find the tangent line of a parametrized curve

**Taurus3** · February 13th 2011, 06:03 PM

ok.

**Prove It** · February 13th 2011, 06:43 PM

1. Evaluate $\displaystyle \gamma'(t)$ at the point $\displaystyle t$ where $\displaystyle \sin{2t} = 0$ and $\displaystyle \cos{2t} = 1$ .

2. Velocity = $\displaystyle \mathbf{v}(t) = \frac{d}{dt}[\mathbf{r}(t)]$ .

To find the unit tangent vector, divide $\displaystyle \mathbf{v}(t)$ by its length.

Acceleration = $\displaystyle \mathbf{a}(t) = \frac{d}{dt}[\mathbf{v}(t)]$ .

**Soroban** · February 13th 2011, 07:03 PM

Hello, Taurus3!

$\text{1. Find the tangent line to given parameterized curve at the given point:}$

. . . . $r(t)\:=\:(2\sin t\cos t,\;\cos 2t),\;\;(0, 1)$

$\text{We have: }\;\begin{Bmatrix}x &=& \sin 2t \\ y &=& \cos 2t \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}\frac{dx}{dt} &=& 2\cos2t \\ \\[-3mm]\frac{dy}{dt} &=& \text{-}2\sin2t \end{Bmatrix}$

$\text{Then: }\;\dfrac{dy}{dx} \;=\;\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} \;=\;\dfrac{\text{-}2\sin2t}{2\cos2t} \;=\;-\tan2t$

$\text{At }(0,1),\:t = 0 \quad\Rightarrow\quad \dfrac{dy}{dx} \:=\:-\tan 0 \:=\:0$

$\text{The line through }(0,1)\text{ with slope 0 is: }\:y = 1$

**Taurus3** · February 13th 2011, 09:54 PM

for the velocity, since x=ln(t) and y= sqrt.(t^2+1), we could write r(t) as (ln(t)+3t)i + (sqrt.(t^2+1))j right?
And v(t) would be r prime t = (1/t + 3)i + t/(sqrt.(t^2+1)) right?

**Prove It** · February 13th 2011, 10:12 PM

Yes.

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