In this section we are going to look at computing the arc
length of a function. Because it’s easy
enough to derive the formulas that we’ll use in this section we will derive one
of them and leave the other to you to derive.
We want to determine the length of the continuous function on the interval . We'll also need to assume that the derivative is continuous on [a, b].
Initially we’ll need to estimate the length
of the curve. We’ll do this by dividing
the interval up into n equal
subintervals each of width and we’ll denote the point on the curve
at each point by Pi. We can then approximate the curve by a series
of straight lines connecting the points.
Here is a sketch of this situation for .
Now denote the length of each of these line segments by and the length of the curve will then be
and we can get the exact length by taking n larger and larger. In other words, the exact length will be,
Now, let’s get a better grasp on the length of each of these
line segments. First, on each segment
let’s define . We can then compute directly the length of
the line segments as follows.
By the Mean
Value Theorem we know that on the interval there is a point so that,
Therefore, the length can now be written as,
The exact length of the curve is then,
However, using the definition of the definite integral,
this is nothing more than,
A slightly more convenient notation (in my opinion anyway)
is the following.
In a similar fashion we can also derive a formula for on . This formula is,
Again, the second form is probably a little more convenient.
Note the difference in the derivative under the square
root! Don’t get too confused. With one we differentiate with respect to x and with the other we differentiate
with respect to y. One way to keep the two straight is to notice
that the differential in the “denominator” of the derivative will match up with
the differential in the integral. This
is one of the reasons why the second form is a little more convenient.
Before we work any examples we need to make a small change
in notation. Instead of having two
formulas for the arc length of a function we are going to reduce it, in part,
to a single formula.
From this point on we are going to use the following formula
for the length of the curve.
Arc Length Formula(s)
Note that no limits were put on the integral as the limits
will depend upon the ds that we’re
using. Using the first ds will require x limits of integration and using the second ds will require y limits
Thinking of the arc length formula as a single integral with
different ways to define ds will be
convenient when we run across arc lengths in future sections. Also, this ds notation will be a nice notation for the next section as well.
Now that we’ve derived the arc length formula let’s work
Example 1 Determine
the length of between .
In this case we’ll need to use the first ds since the function is in the form . So, let’s get the derivative out of the
Let’s also get the root out of the way since there is
often simplification that can be done and there’s no reason to do that inside
Note that we could drop the absolute value bars here since
secant is positive in the range given.
The arc length is then,
As noted in the last example we really do have a choice as
to which ds we use. Provided we can get the function in the form
required for a particular ds we can
use it. However, as also noted above,
there will often be a significant difference in difficulty in the resulting
integrals. Let’s take a quick look at
what would happen in the previous example if we did put the function into the
Example 3 Redo
the previous example using the function in the form instead.
In this case the function and its derivative would be,
The root in the arc length formula would then be.
All the simplification work above was just to put the root
into a form that will allow us to do the integral.
Now, before we write down the integral we’ll also need to
determine the limits. This particular ds requires x limits of integration and we’ve got y limits. They are easy
enough to get however. Since we know x as a function of y all we need to do is plug in the
original y limits of integration
and get the x limits of
integration. Doing this gives,
Not easy limits to deal with, but there they are.
Let’s now write down the integral that will give the
That’s a really unpleasant looking integral. It can be evaluated however using the following
Using this substitution the integral becomes,
So, we got the same answer as in the previous
example. Although that shouldn’t
really be all that surprising since we were dealing with the same curve.
From a technical standpoint the integral in the previous
example was not that difficult. It was
just a Calculus I substitution. However,
from a practical standpoint the integral was significantly more difficult than
the integral we evaluated in Example 2.
So, the moral of the story here is that we can use either formula
(provided we can get the function in the correct form of course) however one
will often be significantly easier to actually evaluate.
Okay, let’s work one more example.
Example 4 Determine
the length of for . Assume that y is positive.
We’ll use the second ds
for this one as the function is already in the correct form for that
one. Also, the other ds would again lead to a particularly
difficult integral. The derivative and
root will then be,
Before writing down the length notice that we were given x limits and we will need y limits for this ds. With the assumption
that y is positive these are easy
enough to get. All we need to do is
plug x into our equation and solve
for y. Doing this gives,
The integral for the arc length is then,
This integral will require the following trig
The length is then,
The first couple of examples ended up being fairly simple
Calculus I substitutions. However, as
this last example had shown we can end up with trig substitutions as well for