Arc length formula — derivation: Visual walkthrough

Step 1 — What are we even measuring?

Intuition WHAT / WHY / PICTURE

WHAT. We have a curve — a wiggly line drawn by some rule $y = f (x)$ . The rule is a machine: feed it a horizontal position $x$ , it hands back a height $y$ . We want the length of the curve itself, the distance a tiny ant would crawl walking along the bumps from the left end $x = a$ to the right end $x = b$ . WHY it's hard. For a straight line we can just measure end-to-end. A curve has no such shortcut — its length is more than the straight distance between its ends, because the ant goes up and over every bump. There is no single memorised formula, so we must build one. PICTURE. Below, the cyan curve is the path. The dashed amber line is the straight "as-the-crow-flies" distance between the ends. Notice the cyan path is clearly longer — that gap is exactly the problem we're solving.

Figure — Arc length formula — derivation

Here $x$ = how far right you are, $y = f (x)$ = the height the rule gives you, $a$ = start, $b$ = end. The letter $L$ will mean the true length of the cyan curve.

Step 2 — Chop the curve into tiny straight pieces

Each dot is a point $(x_{i}, y_{i})$ where $y_{i} = f (x_{i})$ . The $i$ is just a counter: the 1st dot, 2nd dot, and so on up to the $n$ -th.

Step 3 — One tiny triangle, and how long its slanted side is

Intuition WHAT / WHY / PICTURE

WHAT. Focus on the chord between dot $i - 1$ and dot $i$ . Going from one to the next you move right by an amount we call $Δ x_{i}$ and up-or-down by an amount $Δ y_{i}$ . The little symbol $Δ$ (Greek "delta") just means "the change in" — how much the thing grew (or shrank). We name the straight-line length of this chord the symbol $L_{i}$ — that is its definition, the number we are about to compute. WHY the right triangle. Sideways and vertical moves are at $9 0^{\circ}$ to each other, so the chord, the sideways move, and the vertical move form a right-angled triangle. That's the shape Pythagoras was built for. PICTURE. Below, top: a rising chord, horizontal leg labelled $Δ x_{i}$ , vertical leg $Δ y_{i}$ , slanted side (the chord length) labelled $L_{i}$ . Bottom: a falling chord where the curve goes down.

Rebuilding Pythagoras from scratch (no memorising). Put four copies of our triangle around a tilted square whose side is the chord $L_{i}$ . The big outer square has side $Δ x_{i} + Δ y_{i}$ . Its area equals the tilted inner square ( $L_{i}^{2}$ ) plus the four triangles ( $4 \times \frac{1}{2} Δ x_{i} Δ y_{i}$ ):

big square (Δ x_{i} + Δ y_{i})^{2} = tilted square L_{i}^{2} + four triangles 4 \cdot \frac{1}{2} Δ x_{i} Δ y_{i}

Expand the left side: $(Δ x_{i})^{2} + 2Δ x_{i} Δ y_{i} + (Δ y_{i})^{2}$ . The $2Δ x_{i} Δ y_{i}$ cancels the triangles, leaving

L_{i} = (Δ x_{i})^{2} + (Δ y_{i})^{2} .

$(Δ x_{i})^{2}$ — the sideways travel, squared.
$(Δ y_{i})^{2}$ — the vertical travel, squared.
— undoes the squaring to return an actual length. See Pythagorean theorem.

Step 4 — Pull the sideways step $Δ x_{i}$ out of the root

L_{i} = (Δ x_{i})^{2} + (Δ y_{i})^{2} = (Δ x_{i})^{2} (1 + \frac{( Δ y _{i} ) ^{2}}{( Δ x _{i} ) ^{2}}) = 1 + (tilt of the chord \frac{Δ y _{i}}{Δ x _{i}})^{2} Δ x_{i}

We could pull $Δ x_{i}$ out of the root because $Δ x_{i} > 0$ (we always walk left-to-right — this is exactly the "no doubling back, advance rightward" assumption from Step 1 — so $(Δ x_{i})^{2} = Δ x_{i}$ , no minus sign).
The $1$ is the leftover from the sideways leg — it is what stops us forgetting the horizontal travel.
$\frac{Δ y _{i}}{Δ x _{i}}$ = vertical-move ÷ right-move = the slope of the chord (a "secant slope"). If the chord falls, this ratio is negative — but it gets squared next, so again the sign washes out.

Step 5 — Turn the chord's slope into a true derivative (Mean Value Theorem)

Under those conditions, the Mean Value Theorem promises a point $x_{i}^{*}$ strictly between $x_{i - 1}$ and $x_{i}$ with

tangent slope at x_{i}^{*} f^{'} (x_{i}^{*}) = chord’s average slope \frac{f ( x _{i} ) - f ( x _{i - 1} )}{x _{i} - x _{i - 1}} = \frac{Δ y _{i}}{Δ x _{i}} .

$f^{'}$ means "the slope function" — how steep the curve is at a point.
$x_{i}^{*}$ (star) is some spot inside the little interval; MVT says it exists, we don't need to find it.

Substituting into Step 4's result:

L_{i} = 1 + (f^{'} (x_{i}^{*}))^{2} Δ x_{i} .

Step 6 — Add them up, shrink, and read off the integral

L = n \to \infty lim i = 1 \sum n height g (x_{i}^{*}) 1 + (f^{'} (x_{i}^{*}))^{2} strip width Δ x_{i} = \int_{a}^{b} 1 + (f^{'} (x))^{2} d x .

The sample height $g (x_{i}^{*}) = 1 + (f^{'} (x_{i}^{*}))^{2}$ becomes the integrand $g (x)$ .
$Δ x_{i}$ becomes $d x$ , an infinitely thin width.
$\sum$ becomes $\int$ — the same idea "add them all," now continuous.

Step 7 — Edge cases: does the picture still hold?

Flat curve ( $f^{'} = 0$ ). Vertical leg $Δ y_{i} = 0$ ; the triangle collapses to a horizontal segment. Integrand $= 1 + 0 = 1$ , so $L = \int_{a}^{b} 1 d x = b - a$ — exactly the width. ✓ A flat line's length is its horizontal span.
Vertical wall ( $f^{'} \to \infty$ ). The slope blows up; the $y = f (x)$ formula chokes because a vertical piece has $Δ x_{i} = 0$ (nothing to pull out). Fix: flip to the $x = g (y)$ form and integrate in $y$ instead — the vertical leg becomes the "safe" direction.
Corner (slope jumps). At a sharp kink $f^{'}$ doesn't exist — this violates MVT's differentiability condition at that one point. Since one bad point has zero width, the integral is unbothered — just split the interval at the corner and add the two pieces.

The one-picture summary

Read the film strip left-to-right: curve → chords → one triangle → pull out $d x$ → tangent matches chord → strips sum to an integral. Every arrow is one step above.

Recall Feynman: the whole walkthrough in plain words

You want the length of a bumpy path. You can't measure a curve directly, so you lay tiny straight sticks end-to-end along it (Step 2). Each stick is the long side of a little right triangle whose legs are "a bit rightward" and "a bit up or down"; the corner-square trick (Pythagoras) says the stick's length is $right^{2} + vertical^{2}$ — and because the vertical bit gets squared, it doesn't matter if the curve climbs or falls (Step 3). To collect everything into a sum over rightward steps, you factor the rightward bit out and are left with $1 + slope^{2}$ times the rightward step (Step 4). "Slope of the stick" isn't quite the curve's real slope, but the Mean Value Theorem — as long as the curve is smooth (continuous and differentiable) — guarantees the real tangent somewhere inside is exactly as tilted, so you replace it by $f^{'}$ (Step 5). Now every stick reads $1 + (f^{'})^{2} Δ x$ ; adding infinitely many infinitely short ones is what the integral sign means, giving $L = \int_{a}^{b} 1 + (f^{'})^{2} d x$ (Step 6). Test it on a flat road (length = width), a vertical cliff (integrate in $y$ instead), and a sharp corner (split there) — it survives all three (Step 7). Same $d s = d x^{2} + d y^{2}$ later builds surfaces of revolution, parametric and polar lengths.

Flashcards

Which little shape gives each chord its length?

A right triangle with legs

Δ x_{i}, Δ y_{i}

; the chord

L_{i}

is the hypotenuse

(Δ x_{i})^{2} + (Δ y_{i})^{2}

Why does the arc-length formula work for descending curves where

Δ y_{i} < 0

Because

Δ y_{i}

enters squared, and

(Δ y_{i})^{2} = ∣Δ y_{i} ∣

; the sign washes out, so length is the same climbing or falling.

Why can we pull

Δ x_{i}

out of the square root without a

\pm

Because

Δ x_{i} > 0

(we walk left to right, no doubling back), so

(Δ x_{i})^{2} = Δ x_{i}

What two hypotheses does the Mean Value Theorem need on each subinterval?

f

continuous on the closed

[x_{i - 1}, x_{i}]

and differentiable on the open

(x_{i - 1}, x_{i})

What does MVT replace, and with what?

The chord's average slope

Δ y_{i} /Δ x_{i}

with an exact derivative

f^{'} (x_{i}^{*})

at an interior point.

What turns the final sum into an integral?

It is a Riemann sum

\sum g (x_{i}^{*}) Δ x_{i}

; its limit as

Δ x_{i} \to 0

\int_{a}^{b} g d x

What is the arc-length element

d s

d s = d x^{2} + d y^{2} = 1 + (d y / d x)^{2} d x

; then

L = \int d s

For a vertical piece where

f^{'} \to \infty

, what do we do?

Integrate in

y

using

L = \int_{c}^{d} 1 + (d x / d y)^{2} d y

For a curve with a corner, how do we handle the non-differentiable point?

Split the interval at the corner and add the two arc lengths.

Connections

Arc length formula — derivation — the parent this page visualises.
Pythagorean theorem — the engine for each chord's length (rebuilt from scratch in Step 3).
Mean Value Theorem — swaps secant slope for a true derivative (Step 5).
Riemann sums and the definite integral — turns the sum into the integral (Step 6).
Surface area of revolution, Parametric arc length, Polar arc length — where $d s$ goes next.