Sketching Graphs

  • Thread starter Chewy0087
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  • #1
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Homework Statement



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


Homework Equations


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
 
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Answers and Replies

  • #2
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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].
 
  • #3
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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
 
  • #4
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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).
 
  • #5
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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 ><
 
  • #6
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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.
 
  • #7
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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?
 
  • #8
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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.
 
  • #9
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Got it, that's great thanks!
 
  • #10
42
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No problem. For reference here's a pic of the actual curve : )

curve.jpg
 
  • #11
370
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Cheers =D - btw you should use google chrome :PP
 

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