}{dx} = 	an(\frac{\pi}{3}) \sec^2(\frac{\pi}{3}) = \sqrt{3} \cdot (2)^2 = 4\sqrt{3}$. (iv) $ightarrow$ $A$.Thus, $(i)$ $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) $x = a \log \sin \theta, y = a \tan \theta$	$(D)$ $\frac{1}{\sqrt{3}}$
	$(E)$ $\frac{1}{3\sqrt{3}}$

Similar Questions

What is the slope of the tangent to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $t = 2$?

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

If $x = \sin(t)$, $y = \cos(pt)$, then $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = $

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then find $\frac{dy}{dx}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module