# 3.3 Arc length and curvature

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• Determine the length of a particle’s path in space by using the arc-length function.
• Explain the meaning of the curvature of a curve in space and state its formula.
• Describe the meaning of the normal and binormal vectors of a curve in space.

In this section, we study formulas related to curves in both two and three dimensions, and see how they are related to various properties of the same curve. For example, suppose a vector-valued function describes the motion of a particle in space. We would like to determine how far the particle has traveled over a given time interval, which can be described by the arc length of the path it follows. Or, suppose that the vector-valued function describes a road we are building and we want to determine how sharply the road curves at a given point. This is described by the curvature of the function at that point. We explore each of these concepts in this section.

## Arc length for vector functions

We have seen how a vector-valued function describes a curve in either two or three dimensions. Recall [link] , which states that the formula for the arc length of a curve defined by the parametric functions $x=x\left(t\right),y=y\left(t\right),{t}_{1}\le t\le {t}_{2}$ is given by

$s={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt.$

In a similar fashion, if we define a smooth curve using a vector-valued function $\text{r}\left(t\right)=f\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{i}+g\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{j},$ where $a\le t\le b,$ the arc length is given by the formula

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

In three dimensions, if the vector-valued function is described by $\text{r}\left(t\right)=f\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{i}+g\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{j}+h\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}$ over the same interval $a\le t\le b,$ the arc length is given by

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

## Arc-length formulas

1. Plane curve : Given a smooth curve C defined by the function $\text{r}\left(t\right)=f\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{i}+g\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{j},$ where t lies within the interval $\left[a,b\right],$ the arc length of C over the interval is
$s={\int }_{a}^{b}\sqrt{\phantom{\rule{0.2em}{0ex}}{\left[{f}^{\prime }\left(t\right)\right]}^{2}+{\left[{g}^{\prime }\left(t\right)\right]}^{2}}dt={\int }_{a}^{b}‖{r}^{\prime }\left(t\right)‖\phantom{\rule{0.2em}{0ex}}dt.$
2. Space curve : Given a smooth curve C defined by the function $\text{r}\left(t\right)=f\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{i}+g\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{j}+h\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{k},$ where t lies within the interval $\left[a,b\right],$ the arc length of C over the interval is
$s={\int }_{a}^{b}\sqrt{\phantom{\rule{0.2em}{0ex}}{\left[{f}^{\prime }\left(t\right)\right]}^{2}+{\left[{g}^{\prime }\left(t\right)\right]}^{2}+{\left[{h}^{\prime }\left(t\right)\right]}^{2}}dt={\int }_{a}^{b}‖{r}^{\prime }\left(t\right)‖\phantom{\rule{0.2em}{0ex}}dt.$

The two formulas are very similar; they differ only in the fact that a space curve has three component functions instead of two. Note that the formulas are defined for smooth curves: curves where the vector-valued function $\text{r}\left(t\right)$ is differentiable with a non-zero derivative. The smoothness condition guarantees that the curve has no cusps (or corners) that could make the formula problematic.

## Finding the arc length

Calculate the arc length for each of the following vector-valued functions:

1. $\text{r}\left(t\right)=\left(3t-2\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(4t+5\right)\phantom{\rule{0.1em}{0ex}}\text{j},1\le t\le 5$
2. $\text{r}\left(t\right)=⟨t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t,t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t,2t⟩,0\le t\le 2\pi$
1. Using [link] , ${r}^{\prime }\left(t\right)=3\text{i}+4\text{j},$ so
$\begin{array}{cc}\hfill s& ={\int }_{a}^{b}‖{r}^{\prime }\left(t\right)‖\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{a}^{5}\sqrt{{3}^{2}+{4}^{2}}dt\hfill \\ & ={\int }_{1}^{5}5\phantom{\rule{0.2em}{0ex}}dt={5t|}_{1}^{5}=20.\hfill \end{array}$
2. Using [link] , ${r}^{\prime }\left(t\right)=⟨\text{cos}\phantom{\rule{0.1em}{0ex}}t-t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t,\text{sin}\phantom{\rule{0.1em}{0ex}}t+t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t,2⟩,$ so
$\begin{array}{cc}\hfill s& ={\int }_{a}^{b}‖{r}^{\prime }\left(t\right)‖\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{2\pi }\sqrt{{\left(\text{cos}\phantom{\rule{0.1em}{0ex}}t-t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\right)}^{2}+{\left(\text{sin}\phantom{\rule{0.1em}{0ex}}t+t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\right)}^{2}+{2}^{2}}dt\hfill \\ & ={\int }_{0}^{2\pi }\sqrt{\left({\text{cos}}^{2}t-2t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t+{t}^{2}{\text{sin}}^{2}t\right)+\left({\text{sin}}^{2}t+2t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t+{t}^{2}{\text{cos}}^{2}t\right)+4}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{2\pi }\sqrt{{\text{cos}}^{2}t+{\text{sin}}^{2}t+{t}^{2}\left({\text{cos}}^{2}t+{\text{sin}}^{2}t\right)+4}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{2\pi }\sqrt{{t}^{2}+5}\phantom{\rule{0.2em}{0ex}}dt.\hfill \end{array}$

Here we can use a table integration formula
$\int \sqrt{{u}^{2}+{a}^{2}}\phantom{\rule{0.2em}{0ex}}du=\frac{u}{2}\sqrt{{u}^{2}+{a}^{2}}+\frac{{a}^{2}}{2}\text{ln}\phantom{\rule{0.1em}{0ex}}|u+\sqrt{{u}^{2}+{a}^{2}}|+C,$

so we obtain
$\begin{array}{cc}\hfill {\int }_{0}^{2\pi }\sqrt{{t}^{2}+5}\phantom{\rule{0.2em}{0ex}}dt& =\frac{1}{2}{\left(t\sqrt{{t}^{2}+5}+5\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}|t+\sqrt{{t}^{2}+5}|\right)}_{0}^{2\pi }\hfill \\ & =\frac{1}{2}\left(2\pi \sqrt{4{\pi }^{2}+5}+5\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2\pi +\sqrt{4{\pi }^{2}+5}\right)\right)-\frac{5}{2}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\sqrt{5}\hfill \\ & \approx 25.343.\hfill \end{array}$

Calculate the arc length of the parameterized curve

$\text{r}\left(t\right)=⟨2{t}^{2}+1,2{t}^{2}-1,{t}^{3}⟩,0\le t\le 3.$

${r}^{\prime }\left(t\right)=⟨4t,4t,3{t}^{2}⟩,$ so $s=\frac{1}{27}\left({113}^{3\text{/}2}-{32}^{3\text{/}2}\right)\approx 37.785$

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