Arc Length

Learning Outcomes

Determine the length of a particle’s path in space by using the arc-length function.

Arc Length for Vector Functions

We have seen how a vector-valued function describes a curve in either two or three dimensions. Recall Alternative Formulas for Curvature, which states that the formula for the arc length of a curve defined by the parametric functions $x=x\,(t),\ y=t\,(t),\ t_{1}\,\leq\,t\,\leq\,t_{2}$ is given by

$s=\displaystyle\int_{t_{1}}^{t^{2}}\ \sqrt{\big(x'\,(t)\big)^{2}+\big(y'\,(t)\big)^{2}}\,dt$

In a similar fashion, if we define a smooth curve using a vector-valued function ${\bf{r}}\,(t)=f\,(t)\,{\bf{I}}+g\,(t)\,{\bf{j}}$ , where $a\,\leq\,t\,\leq\,b$ , the arc length is given by the formula

$s=\displaystyle\int_{a}^{b}\ \sqrt{\big(f'\,(t)\big)^{2}+\big(g'\,(t)\big)^{2}}\,dt$

In three dimensions, if the vector-valued function is described by ${\bf{r}}\,(t)=f\,(t)\,{\bf{I}}+g\,(t)\,{\bf{j}}+h\,(t)\,{\bf{k}}$ over the same interval $a\,\leq\,t\,\leq\,b$ , the arc length is given by

$s=\displaystyle\int_{a}^{b}\ \sqrt{\big(f'\,(t)\big)^{2}+\big(g'\,(t)\big)^{2}+\big(h'\,(t)\big)^{2}}\,dt$

Arc-Length formulas Theorem

Plane curve: Given a smooth curve $C$ defined by the function ${\bf{r}}\,(t)=g\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}$ , where $t$ lies within the interval $[a,\ b]$ , the arc length of $C$ over the interval is

s = \int_{a}^{b} \sqrt{[f^{'} (t)]^{2} + [g^{'} (t)]^{2}} d t = \int_{a}^{b} ‖ r^{'} (t) ‖ d t

$s=\displaystyle\int_{a}^{b}\ \sqrt{\big[f'\,(t)\big]^{2}+\big[g'\,(t)\big]^{2}}\,dt=\displaystyle\int_{a}^{b}\ \left\Vert{\bf{r}}'\,(t)\right\Vert\ dt$ .

Space curve: Given a smooth curve $C$ defined by the function ${\bf{r}}\,(t)=f\,(t)\,{\bf{i}}+g\,(t)\,{\bf{j}}+h\,(t)\,{\bf{k}}$ , where $t$ lies within the interval $[a,\ b]$ , the arc length of $C$ over the interval is

s = \int_{a}^{b} \sqrt{[f^{'} (t)]^{2} + [g^{'} (t)]^{2} + [h^{'} (t)]^{2}} d t = \int_{a}^{b} ‖ r^{'} (t) ‖ d t

$s=\displaystyle\int_{a}^{b}\ \sqrt{\big[f'\,(t)\big]^{2}+\big[g'\,(t)\big]^{2}+\big[h'\,(t)\big]^{2}}\,dt=\displaystyle\int_{a}^{b}\ \left\Vert{\bf{r}}'\,(t)\right\Vert\ dt$

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 ${\bf{r}}\,(t)$ 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.

Example: finding the arc length

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

Show Solution

Using the Arc Length Formulas, ${\bf{r}}'\,(t)=3\,{\bf{i}}+4\,{\bf{j}}$ , so
$\begin{array}{ccc}\hfill {s} &=\hfill&{\displaystyle\int_{a}^{b}\ \left\Vert{\bf{r}}'\,(t)\right\Vert\ dt}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{a}^{5}\ \sqrt{3^{2}+4^{2}}\,dt} \hfill \\ \hfill & =\hfill & {\displaystyle\int_{1}^{5}\ 5\,dt=5t\Big|^5_1=20.}\hfill \\ \hfill \end{array}$
Using the Arc Length Formulas, ${\bf{r}}'\,(t)=\langle{\cos{t}}-t\sin{t},\ \sin{t}+t\cos{t},\ 2\rangle$ , so
$\begin{array}{ccc}\hfill {s} &=\hfill&{\displaystyle\int_{a}^{b}\ \left\Vert{\bf{r}}'\,(t)\right\Vert\ dt}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{2\pi}\ \sqrt{(\cos{t}-t\sin{t})^{2}+(\sin{t}+t\cos{t})^{2}+2^{2}}\,dt} \hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{2\pi}\ \sqrt{(\cos{^{2}t}-2t\sin{t}\cos{t}+t^{2}\, \sin{^{2}t})+(\sin{^{2}t}+2t\sin{t}\cos{t}+t^{2}\, \cos{^{2}t})+4}\,dt}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{2\pi}\ \sqrt{\cos{^{2}t}+\sin{^{2}t}+t^{2}\,(\cos{^{2}t}+\sin{^{2}t})+4}\,dt}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{2\pi}\ \sqrt{t^{2}+5}\,dt}\hfill \\ \hfill \end{array}$

Here we can use a table integration formula

$\displaystyle\int_{}\ \sqrt{u^{2}+a^{2}}\,du=\frac{u}{2}\,\sqrt{u^{2}+a^{2}}+\frac{a^{2}}{2}\ln{\Big|u+\sqrt{u^{2}+a^{2}}\Big|}+C$ ,

so we obtain

$\begin{array}{ccc}\hfill {\displaystyle\int_{0}^{2\pi}\ \sqrt{t^{2}+5}\,dt} &=\hfill&{\frac{1}{2}\big(t\sqrt{t^{2}+5}+5\,ln\ \Big|t+\sqrt{t^{2}+5}\Big|\big)^{2\pi}_{0}}\hfill \\ \hfill & =\hfill & {\frac{1}{2}\,\big(2\pi\sqrt{4{\pi}^{2}+5}+5\ln{(2\pi+\sqrt{4{\pi}^{2}+5})\big)}-\frac{5}{2}\ln{\sqrt{5}}} \hfill \\ \hfill & \approx\hfill & {25.343.}\hfill \\ \hfill \end{array}$

try it

Calculate the arc length of the parameterized curve

${\bf{r}}\,(t)=\langle{2t^{2}}+1,\ 2t^{2}-1,\ t^{3}\rangle,\ 0\,\leq\,t\,\leq\,3.$

Show Solution

r^{'} (t) = ⟨ 4 t, 4 t, 3 t^{2} ⟩

${\bf{r}}'\,(t)=\langle{4t},\ 4t,\ 3t^{2}\rangle$ , so

s = \frac{1}{27} (113^{3 / 2} - 32^{3 / 2}) \approx 37.785

$s=\frac{1}{27}(113^{3/2}-32^{3/2})\approx{37.785}$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 3.9” here (opens in new window).

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

${\bf{r}}\,(t)=R\cos{\Big(\frac{2\pi\,N\,t}{h}\Big)}\,{\bf{i}}+R\sin{\Big(\frac{2\pi\,N\,t}{h}\Big)}\,{\bf{j}}+t\,{\bf{k}},\ 0\,\leq\,t\,\leq\,h,$

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 Arc-Length Formulas. First of all,

${\bf{r}}'\,(t)=-\frac{2\pi\,N\,R}{h}\sin{\Big(\frac{2\pi\,N\,t}{h}\Big)}\,{\bf{i}}+\frac{2\pi\,N\,R}{h}\cos{\Big(\frac{2\pi\,N\,t}{h}\Big)}\,{\bf{j}}+{\bf{k}}.$

Therefore,

$\begin{array}{ccc}\hfill {s} &=\hfill&{\displaystyle\int_{a}^{b}\ \left\Vert{\bf{r}}'\,(t)\right\Vert\ dt}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{h}\ \sqrt{\Bigg(-\frac{2\pi\,N\,R}{h}\sin{\Big(\frac{2\pi\,N\,t}{h}\Big)}\Bigg)^{2}+\Bigg(\frac{2\pi\,N\,R}{h}\cos{\Big(\frac{2\pi\,N\,t}{h}\Big)}\Bigg)^{2}+1^{2}}\,dt} \hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{h}\ \sqrt{\frac{4{\pi}^{2}N^{2}R^{2}}{h^{2}}\,\Bigg(\sin{^{2}\Big(\frac{2\pi{N}t}{h}\Big)}+\cos{^{2}\Big(\frac{2\pi{N}t}{h}\Big)\Bigg)}+1}\,dt}\hfill \\ \hfill & =\hfill & {\displaystyle\int_{0}^{h}\ \sqrt{\frac{4{\pi}^{2}N^{2}R^{2}}{h^{2}}+1}\,dt}\hfill \\ \hfill & =\hfill & {\Big[t\,\sqrt{\frac{4{\pi}^{2}N^{2}R^{2}}{h^{2}}+1}\Big]^h_0}\hfill \\ \hfill & =\hfill & {h\,\sqrt{\frac{4{\pi}^{2}N^{2}R^{2}+h^{2}}{h^{2}}}}\hfill \\ \hfill & =\hfill & {\sqrt{4{\pi}^{2}N^{2}R^{2}+h^{2}}.}\hfill \\ \hfill \end{array}$

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:

$s\,(t)=\displaystyle\int_{a}^{t}\ \sqrt{\big(f'\,(u)\big)^{2}+\big(g'\,(u)\big)^{2}+\big(h'\,(u)\big)^{2}}\,du$

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 $s(t)$ measures distance traveled as a function of time, $s'(t)$ measures the speed of the particle at any given time. Since we have a formula for $s(t)$ , we can differentiate both sides of the equation:

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

If we assume that ${\bf{r}}\,(t)$ defines a smooth curve, then the arc length is always increasing, so $s'\,(t)\,>\,0$ for $t\,>\,a$ . Last, if ${\bf{r}}\,(t)$ is a curve on which $\left\Vert{\bf{r}}'\,(t)\right\Vert=1$ for all $t$ , then

$s\,(t)=\displaystyle\int_{a}^{t}\ \left\Vert{\bf{r}}'\,(u)\right\Vert\ du=\displaystyle\int_{a}^{t}\ 1\,du=t-a,$

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

Arc-Length function Theorem

Let ${\bf{r}}\,(t)$ describe a smooth curve for $t\,\geq\,a$ . Then the arc-length function is given by

s (t) = \int_{a}^{t} ‖ r^{'} (u) ‖ d u

$s\,(t)=\displaystyle\int_{a}^{t}\ \left\Vert{\bf{r}}'\,(u)\right\Vert\ du$ .

Furthermore, $\frac{ds}{dt}=\left\Vert{\bf{r}}'\,(t)\right\Vert\,>\,0$ . If $\left\Vert{\bf{r}}'\,(t)\right\Vert=1$ for all $t\,\geq\,a$ , 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 ${\bf{r}}\,(t)=\langle{3}\cos{t},\ 3\sin{t}\rangle,\ 0\,\leq\,t\,\leq\,2\pi$ that parameterizes a circle of radius 3, we can change the parameter from $t$ to $4t$ , obtaining a new parameterization ${\bf{r}}\,(t)=\langle{3}\cos{4t},\ 3\sin{4t}\rangle$ . The new parameterization still defines a circle of radius $3$ , but now we need only use the values $0\,\leq\,t\,\leq\,\frac{\pi}{2}$ to traverse the circle once.

Suppose that we find the arc-length function $s(t)$ and are able to solve this function for $t$ as a function of $s$ . We can then reparameterize the original function ${\bf{r}}(t)$ by substituting the expression for $t$ back into ${\bf{r}}(t)$ . 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 ${\bf{r}}(t)$ . 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.

Example: finding an arc-length parameterization

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

Show Solution

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

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

${\bf{r}}\,(s)=4\cos{\big(\frac{s}{4}\big)}{\bf{i}}+4\sin{\big(\frac{s}{4}\big)}{\bf{j}}$ .

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

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 ${\bf{r}}\,(t)=\langle{t}+3,\ 2t-4,\ 2t\rangle$ yields

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

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

try it

Find the arc-length function for the helix

${\bf{r}}\,(t)=\langle{3}\cos{t},\ 3\sin{t},\ 4t\rangle{,}\ t\,\geq\,0.$

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

Show Solution

s = 5 t

$s=5t$ , or

t = \frac{s}{5}

$t=\frac{s}{5}$ . Substituting this into

r (t) = ⟨ 3 \cos t, 3 \sin t, 4 t ⟩

${\bf{r}}\,(t)=\langle{3}\cos{t},\ 3\sin{t},\ 4t\rangle$ gives

r (s) = ⟨ 3 \cos (\frac{s}{5}), 3 \sin (\frac{s}{5}), \frac{4 s}{5} ⟩, s \geq 0

${\bf{r}}\,(s)=\Big\langle3\cos{\big(\frac{s}{5}\big)},\ 3\sin{\big(\frac{s}{5}\big)},\ \frac{4s}{5}\Big\rangle,\ s\geq{0}$ .

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 3.10” here (opens in new window).

Module 3: Vector-Valued Functions

Learning Outcomes

Arc Length for Vector Functions

Arc-Length formulas Theorem

Example: finding the arc length

try it

Arc-Length Parameterization

Arc-Length function Theorem

Example: finding an arc-length parameterization

try it

Candela Citations