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Math Help - find the tangent line of a parametrized curve

  1. #1
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    line

    ok.
    Last edited by Taurus3; February 13th 2011 at 08:46 PM.
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    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)].
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    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

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    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?
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    Yes.
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