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Homework Help: Equations systems

  1. Nov 6, 2008 #1
    given x(t)=30t-t[tex]^{3}[/tex]
    y(t)=22t-4t[tex]^{2}[/tex]
    what i need to do in this excercise is fine an equation with x as a function of y x(y) or the opposite y(x)---- 1 function without any variable t.
    i have managed up till now by finding an expression for t and plugging that in instead of t in the 2nd equation, problem here is the extra t's, sqared and cubed, tried dividing both by t-not much help.. any ideas?
     
  2. jcsd
  3. Nov 6, 2008 #2

    HallsofIvy

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    There won't be one way to do that because the two functions are not one-to-one: x(0)= 0 and x([itex]\sqrt{30})= 0[/itex]. There is no single function that will give both of those.
    Similarly, y(0)= y(11/2)= 0.

    What you can do is, for example, is treat the second equation as a quadratic equation in t: 4t2- 22t+ y= 0 and use the quadratic formula to solve for t:
    [tex]t= \frac{22\pm\sqrt{(22)^2- 4(4)y}}{8}= \frac{22\pm\sqrt{484- 16y}}{8}[/tex]
    It's that "[itex]\pm[/itex]" that is the problem. Choose either "+" or "-" and put that value of t into the equation x= 30t- t3 to get x as a function of y.
     
  4. Nov 7, 2008 #3
    Do you mean like composition y(t) [itex]\circ[/itex] x(t) = h(t), or y(x(t))=h(t).

    For y(30t-t3)=22(30t-t3)-4(30t-t3)2

    Can you continue out of here?
     
  5. Nov 8, 2008 #4
    no, cant continue from there, i dont see how you got there either, from what i see you did was placed (30t-t^3) in place of "t", which doesnt seem right to me, i need an equation that will be y as a function of x, that i can plug in a given y/x and find the other.
     
  6. Nov 8, 2008 #5

    HallsofIvy

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    No, that's not what you want to do: you want to convert parametric equations to a Cartesian equation, not find a composition. The simplest thing to do is to solve for t as a function of y, as I showed and then replace t by that expression in the formula for x.
     
  7. Nov 8, 2008 #6
    so what would i do to solve that, cant see what youre doing from here
     
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