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Vector-Valued Function Ellipse

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Find a vector-valued function f that traces out the given curve in the indicated direction.
    (a) Counterclockwise (b) Clockwise.

    4x2+9y2=36


    2. Relevant equations

    x2+y2=r2
    cos2t+sin2t=1

    3. The attempt at a solution
    From what I can determine, this is an ellipse. I believe this is how the answer is found.

    4x2/36+9y2/36=36/36

    x2/9+y2/4=1

    so cos2t+sin2t=1 and x2/9+y2/4=1
    this means x=3cos(t) and y=2sin(t) to balance it equivalently, I think.

    putting that in component form, 3cos(t)i+2sin(t)j for counterclockwise and 3cos(t)i-2sin(t)j for clockwise

    That does equal the correct answer , but I am not sure if I got to it correctly. Can someone please review my work and let me know if I am approaching this correctly, I do not show anything in my notes or in my book on how to perform this kind of equation.
     
  2. jcsd
  3. Sep 18, 2010 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is perfectly correct. Notice that from the definition of sine and cosine in terms of the unit circle, "t" goes around the circle counterclockwise so your 3cos(t)i+ 2sin(t)j does describe the circle counterclockwise as t increases. To go around the circle clockwise, use -t instead: 3cos(-t)+ 2sin(-t)= 3cos(t)- 2sin(-t) because cosine is an even function and sine is an odd function.
     
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