How to Find the Arc Length of a Function

How to Find the Arc Length of a Function
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  • 0:29 Curve Length
  • 4:02 Example Problem
  • 6:30 Lesson Summary
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Lesson Transcript
Instructor: Heather Higinbotham
You don't always walk in a straight line. Sometimes, you want to know the distance between two points when the path is curved. In this lesson, you'll learn about finding the length of a curve.

Use the delta x and delta y in the Pythagorean theorem to find the distance
Arc Length Problem

I really enjoy looking at maps. And I really enjoy plotting my position on a map and watching it change as I move. What do I mean by this? I mean if I'm at Disneyland and I'm trying to get from one end of Disneyland to the other, I like plotting my path on a map and seeing how far I've actually walked.

Curve Length

Let's say that I'm at Disneyland, and I'm following the line y = (2/3)x^(3/2). I'm going from (x=0, y=0) to (x=4, y=8). How can I find out how far exactly I walked?

Let's take a really close look at one small part of my path. Let's zoom in here. Right here, I will have walked delta x in the x direction and delta y in the y direction. This is kind of like a slope. I've got a delta y north and a delta x east. If I want to find the distance between my initial point and my final point, I'm going to use this delta x and delta y in the Pythagorean theorem, so I know that the distance between my initial point and my end point will be equal to the square root of (delta x^2 + delta y^2). Remember from the Pythagorean theorem, if c is the hypotenuse of a right triangle, then a^2 + b^2 = c^2, where a and b are the lengths of the other two legs.

Simplify what is under the square root by factoring
Curve Length Factoring

If I want to know the total distance that I walked in my trek across Disneyland, I'm going to find this distance, this square root of (delta x^2 + delta y^2), for each of these little regions, and I'm going to sum them all up. This (delta x^2 + delta y^2), all inside of a square root, looks a little bit ugly. Before I go any further, I'm going to simplify it. I'm going to factor out a delta x^2 from these two terms. I get (delta x^2)(1 + ((delta y)/(delta x))^2). Because I've got this delta x^2 as a product inside of my square root, I can pull it out of the product. My entire path becomes the sum over all of these little regions that I walked of delta x times the square root of (1 + ((delta y)/(delta x))^2). This would be my distance walked.

Using the Pythagorean theorem over a large distance isn't as accurate as using it over a much smaller distance. In fact, if I take these segment lengths to be zero, that is take the limit as delta x goes to zero of these segments, I end up getting the exact length of my path. I also end up getting the limit as delta x goes to zero the sum over all of my segments of the square root of ((1 + ((delta y)/(delta x))^2)) (delta x). When I take that limit, it's nothing more than a Riemann sum that I've taken the limit of. This is an integral. This equals the integral from the start of my path, a, to the end of my path, b, of the square root of (1 + (y`)^2)dx. This y` is just dy/dx, so that's what happens to this delta y/delta x term as delta x goes to zero. This stops being the slope of the line and starts becoming the slope of the tangent of the line.

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