Why Is PQ Approximated as rΔθ in Small Angles?

AI Thread Summary
The approximation PQ ≈ rΔθ is valid for small angles, where PQ represents the arc length in circular motion. As the angle Δθ approaches zero, the sine of the angle approximates the angle itself, leading to the conclusion that PQ simplifies to rΔθ. This relationship is derived from the distance formula, where Δr is negligible in circular motion. The discussion emphasizes that this approximation becomes exact in the limit as Δθ approaches zero. Understanding this concept is crucial for analyzing the rotation of rigid bodies.
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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

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.
 
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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.
 
Thanks mfb
 
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)).
 
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