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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 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
Considering , , and , which draws the circle of radius quickest?
Which draws the circle of radius slowest?
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 is a continuous
vector-valued function where the curve drawn by is traversed once for , then the arc
length of the curve from to is given by This is all good and well, but the integral
could be quite difficult to compute. On the other hand, if 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
A curve traced out by a vector-valued function is parameterized by arc length if
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 is parameterized by arc length, then the length
of when , is simply . No integral computations need to be done. Also we
should point out that is typically (though not necessarily) the name of the
variable when a function is parameterized by arc length, as often represents
Suppose the curve below has an arc length parameterization given by .
Compute: , , and
Consider the following example:
Let for . Show that is parameterized by arc length.
Here we need to show that We’ll just
compute the right-hand side of the equation above and see what happens. Write with
me, and so
Now our integral becomes:
Hence is parameterized by arc length.
From your own experience and the work above, we think the next theorem should be
A curve traced out by a continuously differentiable vector-valued function is
parameterized by arc length if and only if .
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 .
Which of the following vector-valued functions are parameterized by arc length?
Consider for . Find that makes this parameterized by arc length.
Set and solve for
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
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 in each component. On the other hand, if
we look at , we see Ah, we can now factor a 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 for is identical to the line segment plotted by:
Which of the following are line segments in disguise?
for for for for
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!
Consider for .
Parameterize this curve by arc length.
If we think about we see that the
variable only appears in the expression as . This means as grows, it will
grow identically in each component of . Indeed a quick check with a graph
will show that a graph of produces the same graph as a graph of when .
Ah, so this is a line in disguise! To parameterize a line by arc length you need to
write something like: So let’s find two points on the line. Setting , we see that is on
the line. Setting we see that is also on the line. The unit vector that runs from to
is: Thus as runs from to , draws the same curve as as runs from to
Give an arc length parameterization of for . for
Try your hand at this one now:
Give an arc length parameterization of for .
Check the values of and
Sometimes the curve we are given is a circle in disguise.
Consider for . Parameterize this curve by arc length.
Here, we should recognize
this curve a unit circle, being drawn in a counterclockwise fashion, starting
(when ) at the point . Ah! So an arc length parameterization is given by
Consider for . Parameterize this curve by arc length. for
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 thousand miles. Make the
The Earth will be at the origin.
At the starting time, , the Moon will be at the point in the -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 , the distance traveled in thousands of miles.
First compute the
circumference of the Moon’s orbit in thousands of miles: Now we write:
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
Regardless, if you want an arc length parameterization of starting at here is the
Now write and solve for . In this case you will have
The function will be parameterized by arc length.
Try your hand at it.
Parameterize for by arc length.
First we’ll compute the magnitude of . Write with
me: So now:
So now set : Now compute , in essence just solve for : Our curve, now parameterized
by arc length is