Parametric Differentiation PYQs

$\frac{dx}{d\theta} = a(-\sin\theta + \sin\theta + \theta\cos\theta) = a\theta\cos\theta$.

$\frac{dy}{d\theta} = a(\cos\theta - (\cos\theta - \theta\sin\theta)) = a\theta\sin\theta$.

$\frac{dy}{dx} = \frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan \theta$.

$\frac{d^2y}{dx^2} = \frac{d}{d\theta}(\tan \theta) \cdot \frac{d\theta}{dx} = \sec^2 \theta \cdot \frac{1}{a\theta\cos\theta} = \frac{\sec^3 \theta}{a\theta}$.

$\frac{dx}{d\theta} = a(1-\cos\theta) = 2a\sin^2(\theta/2)$.

$\frac{dy}{d\theta} = a\sin\theta = 2a\sin(\theta/2)\cos(\theta/2)$.

$\frac{dy}{dx} = \frac{2a\sin(\theta/2)\cos(\theta/2)}{2a\sin^2(\theta/2)} = \cot(\theta/2)$.

$\frac{dx}{d\theta} = a \sec\theta \tan\theta, \frac{dy}{d\theta} = b \sec^2\theta$.

$\frac{dy}{dx} = \frac{b \sec^2\theta}{a \sec\theta \tan\theta} = \frac{b}{a} \frac{\sec\theta}{\tan\theta} = \frac{b}{a} \csc\theta$.

$\frac{d^2y}{dx^2} = \frac{b}{a} (-\csc\theta \cot\theta) \cdot \frac{1}{a \sec\theta \tan\theta} = -\frac{b}{a^2} \frac{\cos\theta}{\sin^2\theta} \cdot \frac{\cos^2\theta}{\sin\theta} = -\frac{b \cos^3\theta}{a^2 \sin^3\theta} = -\frac{b}{a^2} \cot^3\theta$.

$\frac{dx}{dt} = \cos t, \frac{dy}{dt} = p \cos pt$.

$\frac{dy}{dx} = \frac{p \cos pt}{\cos t} \Rightarrow y' \cos t = p \cos pt$.

Differentiate w.r.t $x$: $y'' \cos t + y'(-\sin t) \frac{dt}{dx} = -p^2 \sin pt \frac{dt}{dx}$.

$y'' \cos t - y' \sin t \frac{1}{\cos t} = -p^2 y \frac{1}{\cos t}$.

Multiply by $\cos t$: $y'' \cos^2 t - y' \sin t = -p^2 y$.

$(1-\sin^2 t) y'' - x y' = -p^2 y \Rightarrow (1-x^2) y'' - x y' + p^2 y = 0$.

$\frac{dx}{dt} = a[-2\sin 2t + 2\sin 2t + 4t \cos 2t] = 4at \cos 2t$.

$\frac{dy}{dt} = a[2\cos 2t - (2\cos 2t - 4t \sin 2t)] = 4at \sin 2t$.

$\frac{dy}{dx} = \frac{4at \sin 2t}{4at \cos 2t} = \tan 2t$.

$\frac{d^2y}{dx^2} = \frac{d}{dt}(\tan 2t) \cdot \frac{dt}{dx} = 2\sec^2 2t \cdot \frac{1}{4at \cos 2t} = \frac{\sec^3 2t}{2at}$.

$x' = \cos t, y' = 2 \cos 2t \Rightarrow y' = \frac{2 \cos 2t}{\cos t} = \frac{2(1-2\sin^2 t)}{\cos t} = \frac{2(1-2x^2)}{\sqrt{1-x^2}}$.

Differentiating again leads to the standard form $(1-x^2)y'' - xy' + 4y = 0$.

Let $u = 2\cos^2 x$. $\frac{du}{dx} = 4\cos x (-\sin x) = -2\sin 2x$.

Let $v = \cos 2x$. $\frac{dv}{dx} = -2\sin 2x$.

$\frac{du}{dv} = \frac{-2\sin 2x}{-2\sin 2x} = 1$.