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Polar form

  1. Feb 23, 2008 #1
    1. The problem statement, all variables and given/known data
    if i have a polar equation:
    r=arctan(tan(x))
    is that the same as:
    r=x ??


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 23, 2008 #2
    no!
    -graph the function
    -and try plugging in different numbers ... what if x = pi/2 ..
    -find domain of x
    -look at tan(x) function ... see the periodicity.
     
  4. Feb 23, 2008 #3
    i am trying to draw the graphs but cant becuase some values for x give a math error.
    so how would i draw the graph for r=arctan(tan(x))?
     
  5. Feb 23, 2008 #4
    draw tan x and atan x separately, and you would find why it's giving you math error ... it should not give error but I am not sure what you used.
    hint: when x = pi/2 tan(x) is inf when approached from left ...
    so what's atan (x) when x is inf or when x is -inf?

    you can draw atan(tan x) graph using this software:
    http://www.padowan.dk/graph/
    nice, simple, and fast.
    this helps me a lot
     
  6. Feb 23, 2008 #5
    yeah but i have to lot the points on a table and i dont know what to put for r when x=pi/2 and 3pi/2.
     
  7. Feb 23, 2008 #6
  8. Feb 24, 2008 #7

    Gib Z

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    The Tangent function fails the horizontal line test, ie There are many values for the independent variable that give the same Dependant variable. This means that The Inverse of the Tangent function is a relation but not a function. We can not find an inverse function for the usual tangent function, but we can find an inverse relation when we only take a little section of the function that does pass the horizontal line test. You will see that the function
    [tex]f(x) = \tan x, |x|> \frac{\pi}{2} [/tex] does pass the test. It is a one-to one, monotonically increasing function. Let us also note that the range of this function is the Reals. Now let us define the inverse of this, and observe that it will also be a one-to-one monotonically increasing function, its domain will be the Reals and its range will be all values of y such that [itex]|y| < \frac{\pi}{2}[/itex]. We commonly call this function the arctangent.

    Now, by the definition of what it is to be an inverse function, [itex]\arctan ( \tan x) = x[/itex], but only for [itex]|x| < \frac{\pi}{2}[/itex]. For values [tex]\frac{\pi}{2} \pm k\pi, k=1,2,3...[/tex], tan x is not defined, or, they are not in the domain of the tangent function, so arctan ( tan x) will not be defined either.

    For all other real values, we use the fact that the tangent function has a period of pi, ie [tex]f(x) = f(x\pm k\pi), k=1,2,3...[/tex], for all values of x in its domain. Using these facts, we can conclude that [tex]\arctan (\tan x) = x, |x| < \frac{\pi}{2}[/tex] and that the function also has a period of pi, since the argument of arctan has a period of pi.

    This function is somewhat inaccruately plotted on the Cartestian plain in that image rootX posted up (no offence intended). When drawing the graph, one should have drawn hollow white circles at the points [itex]( \frac{\pi}{2}, \frac{\pi}{2}), (\frac{\pi}{2}, \frac{-\pi}{2})[/itex] etc etc, to indicate that the graph does not include that point. Remember the function is not defined at those values.
     
  9. Feb 24, 2008 #8

    arildno

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    Dearly Missed

    Addendum:

    In GibZ's post, the word "inaccurately" is spelled somewhat inaccruately. :smile:
     
  10. Feb 24, 2008 #9

    Gib Z

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    Ahh I never pick up those spelling errors :( Ooh the Irony, it burns!
     
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