If the solve the problem


If $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$ prove that

$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\cos \theta-\theta \sin \theta), \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\sin \theta+\theta \cos \theta)$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\sec ^{3} \theta}{\mathrm{a} \theta}$


Basic idea:

Second order derivative is nothing but derivative of derivative i.e. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}\right)$

$\sqrt{T h e}$ idea of chain rule of differentiation: If $\mathrm{f}$ is any real-valued function which is the composition of two functions $u$ and $v$, i.e. $f=v(u(x))$. For the sake of simplicity just assume $t=u(x)$

Then $f=v(t) .$ By chain rule, we can write the derivative of $f$ w.r.t to $x$ as:

$\frac{d f}{d x}=\frac{d v}{d t} \times \frac{d t}{d x}$

Product rule of differentiation- $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{uv})=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

The idea of parametric form of differentiation:

If $y=f(\theta)$ and $x=g(\theta)$, i.e. $y$ is a function of $\theta$ and $x$ is also some other function of $\theta$.

Then $d y / d \theta=f^{\prime}(\theta)$ and $d x / d \theta=g^{\prime}(\theta)$

We can write : $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d \theta}}}$


$x=a(\cos \theta+\theta \sin \theta) \ldots \ldots$ equation 1

$y=a(\sin \theta-\theta \cos \theta) \ldots \ldots$ equation 2

to prove :

i) $\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\cos \theta-\theta \sin \theta)$

ii) $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\sin \theta+\theta \cos \theta)$

iii) $\frac{d^{2} y}{d x^{2}}=\frac{\sec ^{2} \theta}{a \theta}$

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{d^{2} y}{d x^{2}}$

$\operatorname{As} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

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

$=a(-\sin \theta+\theta \cos \theta+\sin \theta)$

[ differentiated using product rule for $\theta \sin \theta$ ]

$=a \theta \cos \theta \ldots e q n 4$

Again differentiating w.r.t $\theta$ using product rule:-

$\frac{d^{2} x}{d \theta^{2}}=a(-\theta \sin \theta+\cos \theta)$

$\therefore \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\cos \theta-\theta \sin \theta) \ldots$ proved (i)


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

$=a \cos \theta+a \theta \sin \theta-a \cos \theta$

$\therefore \frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a} \theta \sin \theta \ldots \ldots \ldots \ldots$ equation 5

Again differentiating w.r.t $\theta$ using product rule:-

$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\theta \cos \theta+\sin \theta)$

$\therefore \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \theta^{2}}=\mathrm{a}(\sin \theta+\theta \cos \theta) \ldots$ proved (ii)

$\because \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$

Using equation 4 and 5 :

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

As $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\therefore$ again differentiating w.r.t $\mathrm{x}:-$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}} \tan \theta$

$=\sec ^{2} \theta \frac{\mathrm{d} \theta}{\mathrm{dx}}$ [using chain rule]

$\because \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a} \theta \cos \theta \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dx}}=\frac{1}{\mathrm{a} \theta \cos \theta}$

Putting a value in the above equation-

We have :

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\sec ^{2} \theta \times \frac{1}{\mathrm{a} \theta \cos \theta}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\sec ^{3} \theta}{\mathrm{a} \theta} \ldots \ldots$ proved (iii)

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