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Parametric form to algebraic

  1. Oct 10, 2006 #1
    X1 T = 10T

    Y1 T = 100 + (.5 * -9.8T^2)

    X2 T = 100 - 12.3 T

    X2 T = 0

    How do I put this into algebraic form? it seems easy but I just can't get it.

    Do you simply add the X and Y components? If so what do x and y each stand for?? Does it have something to do with sine and cosine? =/
     
    Last edited: Oct 10, 2006
  2. jcsd
  3. Oct 10, 2006 #2
    Are you needing to know how to put the equations together? Because there's a thing called derivatives of parametric equations in calculus.

    for example
    if x = t + t-1 and y = t + 1

    dx/dt = 1 – t -2
    = 1 – 1/t2
    dy/dt = 1
    Then dy/dx = dy/dt x dt/dx

    Then you substitute in the equations and solve
    If that’s what you are needing to solve your question just do the same thing for your parameters given.
     
  4. Oct 10, 2006 #3
    Think about this in terms of slope and the difference between two different points. In other words, as time increases, how much does x increase or decrease? How much does y increase or decrease?
     
  5. Oct 11, 2006 #4

    andrevdh

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    Homework Helper

    I think the equations should read

    [tex]x_1 (t) = 10t[/tex]

    [tex]y_1 (t) = 100 - \frac{1}{2}gt^2[/tex]

    [tex]x_2 (t) = 100 -12.3t[/tex]

    [tex]y_2 (t) = 0[/tex]

    so these equations describe the positional coordinates of two different objects as functions of time. What do you need to determine about the two objects?
     
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