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Find cartesian equation from parametric equation

  1. Mar 24, 2013 #1

    trollcast

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    1. The problem statement, all variables and given/known data
    A curve is defined by the parametric equations:

    $$x=tan(t-1)\ \ \ \ \ \ \ y=cot^2(t+1)$$

    2. Relevant equations
    3. The attempt at a solution

    I think rearranging the first equation for t gives:

    $$t=tan^{-1}(x)+1$$

    However that doesn't help me as I don't know how to simplify the equation I'd get if I sub t into the y equation?
     
  2. jcsd
  3. Mar 24, 2013 #2

    mfb

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    Do you have to simplify the equation? You found an equation y=f(x).
     
  4. Mar 24, 2013 #3

    trollcast

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    I actually just realised that I've misread the equations so I can see how to solve it now.

    The equations should be:

    $$x=tan(t)-1\qquad y=cot^2(t)+1$$

    So the solution would be:

    $$tan(t)=x+1$$
    $$y=\frac{1}{tan^2(t)}$$
    $$y=\frac{1}{(x+1)^2}$$
     
  5. Mar 24, 2013 #4

    Dick

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    What happened to the '+1' in ##y=cot^2(t)+1##?
     
  6. Mar 24, 2013 #5

    trollcast

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    Oops, so it should be:

    $$x=tan(t)-1\qquad y=cot^2(t)+1\\
    tan(t)=x+1\\

    y - 1=\frac{1}{tan^2(t)}\\
    y=\frac{1}{(x+1)^2} +1\\
    y=\frac{(x+1)^2+1}{(x+1)^2}$$
     
  7. Mar 24, 2013 #6

    Dick

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    Sure.
     
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