I Vector math (small angle approximation)

Given the following vectors:

vectors.png




how can i determine that Θ = Δp/p ?
I can understand that p + Δp = p' but nothing arrives from this. Any help is welcome!
 

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Sin Φ ≈ ∅ for small angles.

(Different) Mentor edit: The above should be ##\sin(\theta) \approx \theta##, for ##\theta## in radians.
 
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Delta2

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Sin Φ ≈ ∅ for small angles.
You meant to say that ##\sin\theta\approx \theta## for small angles.
 
You meant to say that ##\sin\theta\approx \theta## for small angles.
Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.
 
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To be clear, the angle is in radian measure.
 

A.T.

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Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.
You also see that from the fact that sin(0) = 0, sin'(0) = 1 and sin''(0) = 0. So sin is equivalent to the identity function up to the 2nd derivative around 0.
 
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Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.
It's instructive to take ##\sin(\theta)## for some small values of ##\theta## (always in radians) to see just how good the approximation is. Even at ##\theta = 0.1##, which is a little larger than what we usually consider "small compared to 1", it's a pretty good approximation.
 
It's instructive to take ##\sin(\theta)## for some small values of ##\theta## (always in radians) to see just how good the approximation is. Even at ##\theta = 0.1##, which is a little larger than what we usually consider "small compared to 1", it's a pretty good approximation.
okay i will try some small numbers. Thank you sir.
 

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