Question

if x=a(cosTheta+theta sintheta),y=a(sintheta-thetacostheta),find dsquarey ÷dxsquare

Answer 1

The answer is in the above page.

Answer 2

Answer:

$\rm \dfrac{d^2y}{dx^2}=\dfrac{sec^3\theta }{a\theta}.$

Step-by-step explanation:

Given:

$\rm x = a(\cos\theta +\theta \sin\theta).\\y=a(\sin\theta -\theta \cos\theta).$

It is clear that x and y, both are the functions of $\theta$, therefore,

$\rm \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}.$

$\rm \dfrac{dy}{d\theta}= \dfrac{d}{d\theta} (a(\sin\theta -\theta \cos\theta)\\=a\left (\dfrac{d}{d\theta}(\sin\theta) -\dfrac{d}{d\theta}(\theta \cos\theta)\right )\\=a\left( \cos\theta - \cos\theta-\theta(-\sin\theta)\right )\\=a\left( \theta\sin\theta\right ).$

$\rm \dfrac{dx}{d\theta}= \dfrac{d}{d\theta} (a(\cos\theta +\theta \sin\theta)\\=a\left (\dfrac{d}{d\theta}(\cos\theta)+\dfrac{d}{d\theta}(\theta \sin\theta)\right )\\=a\left( -\sin\theta+\sin\theta+\theta(\cos\theta)\right )\\=a\left( \theta\cos\theta\right ).$

Thus,

$\rm \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{a\left( \theta\sin\theta\right )}{a\left( \theta\cos\theta\right )}=\tan\theta\\\\\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left ( \dfrac{dy}{dx}\right )\\=\dfrac{d}{dx}\left ( \tan\theta\right )\\=\dfrac{d(\tan\theta)}{d\theta}\cdot \dfrac{d\theta}{dx}\\=\sec^2\theta\times \dfrac{1}{\dfrac{dx}{d\theta }} \\=\sec^2\theta \times \dfrac{1}{a\theta \cos\theta }\\=\dfrac{sec^3\theta }{a\theta}.$