Arc length of a smooth graph

Understand the idea

Arc length accumulates small distances along a curve. For a graph $\displaystyle y=f(x)$, a short horizontal change and its corresponding vertical change form the legs of a small right triangle. The resulting length element explains the square root and the added $\displaystyle 1$ in the graph formula.

For a continuously differentiable graph on a closed interval, integrate the nonnegative length element over that interval. Differentiate the function before forming the integrand, square the entire derivative, and retain the square root. The answer measures distance along the graph rather than signed area beneath it.

A geometric length is nonnegative. Compare a result with the straight-line distance between the endpoints as a reasonableness check. For one-dimensional motion, distance traveled is instead the integral of absolute velocity; for a general parametrized plane curve, both coordinate derivatives contribute to speed.

Worked example

Find the arc length of $\displaystyle y=\tfrac23x^{3/2}$ from $\displaystyle x=0$ to $\displaystyle x=3$.

1. Differentiating gives $\displaystyle y'=\sqrt x$, so $\displaystyle 1+(y')^2=1+x$.
2. The length is $\displaystyle \int_0^3\sqrt{1+x}\,dx$.
3. An antiderivative is $\displaystyle \tfrac23(1+x)^{3/2}$.
4. Evaluating gives $\displaystyle \tfrac23(4^{3/2}-1)=\tfrac23(8-1)$.