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Is my proof correct?

  1. Jun 13, 2012 #1
    Hi,

    I am working on a proof . Could you kindly check my work in the attachment?
    Am I addressing the problem correctly? Also, am I missing any steps?

    The problem is stated in the attachment. My work in progress and a relevant diagram are also included.

    Thanks in advance for your help.
     

    Attached Files:

  2. jcsd
  3. Jun 13, 2012 #2

    HallsofIvy

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    Your "proof" doesn't mean anything. You list 7 fairly straight forward statements and then add the conclusion at the end. But it doesn't follow from the previous statement.

    In fact, exactly what is it you are trying to prove? "Show that angle [itex]\beta[/itex] will approximate [itex]1/(AB)^2[/itex] as AB increases in length". Are you saying that in the limit as AB goes to infinity? If so then [itex]1/(AB)^2[/itex] will go to 0. Is that what you are saying? That the angle goes to 0? Your "proof" cannot be right because it says nothing about AB "increasing". It is purely static.
     
  4. Jun 13, 2012 #3
    Hi,

    Thank you for answering. I see your point. A proof should reflect an increase in the length of "AB". What I would like to show is that angle Beta decreases (towards zero) at a quadratic rate. Apparently the decrease in that angle should be inversely proportional to (AB)^2. Any further suggestions will be appreciated.
     
  5. Jun 18, 2012 #4
    Something is going wrong with your proof. When AB grows, you have [itex] \cos(\alpha + \beta) \rightarrow 1 + \mathcal{O}(\beta) [/itex], [itex] \sin^2(\beta) \rightarrow \beta^2 [/itex], so you wouldn't get the scaling you're asked for. This problem is very easy to do using Taylor series: all you need to do is to write [itex] \beta = \theta-\alpha [/itex]and then expand the angles into series -- you should get something like [itex] \beta \sim \frac{AC}{AB} - \frac{AC}{AF} [/itex]. Perhaps there is some geometric way of getting a similar relation?
     
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