Question
If $\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1+\cos\theta)$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
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Answer
$\text{x} =\text{a} (\theta -\sin\theta)\dots\text{ eq. } 1$
$\text{y}=\text{a}(1+\cos\theta)\dots\text{ eq. 2}$
To find: $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$ using parametric form and differentiate it again.
$\frac{\text{dx}\text{}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}(\theta-\sin\theta)=\text{a}(1-\cos\theta)\dots\ \text{eq. 3}$
Similarly,
$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}(1+\cos\theta)=-\text{a}\sin\theta\dots\text{ eq. }4$
$\Big[\because\frac{\text{d}}{\text{dx}}\cos\text{x}=-\sin\text{x},\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}\Big]$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}=\frac{-\sin\theta}{(1-\cos\theta)}\dots\ \text{eq. }5$
Differentiating again w.r.t. x:
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=-\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\theta}{1-\cos\theta}\Big)$
Using product rule and chain rule of differentiation together:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{-\frac{1}{1-\cos\theta}\frac{\text{d}}{\text{d}\theta}\sin\theta-\sin\theta\frac{\text{d}}{\text{d}\theta}\frac{1}{(1-\cos\theta)}\Big\}\frac{\text{d}\theta}{\text{dx}}$
Apply chain rule to determine $\frac{\text{d}}{\text{d}\theta}\frac{1}{(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}}=\Big\{\frac{-\cos\theta}{1-\cos\theta}+\frac{\sin^2\theta}{(1-\cos\theta)^2}\Big\}\frac{1}{\text{a}(1-\cos\theta)}$ [using eq.3]
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\bigg\{\frac{-\cos\theta(1-\cos\theta)+\sin^2\theta}{(1-\cos\theta)^2}\bigg\}\frac{1}{\text{a}(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\bigg\{\frac{-\cos\theta+\cos^2\theta+\sin^2\theta}{(1-\cos\theta)^2}\bigg\}\frac{1}{\text{a}(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{\frac{1-\cos\theta}{(1-\cos\theta)^2}\Big\}\frac{1}{\text{a}(1-\cos\theta)}[\because\cos^2\theta+\sin^2\theta=1]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{a(1}-\cos\theta)^2}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{a}\Big(2\sin^2\frac{\theta}{2}\Big)^2}\big[\because-\cos\theta=2\sin^2\frac{\theta}{2}\big]$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{4a}}\text{cosec}^4\frac{\theta}{2}$
$\text{y}=\text{a}(1+\cos\theta)\dots\text{ eq. 2}$
To find: $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$ using parametric form and differentiate it again.
$\frac{\text{dx}\text{}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}(\theta-\sin\theta)=\text{a}(1-\cos\theta)\dots\ \text{eq. 3}$
Similarly,
$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}(1+\cos\theta)=-\text{a}\sin\theta\dots\text{ eq. }4$
$\Big[\because\frac{\text{d}}{\text{dx}}\cos\text{x}=-\sin\text{x},\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}\Big]$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}=\frac{-\sin\theta}{(1-\cos\theta)}\dots\ \text{eq. }5$
Differentiating again w.r.t. x:
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=-\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\theta}{1-\cos\theta}\Big)$
Using product rule and chain rule of differentiation together:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{-\frac{1}{1-\cos\theta}\frac{\text{d}}{\text{d}\theta}\sin\theta-\sin\theta\frac{\text{d}}{\text{d}\theta}\frac{1}{(1-\cos\theta)}\Big\}\frac{\text{d}\theta}{\text{dx}}$
Apply chain rule to determine $\frac{\text{d}}{\text{d}\theta}\frac{1}{(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}}=\Big\{\frac{-\cos\theta}{1-\cos\theta}+\frac{\sin^2\theta}{(1-\cos\theta)^2}\Big\}\frac{1}{\text{a}(1-\cos\theta)}$ [using eq.3]
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\bigg\{\frac{-\cos\theta(1-\cos\theta)+\sin^2\theta}{(1-\cos\theta)^2}\bigg\}\frac{1}{\text{a}(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\bigg\{\frac{-\cos\theta+\cos^2\theta+\sin^2\theta}{(1-\cos\theta)^2}\bigg\}\frac{1}{\text{a}(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{\frac{1-\cos\theta}{(1-\cos\theta)^2}\Big\}\frac{1}{\text{a}(1-\cos\theta)}[\because\cos^2\theta+\sin^2\theta=1]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{a(1}-\cos\theta)^2}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{a}\Big(2\sin^2\frac{\theta}{2}\Big)^2}\big[\because-\cos\theta=2\sin^2\frac{\theta}{2}\big]$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{4a}}\text{cosec}^4\frac{\theta}{2}$
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