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For the following parametric equations, find the horizontal tangent line and the vertical tangent line if:

$x=3 t^{2}-6 t\quad, \quad y=\sqrt{t} \quad,\quad t>0$

For HTL: $\frac{d y}{d x}=0 \quad \text { or } \quad \frac{d y}{d t}=0$

For VTL: $\frac{d y}{d x}=\infty \quad \frac{d x}{d t}=0$

$\frac{d x}{d t}=6 t-6$

$\frac{d y}{d t}=\frac{1}{2 \sqrt{t}} \neq 0 \Rightarrow \text { no } H \cdot T \cdot L$

$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{1 / 2 \sqrt{t}}{6 t-6}=\frac{1}{2 \sqrt{t}(6 t-6)}$

V.T.L at $\frac{d x}{d t}=0 \Rightarrow 6 t-6=0 \Rightarrow 6 t=6 \Rightarrow t=1$

$\Rightarrow x=3(1)^{2}-6(1)=-3$

$y=\sqrt{1}=1 \quad \text { V.T.L at }(-3,1)$

A) Find the tangent the cycloid $x=r(\theta-\sin \theta), y=r(1-\cos \theta)$ at the point where $\theta=\frac{\pi}{3}$

B) At what point is the tangent horizontal? when is it vertical?

(a) $\frac{d x}{d \theta}=\frac{d(r(\theta-\sin \theta))}{d \theta}=\frac{d(r{\theta}-r \sin \theta)}{d \theta}=r-r \cos \theta=r(1-\cos \theta)$

$\frac{d y}{d \theta}=\frac{d(r(1-\cos \theta))}{d \theta}=\frac{d(r-r \cos \theta)}{d \theta}=-r(-\sin \theta)=+r \sin \theta$

the slop of the tangent $=\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$

$=\frac{r \sin \theta }{r(1-\cos \theta)}=\frac{\sin \theta}{1-\cos \theta}$

at $\theta=\frac{\pi}{3} \Rightarrow slop =\frac{d y}{d x}=\frac{\sin \theta}{1-\cos \theta}=\frac{\sin \left(\frac{\pi}{3}\right)}{1-\cos \left(\frac{\pi}{3}\right)}$

$=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=\frac{\sqrt{3} / 2}{1 / 2}=\sqrt{3}$

$x \text { at } \theta=\frac{\pi}{3} \Rightarrow x=r\left(\frac{\pi}{3}-\sin \frac{\pi}{3}\right)=r\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)$

$y \text { at } \theta=\frac{\pi}{3} \Rightarrow y=r\left(1-\cos \frac{\pi}{3}\right)=r\left(1-\frac{1}{2}\right)=r\left(\frac{1}{2}\right)=\frac{1}{2} r$

tangent equation: $y-y_{0}=slope\left(x-x_{0}\right)$

$y-\frac{r}{2}=\sqrt{3}(x-(r(\frac{\pi}{3}-\frac{\sqrt{3}}{2}))$

$y-\frac{r}{2}=\sqrt{3}\left(x-\frac{r \pi}{3}+\frac{r \sqrt{3}}{2}\right)$

(b) the tangent is horizontal when $\frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{\sin \theta}{1-\cos \theta}=0 \Rightarrow \sin \theta=0 \quad \& \:1-\cos \theta \neq 0$

$\theta=\pi, 3 \pi, 5 \pi, \dots$

$\theta=(2 n-1) \pi$

tangent is horizontal at point (x , y)

$x=r{(\theta-\sin \theta)}=r \theta-r \sin \theta=r \theta=r \cdot(2 n-1) \pi$

$y=r(1-\cos \theta)=r-r \cos \theta=r-r(-1)=r+r=2 r$

$∴ \text { Tangent is horizontatal }(r(2 n-1) \pi, 2 r)$

the tangent line is vertical when $\frac{d y}{d x} \rightarrow \infty$

$\frac{d x}{d \theta}=0 \quad \& \quad \frac{d y}{d \theta} \neq 0$

$\frac{d x}{d \theta}=0 \Rightarrow r(1-\cos \theta)=0$

$\frac {r} {r} (1-\cos a)=\frac{0}{r}$

$\begin{array}{l}{1-\cos \theta=0} \\ {\cos \theta=1}\end{array}$

$\theta=\cos ^{-1}(1)=0,2 \pi , 4 \pi, 6 \pi, \dots$

$\theta=2 n \pi$

$\lim _{x \rightarrow 2 n \pi} \frac{d y}{d x}=\lim _{\theta \rightarrow 2n \pi} \frac{\sin \theta}{1-\cos \theta}=\frac{0}{1-1}=\frac{0}{0}$

use L'Hopital Role

$\lim _{\theta \rightarrow 2 n \pi} \frac{\sin \theta}{1-\cos \theta}=\lim _{\theta \rightarrow 2n \pi} \frac{\cos \theta}{\sin \theta}=\frac{\cos (2 n \pi)}{\sin (2 n \pi)}=\frac{1}{0}=\infty$

$x=r(0-\sin \theta)=r(2 n \pi-\sin (2 n \pi))=2n \pi r$