Parameterizing by arc length

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We find a new description of curves that trivializes arc length computations.

For any given a curve in space, there are many different vector-valued functions that draw this curve. For example, consider a circle of radius $3$ centered at the origin. Each of the following vector-valued functions will draw this circle: $\begin{align*} \vec{f}(\theta) &=\vector{3\cos(\theta),3\sin(\theta)} & &\text{for $0\le \theta< 2\pi$} \\ \vec{g}(t) &= \vector{3\cos(2\pi t),3\sin(2\pi t)} & &\text{for $0\le t< 1$} \\ \vec{h}(s) &= \vector{3\cos(s/3),3\sin(s/3)} & &\text{for $0\le s< 6\pi$} \end{align*}$ Each of these functions is a different parameterization of the circle. This means that while these vector-valued functions draw the same circle, they do so at different rates.

Considering $\vec {f}$ , $\vec {g}$ , and $\vec {h}$ , which draws the circle of radius $3$ quickest?

$\vec {f}(\theta ) =\vector {3\cos (\theta ),3\sin (\theta )}$ $\vec {g}(t) = \vector {3\cos (2\pi t),3\sin (2\pi t)}$ $\vec {h}(s) = \vector {3\cos (s/3),3\sin (s/3)}$

Which draws the circle of radius $3$ slowest?

$\vec {f}(\theta ) =\vector {3\cos (\theta ),3\sin (\theta )}$ $\vec {g}(t) = \vector {3\cos (2\pi t),3\sin (2\pi t)}$ $\vec {h}(s) = \vector {3\cos (s/3),3\sin (s/3)}$

In this section, we are going to be interested in parameterizations of curves where there is a one-to-one ratio between the parameter (the variable) and distance drawn (the arc length) from the start of the curve. Recall that if $\vec {f}$ is a continuous vector-valued function where the curve drawn by $\vec {f}(t)$ is traversed once for $a\le t\le b$ , then the arc length of the curve from $\vec {f}(a)$ to $\vec {f}(b)$ is given by $\text {arc length} = \int _a^b |\vec {f}'(t)|\d t.$ This is all good and well, but the integral $\int _a^b |\vec {f}'(t)|\d t$ could be quite difficult to compute. On the other hand, if $\vec {f}$ were an arc length parameterization, this would be simple to compute, because then the arc length is in a one-to-one ratio with the variables. Hence $\int _a^b |\vec {f}'(t)|\d t = b-a.$ Let’s state this as a definition.

A curve traced out by a vector-valued function $\vec {g}(s)$ is parameterized by arc length if $s = \int _0^s |\vec {g}'(t)|\d t.$ Such a parameterization is called an arc length parameterization.

It is nice to work with functions parameterized by arc length, because computing the arc length is easy. If $g$ is parameterized by arc length, then the length of $g(s)$ when $a\le s\le b$ , is simply $b-a$ . No integral computations need to be done. Also we should point out that $s$ is typically (though not necessarily) the name of the variable when a function is parameterized by arc length, as $s$ often represents “distance.”

Suppose the curve below has an arc length parameterization given by $\vec {p}(s)$ .

Compute: $\vec {p}(2)$ , $\vec {p}(4)$ , and $\vec {p}(7.5)$ $\begin{align*} \vec{p}(2) &= \vector{\answer{2},\answer{2}}\\ \vec{p}(4) &= \vector{\answer{2},\answer{0}}\\ \vec{p}(7.5) &= \vector{\answer{0},\answer{1/2}} \end{align*}$

Consider the following example:

Let $\vec {f}(t) = \vector {\cos (t), \sin (t)}$ for $0\le t< 2\pi$ . Show that $\vec {f}$ is parameterized by arc length.

Here we need to show that $s = \int _0^s |\vec {f}'(t)| \d t.$ We’ll just compute the right-hand side of the equation above and see what happens. Write with me, $\vec {f}'(t) = \vector {\answer [given]{-\sin (t)},\answer [given]{\cos (t)}}$ and so $\begin{align*} |\vec{f}'(t)| &= \sqrt{\vec{f}'(t)\dotp \vec{f}'(t)}\\ &= \sqrt{\vector{\answer[given]{-\sin(t)},\answer[given]{\cos(t)}}\dotp\vector{\answer[given]{-\sin(t)},\answer[given]{\cos(t)}}}\\ &= \answer[given]{1}. \end{align*}$ Now our integral becomes: $\begin{align*} \int_0^s |\vec{f}'(t)| \d t &= \int_0^s \d t\\ &= \answer[given]{s}. \end{align*}$ Hence $\vec {f}$ is parameterized by arc length.

From your own experience and the work above, we think the next theorem should be quite sensible.

A curve traced out by a continuously differentiable vector-valued function $\vec {f}:\R \to \R ^2$ is parameterized by arc length if and only if $|\vec {f}'| = 1$ .

If we imagine our vector-valued function as giving the position of a particle, then this theorem says that the path is parameterized by arc length exactly when the particle is moving at a speed of $1$ .

Which of the following vector-valued functions are parameterized by arc length?

$t\vector {11/61,60/61}$ $\vector {3\sin (t/3),3\cos (t/3)}$ $t\vector {16/113,112/113}$ $\vector {7,t17/145,t144/145}$ $\vector {3\cos (t),3\sin (t)}$

Consider $\vec {f}(t) = \vector {3\sin (a t),t/2,3\cos (a t)}$ for $0\le t$ . Find $a$ that makes this parameterized by arc length.

Set $|\vec {f}'(t)| = 1$ and solve for $a$ .

$a = \pm \answer {\frac {1}{2\sqrt {3}}}$

Often given a curve one wishes to have an arc length parameterization of the curve. We proceed by discussing several special cases, and then by giving a general method.

Disguised lines

Sometimes you have a vector-valued function that is merely a line in disguise. How could this be? Well consider the vector-valued function: $\vec {f}(t) = \vector {2-3\sin (t),1+4\sin (t)}\quad \text {for $-\pi /2\le t\le \pi /2$}$ This doesn’t look very much like a line, for one thing it has the function $\sin (t)$ in each component. On the other hand, if we look at $\vec {f}'$ , we see $\vec {f}'(t) = \vector {-3\cos (t),4\cos (t)}$ Ah, we can now factor a $\cos (t)$ out of each component to get: $\underbrace {\cos (t)}_{\text {scalar function}}\cdot \overbrace {\vector {-3,4}}^{\text {constant vector}}$ this is a scalar-function times a constant vector. The fact that we can “pull-out” the scalar function, and are left with a constant vector tells us that the line segment plotted by $\vec {f}$ for $-\pi /2\le t\le \pi /2$ is identical to the line segment plotted by:

$\begin{align*} \vec{g}(s) &=\vector{2,1} + s \vector{-3,4}\quad \text{for $-1\le s\le 1$}\\ &=\vector{2-3s,1+4s} \end{align*}$

Which of the following are line segments in disguise?

$\vector {2+e^t,4,-2-e^t}$ for $0\le t\le 2$ $\vector {3+e^t,-1+2e^t,2-e^{2t}}$ for $0\le t\le 1$ $\vector {5-3\cos (t),4+2\cos (t),1+\cos (t)}$ for $0\le t\le 2\pi$ $\vector {-1 -\sin (t),3+\sin (t),2-\cos (t)}$ for $0\le t\le \pi /2$ $\vector {2+5t,3t^2}$ for $-1\le t\le 1$ $\vector {3-t^3,1+2t^3}$ for $-1\le t\le 1$

Once we identify a vector-valued function as a disguised line, we can rewrite it as $\text {point}+ s \cdot \left (\text {unit vector}\right )$ and we have an arc length parameterization. Note, we need a unit vector to ensure that the magnitude of the derivative is one!

Consider $\vec {f}(t) = \vector {3t^2,4t^2}$ for $0\le t\le 1$ . Parameterize this curve by arc length.

If we think about $\vec {f}$ we see that the variable $t$ only appears in the expression as $t^2$ . This means as $t$ grows, it will grow identically in each component of $\vec {f}$ . Indeed a quick check with a graph will show that a graph of $\vector {3t,4t}$ $\graph {(3t, 4t)}$ produces the same graph as a graph of $\vector {3t^2,4t^2}$ $\graph {(3t^2,4t^2)}$ when $0\le t\le 1$ . Ah, so this is a line in disguise! To parameterize a line by arc length you need to write something like: $\text {point}+ s \cdot \left (\text {unit vector}\right )$ So let’s find two points on the line. Setting $t=0$ , we see that $(0,0)$ is on the line. Setting $t = 1$ we see that $(3,4)$ is also on the line. The unit vector that runs from $(0,0)$ to $(3,4)$ is: $\vector {\answer [given]{3/5},\answer [given]{4/5}}$ Thus as $s$ runs from $\answer [given]{0}$ to $\answer [given]{5}$ , $\vec {g}(s) = \vector {\answer [given]{3s/5},\answer [given]{4s/5}}$ draws the same curve as $\vec {f}$ as $t$ runs from $\answer [given]{0}$ to $\answer [given]{1}$ .

Give an arc length parameterization of $\vec {f}(t) = \vector {3-4t^3,2+t^3,5-t^3}$ for $0\le t\le 1$ . $\vec {g}(s) = \vector {\answer {3-4s/\sqrt {18}},\answer {2+s/\sqrt {18}},\answer {5-s/\sqrt {18}}}$ for $\answer {0} \le s \le \answer {\sqrt {18}}$

Try your hand at this one now:

Give an arc length parameterization of $\vec {f}(t) = \vector {1-e^t,3+e^t,5}$ for $0\le t\le 1$ .

Check the values of $\vec {f}(0)$ and $\vec {f}(1)$ .

$\vec {g}(s) = \vector {\answer {-s/\sqrt {2}},\answer {4+s/\sqrt {2}},\answer {5}}$ for $\answer {0} \le s \le \answer {\sqrt {2(e-1)^2}}$

Disguised circles

Sometimes the curve we are given is a circle in disguise.

Consider $\vec {f}(t) = \vector {\sin (2\pi t^2),\cos (2\pi t^2)}$ for $0\le t\le 1$ . Parameterize this curve by arc length.

Here, we should recognize this curve a unit circle, being drawn in a counterclockwise fashion, starting (when $t=0$ ) at the point $\left (\answer [given]{0},\answer [given]{1}\right )$ . Ah! So an arc length parameterization is given by $\vec {g}(s) = \vector {\answer [given]{\sin (s)},\answer [given]{\cos (s)}}.$

Consider $\vec {f}(t) = \vector {5\cos (t),5\sin (t)}$ for $0\le t< 2\pi$ . Parameterize this curve by arc length. $\vec {g}(s) = \vector {\answer {5\cos (s/5)},\answer {5\sin (s/5)}}$ for $\answer {0}\le s < \answer {10\pi }$

The Moon travels in a orbit around the Earth that can be approximated by a circle. The distance from the Earth to the Moon is around $240$ thousand miles. Make the following assumptions:

The Earth will be at the origin.
At the starting time, $t=0$ , the Moon will be at the point $(240,0)$ in the $(x,y)$ -plane.
The Moon will travel in a counterclockwise direction around the Earth.

Give a parameterization of the Moon’s orbit that will model the Moon’s position in terms of $s$ , the distance traveled in thousands of miles.

First compute the circumference of the Moon’s orbit in thousands of miles: $\text {circumference} = \answer [given]{480\pi }$ Now we write: $\vec {m}(s) = \vector {\answer [given]{240\cos (s/240)},\answer [given]{240\sin (s/240)}}$

A general method

While we are about to present a general method for finding representations of functions parameterized by arc length, one must not overestimate its strength.

Regardless, if you want an arc length parameterization of $\vec {f}(t)$ starting at $t=a$ here is the idea:

(a): Compute $L(t) = \int _a^t |\vec {f}'(u)| \d u$
(b): Now write $s = L(t)$ and solve for $t$ . In this case you will have $t = L^{-1}(s)$
(c): The function $\vec {g}(s) = \vec {f}(L^{-1}(s))$ will be parameterized by arc length.

Try your hand at it.

Parameterize $\vec {f}(t) = \vector {\cos (t),\sin (t),t}$ for $t\ge 0$ by arc length.

First we’ll compute the magnitude of $\vec {f}$ . Write with me: $|\vec {f}'(t)| = \sqrt {\answer [given]{2}}$ So now: $\begin{align*} L(t) &= \int_0^t \sqrt{2} \d u\\ &= \eval{\answer[given]{u\sqrt{2}}}_0^t\\ &= \answer[given]{t\sqrt{2}}. \end{align*}$ So now set $s = L(t)$ : $s = \answer [given]{t\sqrt {2}}$ Now compute $L^{-1}(s)$ , in essence just solve for $t$ : $t = \answer [given]{s/\sqrt {2}}$ Our curve, now parameterized by arc length is $\vec {g}(s) = \vector {\answer [given]{\cos (s/\sqrt {2})},\answer [given]{\sin (s/\sqrt {2})},\answer [given]{s/\sqrt {2}}}.$