Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Strange Trig Function solutions

  1. May 19, 2007 #1
    1. The problem statement, all variables and given/known data

    This is part of a problem for a nonlinear diff class... But it's the basic stuff that's tripping me up.

    Find all the max/min and concavity for

    [tex]v(x) = -cos(x)-Lx+1,\ \ \ 0<L<1 [/tex]

    3. The attempt at a solution

    Here's what I do:

    [tex] v'(x) = sin(x)-L [/tex]
    [tex]v''(x)=cos(x)[/tex]

    Set the first derivative to 0:

    [tex]sin(x)-L=0 \Rightarrow x = arcsin(L)[/tex]

    Here's where I'm confused. I say

    [tex]x = arcsin(L) + 2\pi n, \ \ \ \mbox{where }n \mbox{ is any integer}[/tex]

    But Maple says:

    [tex]x = arcsin(L)+2\pi n, \ \ \ \mbox{where }n \mbox{ is any integer, OR:}[/tex]
    [tex]x = arcsin(L) - 2arcsin(L)+2\pi n+\pi \ \ \ \mbox{where }n \mbox{ is any integer}[/tex]

    Where does the -2arcsin(L) and the +pi come from?????

    I really want to understand this once and for all.... I never took a trig class, and while I get by fine 99% of the time, I hit a brick wall when I come across this kind of stuff.

    Thanks!
     
  2. jcsd
  3. May 19, 2007 #2
    I think I almost understand it... The second set of values of x are the ones where the arcsin is in the II quadrant. Correct?

    And obviously it makes more sense to write:

    [tex] x = \pi - arcsin(L) + 2\pi n[/tex]

    for the 2nd set of x values.

    Yeah, I think I get it now. I was just looking at the graph of arcsine, and it doesn't seem to make sense from that.

    Is there a way to see that those are the solutions by looking at the arcsine graph?

    Thanks.
     
  4. May 19, 2007 #3
    I hope you understand the first solution.
    The second part of the solution is better understood if you manipulate the terms a bit , it can be written as [itex]x = (2n+1)\pi-arcsin(L)[/itex].
    Noting that [itex]sin(n\pi-x)=sin(x) [/itex] for all odd n, you can now see how this is also part of the solution.

    If you still don't follow, perhaps substitution of values for L into the solutions can give a clearer picture.
    For the graphical picture, you should use the sin(x) graph. Fix some L value, find arcsin(L), and see what other values of x, do you get sin(x)=L, that's all.
     
    Last edited: May 19, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook