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If $x$ and $y$ are connected parametrically by the equations given in $x = a ( \theta - \sin \theta ), y = a ( 1 +\cos \theta) $ without eliminating the parameter, Find $\large\frac{dy}{dx}$

$\begin{array}{1 1} \cot\large\frac{\theta}{2} \\ -\cot\large\frac{\theta}{2} \\ -\cot\large\frac{\theta}{4} \\ -\tan\large\frac{\theta}{2}\end{array} $

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Toolbox:
  • By chain rule we have $\large\frac{dy}{dx}=\large\frac{dy}{d\theta}$$\times \large\frac{d\theta}{dx}$
Step 1:
$x=a(\theta-\sin\theta)$
Differentiating with respect to $\theta$
$\large\frac{dx}{d\theta}$$=a(1-\cos \theta)$
$f'(\theta)=a(1-\cos \theta)$
Step 2:
$y=a(1+\cos\theta)$
Differentiating with respect to $\theta$
$\large\frac{dy}{d\theta}$$=a(0+(-\sin \theta))$
$\large\frac{dy}{d\theta}$$=a(-\sin \theta)$
$\large\frac{dy}{d\theta}$$=-a\sin \theta$
Step 3:
$\large\frac{dy}{dx}=\frac{g'(\theta)}{f'(\theta)}$
$\quad\;=\large\frac{\Large\frac{dy}{d\theta}}{\Large\frac{dx}{d\theta}}$
$\Rightarrow \large\frac{dy}{d\theta}$$\times \large\frac{d\theta}{dx}$
$\Rightarrow -a\sin \theta\times\large\frac{1}{a(1-\cos\theta)}$
We know that $\sin 2\theta=2\sin\theta\cos\theta$
$\sin\theta=2\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
$1-\cos\theta=2\sin^2\large\frac{\theta}{2}$
$\large\frac{dy}{dx}=\frac{-a(2\sin\Large\frac{\theta}{2}\cos\Large\frac{\theta}{2})}{a(2\sin^2\Large\frac{\theta}{2})}$
$\Rightarrow -\Large\frac{\cos\Large\frac{\theta}{2}}{\sin\Large\frac{\theta}{2}}$
$\Rightarrow -\cot\large\frac{\theta}{2}$
answered May 10, 2013 by sreemathi.v
 

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