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Homework Help: When is arc length ≈ chord length

  1. Aug 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Maybe this is precalculus? Either way, here is a question that I am curious about. Take a circle of radius R and sweep out an arc length SAB with endpoints 'A' and 'B' over angle theta. For a short enough arc length, I believe that we could approximate SAB by the chord length AB.

    I am trying quantify "when" the ratio SAB/R is such that the approximation is a good one. I guess a good start is to establish some relationships. From the picture below, we see that the arc length is given by SAB=R*theta and the chord length is given by AB = 2*R*sin(theta/2).

    So I believe we should now ask when does R*theta ≈ 2*R*sin(theta/2).

    I know from other problems we often employ the approximation that if an angle 'X' is "small enough", then sin(X)≈X. It looks like this would help here since if we let sin(theta/2) = theta/2, then the approximation above becomes an identity. I am just having trouble figuring out how to relate this all back to the ratio SAB/R ? What if we said that we already know that for some critical value of the angle X we can approximate sin(X) = X. We will call that "known" value Xcr. So if theta/2 < Xcr then SAB≈AB. So
    [itex]\theta/2 < X_{cr}\Rightarrow \theta < 2*X_{cr}[/itex] and from the arc length relationship SAB = r*theta we can assert that when [itex]S_{AB}/R < 2*X_{cr}[/itex], the approximation is good.

    Can someone let me know if they think my logic is flawed? I have never done something like this from scratch before :redface:

    Thanks!

    chord.jpg
     
  2. jcsd
  3. Aug 5, 2010 #2

    Hurkyl

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    Your argument looks like a typical reasonable heuristic one.


    However, do note that you are paying attention to relative error. If x and sin(x) are "close", then relatively speaking, Rx and Rsin(x) are equally "close".

    However, if absolute error matters, your condition on a good approximation will depend on both R and x.
     
  4. Aug 5, 2010 #3
    Hi Hurkyl :smile: I am wondering, isn't this the same as saying that that my condition of a good approximation depends on how close the value of the ratio S/R is to 2*Xcr ? Seeing as the angle is given by S/R.

    Thanks!
     
    Last edited: Aug 6, 2010
  5. Aug 6, 2010 #4
    Generally, we assume sinx~x for x<10degrees. So maybe that would be of some help to you
     
  6. Aug 6, 2010 #5
    Yes you can ! :)

    but it all depends on how accurate you want to be and/or if you don't have an easy alternative.

    *basically its a good approximation, you don't need to guess just crunch the numbers and check by how much percent sin(x) and x differ for a range of values.
     
  7. Aug 6, 2010 #6

    Hurkyl

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    Nope.

    While the value SAB/R tells you everything about the relative error when approximating SAB with AB, it doesn't tell you anything about the absolute error.
     
  8. Aug 6, 2010 #7
    OK. So perhaps you are saying that for some fixed R, the choice of X will fix my absolute error? That is, if given the choice between 2 different values of X1 and X2, then the one that makes sin(X) closest to X is the one with the lowest absolute error.

    Sorry, I am just trying to get a feel for what you are saying.
     
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