Arc Length for Parametric Curves
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.4: “Arc Length of a Curve and Surface Area” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-4-arc-length-of-a-curve-and-surface-area |
| Textbook used in class | Stewart, Calculus, Section 8.1 / 10.2: “Arc Length” |
Opening Scenario
A projectile moves along a curved path in the plane. Its horizontal position $x(t)$ and vertical position $y(t)$ are both functions of time $t$. The path is not described by any single equation $y = f(x)$ (it may loop or reverse), but it is described by the pair $(x(t), y(t))$. How long is the path traced from time $t = a$ to $t = b$?
The Pythagorean theorem still applies. Each infinitesimal time step $dt$ produces a horizontal displacement $dx = x'(t)\,dt$ and a vertical displacement $dy = y'(t)\,dt$. The diagonal length of that tiny piece of path is $\sqrt{(dx)^2 + (dy)^2} = \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$. Integrate over $[a,b]$ to get the total length.
Quick Reference
Arc length for a parametric curve. Let $x = x(t)$, $y = y(t)$ for $t \in [a,b]$. If $x'$ and $y'$ are continuous on $[a,b]$ and the curve does not retrace itself (smooth, no backtracking), then the arc length is: $$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt.$$
The expression under the radical is the squared speed: $\sqrt{(x')^2 + (y')^2}$ is the speed of the point $(x(t), y(t))$ at time $t$.
Key Concepts
1. Where the Formula Comes From
Partition $[a,b]$ into $n$ small subintervals. On the $i$-th piece from $t_{i-1}$ to $t_i$ (with $\Delta t$ small), the curve moves approximately from $(x(t_{i-1}), y(t_{i-1}))$ to $(x(t_i), y(t_i))$. The straight-line distance 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 over all pieces and taking $n \to \infty$ gives the integral.
2. The Smoothness Condition
The formula assumes the curve is smooth: $x'$ and $y'$ are continuous, and $x'(t)^2 + y'(t)^2 > 0$ (i.e., the curve has a well-defined direction at every point). If the curve retraces itself over part of $[a,b]$, the integral counts each traced segment twice (or more), giving the total path length rather than the net displacement.
3. Connection to y = f(x) Case
If $y = f(x)$ and the curve is traversed left to right, set $t = x$: then $x(t) = t$, $y(t) = f(t)$, $x'(t) = 1$, $y'(t) = f'(t)$.
$$L = \int_a^b \sqrt{1^2 + [f'(t)]^2}\,dt = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx.$$
The $y = f(x)$ arc length formula is the special case where $x$ is the parameter.
Worked Example
Find the arc length of the curve $x = t^2$, $y = \dfrac{2t^3}{3}$ from $t = 0$ to $t = 1$.
$x'(t) = 2t$, $\quad y'(t) = 2t^2$.
$(x')^2 + (y')^2 = 4t^2 + 4t^4 = 4t^2(1 + t^2)$.
$\sqrt{4t^2(1+t^2)} = 2t\sqrt{1+t^2}$ (since $t \geq 0$).
$$L = \int_0^1 2t\sqrt{1+t^2}\,dt.$$
Substitution: $u = 1 + t^2$, $du = 2t\,dt$.
When $t = 0$: $u = 1$. When $t = 1$: $u = 2$.
$$L = \int_1^2 \sqrt{u}\,du = \left[\frac{2}{3}u^{3/2}\right]_1^2 = \frac{2}{3}(2\sqrt{2} - 1).$$
Boxed answer: $L = \dfrac{2}{3}(2\sqrt{2}-1) \approx \dfrac{2}{3}(1.828) \approx 1.219$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Treating $t$ as $x$ and omitting $x'$ | Writing $\int\sqrt{1+(y')^2}\,dt$ | Parametric formula is $\int\sqrt{(x')^2+(y')^2}\,dt$; both derivatives appear |
| Failing to simplify before integrating | Leaving $\sqrt{4t^2+4t^4}$ unexpanded | Factor: $\sqrt{4t^2(1+t^2)} = 2t\sqrt{1+t^2}$ for $t \geq 0$ |
| Confusing arc length with displacement | Using $\sqrt{(x(b)-x(a))^2+(y(b)-y(a))^2}$ for arc length | That formula gives the straight-line distance from start to end, not the curve length |
Common Misconceptions
the parametric arc length formula is $L = \int_a^b \sqrt{1+[y'(t)]^2}\,dt$, with only the $y$-derivative appearing.
This is the concept-image-conflicts-definition error. The formula $\int\sqrt{1+(dy/dx)^2}\,dx$ applies when $x$ is the parameter. For a general parametric curve, both components of velocity appear: $L = \int_a^b\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt$. When $x(t) = t$, then $x'(t) = 1$ and the formula reduces to $\int\sqrt{1+(y')^2}\,dt$, recovering the familiar form. Omitting $[x'(t)]^2$ discards the horizontal contribution to arc length and produces an underestimate unless the curve moves only vertically.
arc length is a fixed number that does not depend on how the curve is parameterized.
This is the rate-as-fixed-number error. The arc length is indeed the same regardless of the parameterization, but the integrand $\sqrt{(x')^2+(y')^2}$ changes with the parameterization because the speed $\sqrt{(x')^2+(y')^2}$ changes. A curve traced twice as fast has $x'$ and $y'$ doubled, so the integrand doubles, but the interval of integration halves, and the total length is preserved. Treating the speed as a fixed constant leads to incorrect integrals when the parameterization is not unit speed.
Leveled Practice
Level 1 -- Setup
Problem 1. Write (but do not evaluate) the arc length integral for $x = \cos t$, $y = \sin t$, $0 \leq t \leq 2\pi$.
Show answer
$x'(t) = -\sin t$, $y'(t) = \cos t$.
$(x')^2 + (y')^2 = \sin^2 t + \cos^2 t = 1$.
$L = \displaystyle\int_0^{2\pi} \sqrt{1}\,dt = 2\pi$.
This is the circumference of the unit circle -- a reassuring check.
Level 2 -- Evaluation
Problem 2. Find the arc length of $x = 3t^2$, $y = 2t^3$ from $t = 0$ to $t = \sqrt{3}$.
Show answer
$x' = 6t$, $y' = 6t^2$.
$(x')^2+(y')^2 = 36t^2 + 36t^4 = 36t^2(1+t^2)$.
$\sqrt{\cdot} = 6t\sqrt{1+t^2}$.
$L = \displaystyle\int_0^{\sqrt{3}} 6t\sqrt{1+t^2}\,dt$.
Sub $u = 1+t^2$, $du = 2t\,dt$: limits $u = 1$ to $u = 4$.
$L = \displaystyle\int_1^4 3\sqrt{u}\,du = \left[2u^{3/2}\right]_1^4 = 2(8) - 2(1) = 14$.
Boxed answer: $L = 14$.
Mastery Checklist
Mental Model
Think of a bug crawling along the curve $(x(t), y(t))$. At each instant the bug has a velocity vector $(x'(t), y'(t))$ and the bug’s speed is $\sqrt{(x')^2+(y')^2}$. The total distance traveled is (speed) times (time), integrated: $L = \int_a^b \text{speed}\,dt$. This is exactly the arc length formula. Arc length for parametric curves is just the integral of speed.
Connections
Looking back
- Arc length formula (Section 8.1): The $y = f(x)$ formula is the special case with $x(t) = t$.
- Arc length function (Section 8.1): $s(t) = \int_a^t \sqrt{(x')^2+(y')^2}\,du$ extends the arc length function to parametric curves.
Looking ahead
- Surface area of revolution (Section 8.2): The arc length element $ds = \sqrt{(x')^2+(y')^2}\,dt$ appears in $S = 2\pi\int r\,ds$ for surfaces generated by rotating parametric curves.
- Polar arc length (Section 10.4): Polar curves $r = f(\theta)$ are parametric; their arc length follows the same formula.
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