Arc Length and Curvature: Learn It 2

Giselle Miller

Arc Length and Curvature: Learn It 2

Arc-Length Parameterization

Now that we have a formula for the arc length of a curve, let’s examine what an arc-length function is and how we can use it to reparameterize curves.

The Arc-Length Function

If a vector-valued function represents the position of a particle in space as a function of time, then the arc-length function measures how far that particle travels as a function of time. The formula follows directly from the arc length formula:

[latex]s(t)=\int_{a}^{t} \sqrt{(f'(u))^{2}+(g'(u))^{2}+(h'(u))^{2}}du[/latex]

For curves in two dimensions, only two terms appear under the square root. We use the independent variable [latex]u[/latex] to distinguish between time and the variable of integration.

Since [latex]s(t)[/latex] measures distance traveled as a function of time, [latex]s'(t)[/latex] measures the speed of the particle at any given time. Differentiating both sides:

[latex]\begin{array}{ccc}\hfill {s'(t)} & =\hfill & {\frac{d}{dt}\bigg[\displaystyle\int_{a}^{t}\ \sqrt{\big(f'(u)\big)^{2}+\big(g'(u)\big)^{2}+\big(h'(u)\big)^{2}}du\bigg]}\hfill \\ \hfill & =\hfill & {\frac{d}{dt}\bigg[\displaystyle\int_{a}^{t}\ \left\Vert{\bf{r}}'(u)\right\Vert\ du\bigg]} \hfill \\ \hfill & =\hfill & {\left\Vert{\bf{r}}'(t)\right\Vert.} \hfill \\ \hfill \end{array}[/latex]

If [latex]{\bf{r}}(t)[/latex] defines a smooth curve, then arc length is always increasing, so [latex]s'(t)>0[/latex] for [latex]t>a[/latex].

There’s a special case worth noting: if [latex]{\bf{r}}(t)[/latex] is a curve where [latex]\left\Vert{\bf{r}}'(t)\right\Vert=1[/latex] for all [latex]t[/latex], then:

[latex]s(t)=\int_{a}^{t} \left\Vert{\bf{r}}'(u)\right\Vert du=\int_{a}^{t} 1 du=t-a[/latex]

This means [latex]t[/latex] represents the arc length as long as [latex]a=0[/latex].

arc-length function

Let [latex]{\bf{r}}(t)[/latex] describe a smooth curve for [latex]t\geq a[/latex]. The arc-length function is:

[latex]s(t)=\int_{a}^{t} \left\Vert{\bf{r}}'(u)\right\Vert du[/latex]

Additionally:

[latex]\frac{ds}{dt}=\left\Vert{\bf{r}}'(t)\right\Vert>0[/latex]

If [latex]\left\Vert{\bf{r}}'(t)\right\Vert=1[/latex] for all [latex]t\geq a[/latex], then the parameter [latex]t[/latex] represents the arc length from the starting point at [latex]t=a[/latex].

Reparameterizing by Arc Length

A useful application of this theorem is finding an alternative parameterization of a curve called an arc-length parameterization. Recall that any vector-valued function can be reparameterized via a change of variables.

For example, the function [latex]{\bf{r}}(t)=\langle 3\cos t, 3\sin t\rangle[/latex] for [latex]0\leq t\leq 2\pi[/latex] parameterizes a circle of radius [latex]3[/latex]. Changing the parameter from [latex]t[/latex] to [latex]4t[/latex] gives [latex]{\bf{r}}(t)=\langle 3\cos 4t, 3\sin 4t\rangle[/latex]. This still defines a circle of radius [latex]3[/latex], but now we only need [latex]0\leq t\leq\frac{\pi}{2}[/latex] to traverse the circle once.

To create an arc-length parameterization, we find the arc-length function [latex]s(t)[/latex] and solve for [latex]t[/latex] as a function of [latex]s[/latex]. Then we substitute this expression for [latex]t[/latex] back into the original function [latex]{\bf{r}}(t)[/latex]. The result is a vector-valued function written in terms of the parameter [latex]s[/latex].

How to: Find an Arc-Length Parameterization

Find the arc-length function: Calculate [latex]s(t)=\int_{a}^{t} \left\Vert{\bf{r}}'(u)\right\Vert du[/latex]
Solve for [latex]t[/latex]: Express [latex]t[/latex] as a function of [latex]s[/latex]
Substitute: Replace [latex]t[/latex] in the original function [latex]{\bf{r}}(t)[/latex] with your expression in terms of [latex]s[/latex]
Adjust the domain: Convert the restriction on [latex]t[/latex] to a restriction on [latex]s[/latex]

One advantage of arc-length parameterization is that the distance traveled along the curve starting from [latex]s=0[/latex] equals the parameter [latex]s[/latex]. This makes calculations more intuitive and simplifies formulas in contexts like curvature and line integrals.

Find the arc-length parameterization for each of the following curves:

First we find the arc-length function using the Arc Length Function equations:
[latex]\begin{array}{ccc}\hfill {s(t)} & =\hfill & {\displaystyle\int_{a}^{t}\ \left\Vert{\bf{r}}'(u)\right\Vert\ du}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{t}\ \left\Vert\langle{-4}\sin{u},\ 4\cos{u}\rangle\right\Vert du} \hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{t}\ \sqrt{(-4\sin{u})^{2}+(4\cos{u})^{2}}du}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{t}\ \sqrt{16\sin{^{2}u}+16\cos{^{2}u}}du}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{t}\ 4du=4t,}\hfill \\ \hfill \end{array}[/latex]

which gives the relationship between the arc length [latex]s[/latex] and the parameter [latex]t[/latex] as [latex]s=4t[/latex]; so, [latex]t=s/4[/latex]. Next we replace the variable [latex]t[/latex] in the original function [latex]{\bf{r}}(t)=4\cos{t{\bf{i}}}+4\sin{t{\bf{j}}}[/latex] with the expression [latex]s/4[/latex] to obtain

[latex]{\bf{r}}(s)=4\cos{\big(\frac{s}{4}\big)}{\bf{i}}+4\sin{\big(\frac{s}{4}\big)}{\bf{j}}[/latex].

This is the arc-length parameterization of [latex]{\bf{r}}(t)[/latex]. Since the original restriction on [latex]t[/latex] was given by [latex]t\geq 0[/latex], the restriction on [latex]s[/latex] becomes [latex]s/4\geq 0[/latex], or [latex]s\geq 0[/latex].
The arc-length function is given by the Arc Length Function equations:
[latex]\begin{array}{ccc}\hfill {s(t)} & =\hfill & {\displaystyle\int_{a}^{t}\ \left\Vert{\bf{r}}'(u)\right\Vert\ du}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{3}^{t}\ \left\Vert\langle{1},\ 2,\ 2\rangle\right\Vert\ du} \hfill \\ \hfill & =\hfill & {\displaystyle\int_{3}^{t}\ \sqrt{1^{2}+2^{2}+2^{2}}du}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{3}^{t}\ 3du}\hfill \\ \hfill & =\hfill & {3t-9.}\hfill \\ \hfill \end{array}[/latex]

Therefore, the relationship between the arc length [latex]s[/latex] and the parameter [latex]t[/latex] is [latex]s=3t-9[/latex], so [latex]t=\frac{s}{3}+3[/latex]. Substituting this into the original function [latex]{\bf{r}}(t)=\langle{t}+3,\ 2t-4,\ 2t\rangle[/latex] yields

[latex]{\bf{r}}(s)=\Big\langle\big(\frac{s}{3}+3\big)+3,\ 2\big(\frac{s}{3}+3\big)-4,\ 2\big(\frac{s}{3}+3\big)\Big\rangle=\Big\langle\frac{s}{3}+6,\ \frac{2s}{3}+2,\ \frac{2s}{3}+6\Big\rangle[/latex].

This is an arc-length parameterization of [latex]{\bf{r}}(t)[/latex]. The original restriction on the parameter [latex]t[/latex] was [latex]t\geq 3[/latex], so the restriction on [latex]s[/latex] is [latex](s/3)+3\geq 3[/latex], or [latex]s\geq 0[/latex].