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Unit Vectors

  1. Sep 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the unit vectors that are parallel to the tangent line to the curve y=2sinx at the point ([tex]\pi[/tex]/6, 1). Thereafter, find the unit vectors that are perpendicular to the tangent line.


    2. Relevant equations

    3. The attempt at a solution
    I took the derivative of y=2sinx and got y'=2cosx. Then subbed in [tex]\pi[/tex]/6 and got slope=[tex]\sqrt{3}[/tex]. After this, I was totally confused about what to do next since I don't know how to put the function with respect to i, j. Thanks in advance.
     
  2. jcsd
  3. Sep 12, 2008 #2

    Defennder

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    First thing you should note is that dy/dx tells you how the unit vector x-y components are related to each other. Draw the tangent line at pi/6, and a really small right-angle triangle with the vertical length denoted dy and horizontal length dx. So, see what to do next?
     
  4. Sep 12, 2008 #3
    you have slop which is rise over run . so rise will be j-hat and run will be i-hat.you don't have to find k-hat.just simplify square root of 3.
     
  5. Sep 12, 2008 #4
    Oh yes, I think i got it now. Just to confirm is it (i+root3j) and (-i-root3j)? Thank you so much guys.
     
    Last edited: Sep 12, 2008
  6. Sep 12, 2008 #5

    Defennder

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    Don't see how you got that answer. Might want to re-check your working.
     
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