Parameterizing by arc length

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

Recall that if $\vec {f}$ is a vector-valued function where

$\vec {f}'$ is continuous.
The curve defined by $\vec {f}(t)$ is traversed once for $a\le t\le b$ .

The arc length of the curve from $\vec {f}(a)\quad \text {to}\quad \vec {f}(b)$ is given by $\text {arc length} = \int _a^b |\vec {f}'(t)|\d t.$ This is all good and well; however, the integral $\int _a^b |\vec {f}'(t)|\d t$ could be quite difficult to compute. In this section, we see a new description of the curve drawn by $\vec {f}(t)$ , we’ll call it $\vec {g}(s)$ where the same curve is drawn by both $\vec {f}$ and $\vec {g}$ and we have that $s = \int _0^s |\vec {g}'(t)|\d t.$ This 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. 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 vector-valued function $\vec {f}:\R \to \R ^2$ is parameterized by arc length if and only if $|\vec {f}'| = 1$ .

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

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

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: $\vec {g}(s) = \vector {2-3s,1+4s}\quad -1\le s\le 1$

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 $\text {point}+ s \cdot \text {unit vector}$ and we have an arc length parameterization.

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: $\graph {(3t, 4t)}$ $\graph {(3t^2,4t^2)}$ produce the same graph. Now, by our theorem above, $\vec {g}$ is parameterized by arc length if and only if $|\vec {g}'(t)| = \answer [given]{1}$ Hence we need to find a unit vector in the the same direction as the line drawn by $\vec {f}$ . Write with me, $\begin{align*} \frac{\vec{f}'(t)}{|\vec{f}'(t)|} &= \frac{\vector{\answer[given]{6t},\answer[given]{8t}}}{\answer[given]{10t}}\\ &=\vector{\answer[given]{3/5},\answer[given]{4/5}}. \end{align*}$ Hence $\vec {g}(s) = s\vector {\answer [given]{3/5},\answer [given]{4/5}}$ draws the same curve as $\vec {f}$ and is parameterized by arc length.

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

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 $0\le t$ .

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

$\vec {g}(s) = \vector {\answer {-s/\sqrt {2}},\answer {4+s/\sqrt {2}},\answer {5}}$

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

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

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

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 $\vec {f}(t) = \vector {6t^3,8t^3,3}$ for $t\ge 0$ by arc length.

Write with me, $\begin{align*} |\vec{f}'(t)| &= \sqrt{\answer[given]{18^2t^4+24^2t^4}}\\ &=\answer[given]{30t^2} \end{align*}$ $\begin{align*} L(t) &= \int_0^t 30 u^2 \d u\\ &= \eval{\answer[given]{10 u^3}}_0^t\\ &= \answer[given]{10t^3}. \end{align*}$ So now set $s = \answer [given]{10t^3}$ and solve for $t$ . $t = \sqrt [3]{\answer [given]{s/10}}$ Our function parameterized by arc length is $\vec {g}(s) = \vector {\answer [given]{6s/10},\answer [given]{8s/10},\answer [given]{3}}.$