Question

If $x = a(1-cost)$, $y = a(t + sint)$, then match each entry in List-I to the correct entries in List-II.

	List-I		List-II
(P)	$\frac{dy}{dx}$	(1)	$tan \frac{t}{2}$
(Q)	$\frac{d^2y}{dx^2}$	(2)	$\frac{1}{4a} sec^4 \frac{t}{2}$
(R)	$\frac{dx}{dy}$	(3)	$cot \frac{t}{2}$
(S)	$\frac{d^2x}{dy^2}$	(4)	$\frac{-1}{4a} (sec \frac{t}{2} cosec^3 \frac{t}{2})$
		(5)	$sin (\frac{t}{2})$

• A. (P) $\leftrightarrow$ (3), (Q) $\leftrightarrow$ (4), (R) $\leftrightarrow$ (1), (S) $\leftrightarrow$ (2)
• B. (P) $\leftrightarrow$ (1), (Q) $\leftrightarrow$ (2), (R) $\leftrightarrow$ (3), (S) $\leftrightarrow$ (4)
• C. (P) $\leftrightarrow$ (3), (Q) $\leftrightarrow$ (1), (R) $\leftrightarrow$ (4), (S) $\leftrightarrow$ (2)
• D. (P) $\leftrightarrow$ (4), (Q) $\leftrightarrow$ (3), (R) $\leftrightarrow$ (2), (S) $\leftrightarrow$ (1)

Answer 1

Answer: (A)

(P) $\leftrightarrow$ (3), (Q) $\leftrightarrow$ (4), (R) $\leftrightarrow$ (1), (S) $\leftrightarrow$ (2)

Answer 2

Given $x = a(1-\cos t)$ and $y = a(t+\sin t)$. We find $\frac{dx}{dt} = a \sin t$ and $\frac{dy}{dt} = a(1+\cos t)$.

(P) $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a(1+\cos t)}{a \sin t} = \frac{2\cos^2(t/2)}{2\sin(t/2)\cos(t/2)} = \cot(t/2)$. Matches (3).

(R) $\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{a \sin t}{a(1+\cos t)} = \frac{2\sin(t/2)\cos(t/2)}{2\cos^2(t/2)} = \tan(t/2)$. Matches (1).

(Q) $\frac{d^2y}{dx^2} = \frac{d}{dx}(\cot(t/2)) = \frac{d}{dt}(\cot(t/2)) \cdot \frac{dt}{dx}$. $\frac{d}{dt}(\cot(t/2)) = -\frac{1}{2}\csc^2(t/2)$. $\frac{dt}{dx} = \frac{1}{a \sin t} = \frac{1}{2a\sin(t/2)\cos(t/2)}$. So, $\frac{d^2y}{dx^2} = -\frac{1}{2}\csc^2(t/2) \cdot \frac{1}{2a\sin(t/2)\cos(t/2)} = -\frac{1}{4a\sin^3(t/2)\cos(t/2)} = -\frac{1}{4a}\csc^3(t/2)\sec(t/2)$. Matches (4).

(S) $\frac{d^2x}{dy^2} = \frac{d}{dy}(\tan(t/2)) = \frac{d}{dt}(\tan(t/2)) \cdot \frac{dt}{dy}$. $\frac{d}{dt}(\tan(t/2)) = \frac{1}{2}\sec^2(t/2)$. $\frac{dt}{dy} = \frac{1}{a(1+\cos t)} = \frac{1}{2a\cos^2(t/2)}$. So, $\frac{d^2x}{dy^2} = \frac{1}{2}\sec^2(t/2) \cdot \frac{1}{2a\cos^2(t/2)} = \frac{1}{4a\cos^4(t/2)} = \frac{1}{4a}\sec^4(t/2)$. Matches (2).

Matches: (P)-(3), (Q)-(4), (R)-(1), (S)-(2).