find the equation of tangent and normal to the curve x=1-cos theta, y= theta-sin theta at theta=pi/4.

x = 1 - cosθ  and y = θ - sinθ

find dy/dx 

dy/dθ  =  1  - cosθ

dy/dx  =  sinθ

therefore dy/dx  =  (dy/dθ) /(dx/dθ)  =  (1 - cosθ)/sinθ

slope dy/dx  at θ  =  pi/4  =  [1 - cos(π/4)] / sin(π/4)  =  (√2)   -  1

We need to find the coordinates of the point θ = π/4  in cartesian coordinates.

y1  =  π/4  - sin(π/4)  =  π/4  -  1/(√2)

x1  =  1  - cos(π/4)  =  1  -  1/(√2)

 

Therfore the equation of tangent

Y  -  y1  =  dy/dx(X  -  x1)

Y  - (π/4  -  1/(√2))  =  [(√2)   -  1] [X  -  (1  -  1/(√2)]

You can rearrange it according to your aswer.

Y  - (√2)   -  1)X  =  π/4  + 2  -  √2

 

For equation of normal. use negative reciprocal of slope -dx/dy  =  -1/[(√2)   -  1]

-dx/dy  =  -[(√2)  +  1] I have rationalized the denominator.

Using  Y  -  y1  =  -dx/dy(X  -  x1)

gives

Y +  [√2)   +   1]X  =  π/4

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