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For the curve $x=e^{t}\cos t: y=e^{t}\sin t$ the tangent line is parallel to the $x$-axis when $t$ is equal to

\[\begin{array}{1 1}(1)-\frac{\pi}{4}&(2)\frac{\pi}{4}\\(3)0&(4)\frac{\pi}{2}\end{array}\]

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$x=e^t \cos t$
$y=e^t \sin t$
$\large\frac{dx}{dt} $$=e^{t} x - \sin t+ \cos t .e^t. \qquad \large\frac{dy}{dt}$$=e^{t}. \cos t + \sin t e^{t}$
$\large\frac{dx}{dt} $$=e^{t}(x -\cos t - \sin t) \qquad \large\frac{dy}{dt}$$=e^{t}. (\cos t + \sin t )$
$\large\frac{dy}{dx} =\large\frac{dy/dt}{dx/dt}=\frac{e^t (\cos t + \sin t)}{e^t (\cos t + \sin t)}$
$\large\frac{dy}{dx} =\large\frac{ \cos t + \sin t }{\cos t - \sin t}$
$=> \cos t + \sin t =0$
$\sin t = -\cos t$
$=> \large\frac{\sin t }{\cos t}$$=-1$
$\tan t=-1$
$t= -\large\frac{ \pi}{4}$
Hence 1 is the correct answer
answered May 19, 2014 by meena.p
 

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