Parametrizations and regular curves

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I'd like to have someone explain to me the purpose of the variable change in the proof of the following theorem:

A parametrized curve has a unit-speed reparametrization if and only if it is regular.

Proof: Suppose first that a parametrized curve γ: (a,b) → R^n has a unit-speed reparametrization μ, with reparametrization map φ. Letting t = φ(s) we have μ(s) = γ(t) and so

dμ/ds = (dγ/dt)(dt/ds)

Therefore, ||dμ/ds|| = ||dγ/dt|| ⋅ |dt/ds|

Since μ has unit speed, ||dμ/ds|| = 1, so dγ/dt cannot be zero.
===============

I don't see why it was necessary? Couldn't we have just concluded the exact same thing without the variable change? Another thing, I notice that they use absolute value for |dt/ds| this is because t = φ(s) is a single variable function right? So it's like a 'constant'? Was it just to make this more explicit? Because the norm properties don't have ||fg|| = ||f|| ||g||.

Thanks for the help!
 

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  • #2
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Anyone?
 
  • #3
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Can anyone help me out on this?
 
  • #4
lavinia
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===============

I don't see why it was necessary? Couldn't we have just concluded the exact same thing without the variable change?
How would you make the conclusion without the chain rule?

Another thing, I notice that they use absolute value for |dt/ds| this is because t = φ(s) is a single variable function right? So it's like a 'constant'?
You are just taking the norm of a scalar time a vector.
 
  • #5
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How would you make the conclusion without the chain rule?
I'm saying we're not supposed to use the Chain Rule. The Chain Rule is obviously necessary, but we could apply the chain rule without having to make such a variable change.


You are just taking the norm of a scalar time a vector.
I know the norm property for this, but I don't see how dt/ds is a scalar? Could you explain?
 
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When you say scalar, I suppose you mean that it is a function of one variable plus its a vector valued function - its a 1-tuple. So its norm just simply behaves as a normal absolute value.
 
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lavinia
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When you say scalar, I suppose you mean that it is a function of one variable plus its a vector valued function - its a 1-tuple. So its norm just simply behaves as a normal absolute value.
right
 
  • #8
lavinia
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When you say scalar, I suppose you mean that it is a function of one variable plus its a vector valued function - its a 1-tuple. So its norm just simply behaves as a normal absolute value.
If the curve is not regular then no reparameterization can regularize it. This follows from the chain rule. I suspect any other proof is equivalent.

If the curve is reparameterized by arc length then this reparameterized curve is regular. But the original curve may not be. To show that it is - use the chain rule.
 
  • #9
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If the curve is not regular then no reparameterization can regularize it. This follows from the chain rule. I suspect any other proof is equivalent.

If the curve is reparameterized by arc length then this reparameterized curve is regular. But the original curve may not be. To show that it is - use the chain rule.
I don't think you understood my problem. I know you need to use the chain rule. But I don't see the purpose of the variable change in the proof since you can apply the chain rule without making the variable change. So my question was what was the purpose of it...
 
  • #10
lavinia
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I don't think you understood my problem. I know you need to use the chain rule. But I don't see the purpose of the variable change in the proof since you can apply the chain rule without making the variable change. So my question was what was the purpose of it...
How do you apply the Chain rule without the change of variables? Can you show me?
 

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