Topic 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled: compute the length of a curve y = f(x) and the distance a particle travels using the arc length integral (BC).

What this topic is asking

The College Board (Topic 8.13, BC only) applies the definite integral to a new geometric quantity: the length of a curve. Instead of accumulating area, you accumulate distance along the curve itself. The same integral also gives the distance traveled by a particle whose path is the curve.

Where the formula comes from

A worked exact arc length

When to reach for the calculator

The worked example was engineered so that $1 + (dy/dx)^2$ became a perfect square; that almost never happens for a randomly chosen curve. For something like $y = x^2$, the integrand is $\sqrt{1 + 4x^2}$, whose antiderivative is beyond the AP toolkit. On the calculator-active parts of the exam, you set up the integral by hand and then evaluate it numerically. The expected work is the correct integral with correct limits; the number comes from the calculator's definite-integral function. Recognizing which type of problem you are in (exact-by-hand versus calculator) saves time: if squaring and adding $1$ does not collapse to something integrable, it is a calculator problem.

Distance traveled along a path

The arc-length formula doubles as the distance traveled. If a particle moves along the curve $y = f(x)$, the total distance it covers from $x = a$ to $x = b$ is the arc length $L$. This connects to the Unit 8 motion idea that distance traveled is the integral of speed: along a graph $y = f(x)$, speed in terms of $x$ carries the factor $\sqrt{1 + (dy/dx)^2}$. In Unit 9 the same idea reappears for parametric paths, where speed is $\sqrt{(dx/dt)^2 + (dy/dt)^2}$ and distance traveled is $\int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$; the planar arc-length formula here is the special case where the parameter is $x$ itself.

Setting up correctly under exam pressure

The marks on a free-response arc-length question are concentrated in the setup: the correct integrand $\sqrt{1 + (dy/dx)^2}$ and the correct limits. A reliable routine is to write the formula first, then fill in $\frac{dy}{dx}$, then square. Keep the square root in place rather than expanding prematurely, since the calculator handles it directly. If the curve is given as $x = g(y)$ instead, the mirror formula is $L = \int_c^d \sqrt{1 + (dx/dy)^2}\,dy$, integrating in $y$. Choosing the variable that matches how the curve is described avoids an unnecessary rearrangement.