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Thread: Parametric equation / Cartesian equation

  1. #1
    Member wiseguy's Avatar
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    Parametric equation / Cartesian equation

    My textbook says:




    x = 2 + 5 tan t
    y = 1 + 3 sec t

    By performing appropriate operations on the parametric equations, eliminate the parameter and get a Cartesian equation relating x and y. You should find that one of the Pythagorean properties will help.





    I believe the question is asking for me to take the two equations and make one equation in the f(x)=x format so it may be put into a graphing calculator. If this is true, I am 100% at a loss; I cannot find any reference in the chapter that tells how to do this, and as a last resort, I even checked all of the example problems... none relate to this.
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  2. #2
    A Plied Mathematician
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    Take $\displaystyle \sin^{2}(t)+\cos^{2}(t)=1$, and divide the entire equation through by $\displaystyle \cos^{2}(t)$ to get an identity that might help.
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by wiseguy View Post
    My textbook says:

    x = 2 + 5 tan t
    y = 1 + 3 sec t

    By performing appropriate operations on the parametric equations, eliminate the parameter and get a Cartesian equation relating x and y. You should find that one of the Pythagorean properties will help.


    I believe the question is asking for me to take the two equations and make one equation in the f(x)=x format so it may be put into a graphing calculator.
    Yes, that's what you are asked to do.

    If this is true, I am 100% at a loss; I cannot find any reference in the chapter that tells how to do this, and as a last resort, I even checked all of the example problems... none relate to this.
    You have the following system of equations:

    $\displaystyle x=2+5\frac{\sin t}{\cos t}$ and $\displaystyle y=1+3\frac{1}{\cos t}$

    Given the second equation, you can express $\displaystyle \cos t$ in terms of $\displaystyle y$.

    You can then replace $\displaystyle \scriptstyle \cos t$ (and even $\displaystyle \scriptstyle \sin t=\pm \sqrt{1-\cos^2 t}$) in the first equation by terms in which $\displaystyle t$ does not occur anymore. Finally, solve that remaining equation (in x and y) for y.
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  4. #4
    Member wiseguy's Avatar
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    Okay, I somewhat overlooked that I may of taken the three variable system of equations and solved for t, then y.


    x=2+5tan(t)
    (x-2)/5=tanT
    arctan((x-2)/5)=T

    y=1+3secT
    1/3=1/cosT
    3/1=cosT
    arccos3=T

    arctan((x-2)/5)=arccos3
    (arctan((x-2)/5)/(arccos3)=y


    Am I headed in the right direction with this work?
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  5. #5
    Senior Member eumyang's Avatar
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    (Ignore this, sorry. Something went wrong in my work.)
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  6. #6
    A Plied Mathematician
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    I think you might be able to get that approach to work. But if you do what I suggested, that is, see that $\displaystyle \tan^{2}(t)+1=\sec^{2}(t),$ then you simply solve each of your two equations for the relevant trig function, and drop them into the equation I just wrote down. Voila, you're done.
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  7. #7
    Member wiseguy's Avatar
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    x=2+5sinT/cosT
    (x-2)/5sinT=1/cosT;
    y=1+3(1/cosT)
    (y-1)/3=1/cosT;
    (x-2)/5sinT=(y-1)/3

    3(x-2)/5sinT+1=y


    I don't know if I screwed that up or not...

    I'm a little confused about using the tan^2 identity :/ would be nice to impress the teacher, lol.
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  8. #8
    Senior Member eumyang's Avatar
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    What Ackbeet was suggesting was what I erased -- I ended up getting two possible equations when I tried to solve for y, and I thought I was missing something. I'll just show you the beginning part:

    Solve each equation for the corresponding trig function:
    $\displaystyle \begin{aligned}
    tan\, t &= \frac{x - 2}{5} \\
    sec\, t &= \frac{y - 1}{3}
    \end{aligned}$

    Use the pythagorean identity $\displaystyle 1 + tan^2\, t = sec^2\, t$ that Ackbeet mentioned and plug in :
    $\displaystyle \begin{aligned}
    1 + tan^2\, t &= sec^2\, t \\
    1 + \left( \frac{x - 2}{5} \right)^2 &= \left( \frac{y - 1}{3} \right)^2 \\
    \end{aligned}$

    EDIT: Duh, this is an equation for a hyperbola. Who says we need to solve for y? I'm such an idiot. So just put it in standard form:
    $\displaystyle \begin{aligned}
    1 + \left( \frac{x - 2}{5} \right)^2 &= \left( \frac{y - 1}{3} \right)^2 \\
    \frac{(y - 1)^2}{9} - \frac{(x - 2)^2}{25} &= 1
    \end{aligned}$
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  9. #9
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    Hello, wiseguy!

    $\displaystyle \begin{array}{cccc}x &=& 2 + 5 \tan t & [1]\\
    y &=& 1 + 3\sec t & [2] \end{array}$

    Eliminate the parameter and get a Cartesian equation in $\displaystyle x$ and $\displaystyle y$.

    $\displaystyle \begin{array}{cccccc}\text{Rewrite [2]:} & \dfrac{y-1}{3} &=& \sec t & [3] \\ \\[-3mm] \text{Rewrite [1]:} & \dfrac{x-2}{5} &=& \tan t & [4] \end{array}$


    . . $\displaystyle \begin{array}{cccccc}
    \text{Square [3]:} & \left(\dfrac{y-1}{3}\right)^2 &=& \sec^2t & [5] \\ \\[-3mm]
    \text{Square [4]:} & \left(\dfrac{x-2}{5}\right)^2 &=& \tan^2t & [6] \end{array}$


    $\displaystyle \text{Subtract [5] - [6]: }\;\left(\dfrac{y-1}{3}\right)^2 - \left(\dfrac{x-2}{5}\right)^2 \;=\;\underbrace{\sec^2t - \tan^2t}_{\text{This is 1}} $


    $\displaystyle \text{Therefore: }\;\dfrac{(y-1)^2}{9} - \dfrac{(x-2)^2}{25} \;=\;1$

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  10. #10
    Member wiseguy's Avatar
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    Okay, so it doesn't matter that the equation needs to be solved for Y, $\displaystyle (x-h)^2/a^2+(y-k)2 /b^2 = 1$ works. Maybe I'll be the only one in the class who figured this tidbit out and I'll impress the instructor
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