
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:

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 This is all good and well, but the integral 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 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 $\vec {p}(7.5)$

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 $a$.

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: 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 Ah, we can now factor a $\cos (t)$ out of each component to get: 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:

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 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 $0\le t\le 1$. for

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 $\vec {f}(1)$.
for

### 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. for

### 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
(b)
Now write and solve for $t$. In this case you will have
(c)
The function will be parameterized by arc length.