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Converting Position Vector vs Time to Cartesian Coordinates

  1. Oct 22, 2011 #1
    1. The position vector of a particle at time t ≥ 0 is given by r = sin(t)*i + cos(2t)*j. Find the cartesian equation for the path of the particle.



    2. I was told that the answer is:
    y = 1 - 2x^2
    But I don't know how to obtain that solution.




    3. r = sin(t)*i + cos(2t)*j
    At first I thought I would merely plug in the values:
    x = i, y = j and √(x^2 + y^2) = r, but that wasn't working out:
    √(x^2 + y^2) = x sin(t) + y cos(2t)
    x^2 + y^2 = x^2(sin(t))^2 + 2xysin(t)cos(2t) + y^2(cos(2t))^2
    Solve with CAS Calculator:
    t = -pi/2
    Substitute that back in:
    x^2 + y^2 = x^2(-1)^2 + 2xy(-1)(0) + y^2(-1)^2
    x^2 + y^2 = x^2 + y^2
    0 = 0
    Now I am lost. Please help me!
     
  2. jcsd
  3. Oct 22, 2011 #2

    dynamicsolo

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    Use the "double-angle formula" for cosine -- there are three forms of it, but one of the expressions for y = cos(2t) will have a simple connection to x = sin t in the vector equation for r .

    (The double-angle formulas for sine and cosine from trigonometry are ones that it are useful to know. They see a lot of application...)
     
  4. Oct 22, 2011 #3
    cos(2t) = 1 - 2(sin(t))^2
    So r = sin(t)*i + (1 - 2(sin(t))^2)*j
    But I'm still confused because I don't know how to get rid of the i's and j's...
    Wait I think I get it:
    x = sin(t)
    y = 1 - 2(sin(t))^2
    y = 1 - 2x^2
    Is that right?
     
  5. Oct 22, 2011 #4

    SteamKing

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    Let x = sin (t). What is y in terms of x?
     
  6. Oct 22, 2011 #5
    Thanks a lot for your help, dynamicsolo and SteamKing. Now I know how to solve these types of problems for the future.
    I let x = the value that is being multiplied by i and y = the value that is being multiplied by j and then ignore the r.
    Then I express y in terms of the value being multiplied by i and then make the substitution to get y in terms of x.
    Thank you once again!
     
  7. Oct 22, 2011 #6

    dynamicsolo

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    Yes. You don't have to "get rid of" the i and j: they are just the unit vectors connected with the x- and y-components of the vector equation. We would write

    r(t) = x(t) i + y(t) j = sin(t) i + cos(2t) j .

    You are just being asked to look for the equation relating y(t) to x(t) .
     
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