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Homework Help: Probability and expectation values

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations

    I have that [tex]p(\theta)d\theta = \frac{1}{\pi}d\theta[/tex], this is definitely correct.

    Also y=r*sin(theta) so dy=r*cos(theta)*d(theta).

    Substituting d(theta) in above and simplifying, I have:

    [tex] p(y)dy = \frac{1}{pi} \frac{dy}{\sqrt{r^2-y^2}} [/tex]

    The problem is, integrating this from y=0 to y=r, I obtain 1/2, and not 1, as expected. I don't understand why this is.... anyone?
     
    Last edited: Sep 19, 2011
  2. jcsd
  3. Sep 19, 2011 #2
    I'm not totally sure, but I think the problem is that in your integration your considering only half of the posibilities, I mean, you are only considering one half of the semi circle that can be described by the needle.

    The min and max angles are \pi and 0. since y=r sin(\theta), you have y=0 for theta=0 or pi. This means that, when integrating between 0 and r, you need to multiply by a factor 2 to consider the whole spectrum.

    Hope it helps...
     
  4. Sep 19, 2011 #3
    Thanks, I thought exactly what you wrote. Still, is there any reason mathematically why this is so? When I do the same problem, but for the x-projection, this isn't an issue since the integration runs from -r to r.
     
  5. Sep 19, 2011 #4
    Yes... the mathemathical reason is that the satisfied relation for trigonometric functions is [itex]\cos^2(\theta)=1-\sin^2(\theta)[/itex], but when you replace [itex]\cos(\theta)=\sqrt{1-\sin^2(theta)}[/itex] you are leaving out all negative values of cos(theta), i.e. all angles between pi/2 and pi.... and so... you are considering only half of the posibilities.
     
  6. Sep 19, 2011 #5
    Perfect. Thanks!
     
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