The Arc Length of a Smooth, Planar Curve and Distance Traveled: Study Guide (Calculus BC 8.13)

BC-only material. Arc length is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

If any of these is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Pythagoras’ theorem and the distance formula	Before calculus	The length of each short straight piece
Derivatives and the chain rule	Units 2 and 3	Finding f′(x) for the integrand
Mean Value Theorem	5.1	Turning a chord’s slope into a value of f′
Riemann sums and definite integrals	6.2 to 6.6	Turning a sum of pieces into an integral
Fundamental Theorem and substitution	6.7, 6.9	Evaluating the integral exactly, when that is possible

Notation on this page: ∫ from a to b of f(x) dx is the definite integral with lower limit a and upper limit b, and [F(x)] from a to b means F(b) − F(a). L stands for length.

What arc length means

Earlier in Unit 8 you used definite integrals to add up thin pieces of area and volume. This topic adds up thin pieces of length.

Imagine a piece of string lying exactly along the graph of y = f(x) from x = a to x = b. Pull the string straight and measure it. That measurement is the arc length of the curve. It is a length, so its units are the units of the axes (metres, kilometres, and so on), not square units.

You already know one case. For a straight line, the distance formula gives the length directly. Calculus is needed when the curve bends.

Building the formula from straight pieces

Step 1: cut the curve into pieces. Split [a, b] into n subintervals of width Δx. Join the points on the curve above the division points with straight segments (chords). The chain of chords is a polygon that follows the curve.

Step 2: measure one chord with Pythagoras. If a chord covers a horizontal change Δx and a vertical change Δy, its length is

Δs = √((Δx)² + (Δy)²) = √(1 + (Δy/Δx)²) · Δx

Step 3: replace the slope of the chord by a derivative. If f is differentiable, the Mean Value Theorem says the chord’s slope Δy/Δx equals f′(c) for some c inside that subinterval. So each chord has length √(1 + (f′(c))²) · Δx.

Step 4: add and take the limit. The total length of the chords is a Riemann sum:

Σ √(1 + (f′(cᵢ))²) Δx

As n → ∞ the chords hug the curve and the sum becomes a definite integral.

You can watch the chord totals approach the true length for the curve in Figure 1 (Worked example 1 finds the exact value):

Number of chords	3	6	30	Exact (integral)
Total chord length	4.647	4.660	4.666	14/3 ≈ 4.667

Chords always come out a little short, because a straight segment is the shortest path between its end points.

The arc length formula

Length of y = f(x) from x = a to x = b: L = ∫ from a to b of √(1 + (f′(x))²) dx

Condition: the curve must be smooth. Here this means f′ exists and is continuous on [a, b]. Then the integrand is continuous and the integral exists. If the graph has a corner, such as y = |x − 2| at x = 2, split the interval at the corner and add the lengths of the pieces. (For y = |x − 2| on [0, 4] each piece is a straight segment of length 2√2, so L = 4√2 ≈ 5.657.)

Three quick facts that help you check answers.

• The integrand √(1 + (f′(x))²) is always at least 1, so L ≥ b − a. Equality holds only for a horizontal line.
• For a straight line y = mx + c, the integrand is the constant √(1 + m²), so L = (b − a)√(1 + m²). This is the distance formula again.
• L is never less than the straight-line distance (the chord) between the two end points.

A useful shorthand: ds = √(1 + (dy/dx)²) dx is the length of a tiny piece of curve. The integral adds up all the ds pieces.

Curves written as x = g(y)

Sometimes a curve is easier to describe with x as a function of y, for example x = 4 − y². Swap the roles of the variables:

Length of x = g(y) from y = c to y = d: L = ∫ from c to d of √(1 + (g′(y))²) dy

This needs g′ continuous on [c, d]. The limits are y-values, and the variable of integration is y. Use this form when the curve fails the vertical line test, or when dy/dx would be undefined at an end point (as for y = √(4 − x) at x = 4) but dx/dy is not.

Arc length as distance traveled

If a point (a hiker, a bead on a wire, a car on a road) moves along the curve y = f(x) from x = a to x = b, and never turns back, then the distance it travels along the path is the arc length. It does not matter how fast it moves. Only the shape of the path matters.

Compare three kinds of “distance” you meet in calculus:

Situation	Distance traveled	Topic
Motion along a straight line with velocity v(t)	∫ from t₁ to t₂ of |v(t)| dt	8.2
Moving along a curve y = f(x) from x = a to x = b	∫ from a to b of √(1 + (f′(x))²) dx	8.13 (this topic)
Motion in the plane given as x(t), y(t)	∫ of the speed √((x′(t))² + (y′(t))²) dt	9.3 and 9.6 (later)

Two other quantities are easy to confuse with distance traveled. The straight-line distance between the end points is the chord length, and it is the shortest possible. The vertical change f(b) − f(a) ignores horizontal movement completely.

Worked example 1: an exact length without a calculator

Question. Find the exact length of the curve y = (2/3)(x − 1)^(3/2) for 1 ≤ x ≤ 4. Check that your answer is sensible.

1. Differentiate. By the chain rule, f′(x) = (2/3) · (3/2)(x − 1)^(1/2) · 1 = √(x − 1).
2. Check smoothness. √(x − 1) is continuous on [1, 4], so the formula applies.
3. Build the integrand. 1 + (f′(x))² = 1 + (x − 1) = x. So √(1 + (f′(x))²) = √x.
4. Set up. L = ∫ from 1 to 4 of √x dx.
5. Integrate. An antiderivative of x^(1/2) is (2/3)x^(3/2). So L = [(2/3)x^(3/2)] from 1 to 4 = (2/3)(8) − (2/3)(1) = 14/3 ≈ 4.667.

Checks.

• Antiderivative: d/dx[(2/3)x^(3/2)] = x^(1/2) = √x. ✓
• Chord: the end points are (1, 0) and (4, 2√3), since f(4) = (2/3)(3)^(3/2) = 2√3 ≈ 3.464. The chord length is √(3² + (2√3)²) = √21 ≈ 4.583. The curve must be a little longer, and 4.667 > 4.583. ✓
• At least b − a: 4.667 ≥ 4 − 1 = 3. ✓

Why this worked. The function was chosen so that 1 + (f′(x))² became a simple expression. That is typical of no-calculator arc length questions. If the square root does not simplify, you will need a calculator.

Worked example 2: distance traveled over a ridge (calculator)

Context. A walking trail crosses a ridge. Seen from the side, the trail follows y = 0.06(25 − x²) for −5 ≤ x ≤ 5, where x and y are in kilometres (x is horizontal distance, y is height above the valley floor). A tunnel would instead run straight along y = 0 from x = −5 to x = 5.

(a) How far does a walker travel along the trail? (b) How much further is this than the tunnel route? (c) At a steady 4 km/h along either route, how much extra time does the trail take?

(a)

1. Differentiate. f(x) = 1.5 − 0.06x², so f′(x) = −0.12x. This is continuous on [−5, 5], so the curve is smooth.
2. Set up. L = ∫ from −5 to 5 of √(1 + (−0.12x)²) dx = ∫ from −5 to 5 of √(1 + 0.0144x²) dx.
3. Evaluate with a calculator. L ≈ 10.571 km.

Write the integral before you press any buttons. On a free-response question, the setup earns credit in its own right.

(b) Extra distance ≈ 10.571 − 10 = 0.571 km, about 571 m.

(c) Trail time = 10.571 ÷ 4 ≈ 2.643 hours; tunnel time = 10 ÷ 4 = 2.5 hours. Extra time ≈ 0.143 hours ≈ 8.6 minutes. (To avoid rounding errors, divide the stored value of L, not the rounded one.)

Checks.

• Units: x and y are both in km, so the arc length is in km. ✓
• Bounds: |f′(x)| ≤ 0.6 on the interval, so the integrand lies between 1 and √1.36 ≈ 1.166. Then 10 ≤ L ≤ 11.662, and 10.571 fits. ✓
• Symmetry: the curve is symmetric about x = 0, and 2 × ∫ from 0 to 5 of √(1 + 0.0144x²) dx gives the same value (each half is about 5.286 km). ✓

Two tempting wrong integrals. ∫ from −5 to 5 of f(x) dx = 10 is an area (in km²), not a length. ∫ from −5 to 5 of |f′(x)| dx = 3 is the total up-and-down change (1.5 km up, then 1.5 km down), which ignores the horizontal distance.

Common misconceptions

• Forgetting the “1” or the square root. √(1 + (f′)²) is not √((f′)²) = |f′|, and it is not 1 + (f′)².
• Using f instead of f′. ∫ √(1 + (f(x))²) dx is not the arc length. For Worked example 1 it gives about 5.498 instead of 4.667.
• Splitting the square root. √(1 + (f′)²) is not 1 + f′. A square root does not split over a sum.
• Mixing variables. If you integrate with respect to y, the limits must be y-values and the derivative must be dx/dy.
• Using the formula across a corner. At a corner or a vertical tangent f′ is not continuous. Split the interval, or switch to x = g(y).
• Thinking the length can be less than the chord. If your answer is shorter than the straight-line distance between the end points, something is wrong.
• Confusing distance traveled with net change. The distance along the path is not |f(b) − f(a)|, and it is not the chord either.
• Rounding too early. Store the calculator value and round only the final answer to 3 decimal places.

Where this leads

This is the last topic of Unit 8. Unit 9 (BC only) starts with Topic 9.1, defining and differentiating parametric equations. In Topic 9.3 the same idea of adding short chord lengths gives the length of a parametric curve, using √((dx/dt)² + (dy/dt)²) dt, and in Topic 9.6 the integral of speed gives the distance traveled by a particle moving in the plane. The previous topic was Topic 8.12, the washer method around other axes. Use the Calculus BC roadmap to see the order.

Try the practice questions now, then use the revision notes and the checklist to consolidate.