Difference between revisions of "Regular curve"

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Requires knowledge of [[Curve]] and [[Parametrisation]]
  
 
==Definition==
 
==Definition==
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==Important point==
 
==Important point==
The curve <math>\gamma(t)\mapsto(t,t^2)</math> is regular however <math>\gamma'(t)\mapsto(t^3,t^6)</math> is not - it is not technically a [[Reparametrisation|reparametrisation]]  
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{{Begin Theorem}}
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The curve <math>\gamma(t)\mapsto(t,t^2)</math> is regular however <math>\tilde{\gamma}(t)\mapsto(t^3,t^6)</math> is not - it is not technically a [[Reparametrisation|reparametrisation]]  
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{{Begin Proof}}
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Take the regular curve {{M|\gamma}}, and the "reparametrisation" <math>\phi(t)\mapsto t^3=\tilde{t}</math> - this is indeed bijective and smooth, however its inverse <math>\phi^{-1}(\tilde{t})=\tilde{t}^\frac{1}{3}</math> is not smooth.
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Thus <math>\phi</math> is not a diffeomorphism. Thus <math>\tilde{\gamma}(\tilde{t})=(\tilde{t}^3,\tilde{t}^6)</math> is not a reparametrisation.
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{{End Proof}}
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{{End Theorem}}
  
 
==Any reparametrisation of a regular curve is regular==
 
==Any reparametrisation of a regular curve is regular==
{{Todo|reparametrisation theorem}}
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{{Begin Theorem}}
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Theorem: Any reparametrisation of a regular curve is regular
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{{Begin Proof}}
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Consider two [[Parametrisation|parameterised curves]] {{M|\gamma}} and {{M|\tilde{\gamma} }} where {{M|\gamma}} is regular.
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We wish to show that {{M|\tilde{\gamma} }} is regular. By being a reparametrisation we know <math>\exists\phi</math> which is a diffeomorphism such that: <math>\tilde{\gamma}(\tilde{t})=\gamma(\phi(\tilde{t}))</math>
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Then taking the equation: <math>t=\phi(\psi(t))</math> and differentiating with respect to {{M|t}} we see:
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<math>1=\frac{d\phi}{d\tilde{t}}\frac{d\psi}{dt}</math> - this means both {{M|\frac{d\phi}{d\tilde{t} } }} and {{M|\frac{d\psi}{dt} }} are non-zero
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Next consider <math>\tilde{\gamma}(\tilde{t})=\gamma(\phi(\tilde{t}))</math> differentiating this with respect to {{M|\tilde{t} }} yields:
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<math>\frac{d\tilde{\gamma}}{d\tilde{t}}=\frac{d\gamma}{dt}\frac{d\phi}{d\tilde{t}}</math> but:
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* <math>\frac{d\gamma}{dt}\ne 0</math> as the curve {{M|\gamma}} is regular
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* <math>\frac{d\phi}{d\tilde{t}}\ne 0</math> from the above.
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This completes the proof.
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{{End Proof}}
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{{End Theorem}}
  
 
==See also==
 
==See also==
* [[Parameterisation]]
 
 
* [[Curve]]
 
* [[Curve]]
  
 
{{Definition|Geometry of Curves and Surfaces|Differential Geometry}}
 
{{Definition|Geometry of Curves and Surfaces|Differential Geometry}}

Latest revision as of 22:12, 29 March 2015

Requires knowledge of Curve and Parametrisation

Definition

A curve [math]\gamma:\mathbb{R}\rightarrow\mathbb{R}^3[/math] usually (however [math]\gamma:A\subseteq\mathbb{R}\rightarrow\mathbb{R}^n[/math] more generally) is called regular if all points ([math]\in\text{Range}(\gamma)[/math]) are regular

Definition: Regular Point

A point [math]\gamma(t)[/math] is called regular of [math]\dot\gamma\ne 0[/math] otherwise it is a Singular point

Important point

The curve [math]\gamma(t)\mapsto(t,t^2)[/math] is regular however [math]\tilde{\gamma}(t)\mapsto(t^3,t^6)[/math] is not - it is not technically a reparametrisation


Take the regular curve [ilmath]\gamma[/ilmath], and the "reparametrisation" [math]\phi(t)\mapsto t^3=\tilde{t}[/math] - this is indeed bijective and smooth, however its inverse [math]\phi^{-1}(\tilde{t})=\tilde{t}^\frac{1}{3}[/math] is not smooth.

Thus [math]\phi[/math] is not a diffeomorphism. Thus [math]\tilde{\gamma}(\tilde{t})=(\tilde{t}^3,\tilde{t}^6)[/math] is not a reparametrisation.


Any reparametrisation of a regular curve is regular

Theorem: Any reparametrisation of a regular curve is regular


Consider two parameterised curves [ilmath]\gamma[/ilmath] and [ilmath]\tilde{\gamma} [/ilmath] where [ilmath]\gamma[/ilmath] is regular.

We wish to show that [ilmath]\tilde{\gamma} [/ilmath] is regular. By being a reparametrisation we know [math]\exists\phi[/math] which is a diffeomorphism such that: [math]\tilde{\gamma}(\tilde{t})=\gamma(\phi(\tilde{t}))[/math]


Then taking the equation: [math]t=\phi(\psi(t))[/math] and differentiating with respect to [ilmath]t[/ilmath] we see: [math]1=\frac{d\phi}{d\tilde{t}}\frac{d\psi}{dt}[/math] - this means both [ilmath]\frac{d\phi}{d\tilde{t} } [/ilmath] and [ilmath]\frac{d\psi}{dt} [/ilmath] are non-zero


Next consider [math]\tilde{\gamma}(\tilde{t})=\gamma(\phi(\tilde{t}))[/math] differentiating this with respect to [ilmath]\tilde{t} [/ilmath] yields:

[math]\frac{d\tilde{\gamma}}{d\tilde{t}}=\frac{d\gamma}{dt}\frac{d\phi}{d\tilde{t}}[/math] but:

  • [math]\frac{d\gamma}{dt}\ne 0[/math] as the curve [ilmath]\gamma[/ilmath] is regular
  • [math]\frac{d\phi}{d\tilde{t}}\ne 0[/math] from the above.

This completes the proof.


See also