When is arc length ≈ chord length

1. Aug 5, 2010

Saladsamurai

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
$\theta/2 < X_{cr}\Rightarrow \theta < 2*X_{cr}$ and from the arc length relationship SAB = r*theta we can assert that when $S_{AB}/R < 2*X_{cr}$, 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

Thanks!

2. Aug 5, 2010

Hurkyl

Staff Emeritus
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.

3. Aug 5, 2010

Saladsamurai

Hi Hurkyl 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
4. Aug 6, 2010

sagardip

Generally, we assume sinx~x for x<10degrees. So maybe that would be of some help to you

5. Aug 6, 2010

gomunkul51

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.

6. Aug 6, 2010

Hurkyl

Staff Emeritus
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.

7. Aug 6, 2010

Saladsamurai

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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