We can use the procedure of “Slice, Approximate, Integrate” to find the length of curves.
Find the length of this segment of the curve.
Step 1: Slice Since the curve is described as a function of , we begin by slicing the curve into many segments with respect to .
Approximate: In order to find the approximate length of the curve, we must approximate each slice by a type of curve whose length we know how to compute.
We really only know how to compute the arclength of one type of curve - a line segment! In fact, if the endpoints of a line segment are and then the Pythagorean Theorem gives the distance between the points.
We thus approximate each slice as a line segment.
We thus have that the length of a single segment is and can write
to indicate that the approximate length of the curve is found by adding together all of the lengths of the line segments.
Step 3: Integrate As usual, we want to let the slice width become arbitrarily small, and since we have sliced with respect to , we eventually want to integrate with respect to . While the expression for the approximate length is conceptually useful, it does not pass to an integral easily in its current form! To write in a manageable form, we can do some algebra:
Now, note that for the two endpoints of the slice:
Assuming that the curve is differentiable along each slice, as the slice becomes arbitrarily small, the quantity :
We can now write the exact length as
Here, , so .
Thus, the arclength is
Using the trigonometric identity , we have:The exact length is thus . Using a calculator to find the length to decimal places gives: .
The length of a curve is given by the accumulated length determined by the instantaneous horizontal change and the instantaneous vertical change.
Let’s see a few examples.
Just as it was sometimes advantageous to integrate with respect to in our area and volume calculations, it can also help us sometimes in arclength calculations. Unlike the area and volume problems, where the geometry of the region often suggested a preferred variable of integration, these problems require us only to consider how we describe the curve in question (and whether we want to work with its given description!) when choosing the variable of integration.
Sometimes, the integrals that arise can be tricky to compute analytically and require careful differentiation and algebra.
Finally, most of the integrands arising in length calculations do not have elementary antiderivatives, so oftentimes you will only be able to set them up and estimate them numerically.
To summarize some important points from this section:
- The integrand is a square root of a linear function of . If so, evaluate it by inspection (or by using a substitution if necessary).
- The expression under the square root is actually a perfect square in disguise. This often requires careful algebra and differentiation.
As usual, practice is important. Make sure you work through the problems slowly. A small mistake early on in the problem can produce catastrophic results.
“Did you hear about the mathematician who took up gardening? He only grows vegetables with square roots!” - Anonymous