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Math Help - vector equations

  1. #1
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    vector equations

    each curve represents the motion of a particle moving in R2

    x(t) = (2cos t, 3sin t) , 0 <= t <= 2Pi

    x(t) = (3 sec t, 2 tan t) , (-1/3)pi <= t <= (1/3)pi

    can someone sketch both these curves showing the direction of motion for me???

    and find the velocity and speed of each of the particles with a step by step explanation

    the speed of the first one i got as:

    sqrt ( 4(sin t)^2 + 6 (cos t)^2 )
    is there any way to further simplify this or should i leave it like this???

    i also have to locate the points (x,y) where the velocity is horizontal, vertical or 0
    i think this means when the derivative or x-component is 0, when derivative or y-component is 0, and when both are zero?
    am i correct?
    and can you list those points for both equations as well so i can verify them
    Last edited by razorfever; February 12th 2009 at 04:08 PM.
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  2. #2
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    Hello, razorfever!

    I'll get you started on this one.

    But they really shouldn't use x that way . . .


    Each curve represents the motion of a particle moving in Rē.

    . . f(t) \:= \:(2\cos t,\:3\sin t),\quad0 \leq t \leq 2\pi
    We have the parametric equations: . \begin{array}{cccc} x\:=\: 2\cos t & {\color{blue}(1)}\\ y\:=\:3\sin t & {\color{blue}(2)}\end{array}

    We can eliminate the parameter . . .

    We have: . \begin{array}{ccccccccc}{\color{blue}(1)} & x \:=\: 2\cos t & \Rightarrow & \dfrac{x}{2} \:=\: \cos t & {\color{blue}(3)}\\ \\[-3mm]<br />
{\color{blue}(2)} & y \:=\: 3\sin t & \Rightarrow & \dfrac{y}{3} \:=\: \sin t & {\color{blue}(4)}\end{array}


    . . \begin{array}{cccccc}\text{Square {\color{blue}(3)}:} & \dfrac{x^2}{4} \:=\:\cos^2\!t \\ \\[-3mm]<br />
\text{Square {\color{blue}(4)}:} & \dfrac{y^2}{9} \:=\:\sin^2\!t \end{array}

    Add: . \frac{x^2}{4} + \frac{y^2}{9} \;=\;\underbrace{\cos^2\!t + \sin^2\!t}_{\text{This is 1}}

    And we have the ellipse: . \frac{x^2}{4} + \frac{y^2}{9} \:=\:1



    Calcuate the points (x,y) for various values of t.

    . . . \begin{array}{|c||c|c|}<br />
t & x & y \\ \hline\hline<br />
0 & 2 & 0 \\ \frac{\pi}{2} & 0 & 3 \\ \pi & \text{-}2 & 0 \\ \frac{3\pi}{2} & 0 & \text{-}3 \\ 2\pi & 2 & 0 \end{array}


    We see that the ellipse starts at (2,0) and is generated in the CCW direction.


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