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.

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:

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

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.

1 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!

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
\[ \answer {0} \le s \le \answer {\sqrt {2(e-1)^2}} \]

2 Disguised circles

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

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
\[ \answer {0}\le s < \answer {10\pi } \]

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.