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8.13 - The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC ONLY)

Note: this is a BC Calculus topic. AB students can skip this section.

Review: The Distance Formula

The distance formula between two points comes straight from the Pythagorean Theorem.
For two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2):
d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Or in change notation:
d=(Δx)2+(Δy)2d = \sqrt{(\Delta x)^2 + (\Delta y)^2}
This is just c=a2+b2c = \sqrt{a^2 + b^2} from the Pythagorean Theorem.
8.13 - The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC ONLY)

The Concept of Arc Length

Say we want the length of a curve between two points. We can’t just draw a straight line, because we need to account for the curve itself.

Approximating with Line Segments

The key insight is to approximate the curve with many small line segments.
One Line Segment (a Poor Approximation)
8.13 - The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC ONLY)
A single line segment from start to finish gives a terrible approximation.
Two Line Segments (Better)
8.13 - The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC ONLY)
Two segments follow the curve more closely.
Four Line Segments (Even Better)
8.13 - The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC ONLY)
With four segments, we’re getting pretty close to the actual curve length.

The Calculus Approach

Key idea: as the number of line segments approaches infinity, the sum of their lengths approaches the actual arc length.
If we label the points x0,x1,x2,x3,…,xnx_0, x_1, x_2, x_3, \ldots, x_n, where x0=ax_0 = a is the start and xn=bx_n = b is the end, each tiny segment has a length given by the distance formula. As n→∞n \to \infty, the sum of those segments becomes an integral.

The Arc Length Formula

For a smooth, continuous, differentiable function y=f(x)y = f(x) on the interval [a,b][a, b]:
L=∫ab1+[f′(x)]2 dx\boxed{L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx}

Important Notes

  • The function has to be smooth and continuous, with no corners or discontinuities
  • The curve has to be differentiable throughout the interval
  • Look carefully: it’s [f′(x)]2[f'(x)]^2, not [f(x)]2[f(x)]^2
  • If the curve is written as x=g(y)x = g(y) for c≤y≤dc \leq y \leq d, swap the roles: L=∫cd1+[g′(y)]2 dyL = \int_c^d \sqrt{1 + [g'(y)]^2}\,dy

Why This Formula?

The 1+[f′(x)]2\sqrt{1 + [f'(x)]^2} comes from applying the distance formula to infinitesimally small segments. Each tiny piece of the curve behaves like a miniature tangent line with slope f′(x)f'(x).

Example 1: Setting Up Arc Length

Find the arc length of y=16x3+12xy = \frac{1}{6}x^3 + \frac{1}{2x} from x=1x = 1 to x=2x = 2.
Step 1: Find the derivative.
y′=ddx[16x3+12x−1]=16(3x2)+12(−1)x−2y' = \frac{d}{dx}\left[\frac{1}{6}x^3 + \frac{1}{2}x^{-1}\right] = \frac{1}{6}(3x^2) + \frac{1}{2}(-1)x^{-2}y′=x22−12x2y' = \frac{x^2}{2} - \frac{1}{2x^2}
Step 2: Square the derivative.
[y′]2=(x22−12x2)2=x44−2⋅x22⋅12x2+14x4[y']^2 = \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2 = \frac{x^4}{4} - 2 \cdot \frac{x^2}{2} \cdot \frac{1}{2x^2} + \frac{1}{4x^4}=x44−12+14x4= \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}
Step 3: Add 1.
1+[y′]2=x44+12+14x4=14(x4+2+1x4)1 + [y']^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \frac{1}{4}\left(x^4 + 2 + \frac{1}{x^4}\right)
Notice this is a perfect square:
=14(x2+1x2)2= \frac{1}{4}\left(x^2 + \frac{1}{x^2}\right)^2
Step 4: Set up the integral.
L=∫1214(x2+1x2)2 dx=∫1212(x2+1x2)dxL = \int_1^2 \sqrt{\frac{1}{4}\left(x^2 + \frac{1}{x^2}\right)^2} \, dx = \int_1^2 \frac{1}{2}\left(x^2 + \frac{1}{x^2}\right) dx
Step 5: Integrate.
=12∫12(x2+x−2)dx=12[x33−1x]12= \frac{1}{2} \int_1^2 \left(x^2 + x^{-2}\right) dx = \frac{1}{2} \left[\frac{x^3}{3} - \frac{1}{x}\right]_1^2=12[(83−12)−(13−1)]=12[83−12−13+1]= \frac{1}{2} \left[\left(\frac{8}{3} - \frac{1}{2}\right) - \left(\frac{1}{3} - 1\right)\right] = \frac{1}{2} \left[\frac{8}{3} - \frac{1}{2} - \frac{1}{3} + 1\right]=12[73+12]=12⋅176=1712= \frac{1}{2} \left[\frac{7}{3} + \frac{1}{2}\right] = \frac{1}{2} \cdot \frac{17}{6} = \frac{17}{12}
Answer: L=1712L = \frac{17}{12} units

Example 2: Perfect Square Pattern

Find the arc length of y=112(x2+8)3/2y = \frac{1}{12}(x^2 + 8)^{3/2} from x=1x = 1 to x=4x = 4.
Step 1: Find y′y'.
y′=112⋅32(x2+8)1/2⋅2x=324x(x2+8)1/2=x4x2+8y' = \frac{1}{12} \cdot \frac{3}{2}(x^2 + 8)^{1/2} \cdot 2x = \frac{3}{24}x(x^2 + 8)^{1/2} = \frac{x}{4}\sqrt{x^2 + 8}
Step 2: Square it.
[y′]2=x216(x2+8)[y']^2 = \frac{x^2}{16}(x^2 + 8)
Step 3: Set up the integral.
L=∫141+x2(x2+8)16 dx=∫141+x4+8x216 dxL = \int_1^4 \sqrt{1 + \frac{x^2(x^2 + 8)}{16}} \, dx = \int_1^4 \sqrt{1 + \frac{x^4 + 8x^2}{16}} \, dx=∫1416+x4+8x216 dx= \int_1^4 \sqrt{\frac{16 + x^4 + 8x^2}{16}} \, dx
Step 4: Factor inside the square root.
=∫14x4+8x2+1616 dx= \int_1^4 \sqrt{\frac{x^4 + 8x^2 + 16}{16}} \, dx
Notice that x4+8x2+16=(x2+4)2x^4 + 8x^2 + 16 = (x^2 + 4)^2, a perfect square.
=∫14(x2+4)216 dx=∫14x2+44 dx= \int_1^4 \sqrt{\frac{(x^2 + 4)^2}{16}} \, dx = \int_1^4 \frac{x^2 + 4}{4} \, dx
Step 5: Integrate.
=14∫14(x2+4) dx=14[x33+4x]14= \frac{1}{4} \int_1^4 (x^2 + 4) \, dx = \frac{1}{4} \left[\frac{x^3}{3} + 4x\right]_1^4=14[(643+16)−(13+4)]=14[643+16−13−4]= \frac{1}{4} \left[\left(\frac{64}{3} + 16\right) - \left(\frac{1}{3} + 4\right)\right] = \frac{1}{4} \left[\frac{64}{3} + 16 - \frac{1}{3} - 4\right]=14[633+12]=14[21+12]=334= \frac{1}{4} \left[\frac{63}{3} + 12\right] = \frac{1}{4}[21 + 12] = \frac{33}{4}
Answer: L=334L = \frac{33}{4} units

Example 3: Using a Calculator

Find the arc length of y=sin⁡(x)y = \sin(x) from x=0x = 0 to x=πx = \pi.
Step 1: Find the derivative.
y′=cos⁡(x)⇒[y′]2=cos⁡2(x)y' = \cos(x) \quad \Rightarrow \quad [y']^2 = \cos^2(x)
Step 2: Set up the integral.
L=∫0π1+cos⁡2(x) dxL = \int_0^\pi \sqrt{1 + \cos^2(x)} \, dx
This integral can’t be evaluated with elementary functions, so we use a calculator.
Step 3: Calculator evaluation.
Make sure your calculator is in radian mode.
On a TI-84:
MATH → 9:fnInt
fnInt(√(1 + cos²(x)), x, 0, π)
Or more explicitly:
fnInt(√(1 + (cos(x))²), x, 0, π)
Answer: L≈3.820L \approx 3.820 units

Example 4: Another Perfect Square

Find the arc length of y=23x3/2−12x1/2y = \frac{2}{3}x^{3/2} - \frac{1}{2}x^{1/2} from x=1x = 1 to x=9x = 9.
Step 1: Find y′y'.
y′=23⋅32x1/2−12⋅12x−1/2=x1/2−14x−1/2=x−14xy' = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} - \frac{1}{2} \cdot \frac{1}{2}x^{-1/2} = x^{1/2} - \frac{1}{4}x^{-1/2} = \sqrt{x} - \frac{1}{4\sqrt{x}}
Step 2: Square it.
[y′]2=(x−14x)2=x−2⋅x⋅14x+116x=x−12+116x[y']^2 = \left(\sqrt{x} - \frac{1}{4\sqrt{x}}\right)^2 = x - 2 \cdot \sqrt{x} \cdot \frac{1}{4\sqrt{x}} + \frac{1}{16x} = x - \frac{1}{2} + \frac{1}{16x}
Step 3: Add 1.
1+[y′]2=x+12+116x1 + [y']^2 = x + \frac{1}{2} + \frac{1}{16x}
This looks like it should factor, and it does:
=(x+14x)2= \left(\sqrt{x} + \frac{1}{4\sqrt{x}}\right)^2
Check: (x)2+2x⋅14x+116x=x+12+116x(\sqrt{x})^2 + 2\sqrt{x} \cdot \frac{1}{4\sqrt{x}} + \frac{1}{16x} = x + \frac{1}{2} + \frac{1}{16x} ✓\checkmark
Step 4: Set up and evaluate.
L=∫19(x+14x)2 dx=∫19(x+14x)dxL = \int_1^9 \sqrt{\left(\sqrt{x} + \frac{1}{4\sqrt{x}}\right)^2} \, dx = \int_1^9 \left(\sqrt{x} + \frac{1}{4\sqrt{x}}\right) dx=∫19(x1/2+14x−1/2)dx=[2x3/23+14⋅2x1/2]19= \int_1^9 \left(x^{1/2} + \frac{1}{4}x^{-1/2}\right) dx = \left[\frac{2x^{3/2}}{3} + \frac{1}{4} \cdot 2x^{1/2}\right]_1^9=[2x3/23+x1/22]19=[2(27)3+32]−[23+12]= \left[\frac{2x^{3/2}}{3} + \frac{x^{1/2}}{2}\right]_1^9 = \left[\frac{2(27)}{3} + \frac{3}{2}\right] - \left[\frac{2}{3} + \frac{1}{2}\right]=[18+32]−76=392−76=117−76=1106=553= \left[18 + \frac{3}{2}\right] - \frac{7}{6} = \frac{39}{2} - \frac{7}{6} = \frac{117 - 7}{6} = \frac{110}{6} = \frac{55}{3}
Answer: L=553L = \frac{55}{3} units

Arc Length as Distance Traveled

This topic’s title pairs arc length with distance traveled. If an object moves along the graph of y=f(x)y = f(x) from x=ax = a to x=bx = b without turning back, the distance it travels along the path is the arc length:
Distance traveled=L=∫ab1+[f′(x)]2 dx\text{Distance traveled} = L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx
Compare this with 8.2. There the object moves along a straight line, so distance traveled is ∫∣v(t)∣ dt\int |v(t)|\,dt. Here the path itself curves, so you measure the length of the path. (In Unit 9 you’ll get the same distance from the velocity in both directions, ∫(x′(t))2+(y′(t))2 dt\int \sqrt{(x'(t))^2 + (y'(t))^2}\,dt.)

Example 5: Distance Along a Curved Path

A bug crawls along the curve y=x3/2y = x^{3/2} from (0,0)(0, 0) to (4,8)(4, 8), with distances in centimeters. How far does it crawl?
Step 1: Find the derivative.
y′=32x1/2⇒[y′]2=94xy' = \frac{3}{2}x^{1/2} \quad \Rightarrow \quad [y']^2 = \frac{9}{4}x
Step 2: Set up the integral.
L=∫041+94x dxL = \int_0^4 \sqrt{1 + \frac{9}{4}x} \, dx
Step 3: Integrate with u-substitution. Let u=1+94xu = 1 + \frac{9}{4}x, so du=94 dxdu = \frac{9}{4}\,dx and dx=49 dudx = \frac{4}{9}\,du. When x=0x = 0, u=1u = 1. When x=4x = 4, u=10u = 10.
L=49∫110u1/2 du=49⋅23[u3/2]110=827(1010−1)L = \frac{4}{9}\int_1^{10} u^{1/2} \, du = \frac{4}{9} \cdot \frac{2}{3}\left[u^{3/2}\right]_1^{10} = \frac{8}{27}\left(10\sqrt{10} - 1\right)
Answer: L=827(1010−1)≈9.073L = \frac{8}{27}\left(10\sqrt{10} - 1\right) \approx 9.073 cm
Sanity check: the straight line from (0,0)(0, 0) to (4,8)(4, 8) is 42+82=80≈8.944\sqrt{4^2 + 8^2} = \sqrt{80} \approx 8.944 cm. The curved path is a little longer, as it should be. Arc length is never shorter than the straight-line distance between the endpoints.

Common Algebraic Technique: Factoring Out Constants

When 1+[f′(x)]21 + [f'(x)]^2 comes out messy, you can often factor a constant out from under the square root.

Example Pattern

If you have
1+x2⋅g(x)16\sqrt{1 + \frac{x^2 \cdot g(x)}{16}}
factor out 116\frac{1}{16}:
=116(16+x2⋅g(x))=1416+x2⋅g(x)= \sqrt{\frac{1}{16}\left(16 + x^2 \cdot g(x)\right)} = \frac{1}{4}\sqrt{16 + x^2 \cdot g(x)}
This often reveals a perfect square hiding inside.

Common Mistakes

Mistake 1: Using f(x)f(x) Instead of f′(x)f'(x)

Wrong: ∫ab1+[f(x)]2 dx\int_a^b \sqrt{1 + [f(x)]^2} \, dx
Right: ∫ab1+[f′(x)]2 dx\int_a^b \sqrt{1 + [f'(x)]^2} \, dx
You have to take the derivative first.

Mistake 2: Forgetting to Square the Derivative

Wrong: ∫ab1+f′(x) dx\int_a^b \sqrt{1 + f'(x)} \, dx
Right: ∫ab1+[f′(x)]2 dx\int_a^b \sqrt{1 + [f'(x)]^2} \, dx

Mistake 3: Not Checking for Perfect Squares

A lot of arc length problems are designed so that 1+[f′(x)]21 + [f'(x)]^2 turns out to be a perfect square. Look for that pattern before you start trying to integrate.

Mistake 4: Calculator in Degrees Instead of Radians

For trig arc length problems, always double-check that your calculator is in radian mode.

When Does the Arc Length Formula Apply?

The formula requires that the function is:
  • Continuous on [a,b][a, b]
  • Differentiable on [a,b][a, b]
  • Smooth, with no sharp corners
If any of those fail, the formula doesn’t apply.

Why Does This Formula Work? (Optional Proof Sketch)

Note: you don’t need this derivation for the AP exam, but it gives you the mathematical insight.

Step 1: Sum of Line Segments

For nn segments from x0x_0 to xnx_n, the length of segment ii is
Δsi=(Δxi)2+(Δyi)2\Delta s_i = \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2}
Total length approximation:
L≈∑i=1n(Δxi)2+(Δyi)2L \approx \sum_{i=1}^{n} \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2}

Step 2: Factor Out Δx\Delta x

=∑i=1n(Δxi)2(1+(Δyi)2(Δxi)2)=∑i=1nΔxi1+(ΔyiΔxi)2= \sum_{i=1}^{n} \sqrt{(\Delta x_i)^2\left(1 + \frac{(\Delta y_i)^2}{(\Delta x_i)^2}\right)} = \sum_{i=1}^{n} \Delta x_i \sqrt{1 + \left(\frac{\Delta y_i}{\Delta x_i}\right)^2}

Step 3: Apply the Mean Value Theorem

By the Mean Value Theorem, for each segment there’s some cic_i where
ΔyiΔxi=f′(ci)\frac{\Delta y_i}{\Delta x_i} = f'(c_i)
So
L≈∑i=1n1+[f′(ci)]2 ΔxiL \approx \sum_{i=1}^{n} \sqrt{1 + [f'(c_i)]^2} \, \Delta x_i

Step 4: Take the Limit

As n→∞n \to \infty and Δxi→0\Delta x_i \to 0, this Riemann sum becomes an integral:
L=lim⁡n→∞∑i=1n1+[f′(ci)]2 Δxi=∫ab1+[f′(x)]2 dxL = \lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{1 + [f'(c_i)]^2} \, \Delta x_i = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx
That’s the definition of the definite integral.

Strategy Summary

Step 1: Verify the function is smooth and continuous on [a,b][a, b]
Step 2: Find f′(x)f'(x)
Step 3: Square it to get [f′(x)]2[f'(x)]^2
Step 4: Form 1+[f′(x)]21 + [f'(x)]^2
Step 5: Check whether it’s a perfect square, or can be factored to reveal one
Step 6: Set up L=∫ab1+[f′(x)]2 dxL = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx
Step 7: Evaluate by hand if you can, or with a calculator if you can’t

Key Takeaways

  • Arc length formula: L=∫ab1+[f′(x)]2 dxL = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx
  • Take the derivative first. It’s [f′(x)]2[f'(x)]^2, not [f(x)]2[f(x)]^2
  • Square the derivative. Don’t forget this step
  • Look for perfect squares. 1+[f′(x)]21 + [f'(x)]^2 often factors nicely
  • Smooth and continuous is required. No corners, no breaks
  • Use a calculator for the messy ones when the integral can’t be done by hand
  • Check radian mode for trig functions
  • The proof uses limits. Arc length comes from summing infinitely many tiny line segments, though this isn’t tested on the AP exam
  • Distance traveled along a curve is the arc length of the path: ∫ab1+[f′(x)]2 dx\int_a^b \sqrt{1 + [f'(x)]^2} \, dx

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