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Arc Length of a
Smooth, Planar Curve

Learn to compute the exact length of a curve defined by \( y = f(x) \) or parametric equations, and connect it to distance traveled.

AP Calculus BC

Why Arc Length Matters

Straight-line distance between two points is easy: just use the distance formula. But what if a particle moves along a curved path? The actual distance it travels is the arc length of that curve: the total length measured along the curve itself.

Engineering
Length of cable, road, or bridge arc
Physics
Distance a particle actually travels
Geometry
Circumference is arc length of a circle

From Straight Lines to Curves

x y a b Approximate with line segments Actual curve

The Differential Element \( ds \)

\(dx\)
\(dy\)
\(ds\)

The Arc Length Formula

1
Smooth curve: \( f'(x) \) exists and is continuous on \([a,b]\).
2
Square the derivative: compute \( [f'(x)]^2 \).
3
Add 1, take the square root, then integrate from \( a \) to \( b \).

Parametric Arc Length

This is the natural generalization: instead of factoring out \( dx \), we express both \( dx \) and \( dy \) in terms of the parameter \( t \). The integrand is the speed \( \left\|\mathbf{r}'(t)\right\| \) of the parametric curve.

Arc Length vs. Displacement

Start End Displacement (straight line) Arc length (actual path)
Arc Length / Distance
Total path traveled. Uses \( \int \left\|\mathbf{r}'(t)\right\| dt \). Always \( \geq 0 \).
Displacement
Net change in position. Uses \( \int \mathbf{r}'(t)\, dt \). Can be smaller than arc length.

Speed and Distance Traveled

The total distance traveled from \( t = \alpha \) to \( t = \beta \) is: $$ \text{Distance} = \int_\alpha^\beta v(t)\, dt = \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt $$ This is exactly the parametric arc length formula: distance traveled equals arc length of the trajectory.

How to Compute Arc Length

1
Identify the form: Is it \( y = f(x) \), \( x = g(y) \), or parametric \( (x(t), y(t)) \)?
2
Compute the derivative(s): find \( f'(x) \), or \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
3
Build the integrand: \( \sqrt{1 + [f'(x)]^2} \) or \( \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \).
4
Simplify under the radical: look for perfect squares or clean expressions.
5
Evaluate the integral: use antiderivatives, substitution, or a calculator if allowed.

Example 1: \( y = x^{3/2} \) on \([0, 4]\)

x y 1 2 3 4 8 4
1
\( f(x) = x^{3/2} \), so \( f'(x) = \frac{3}{2}x^{1/2} \).
2
\( [f'(x)]^2 = \frac{9}{4}x \). Integrand: \( \sqrt{1 + \frac{9}{4}x} \).
3
Let \( u = 1 + \frac{9}{4}x \), so \( du = \frac{9}{4}\,dx \). Evaluate: $$ L = \frac{4}{9} \cdot \frac{2}{3}\left[\left(1 + \tfrac{9}{4}x\right)^{3/2}\right]_0^4 = \frac{8}{27}\left(10^{3/2} - 1\right) $$

Example 1: Result

The curve \( y = x^{3/2} \) from \( x = 0 \) to \( x = 4 \) has length approximately 9.073 units. Notice this is much longer than the straight-line distance \( \sqrt{4^2 + 8^2} = \sqrt{80} \approx 8.944 \): the curve is always at least as long as the chord.

Example 2: Parametric Circle

\(r\)
\(t=0\)
\(t=\pi/2\)
1
\( x = r\cos t,\; y = r\sin t,\; t \in [0, 2\pi] \).
2
\( \frac{dx}{dt} = -r\sin t,\;\frac{dy}{dt} = r\cos t \).
3
\( L = \int_0^{2\pi}\sqrt{r^2\sin^2 t + r^2\cos^2 t}\,dt = \int_0^{2\pi} r\,dt = 2\pi r \). ★

Example 3: \( y = \frac{x^2}{4} - \frac{\ln x}{2} \) on \([1, e]\)

1
Differentiate: \( f'(x) = \frac{x}{2} - \frac{1}{2x} = \frac{x^2 - 1}{2x} \).
2
Square it: \( [f'(x)]^2 = \frac{(x^2-1)^2}{4x^2} = \frac{x^4 - 2x^2 + 1}{4x^2} \).
3
Add 1: \( 1 + [f'(x)]^2 = \frac{4x^2 + x^4 - 2x^2 + 1}{4x^2} = \frac{(x^2+1)^2}{4x^2} \).
4
Simplify radical: \( \sqrt{\frac{(x^2+1)^2}{4x^2}} = \frac{x^2+1}{2x} \) (since \( x > 0 \)).
5
Integrate: $$ L = \int_1^e \frac{x^2+1}{2x}\,dx = \int_1^e \left(\frac{x}{2} + \frac{1}{2x}\right)dx = \left[\frac{x^2}{4} + \frac{\ln x}{2}\right]_1^e = \frac{e^2 + 1}{4} $$

The Perfect Square Trick

In Example 3, the expression under the radical simplified to a perfect square. This is not a coincidence: AP exam problems are carefully crafted so the radical simplifies cleanly. Watch for these patterns:

Pattern A
\( f(x) = \frac{x^n}{a} + \frac{1}{bx^n} \): the cross terms cancel neatly.
Pattern B
\( f(x) = \frac{x^2}{a} \pm \frac{\ln x}{b} \): the derivative yields a ratio of polynomials.
Pattern C
Parametric with trig: \( \sin^2 \theta + \cos^2 \theta = 1 \) cleans up immediately.

Common Mistakes to Avoid

✕ Forgetting the +1
Writing \( \int \sqrt{[f'(x)]^2}\,dx \) instead of \( \int \sqrt{1 + [f'(x)]^2}\,dx \). The 1 comes from \( dx^2 \).
✕ Using displacement formula
Computing \( \int v_x\,dt \) (net displacement) instead of \( \int \|v\|\,dt \) (arc length). Speed is the magnitude, not a component.
✕ Wrong variable of integration
For \( x = g(y) \), use \( \sqrt{1 + [g'(y)]^2}\,dy \), not \( dx \). Match the variable to the form you are given.

Summary of Formulas

Formula When to Use
\( y = f(x) \) \( \int_a^b \sqrt{1 + [f'(x)]^2}\,dx \) Explicit function of \( x \)
\( x = g(y) \) \( \int_c^d \sqrt{1 + [g'(y)]^2}\,dy \) Easier as function of \( y \)
Parametric \( \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt \) Parametric or distance traveled
Speed \( v(t) = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \) Instantaneous speed at time \( t \)

Question 1 of 5

What is the arc length of \( y = \frac{2}{3}x^{3/2} \) from \( x = 0 \) to \( x = 3 \)?

Question 2 of 5

A particle moves with \( x(t) = \cos t \), \( y(t) = \sin t \) for \( t \in [0, \pi] \). What is the distance traveled?

Question 3 of 5

Which expression gives the arc length of \( y = \ln(\cos x) \) on \( [0, \frac{\pi}{4}] \)?

Question 4 of 5

The speed of a particle at time \( t \) is given by \( v(t) = \sqrt{9t^4 + 4t^2} \). Which could be its position functions?

Question 5 of 5

What is the arc length of \( y = \frac{x^2}{4} - \frac{\ln x}{2} \) from \( x = 1 \) to \( x = e \)?