# 3.3 Arc length and curvature  (Page 3/18)

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## Finding an arc-length parameterization

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

1. $\text{r}\left(t\right)=4\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{i}+4\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{j},t\ge 0$
2. $\text{r}\left(t\right)=⟨t+3,\phantom{\rule{0.2em}{0ex}}2t-4,2t⟩,t\ge 3$
1. First we find the arc-length function using [link] :
$\begin{array}{cc}\hfill s\left(t\right)& ={\int }_{a}^{t}‖{r}^{\prime }\left(u\right)‖\phantom{\rule{0.2em}{0ex}}du\hfill \\ & ={\int }_{0}^{t}‖⟨-4\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}u,4\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}u⟩‖\phantom{\rule{0.2em}{0ex}}du\hfill \\ & ={\int }_{0}^{t}\sqrt{{\left(-4\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}u\right)}^{2}+{\left(4\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}u\right)}^{2}}\phantom{\rule{0.2em}{0ex}}du\hfill \\ & ={\int }_{0}^{t}\sqrt{16\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{2}u+16\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}u}\phantom{\rule{0.2em}{0ex}}du\hfill \\ & ={\int }_{0}^{t}4\phantom{\rule{0.2em}{0ex}}du=4t,\hfill \end{array}$

which gives the relationship between the arc length s and the parameter t as $s=4t;$ so, $t=s\text{/}4.$ Next we replace the variable t in the original function $\text{r}\left(t\right)=4\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{i}+4\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{j}$ with the expression $s\text{/}4$ to obtain
$\text{r}\left(s\right)=4\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\frac{s}{4}\right)\phantom{\rule{0.1em}{0ex}}\text{i}+4\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\frac{s}{4}\right)\phantom{\rule{0.1em}{0ex}}\text{j}\text{.}$

This is the arc-length parameterization of $\text{r}\left(t\right).$ Since the original restriction on t was given by $t\ge 0,$ the restriction on s becomes $s\text{/}4\ge 0,$ or $s\ge 0.$
2. The arc-length function is given by [link] :
$\begin{array}{cc}\hfill s\left(t\right)& ={\int }_{a}^{t}‖{r}^{\prime }\left(u\right)‖\phantom{\rule{0.2em}{0ex}}du\hfill \\ & ={\int }_{3}^{t}‖⟨1,2,2⟩‖\phantom{\rule{0.2em}{0ex}}du\hfill \\ & ={\int }_{3}^{t}\sqrt{{1}^{2}+{2}^{2}+{2}^{2}}\phantom{\rule{0.2em}{0ex}}du\hfill \\ & ={\int }_{3}^{t}3\phantom{\rule{0.2em}{0ex}}du\hfill \\ & =3t-9.\hfill \end{array}$

Therefore, the relationship between the arc length s and the parameter t is $s=3t-9,$ so $t=\frac{s}{3}+3.$ Substituting this into the original function $\text{r}\left(t\right)=⟨t+3,\phantom{\rule{0.2em}{0ex}}2t-4,2t⟩$ yields
$\text{r}\left(s\right)=⟨\left(\frac{s}{3}+3\right)+3,\phantom{\rule{0.2em}{0ex}}2\left(\frac{s}{3}+3\right)-4,2\left(\frac{s}{3}+3\right)⟩=⟨\frac{s}{3}+6,\phantom{\rule{0.2em}{0ex}}\frac{2s}{3}+2,\frac{2s}{3}+6⟩.$

This is an arc-length parameterization of $\text{r}\left(t\right).$ The original restriction on the parameter $t$ was $t\ge 3,$ so the restriction on s is $\left(s\text{/}3\right)+3\ge 3,$ or $s\ge 0.$

Find the arc-length function for the helix

$\text{r}\left(t\right)=⟨3\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t,3\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t,4t⟩,t\ge 0.$

Then, use the relationship between the arc length and the parameter t to find an arc-length parameterization of $\text{r}\left(t\right).$

$s=5t,$ or $t=s\text{/}5.$ Substituting this into $\text{r}\left(t\right)=⟨3\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t,3\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t,4t⟩$ gives

$\text{r}\left(s\right)=⟨3\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\frac{s}{5}\right),\phantom{\rule{0.2em}{0ex}}3\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\frac{s}{5}\right),\frac{4s}{5}⟩,s\ge 0.$

## Curvature

An important topic related to arc length is curvature. The concept of curvature provides a way to measure how sharply a smooth curve turns. A circle has constant curvature. The smaller the radius of the circle, the greater the curvature.

Think of driving down a road. Suppose the road lies on an arc of a large circle. In this case you would barely have to turn the wheel to stay on the road. Now suppose the radius is smaller. In this case you would need to turn more sharply to stay on the road. In the case of a curve other than a circle, it is often useful first to inscribe a circle to the curve at a given point so that it is tangent to the curve at that point and “hugs” the curve as closely as possible in a neighborhood of the point ( [link] ). The curvature of the graph at that point is then defined to be the same as the curvature of the inscribed circle.

## Definition

Let C be a smooth curve in the plane or in space given by $\text{r}\left(s\right),$ where $s$ is the arc-length parameter. The curvature     $\kappa$ at s is

$\kappa =‖\frac{d\phantom{\rule{0.2em}{0ex}}\text{T}}{ds}‖=‖{T}^{\prime }\left(s\right)‖.$

The formula in the definition of curvature is very useful in terms of calculation. In particular, recall that $\text{T}\left(t\right)$ represents the unit tangent vector to a given vector-valued function $\text{r}\left(t\right),$ and the formula for $\text{T}\left(t\right)$ is $\text{T}\left(t\right)=\frac{{r}^{\prime }\left(t\right)}{‖{r}^{\prime }\left(t\right)‖}.$ To use the formula for curvature, it is first necessary to express $\text{r}\left(t\right)$ in terms of the arc-length parameter s , then find the unit tangent vector $\text{T}\left(s\right)$ for the function $\text{r}\left(s\right),$ then take the derivative of $\text{T}\left(s\right)$ with respect to s. This is a tedious process. Fortunately, there are equivalent formulas for curvature.

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