Length uses slope, not vertical change alone

Try

The graph y=3x+2 rises by 12 units while x increases from 0 to 4. Its length is not 12: horizontal movement also contributes. Use the right triangle formed by the endpoint displacement to calculate length 4√10. Predict how the graph-length integral reproduces that value.

Predict

For y=x² on [0,1], endpoint displacement has components (1,1). The straight chord has length √2, while the curved arc is longer. Predict why integrating either f(x) or f′(x) cannot directly give arc length: those integrals measure signed area and vertical change respectively.

Experiment

Parameterize a graph by its horizontal coordinate, x=t and y=f(t). The velocity vector is (1,f′(t)), so speed is √(1+[f′(t)]²).

Graph and interval	Length integrand	Exact result
y=3x+2, [0,4]	√10	4√10
y=(2/3)x^(3/2), [0,3]	√(1+x)	14/3
y=x², [0,1]	√(1+4x²)	the stated definite integral

For the second row, derive f′=√x and integrate the resulting simple radical. For the third, leave the exact integral intact until choosing a valid symbolic or numerical method. An unevaluated definite integral is not inherently an approximation.

Understand

For a continuously differentiable graph y=f(x) on [a,b], length is ∫ₐᵇ√(1+[f′(x)]²)dx. A derivation joins nearby graph points by chords, whose lengths are √[(Δx)²+(Δy)²]. Factoring the positive Δx gives √[1+(Δy/Δx)²]Δx. Under the smoothness conditions, refinement converts the slope quotients into derivatives and the sum into the integral.

The constant 1 under the radical records horizontal movement per horizontal input. Squaring f′ makes upward and downward motion contribute positively. There is no cancellation of length when the graph turns downward. The integrand is at least 1, so graph length is at least the horizontal interval length, a useful quick check.

Choosing y as the parameter can be simpler. For x=g(y), length is ∫√(1+[g′(y)]²)dy over the stated y interval. The graph x=y² on [0,1] in y gives the same expression as the parameterization x=t²,y=t. Use a coordinate representation that covers the intended arc once and avoids unnecessary inverse branches.

Corners can be handled piecewise. The graph y=|x| on [−1,1] is not continuously differentiable across zero, but each side is a straight segment of length √2. Their sum is 2√2. A failed global theorem hypothesis does not imply that the curve has no length; it means the argument should be split or use a more general result.

Not every continuous graph has finite length, and extreme oscillation can defeat a naive chord-refinement intuition. The stated smoothness or piecewise-smoothness assumptions provide the setting for these formulas. This unit does not replace them with a claim that any drawn curve has the same simple integral interpretation.

Computational comparisons should separate an exact integral representation, a numerically approximated value and an observed polygonal length. They answer related questions but carry different guarantees. If a numerical routine encounters a near-vertical derivative or a singular endpoint, reparameterization or an improper-limit treatment may be needed.

Apply

For each task, state which coordinate parameterizes the arc and whether it traces the arc once. Distinguish vertical displacement, straight chord distance and curved length. In the corner example, justify each smooth piece and add their positive lengths.

Continue

Next rotate a curve to create a surface. Its area uses circumference times slant length, so the arc-length factor carries directly into the new integral.