Application of Derivatives: Tangents and Normals

Find the equation of tangent to $x=\cos \theta$, $y=\sin \theta$ $\;\;$ $\theta \in \left[0,2\pi\right)$ at $\theta=\dfrac{\pi}{4}$


$\dfrac{dx}{d\theta}=-\sin \theta$

$\dfrac{dy}{d\theta}=\cos \theta$

$\therefore$ $\;\;$ $\dfrac{dy}{dx}=\dfrac{dy / d\theta}{dx / d\theta} = -\dfrac{\cos \theta}{\sin \theta}=-\cot \theta$

$\therefore$ $\;\;$ $\dfrac{dy}{dx}\bigg|_{\theta=\frac{\pi}{4}} = -\cot \left(\dfrac{\pi}{4}\right)=-1$

When $\theta = \dfrac{\pi}{4}$,

$x = \cos \left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$

$y = \sin \left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$

$\therefore$ $\;\;$ Equation of tangent at $\theta=\dfrac{\pi}{4}$ is

$y-\dfrac{1}{\sqrt{2}}=-1 \left(x-\dfrac{1}{\sqrt{2}}\right)$

i.e. $x\sqrt{2} + y \sqrt{2}=2$