Regular curve

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

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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

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Take the regular curve $\gamma$, 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

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Theorem: Any reparametrisation of a regular curve is regular

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Consider two parameterised curves $\gamma$ and $\tilde{\gamma}$ where $\gamma$ is regular.

We wish to show that $\tilde{\gamma}$ 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 $t$ we see: $$1=\frac{d\phi}{d\tilde{t}}\frac{d\psi}{dt}$$ - this means both $\frac{d\phi}{d\tilde{t} }$ and $\frac{d\psi}{dt}$ are non-zero

Next consider $$\tilde{\gamma}(\tilde{t})=\gamma(\phi(\tilde{t}))$$ differentiating this with respect to $\tilde{t}$ 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 $\gamma$ is regular
• $$\frac{d\phi}{d\tilde{t}}\ne 0$$ from the above.

This completes the proof.

See also

• Curve