For arc length, we cannot just use one application of the distance formula like we can in straight line distance between two points. Instead, you divide the curve into infinitely tiny pieces that can be approximated as straight lines.

• Each tiny segment forms a tiny right triangle with dx, dy, and ds
  • dy/dx tells us the slope of each tiny segment, which allows us to express dy in terms of dx and integrate
  • adding all tiny distances ds using an integral allows us to find the total arc length

Therefore, by Pythagorean Theorem:

$ds^2=dx^2+dy^2$ where ds is a tiny piece of arc length

we get $ds=√(dx^2+dy^2)$

What is it?

Like in Calculus 1, we understand that a tiny piece of a curve looks more and more straight zooming in. Knowing this, we approximate the curve then using tiny line segments where

• horizontal change dx - tiny displacement in x
• vertical change dy - tiny displacement in t

We don’t just use Δx and Δy because the path is already straight, and one application of the distance formula gets you the distance

• Curved paths have no single triangle whose hypotenuse follows the entire curve, so we chop the curve into tiny pieces that look straight with sides dx and dy in the Pythagorean theorem and add up all the tiny lengths to find length

Δx and Δy = total change over the WHOLE path

dx and dy = tiny changes for one tiny piece of the path

The Integral

We get $dy/dx$ because we want integrals in terms of dx. We need a way to replace dy with something involving dx, which is where the derivative comes in.

We know vertical change = slope x horizontal change: $dy=dy/dx * dx$ which we just substitute into the distance formula: