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If $x$ and $y$ are connected parametrically by the equations given in $x = a \big(\cos t +\ log \tan \large\frac{t}{2}\big) $ and $ y = a \sin t $ without eliminating the parameter, Find $ \frac{\large dy}{\large dx}$

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Toolbox:
  • By chain rule we have $\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
  • $\sin\theta=2\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
Step 1:
$x=a(\cos t+\log\tan\large\frac{t}{2})$
Differentiating with respect to $t$
$\large\frac{dx}{dt}=$$a\begin{bmatrix}\normalsize-\sin t+\large\frac{1}{\tan\Large\frac{t}{2}}\frac{d}{dx} (\normalsize\tan\large\frac{t}{2})\end{bmatrix}$
$\quad\;\;=a\begin{bmatrix}-\sin t+\large\frac{1}{\tan\large\frac{t}{2}}\normalsize\sec^2\large\frac{t}{2}.\frac{1}{2}\end{bmatrix}$
$\quad\;\;=a\begin{bmatrix}-\sin t+\large\frac{1}{\tan\Large\frac{t}{2}}\frac{1}{\cos^2\Large\frac{t}{2}}.\frac{1}{2}\end{bmatrix}$
$\quad\;\;=a\begin{bmatrix}-\sin t+\large\frac{1}{2\tan\Large\frac{t}{2}}\frac{1}{\cos^2\Large\frac{t}{2}}\end{bmatrix}$
$\quad\;\;=a\begin{bmatrix}-\sin t+\large\frac{1}{2\Large\frac{\Large\sin\Large\frac{t}{2}}{\Large\cos\large\frac{t}{2}}}\frac{1}{\cos^2\Large\frac{ t}{2}}\end{bmatrix}$
$\quad\;\;=a\begin{bmatrix}-\sin t+\large\frac{1}{2\sin\frac{t}{2}\cos\frac{ t}{2}}\end{bmatrix}$
We know that $\sin\theta=2\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
$\quad\;=a\begin{bmatrix}-\sin t+\large\frac{1}{\sin t}\end{bmatrix}$
$\quad\;\;=a\begin{bmatrix}\large\frac{-\sin^2t+1}{\sin t}\end{bmatrix}$
$\quad\;\;=a\begin{bmatrix}\large\frac{1-\sin^2t}{\sin t}\end{bmatrix}$
$\quad\;\;=\begin{bmatrix}\large\frac{a\cos^2t}{\sin t}\end{bmatrix}$
Step 2:
$y=a\sin t$
Differentiating with respect to $t$
$\large\frac{dy}{dt}$$=a\cos t$
Step 3:
$\large\frac{dy}{dx}=\Large\frac{\frac{dy}{dx}}{\frac{dx}{dt}}$
$\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
$\quad\;\;=a\cos t\times \large\frac{1}{\Large\frac{a\cos^2t}{\sin t}}$
$\quad\;\;=a\cos t\times \large\frac{\sin t}{a\cos^2 t}$
$\quad\;\;=\large\frac{\sin t}{\cos t}$$=\tan t.$
answered May 10, 2013 by sreemathi.v
 

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