Short on time? This page is the recap. For the derivation, the diagrams and worked examples, use the full study guide. BC-only material. Prerequisites (chain rule, Mean Value Theorem 5.1, Riemann sums 6.2 to 6.6, FTC 6.7) are on the Calculus BC roadmap.
Recap
- Arc length is the length of a curve, as if you laid a string along it and pulled it straight. Units: the units of the axes, not square units.
- Cut the curve into chords. Each chord has length √(Δx² + Δy²) = √(1 + (Δy/Δx)²) Δx. The Mean Value Theorem turns Δy/Δx into f′(c). Adding and taking the limit gives an integral.
- Smooth means f′ is continuous on the interval. Split at any corner.
- A point moving along the curve from one end to the other, without turning back, travels a distance equal to the arc length.
Key relationships
| Situation | Formula | Note |
|---|---|---|
| y = f(x), a ≤ x ≤ b | L = ∫ from a to b of √(1 + (f′(x))²) dx | f′ continuous on [a, b] |
| x = g(y), c ≤ y ≤ d | L = ∫ from c to d of √(1 + (g′(y))²) dy | Limits are y-values |
| Small piece of length | ds = √(1 + (dy/dx)²) dx | Add the pieces with an integral |
| Straight line y = mx + c | L = (b − a)√(1 + m²) | Same as the distance formula |
| Lower bounds | L ≥ b − a and L ≥ chord length | Equality only for straight lines |
When can you do it by hand?
- Only when 1 + (f′)² simplifies: it becomes a perfect square, or a simple expression such as x.
- Example: y = (2/3)(x − 1)^(3/2) has f′ = √(x − 1), so 1 + (f′)² = x and L on [1, 4] is 14/3.
- Otherwise write the integral and use a calculator.
Mistakes to avoid
- Dropping the 1 or the square root: the integrand is not |f′| and not 1 + (f′)².
- Using f instead of f′ inside the square root.
- Splitting the root: √(1 + (f′)²) ≠ 1 + f′.
- Wrong limits: dy integrals need y-limits.
- Ignoring corners where f′ is not continuous.
- An answer shorter than the chord. That is impossible, so recheck.
- Rounding early. Round only the final value, to 3 decimal places.
Quick self-check
- Find the exact length of y = (4/3)x^(3/2) for 0 ≤ x ≤ 2. (f′ = 2√x, so L = ∫ from 0 to 2 of √(1 + 4x) dx = 13/3)
- Write and evaluate the length of y = sin x for 0 ≤ x ≤ π. (∫ from 0 to π of √(1 + cos²x) dx ≈ 3.820, calculator)
- Use the formula on y = 3x + 1 for 0 ≤ x ≤ 2, then check with the distance formula. (2√10 ≈ 6.325 both ways)
Next: practice questions.