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Trig: Writing the equation for vertical asymptote of a secant function?

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations
    How did they come up with [itex]\frac{1}{2}[/itex]+k for the equation of the vertical asymptote? I understand everything else except this.


    3. The attempt at a solution
    On this particular exercise, I graphed it and saw that each of my vertical dashed lines were all one whole unit apart. I've tried this method with another problem that had the dashed lines separated 2 units apart, took that 2, and multiplied it by the x=[itex]\prod[/itex]/2 + k[itex]\prod[/itex]. The result was [itex]\prod[/itex]+2k[itex]\prod[/itex], which was correct.

    I tried it with other numbers and have gotten the correct answer, but I have a feeling I'm still doing something wrong. Because with this particular one using my method, I input [itex]\prod[/itex]/2 + k[itex]\prod[/itex] as my answer for the vertical asymptote which was incorrect. What am I doing wrong?
     
  2. jcsd
  3. Sep 30, 2011 #2

    Dick

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    Science Advisor
    Homework Helper

    You would be right if the equation were y=4*sec(x). It's not. It's y=4*sec(pi*x). There's already a pi in the equation for y. y=4*sec(pi*x) doesn't have an asymptote at x=pi/2.
     
  4. Sep 30, 2011 #3
    I noticed multiplying the vertical asymptote formula/equation by 1/pi cancels out the pi, resulting in that 1/2+k... but where did they get 1/pi from? Does that have any relation to secant being 1/cos?
     
  5. Sep 30, 2011 #4
    Worked on another problem set up similarly and I think I got it!

    I noticed that simply taking the 'B' (like in the y=Asin[B(x-C)]+D formula), turning it into the reciprocal (1/B), and thennnn multiplying it by [itex]\prod[/itex]/2 + k[itex]\prod[/itex] gets me the right asymptote. This *does* relate to inverse trig functions (ie, sec being the reciprocal of cos), right?
     
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