# 3.3 Arc length and curvature  (Page 2/18)

 Page 2 / 18

We now return to the helix introduced earlier in this chapter. A vector-valued function that describes a helix can be written in the form

$\text{r}\left(t\right)=R\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\frac{2\pi Nt}{h}\right)\phantom{\rule{0.1em}{0ex}}\text{i}+R\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\frac{2\pi Nt}{h}\right)\phantom{\rule{0.1em}{0ex}}\text{j}+t\phantom{\rule{0.2em}{0ex}}\text{k},\phantom{\rule{0.2em}{0ex}}0\le t\le h,$

where R represents the radius of the helix, h represents the height (distance between two consecutive turns), and the helix completes N turns. Let’s derive a formula for the arc length of this helix using [link] . First of all,

${r}^{\prime }\left(t\right)=-\frac{2\pi NR}{h}\text{sin}\left(\frac{2\pi Nt}{h}\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\frac{2\pi NR}{h}\text{cos}\left(\frac{2\pi Nt}{h}\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\text{k}.$

Therefore,

$\begin{array}{cc}\hfill s& ={\int }_{a}^{b}‖{r}^{\prime }\left(t\right)‖\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{h}\sqrt{{\left(-\frac{2\pi NR}{h}\text{sin}\left(\frac{2\pi Nt}{h}\right)\right)}^{2}+{\left(\frac{2\pi NR}{h}\text{cos}\left(\frac{2\pi Nt}{h}\right)\right)}^{2}+{1}^{2}}dt\hfill \\ & ={\int }_{0}^{h}\sqrt{\frac{4{\pi }^{2}{N}^{2}{R}^{2}}{{h}^{2}}\left({\text{sin}}^{2}\left(\frac{2\pi Nt}{h}\right)+{\text{cos}}^{2}\left(\frac{2\pi Nt}{h}\right)\right)+1}dt\hfill \\ & ={\int }_{0}^{h}\sqrt{\frac{4{\pi }^{2}{N}^{2}{R}^{2}}{{h}^{2}}+1}dt\hfill \\ & ={\left[t\sqrt{\frac{4{\pi }^{2}{N}^{2}{R}^{2}}{{h}^{2}}+1}\right]}_{0}^{h}\hfill \\ & =h\sqrt{\frac{4{\pi }^{2}{N}^{2}{R}^{2}+{h}^{2}}{{h}^{2}}}\hfill \\ & =\sqrt{4{\pi }^{2}{N}^{2}{R}^{2}+{h}^{2}}.\hfill \end{array}$

This gives a formula for the length of a wire needed to form a helix with N turns that has radius R and height h.

## Arc-length parameterization

We now have a formula for the arc length of a curve defined by a vector-valued function. Let’s take this one step further and examine what an arc-length function    is.

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 for the arc-length function follows directly from the formula for arc length:

$s\left(t\right)={\int }_{a}^{t}\sqrt{{\left({f}^{\prime }\left(u\right)\right)}^{2}+{\left({g}^{\prime }\left(u\right)\right)}^{2}+{\left({h}^{\prime }\left(u\right)\right)}^{2}}du.$

If the curve is in two dimensions, then only two terms appear under the square root inside the integral. The reason for using the independent variable u is to distinguish between time and the variable of integration. Since $s\left(t\right)$ measures distance traveled as a function of time, ${s}^{\prime }\left(t\right)$ measures the speed of the particle at any given time. Since we have a formula for $s\left(t\right)$ in [link] , we can differentiate both sides of the equation:

$\begin{array}{cc}\hfill {s}^{\prime }\left(t\right)& =\frac{d}{dt}\left[{\int }_{a}^{t}\sqrt{{\left({f}^{\prime }\left(u\right)\right)}^{2}+{\left({g}^{\prime }\left(u\right)\right)}^{2}+{\left({h}^{\prime }\left(u\right)\right)}^{2}}du\right]\hfill \\ & =\frac{d}{dt}\left[{\int }_{a}^{t}‖{r}^{\prime }\left(u\right)‖\phantom{\rule{0.2em}{0ex}}du\right]\hfill \\ & =‖{r}^{\prime }\left(t\right)‖.\hfill \end{array}$

If we assume that $\text{r}\left(t\right)$ defines a smooth curve, then the arc length is always increasing, so ${s}^{\prime }\left(t\right)>0$ for $t>a.$ Last, if $\text{r}\left(t\right)$ is a curve on which $‖{r}^{\prime }\left(t\right)‖=1$ for all t , then

$s\left(t\right)={\int }_{a}^{t}‖{r}^{\prime }\left(u\right)‖\phantom{\rule{0.2em}{0ex}}du={\int }_{a}^{t}1\phantom{\rule{0.2em}{0ex}}du=t-a,$

which means that t represents the arc length as long as $a=0.$

## Arc-length function

Let $\text{r}\left(t\right)$ describe a smooth curve for $t\ge a.$ Then the arc-length function is given by

$s\left(t\right)={\int }_{a}^{t}‖{r}^{\prime }\left(u\right)‖\phantom{\rule{0.2em}{0ex}}du.$

Furthermore, $\frac{ds}{dt}=‖{r}^{\prime }\left(t\right)‖>0.$ If $‖{r}^{\prime }\left(t\right)‖=1$ for all $t\ge a,$ then the parameter t represents the arc length from the starting point at $t=a.$

A useful application of this theorem is to find an alternative parameterization of a given curve, called an arc-length parameterization    . Recall that any vector-valued function can be reparameterized via a change of variables. For example, if we have a function $\text{r}\left(t\right)=⟨3\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t,3\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t⟩,0\le t\le 2\pi$ that parameterizes a circle of radius 3, we can change the parameter from t to $4t,$ obtaining a new parameterization $\text{r}\left(t\right)=⟨3\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}4t,3\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}4t⟩.$ The new parameterization still defines a circle of radius 3, but now we need only use the values $0\le t\le \pi \text{/}2$ to traverse the circle once.

Suppose that we find the arc-length function $s\left(t\right)$ and are able to solve this function for t as a function of s. We can then reparameterize the original function $\text{r}\left(t\right)$ by substituting the expression for t back into $\text{r}\left(t\right).$ The vector-valued function is now written in terms of the parameter s. Since the variable s represents the arc length, we call this an arc-length parameterization of the original function $\text{r}\left(t\right).$ One advantage of finding the arc-length parameterization is that the distance traveled along the curve starting from $s=0$ is now equal to the parameter s. The arc-length parameterization also appears in the context of curvature (which we examine later in this section) and line integrals, which we study in the Introduction to Vector Calculus .

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