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1. Evaluate $\displaystyle \displaystyle \gamma'(t)$ at the point $\displaystyle \displaystyle t$ where $\displaystyle \displaystyle \sin{2t} = 0$ and $\displaystyle \displaystyle \cos{2t} = 1$.
2. Velocity = $\displaystyle \displaystyle \mathbf{v}(t) = \frac{d}{dt}[\mathbf{r}(t)]$.
To find the unit tangent vector, divide $\displaystyle \displaystyle \mathbf{v}(t)$ by its length.
Acceleration = $\displaystyle \displaystyle \mathbf{a}(t) = \frac{d}{dt}[\mathbf{v}(t)]$.
Hello, Taurus3!
$\displaystyle \text{1. Find the tangent line to given parameterized curve at the given point:}$
. . . . $\displaystyle r(t)\:=\:(2\sin t\cos t,\;\cos 2t),\;\;(0, 1)$
$\displaystyle \text{We have: }\;\begin{Bmatrix}x &=& \sin 2t \\ y &=& \cos 2t \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}\frac{dx}{dt} &=& 2\cos2t \\ \\[-3mm]\frac{dy}{dt} &=& \text{-}2\sin2t \end{Bmatrix}$
$\displaystyle \text{Then: }\;\dfrac{dy}{dx} \;=\;\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} \;=\;\dfrac{\text{-}2\sin2t}{2\cos2t} \;=\;-\tan2t$
$\displaystyle \text{At }(0,1),\:t = 0 \quad\Rightarrow\quad \dfrac{dy}{dx} \:=\:-\tan 0 \:=\:0$
$\displaystyle \text{The line through }(0,1)\text{ with slope 0 is: }\:y = 1$