Velocity with respect to arclength is a unit tangent vector?

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Hi all,

I have long had this unsolved question about arclength parameterization in my head and I just can't bend my head around it. I seem not to be able to understand why velocity with arclength as the parameter is automatically a unit tangent vector. My professor proved in class that

s(s) = s = ∫ ||v(s)|| ds with lower and upper bound so and s1

He said that by fundamental theorem of calculus, ds/ds = 1 = ||v(s)||, so the velocity vector with respect to arclength s is always a unit tangent vector.

I understand the mathematical proof the professor did but I really need an intuitive explanation. I have thought about this situation where velocity is represented by distance in one dimension, but it seems that it is not always the case that V(s) will be unit speed. Then why is it the case when velocity is parameterized in terms of arclength so that the speed automatically becomes 1?
 

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  • #2
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where velocity is represented by distance in one dimension, but it seems that it is not always the case that V(s) will be unit speed. Then why is it the case when velocity is parameterized in terms of arclength so that the speed automatically becomes 1?
i do not know but how you define velocity with respect to arc length becomes important
if you are looking at a length -one dimension then the rate of change of displacement will be velocity.....the length taken is of some curve but perhaps now one has to forget about the curve- i.e. a two dimensional displacement.
the observer does not know he is moving on a curve....its my common sense response... may be wrong!
 
  • #3
FactChecker
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Suppose the path was parameterized by the variable t=time. As an object moves along a path, its velocity vector is always tangent to the path it is following. Also, the distance, ds, that it covers in time dt, on the path is instantaneously proportional to the magnitude of the velocity vector.
Now suppose that the path is parameterized by the arc length, s, so the "velocity" with respect to s has magnitude 1. My only concern is that it seems like a strange thing to call a velocity.
 
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Hi FactChekcer, I don't seem to understand the logic leap you made between
Also, the distance, ds, that it covers in time dt, on the path is instantaneously proportional to the magnitude of the velocity vector.
and
Now suppose that the path is parameterized by the arc length, s, so the "velocity" with respect to s has magnitude 1.
What I want to understand is what happens when you set arclength as the parameter and why
so the "velocity" with respect to s has magnitude 1.
 
  • #5
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At any point along the curve, the magnitude of velocity is telling you how fast the point is moving in the direction of the path, adding to s(t). So |v(t)| = ds/dt
 
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Svein
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I have long had this unsolved question about arclength parameterization in my head and I just can't bend my head around it. I seem not to be able to understand why velocity with arclength as the parameter is automatically a unit tangent vector.
That is one of the basic tenets of classical differential geometry (start with https://en.wikipedia.org/wiki/Torsion_of_a_curve).
 

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