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

Answers (1)

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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