# Regular curve

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.