Arc Length Parametric
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.2: “Calculus of Parametric Curves” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-2-calculus-of-parametric-curves |
| Textbook used in class | Stewart, Calculus, Section 10.2: “Calculus with Parametric Curves” |
Opening Scenario
A robot arm sweeps a curved path with position $(x(t), y(t))$. To find how far the tip of the arm travels from $t = 0$ to $t = T$, slice the path into tiny steps: each step has a horizontal component $dx = x'(t)\,dt$ and a vertical component $dy = y'(t)\,dt$. The length of each tiny step is $\sqrt{(dx)^2+(dy)^2} = \sqrt{[x'(t)]^2+[y'(t)]^2}\,dt$. Integrating gives the total arc length.
Quick Reference
Parametric arc length. For $x = x(t)$, $y = y(t)$, $t \in [\alpha, \beta]$, with $x'$ and $y'$ continuous: $$L = \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt.$$
The quantity $\sqrt{(x')^2+(y')^2}$ is the speed of the moving point.
Key Concepts
1. Derivation from the Pythagorean Theorem
Partition $[\alpha,\beta]$ into $n$ subintervals. On the $i$-th piece, the straight-line distance from $(x(t_{i-1}),y(t_{i-1}))$ to $(x(t_i),y(t_i))$ is: $$\sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} = \sqrt{\left(\frac{\Delta x_i}{\Delta t}\right)^2+\left(\frac{\Delta y_i}{\Delta t}\right)^2}\,\Delta t \approx \sqrt{(x'(t_i^*))^2+(y'(t_i^*))^2}\,\Delta t.$$
Summing and taking the limit gives the integral.
2. Connection to Speed
If $t$ represents time, then $(x'(t), y'(t))$ is the velocity vector and $\sqrt{(x')^2+(y')^2}$ is the speed. Arc length equals the integral of speed over time: $$L = \int_\alpha^\beta (\text{speed})\,dt.$$
3. The Curve Must Not Retrace Itself
If the curve doubles back and traces a segment more than once, the integral counts each traversal separately, giving the total path length rather than the length of the traced set. In most textbook problems the curve is smooth and traces each point at most once.
4. Connection to $y = f(x)$ Formula
Set $t = x$: $x(t) = t$, $y(t) = f(t)$, $dx/dt = 1$. Then: $$L = \int_a^b\sqrt{1^2 + (f'(t))^2}\,dt = \int_a^b\sqrt{1+(f'(x))^2}\,dx,$$ recovering the arc length formula from Chapter 8.
Worked Example
Find the arc length of the curve $x = t - \sin t$, $y = 1 - \cos t$, $0 \leq t \leq 2\pi$ (one arch of the cycloid).
$x'(t) = 1-\cos t$, $y'(t) = \sin t$.
$(x')^2+(y')^2 = (1-\cos t)^2 + \sin^2 t = 1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t = 2(1-\cos t)$.
Use the identity $1-\cos t = 2\sin^2(t/2)$: $(x')^2+(y')^2 = 4\sin^2(t/2)$.
$\sqrt{(x')^2+(y')^2} = 2|\sin(t/2)| = 2\sin(t/2)$ for $t \in [0,2\pi]$ (since $\sin(t/2) \geq 0$ there).
$$L = \int_0^{2\pi} 2\sin(t/2)\,dt = 2\left[-2\cos(t/2)\right]_0^{2\pi} = -4[\cos\pi - \cos 0] = -4(-1-1) = 8.$$
Boxed answer: $L = 8$ (in units of $r = 1$; for a cycloid of radius $r$, $L = 8r$).
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Omitting the absolute value of $\sin(t/2)$ | Treating $2\sin(t/2)$ as negative for $t > 2\pi$ | Arc length must be positive; use $|\sin(t/2)|$ or confirm that $\sin(t/2) \geq 0$ on the interval |
| Using $\sqrt{(x'')^2+(y'')^2}$ (second derivatives) | Differentiating again before taking the square root | The arc length integrand uses $x' = dx/dt$ and $y' = dy/dt$, the first derivatives |
| Forgetting to square both components | Writing $\sqrt{x'+y'}$ | The formula is $\sqrt{(x')^2+(y')^2}$; both components are squared |
Common Misconceptions
the arc length integrand is $\sqrt{x'(t) + y'(t)}$ rather than $\sqrt{[x'(t)]^2 + [y'(t)]^2}$.
This is the multiplicative-not-additive error. The arc length element comes from the Pythagorean theorem applied to infinitesimal steps: each step has horizontal component $x'(t)\,dt$ and vertical component $y'(t)\,dt$, and the hypotenuse is $\sqrt{(x')^2 + (y')^2}\,dt$. Adding the components instead of squaring and then taking the square root of their sum ignores the geometry of right triangles and produces an incorrect integrand.
if the curve retraces itself, the arc length integral gives the length of the path set.
This is the iconic-graph error applied to length. Arc length is the total distance traveled by the moving point, not the length of the set of points the curve occupies. If the curve doubles back over a segment, that segment is counted twice in the integral. The integral computes path length, which equals geometric length only when each point of the curve is visited exactly once.
Leveled Practice
Level 1 -- Circle Circumference
Problem 1. Verify that the circumference of the circle $x = r\cos t$, $y = r\sin t$, $0 \leq t \leq 2\pi$ is $2\pi r$.
Show answer
$x'(t) = -r\sin t$, $y'(t) = r\cos t$.
$(x')^2+(y')^2 = r^2\sin^2 t + r^2\cos^2 t = r^2$.
$L = \displaystyle\int_0^{2\pi}\sqrt{r^2}\,dt = r\int_0^{2\pi}dt = 2\pi r$. Confirmed.
Level 2 -- Evaluate
Problem 2. Find the arc length of $x = t^2/2$, $y = (2t+1)^{3/2}/3$ from $t = 0$ to $t = 1$.
Show answer
$x'(t) = t$, $y'(t) = \frac{3}{2}\cdot\frac{2(2t+1)^{1/2}}{3} = (2t+1)^{1/2}$.
$(x')^2+(y')^2 = t^2 + (2t+1) = t^2 + 2t + 1 = (t+1)^2$.
$\sqrt{(t+1)^2} = t+1$ for $t \geq 0$.
$L = \displaystyle\int_0^1 (t+1)\,dt = \left[\dfrac{(t+1)^2}{2}\right]_0^1 = \dfrac{4}{2} - \dfrac{1}{2} = \dfrac{3}{2}$.
Boxed answer: $L = \dfrac{3}{2}$.
Mastery Checklist
Mental Model
Arc length equals the integral of speed. Each infinitesimal time step $dt$ contributes a tiny path segment of length (speed)$\,dt = \sqrt{(x')^2+(y')^2}\,dt$. Integrating accumulates all those tiny diagonal steps. The key computational step is simplifying $(x')^2+(y')^2$ -- this almost always factors or collapses into a perfect square in textbook problems.
Connections
Looking back
- Arc length for $y = f(x)$ (Section 8.1): The parametric formula generalizes the $y = f(x)$ case.
- Pythagorean theorem: Each infinitesimal step is the hypotenuse of a right triangle with legs $dx$ and $dy$.
Looking ahead
- Polar arc length (Section 10.4): Convert polar to parametric; the same formula applies.