Welcome back to class! As we move into Chapter 8: Further Applications of Integration, we face a fundamental geometry question: How long is a curved line?

In the past, finding the length of a straight line segment was easy using the distance formula. But what happens when the path curves? In Section 8-1, we explore the concept of Arc Length. Imagine fitting a piece of string to a curve and then straightening it out against a ruler. While this gives us a rough physical idea, Calculus allows us to be precise.

The Concept: From Polygons to Integrals

To find the length of a curve $C$, we approximate it by connecting a series of points on the curve with straight line segments (a polygon). As the number of segments ($n$) increases and their width approaches zero, the polygon fits the curve more and more perfectly. By taking the limit of the sum of these segment lengths, we derive the integral formula for arc length.

The Arc Length Formula

If a function $f$ is smooth (meaning $f'$ is continuous) on the interval $[a, b]$, the length $L$ of the curve $y = f(x)$ is given by:

$$ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx $$

Or, using Leibniz notation:

$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

Note: This formula looks very similar to the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ because it is essentially adding up infinitely many tiny hypotenuses!

Integration with Respect to y

Sometimes, a curve is better defined as $x = g(y)$. In these cases, we simply switch the variables. If $g'(y)$ is continuous on $[c, d]$, the formula becomes:

$$ L = \int_{c}^{d} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy $$

Example from the Notes

Let's look at Example 1 from the attached slides. We want to find the length of the semicubical parabola $y^2 = x^3$ between points $(1, 1)$ and $(4, 8)$.

  • Step 1: Solve for $y$. Since we are looking at the top half, $y = x^{3/2}$.
  • Step 2: Find the derivative $\frac{dy}{dx}$. $$ \frac{dy}{dx} = \frac{3}{2}x^{1/2} $$
  • Step 3: Square the derivative. $$ \left(\frac{dy}{dx}\right)^2 = \frac{9}{4}x $$
  • Step 4: Set up the integral. $$ L = \int_{1}^{4} \sqrt{1 + \frac{9}{4}x} \, dx $$

From here, you would use $u$-substitution (letting $u = 1 + \frac{9}{4}x$) to solve the definite integral. (Check the PDF for the final numerical answer!)

The Arc Length Function

We also introduce the Arc Length Function, $s(x)$, which measures the distance along a curve from a starting point $a$ to a variable point $x$. This leads to the useful differential relationship:

$$ (ds)^2 = (dx)^2 + (dy)^2 $$

This differential form is a great mnemonic device to remember the formulas!

Please download the full notes below to see the complete derivation, the detailed solution to Example 1, and Example 4 regarding the arc length function.