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. 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.
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.
Start with the actual distance: PQ = sqrt(Δr^{2} + (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)).