Feynman Lectures on Physics - Rotation in 2 dimensions

AI Thread Summary
The discussion focuses on understanding concepts from section 18-2 of The Feynman Lectures on Physics regarding rotation in two dimensions. The user raises two questions about the geometry involved, specifically the right angle at point P and the relationship between QP and rΔθ. Responses clarify that for small angles, sin(Δθ) approximates Δθ, allowing the use of QP = rΔθ instead of r * tan(Δθ). The conversation highlights the simplifications used in calculus and the historical context of mathematical approaches. Overall, the discussion emphasizes the nuances of angular relationships in physics.
ZirconiumAce
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This isn't really a 'problem', I'm just trying to follow Feynman's reasoning in section 18-2 of Volume 1 The Feynman Lectures on Physics. I've attached a png of the paragraph in question.

I have 2 issues with this:

1. If the length OQ = OP, how can there be a right angle at P(x,y)?

2. I don't understand why QP = r \Delta\theta, rather than r * tan (\Delta\theta)

Am I missing something? I've checked the latest errata and this isn't mentioned.

Thanks for your help.
 

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Welcome to PF!

Hi ZirconiumAce! Welcome to PF! :wink:
ZirconiumAce said:
1. If the length OQ = OP, how can there be a right angle at P(x,y)?

2. I don't understand why QP = r \Delta\theta, rather than r * tan (\Delta\theta)

Because that's the way these ∆s work …

we ignore anything that's small compared with a ∆ (in the same way that we ignore dh2, compared with dh, when we do calculus proofs)

So ∆θ = sin∆θ = tan∆θ. :wink:

(sin∆θ = ∆θ - (∆θ)3/3! + …)
 
Ah yes a Taylor series. Mathematics back in the days of the slide rule I suppose. Now we don't bother with simplifications as much. Cheers! How about the right angle?
 
I've got it.
 
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