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# Find the equations of the tangent and normal to the given curves at the indicated points: x = cos t, y = sint at t = pi /4

14) Find the equations of the tangent and normal to the given curves at the indicated points:

e) $x = \cos t , y = \sin t \: \: at \: \: t = \pi /4$

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We know that Slope of the tangent at a point on the given curve is given  by  $\frac{dy}{dx}$
Given the equation of the curve
$x = \cos t , y = \sin t$
Now,
$\frac{dx}{dt} = -\sin t$            and           $\frac{dy}{dt} = \cos t$
Now,
$\left ( \frac{dy}{dx} \right )_{t=\frac{\pi}{4}} = \frac{ \frac{dy}{dt}}{ \frac{dx}{dt}} = \frac{\cos t}{-\sin t} = -\cot t = =- \cot \frac{\pi}{4} = -1$
Hence slope of the tangent is -1
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-1} = 1$
Now, the equation of the tangent at the point $t = \frac{\pi}{4}$ with slope = -1 is
$x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}$          and

$y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}$
equation of the tangent at

$t = \frac{\pi}{4}$ i.e. $\left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right )$ is

$y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = -1(x- \frac{1}{\sqrt2})\\ \sqrt2y + \sqrt2x = 2\\ y + x = \sqrt2$
Similarly, the equation of normal at $t = \frac{\pi}{4}$  with slope = 1 is
$x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}$          and

$y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}$
equation of the tangent at

$t = \frac{\pi}{4}$ i.e. $\left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right )$ is
$\\y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = 1(x- \frac{1}{\sqrt2})\\ \sqrt2y - \sqrt2x = 0\\ y - x = 0\\ x=y$

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