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Sketching Graphs

  1. Jun 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Sketch [tex]\frac{cos x}{x + \frac{\pi}{2}}[/tex]


    2. Relevant equations
    3. The attempt at a solution

    hey guys, I just need help sketching this graph, I can work out the points of intercept with the y axis however having trouble doing more than that.

    On a more general note, I generally find sketching graphs such as this difficult, are there any tips you can give me? I suppose I could try to work out turning points by differentiating however is there a better way?

    Thanks in advance
     
    Last edited: Jun 17, 2009
  2. jcsd
  3. Jun 17, 2009 #2
    It might help to make a substitution [itex]u=x+\frac{1}{2}[/itex], then you can sketch [itex]\frac{sin(u)}{u}[/itex], which is easier, then translate it to get [itex]\frac{cos(x)}{x+\frac{\pi}{2}}[/itex].
     
  4. Jun 17, 2009 #3
    Ah, i get the sin bit, that's clever =P, (sorry we havn't been taught any of this yet, took me ages to see it).

    But as far as sketching sin (u) / u how would you even sketch that? :s
     
  5. Jun 17, 2009 #4
    I guess first think about whether it is even or odd, then think about what happens for large values of x, to get a general idea. Then you could consider what happens for x approaching 0 (try using the small angles approximation for sin, or you could use l'hopital's rule).
     
  6. Jun 17, 2009 #5
    I think I see, yeah =o, one final thing though, how would you translate from sin(u)/u to get to the original? Just move it pi/2 to the left? However i don't understand why having x + pi/2 will translate it like that, because it's it effectively;

    f(x) = sin (u) / u so to translate to cos (x) / x + pi/2 wouldn't it be like...

    f(x - pi/2) or something? I don't get it >.<

    Sureley it's an f(x + pi/2) translation but that would mean that the graph you need...eh i don't get this ><
     
  7. Jun 17, 2009 #6
    If you've just sketched [itex]\frac{sin(u)}{u}[/itex], call this [itex]f(u)[/itex], then you can easily sketch [itex]\frac{sin(u+\frac{\pi}{2})}{u+\frac{\pi}{2}}[/itex], because this is just [itex]f(u+\frac{\pi}{2})[/itex].

    But [itex]sin(u+\frac{\pi}{2})=cos(u)[/itex], so

    [tex]\frac{sin(u+\frac{\pi}{2})}{u+\frac{\pi}{2}} = \frac{cos(u)}{u+\frac{\pi}{2}}[/tex]

    This is just the function you were trying to sketch, but with the variable called u instead of x.
     
  8. Jun 17, 2009 #7
    Oh i see, that's great thanks, however would it not be -sin(u) / u then? because at the start substituting x + pi/2 into x for cos would move it <-- by pi/2 making it -sin then?
     
  9. Jun 17, 2009 #8
    When you substitute u = x + pi/2, you are actually translating the curve to the pi/2 to the right.

    Keeping everything in terms of x, you started with [itex]\frac{cos x}{x + \frac{\pi}{2}}[/itex], call this f(x), then you sketched [itex]\frac{sin(x)}{x}[/itex], which is [itex]\frac{cos (x-\frac{\pi}{2})}{(x-\frac{\pi}{2}) + \frac{\pi}{2}}[/itex], i.e. [itex]f(x-\frac{\pi}{2})[/itex], i.e. [itex]f(x)[/itex] translated [itex]\frac{\pi}{2}[/itex] to the right.

    Overall you took your function, translated it to the right, sketched it, and then translated it back to the left.
     
  10. Jun 17, 2009 #9
    Got it, that's great thanks!
     
  11. Jun 17, 2009 #10
    No problem. For reference here's a pic of the actual curve : )

    curve.jpg
     
  12. Jun 17, 2009 #11
    Cheers =D - btw you should use google chrome :PP
     
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