Find the equation of tangent and normal to the curve $x = 1 - cos θ$ , $y = 1 - sin θ$ at $θ = \frac{π}{4}$

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Solution

$x = 1 - cos θ$

$\frac{d x}{d θ} = sin θ$

$y = θ - sin θ$

$\frac{d y}{d θ} = 1 - cos θ$

$\frac{d y}{d x} = \frac{1 - cos θ}{sin θ}$

$\frac{d y}{d x} \to θ = \frac{π}{4} \Rightarrow \frac{d y}{d x} = \sqrt{2} - 1$

$y_{1} = \frac{π}{4} - sin \frac{π}{4} = \frac{π}{4} - \frac{1}{\sqrt{2}}$

$x_{1} = 1 - cos \frac{π}{4} = 1 - \frac{1}{\sqrt{2}}$

equation of tangent,

$y - y_{1} = \frac{d y}{d x} (x - x_{1})$

$y - (\frac{π}{4} - \frac{1}{\sqrt{2}}) = (\sqrt{2} - 1) [x - (1 - \frac{1}{\sqrt{2}})]$

$y - (\sqrt{2} - 1) x = \frac{π}{4} + 2 - \sqrt{2}$

equation of normal,

$y - y_{1} = - \frac{d x}{d y} (x - x_{1})$

$y + (\sqrt{2} + 1) x = \frac{π}{4}$

Suggest Corrections

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