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Math Help - polar equations with trig intersection points

  1. #1
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    polar equations with trig intersection points

    asd
    Last edited by stargirldrummer187; March 2nd 2008 at 09:56 AM.
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  2. #2
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    Quote Originally Posted by stargirldrummer187 View Post
    I am trying to find the intersecting points of two equations in a polor coordinate grid, by setting them equal.

    In the equations x is meant to be theta.
    And we are in degrees.

    Our first equation is
    r=1+5cos(x)
    and the second is
    r=3sec(x-30)

    I have many trig identities and have tired using them any way I can.
    If you can solve this:
    1+5cos(x)=3sec(x-30)
    Please let me know, it would be much appreciated.
    Thank you,
    from many very lost pre calc students
    It might be easier to switch to Cartesian coordinates. I'll explain soon - have to run and put a fire out right now.

    Nope, false alarm. Same for the fire.

    Any reason to think you have to get exact solutions to the equation?

    Also, have you tried drawing the curves .... you realise that r = 3 \sec (\theta - 30) is a line ...?
    Last edited by mr fantastic; March 1st 2008 at 06:06 PM.
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  3. #3
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    We have somewhat tired converting the equations to cartesian equations with conversion formulas, but it turned out..funny.
    If you can do it, that would be wonderful.

    We also know you can graph them as cartesian equations and get the intersections, but if we were to do that (according to our teacher) we have to have proof of why it works.

    Thank you for your input, We'll look forward to more help.
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    Yeah, we need exact..our teacher is a tad insane.
    Good..but very insane.

    Yes we know it's a line, but how is that going to help us solve the situation?
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  5. #5
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    Quote Originally Posted by stargirldrummer187 View Post
    Yeah, we need exact..our teacher is a tad insane.
    Good..but very insane.

    Yes we know it's a line, but how is that going to help us solve the situation?
    My only other thought at this stage is to get an equation with r.

    r = 3 \sec (\theta - 30)

    \Rightarrow \sqrt{3} r \cos \theta + r \sin \theta = 6 .... (1).


    r = 1 + 5 \cos \theta

    \Rightarrow \cos \theta = \frac{r - 1}{5} .... (2).


    From (2), \sin \theta = \pm \frac{\sqrt{25 - (r - 1)^2}}{5} = \pm \frac{\sqrt{(6 - r)(r + 4)}}{5} .... (3).


    Substitute (2) and (3) into (1) and re-arrange into a quartic equation in r. You might get r from this (I'll cop that I haven't actually tried .....)
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    I'm a little confused as to how you got that last line, but I'll work on it and get back to you.
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  7. #7
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    Quote Originally Posted by stargirldrummer187 View Post
    I'm a little confused as to how you got that last line, but I'll work on it and get back to you.
    From the Pythagorean Identity: \sin^2 \theta = 1 - \cos^2 \theta

    1 - \cos^2 \theta = 1 - \left( \frac{r - 1}{5} \right)^2 = 1 - \frac{r^2 - 2r + 1}{25}

    = \frac{25 - r^2 + 2r - 1}{25} = \frac{-r^2 + 2r + 24}{25} = \frac{-(r^2 - 2r - 24)}{25} etc.

    But on reflection I don't think you're gonna get anything even a little bit easy popping out of the resulting quartic equation in r .....

    I'll keep it in the back of my mind.
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