Ngô Quốc Anh

July 13, 2010

What is a curve parametrized by arc length?

Filed under: Riemannian geometry — Ngô Quốc Anh @ 6:58

Let \alpha :I \to \mathbb R^3 be any curve and t_0 \in I. We define the arc length function from t_0 which will denote by S : I \to \mathbb R by

\displaystyle S(t)=\int_{t_0}^t|\alpha'(s)|ds.

Since s \mapsto |\alpha'(s)| is, in general, continuous, the function S is only C^1 and

S'(t)=|\alpha'(t)|.

If we assume \alpha is regular, i.e. \alpha'(t) \ne 0 for any t \in I, then by the Inverse Function Theorem S is differentiable increasing open function. Then if we put

J=S(I)

then S : I \to J is a diffeomorphism between two open intervals. Let \phi : J \to I be the inverse diffeomorphism and let \beta : J \to \mathbb R^3 be the re-parametrization of \alpha given by

\beta = \alpha \circ \phi.

Then this new curve satisfies

\displaystyle \beta '(s) = \alpha '(\phi (s))\phi '(s) = \frac{{\alpha '(\phi (s))}}{{|\alpha '(\phi (s))|}}

and thus |\beta'(s)|=1 for any s \in J. It follows from the above that any regular curve admits a re-parametrization by arc length. So we have

Definition. A curve \alpha :I \to \mathbb R^3 is said to be parametrized by arc length S if |\alpha'(t)| = 1 for all t \in I.

Roughly speaking, instead of using a time variable t to parametrize a curve we use its arc length S. The new equation for the curve in terms of S is called the curve parametrized by arc length.

Example (logarithmic spiral). The curve \alpha : \mathbb R \to \mathbb R^2 given by

\displaystyle \alpha(t)=(ae^{bt}\cos t, a e^{bt} \sin t)

with a>0 and b<0 is called the logarithmic spiral. The case a=2 and b=-\frac{1}{5} is showed below.

The arc length of the logarithmic spiral is given by

\displaystyle S(t) = \frac{{\sqrt {{a^2}(1 + {b^2}){e^{2bt}}} }}{b}.

Solving this equation gives

\displaystyle t = \frac{1}{b}\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}.

Thus the equation for \alpha parametrized by arc length is given by

\displaystyle \alpha (s) = \left(a{e^{\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}}}\cos \left( {\frac{1}{b}\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}} \right),a{e^{\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}}}\sin \left( {\frac{1}{b}\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}} \right)\right).

A lot of formulas become simpler when your curve is parametrized by arc length. For example, the curvature is calculated by taking the derivative of the tangent vector. When the curve is parametrized by arc length, the tangent vector has constant length one. This implies that the derivative of the tangent vector is always normal to the tangent vector, which is necessary to find the curvature.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: