Difference between revisions of "Regular curve"
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==Important point== | ==Important point== | ||
| − | The curve <math>\gamma(t)\mapsto(t,t^2)</math> is regular however <math>\gamma | + | {{Begin Theorem}} |
| + | 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]] | ||
| + | {{Begin Proof}} | ||
| + | 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. | ||
| + | |||
| + | 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. | ||
| + | {{End Proof}} | ||
| + | {{End Theorem}} | ||
==Any reparametrisation of a regular curve is regular== | ==Any reparametrisation of a regular curve is regular== | ||
| − | {{ | + | {{Begin Theorem}} |
| + | Theorem: Any reparametrisation of a regular curve is regular | ||
| + | {{Begin Proof}} | ||
| + | Consider two [[Parametrisation|parameterised curves]] {{M|\gamma}} and {{M|\tilde{\gamma} }} where {{M|\gamma}} is regular. | ||
| + | |||
| + | 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: | ||
| + | <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: | ||
| + | |||
| + | <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 {{M|\gamma}} is regular | ||
| + | * <math>\frac{d\phi}{d\tilde{t}}\ne 0</math> from the above. | ||
| + | |||
| + | This completes the proof. | ||
| + | {{End Proof}} | ||
| + | {{End Theorem}} | ||
==See also== | ==See also== | ||
Latest revision as of 22:12, 29 March 2015
Requires knowledge of Curve and Parametrisation
Contents
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.
