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Math Help - At what point does this parametric curve cross itself?

  1. #16
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    Re: At what point does this parametric curve cross itself?

    Quote Originally Posted by s3a View Post
    Actually, I found the angle that was roughly 160 degrees, and I converted it to the angle that was roughly 20 degrees, which yields the same answer, but I prefer my way, because it makes me understand what is going on geometrically a little better, but thank you for emphasizing that the answer must be positive. (When drawing the tangents, there are two pairs of angles, where each pair consists of two equivalent angles, and I chose (one of) the acute one(s).)
    But it is more important, I think, to understand that if \arccos(x)<0 then the angle is obtuse.

    So the |\arccos(x)| is the acute supplement.
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  2. #17
    s3a
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    Re: At what point does this parametric curve cross itself?

    Did you mean cos(x) < 0 instead of arccos(x) < 0?
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  3. #18
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    Re: At what point does this parametric curve cross itself?

    Quote Originally Posted by s3a View Post
    Did you mean cos(x) < 0 instead of arccos(x) < 0?
    No I did not. That is the whole point.
    Actually I have suspected that you are confused as to the difference in those two functions.

    Here is an example. Two lines \ell_1: P+t\vec{D}~\&~\ell_2: P+t\vec{E} where \ell_1\not\|\ell_2.

    Then the angle between \ell_1~\&~\ell_2 is \theta  = \arccos \left( {\frac{{D \cdot E}}{{\left\| D \right\| \cdot \left\| E \right\|}}} \right).

    Now if \arccos \left( {\frac{{D \cdot E}}{{\left\| D \right\| \cdot \left\| E \right\|}}} \right)>0 then \theta is acute.
    And if \arccos \left( {\frac{{D \cdot E}}{{\left\| D \right\| \cdot \left\| E \right\|}}} \right)<0 then \theta is obtuse.

    But if your question were to ask for the acute angle betwwen \ell_1~\&~\ell_2 then it is \theta  = \arccos \left( {\frac{{|D \cdot E|}}{{\left\| D \right\| \cdot \left\| E \right\|}}} \right).
    Last edited by Plato; December 6th 2013 at 04:50 PM.
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  4. #19
    s3a
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    Re: At what point does this parametric curve cross itself?

    If I remember correctly, I got a positive angle, alpha, when taking the inverse cosine and then did pi - alpha = theta, where cos(theta) gave me the correct answer.

    Is there something wrong with what I just said (in the post I am writing right now)?
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