# Regular curve

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Requires knowledge of Curve and Parametrisation

## Definition

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

## Definition: Regular Point

A point $\gamma(t)$ is called regular of $\dot\gamma\ne 0$ otherwise it is a Singular point

## Important point

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

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

Thus $\phi$ is not a diffeomorphism. Thus $\tilde{\gamma}(\tilde{t})=(\tilde{t}^3,\tilde{t}^6)$ 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 $\exists\phi$ which is a diffeomorphism such that: $\tilde{\gamma}(\tilde{t})=\gamma(\phi(\tilde{t}))$

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

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

$\frac{d\tilde{\gamma}}{d\tilde{t}}=\frac{d\gamma}{dt}\frac{d\phi}{d\tilde{t}}$ but:

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

This completes the proof.