# Is this arc length?

1. Mar 11, 2013

### autodidude

Could someone please explain why PQ in the diagram below is rΔθ? Isn't rΔθ arc length?

The best reason I can think of is that it's only an approximation for when the angle is very small, so PQ≈arclength=rΔθ. Not 100% sure though.

http://imageshack.us/scaled/landing/199/feynmanangle.jpg [Broken]

The diagram is from the first volume of the Feynman lectures in 18-3, in the section where he talks about rotation of rigid bodies.

Last edited by a moderator: May 6, 2017
2. Mar 11, 2013

### Staff: Mentor

It is an approximation for small $\Delta \theta$, right. I would expect that it is used in a differential somewhere, where the approximation gets exact.

3. Mar 11, 2013

### autodidude

Thanks mfb

4. Mar 11, 2013

### rcgldr

Start with the actual distance:

PQ = sqrt(Δr2 + (r sin(Δθ))2)

If this is circular motion, then Δr = 0, and as Δθ -> 0, then sin(Δθ) -> Δθ, and you end up with lim Δθ -> 0 of sqrt((r sin(Δθ))2) -> sqrt((r Δθ)2) -> r Δθ.

If r is some function of θ, then as long as Δr approaches zero more rapidly than r sin(Δθ), then lim Δθ -> 0 of f(Δr, r sin(Δθ)) -> f(0, r sin(Δθ)).

Last edited: Mar 11, 2013
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