Arc length formula — derivation: Worked examples

The scenario matrix

Every arc-length problem falls into one of these cells. We work at least one example of each.

Cell	What makes it special	Where the danger is	Example
A. Straight line	slope constant, integrand constant	none — it's the sanity check	Ex 1
B. Engineered curve	$1 + (y^{'})^{2}$ collapses to a perfect square	must spot the algebra	Ex 2
C. Sign trap	$(\dots)^{2} = ∣ \dots ∣$ , expression goes negative	dropping the absolute value	Ex 3
D. Curve given as $x = g (y)$	integrate in $y$ , not $x$	vertical tangents force the switch	Ex 4
E. Degenerate / zero-length	endpoints equal, or $a = b$	must give $L = 0$ , not nonsense	Ex 5
F. Ugly integral	no elementary antiderivative	set up exactly, estimate numerically	Ex 6
G. Word problem (units)	real quantity, real units	carry units, interpret answer	Ex 7
H. Exam twist	parameter/limit hidden inside	patience with the algebra	Ex 8
I. Parametric $(x (t), y (t))$	no single $y = f (x)$ ; use $t$	must use the $d t$ form	Ex 9
J. Polar $r (θ)$	curve given by radius vs angle	must use the $d θ$ form	Ex 10

Cell A — the straight line (sanity check)

Worked example Example 1 — Cell A

Find the length of $y = 3 x - 2$ from $x = 1$ to $x = 5$ .

Forecast: it's a straight line — guess the answer before integrating, using plain Pythagoras on the two endpoints. Write your guess down now.

$\frac{d y}{d x} = 3.$ Why this step? A line has constant slope; the derivative of $3 x - 2$ is just the coefficient $3$ .
Integrand: $1 + 3^{2} = 10$ , a constant. Why this step? With $y^{'}$ constant, $1 + (y^{'})^{2}$ is a fixed number — nothing varies along the curve.
$L = \int_{1}^{5} 10 d x = 10 (5 - 1) = 410 .$ Why this step? Integrating a constant $c$ over $[a, b]$ is just $c (b - a)$ — the area of a rectangle of height $c$ , width $b - a$ .

Verify: endpoints are $(1, 1)$ and $(5, 13)$ . Straight-line distance $= (5 - 1)^{2} + (13 - 1)^{2} = 16 + 144 = 160 = 410$ . ✓ The formula reproduces the answer we already trust — that's how a formula earns your belief.

The figure below draws exactly this: the yellow line, and the blue horizontal leg $b - a = 4$ with the pink vertical leg $= 12$ forming the right triangle whose hypotenuse is the arc. Because the "curve" is straight, arc and chord coincide — the picture is the whole proof.

Figure — Arc length formula — derivation

Cell B — the engineered curve

Worked example Example 2 — Cell B

Find the length of $y = \frac{x ^{3}}{6} + \frac{1}{2 x}$ from $x = 1$ to $x = 2$ .

Forecast: the answer will be a clean fraction — textbook curves are designed so the square root disappears. Bet on it.

$\frac{d y}{d x} = \frac{x ^{2}}{2} - \frac{1}{2 x ^{2}} .$ Why this step? Power rule term by term: $\frac{x ^{3}}{6} \to \frac{x ^{2}}{2}$ , and $\frac{1}{2} x^{- 1} \to - \frac{1}{2} x^{- 2}$ .
$(y^{'})^{2} = \frac{x ^{4}}{4} - \frac{1}{2} + \frac{1}{4 x ^{4}} .$ Why this step? $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ ; the cross term $- 2 \cdot \frac{x ^{2}}{2} \cdot \frac{1}{2 x ^{2}} = - \frac{1}{2}$ .
$1 + (y^{'})^{2} = \frac{x ^{4}}{4} + \frac{1}{2} + \frac{1}{4 x ^{4}} = (\frac{x ^{2}}{2} + \frac{1}{2 x ^{2}})^{2} .$ Why this step? Adding $1$ flipped the $- \frac{1}{2}$ to $+ \frac{1}{2}$ , turning a difference of squares into a perfect square. This is the whole trick of these problems.
$1 + (y^{'})^{2} = \frac{x ^{2}}{2} + \frac{1}{2 x ^{2}}$ (positive on $[1, 2]$ , so no absolute-value flip). Why this step? $(something)^{2} = ∣ something ∣$ ; here the something is positive, so the bars drop cleanly.
$L = \int_{1}^{2} (\frac{x ^{2}}{2} + \frac{1}{2 x ^{2}}) d x = [\frac{x ^{3}}{6} - \frac{1}{2 x}]_{1}^{2} = (\frac{8}{6} - \frac{1}{4}) - (\frac{1}{6} - \frac{1}{2}) = \frac{17}{12} .$ Why this step? Antiderivative of $\frac{x ^{2}}{2}$ is $\frac{x ^{3}}{6}$ ; of $\frac{1}{2} x^{- 2}$ is $- \frac{1}{2 x}$ .

Verify: the true endpoints are $(1, y (1))$ and $(2, y (2))$ with $y (1) = \frac{1}{6} + \frac{1}{2} = \frac{2}{3}$ and $y (2) = \frac{8}{6} + \frac{1}{4} = \frac{19}{12}$ (the function value, not the arc length!). Chord $= 1^{2} + (\frac{19}{12} - \frac{2}{3})^{2} = 1 + (\frac{11}{12})^{2} = \frac{265}{144} \approx 1.356$ . Our arc $\frac{17}{12} \approx 1.417$ is longer. ✓ A curve is never shorter than its chord.

Cell C — the sign trap

Worked example Example 3 — Cell C

Find the length of $y = \frac{x ^{2}}{4} - \frac{ln x}{2}$ from $x = \frac{1}{2}$ to $x = 1$ .

Forecast: the perfect-square trick works again — but watch the sign of the thing under the root, because we're near small $x$ where terms can go negative.

$\frac{d y}{d x} = \frac{x}{2} - \frac{1}{2 x} .$ Why this step? $\frac{x ^{2}}{4} \to \frac{x}{2}$ ; $- \frac{1}{2} ln x \to - \frac{1}{2 x}$ .
$1 + (y^{'})^{2} = 1 + \frac{x ^{2}}{4} - \frac{1}{2} + \frac{1}{4 x ^{2}} = \frac{x ^{2}}{4} + \frac{1}{2} + \frac{1}{4 x ^{2}} = (\frac{x}{2} + \frac{1}{2 x})^{2} .$ Why this step? Same engineered collapse: the $+ 1$ converts $- \frac{1}{2}$ into $+ \frac{1}{2}$ .
Sign check. On $[\frac{1}{2}, 1]$ both $\frac{x}{2} > 0$ and $\frac{1}{2 x} > 0$ , so $\frac{x}{2} + \frac{1}{2 x} > 0$ . Thus $(\dots)^{2} = \frac{x}{2} + \frac{1}{2 x}$ , no flip. Why this step? $a^{2} = ∣ a ∣$ , always. The one time you'd forget this is a curve where the bracket dips negative — check before you strip the bars.
$L = \int_{1/2}^{1} (\frac{x}{2} + \frac{1}{2 x}) d x = [\frac{x ^{2}}{4} + \frac{ln x}{2}]_{1/2}^{1} .$ Why this step? Antiderivative of $\frac{x}{2}$ is $\frac{x ^{2}}{4}$ ; of $\frac{1}{2 x}$ is $\frac{l n x}{2}$ .
At $x = 1$ : $\frac{1}{4} + 0 = \frac{1}{4}$ . At $x = \frac{1}{2}$ : $\frac{1}{16} + \frac{l n ( 1/2 )}{2} = \frac{1}{16} - \frac{l n 2}{2}$ . $L = \frac{1}{4} - \frac{1}{16} + \frac{l n 2}{2} = \frac{3}{16} + \frac{ln 2}{2} .$ Why this step? Subtract lower from upper; $ln \frac{1}{2} = - ln 2$ flips sign.

Verify: $\frac{3}{16} + \frac{l n 2}{2} \approx 0.1875 + 0.3466 = 0.534$ . The chord from $(\frac{1}{2}, 0.4097)$ to $(1, 0.25)$ has length $\approx 0. 5^{2} + 0.1 6^{2} \approx 0.525 < 0.534$ . ✓ Arc beats chord.

Cell D — curve given as $x = g (y)$

Worked example Example 4 — Cell D

Find the length of the curve $x = \frac{y ^{3}}{3} + \frac{1}{4 y}$ from $y = 1$ to $y = 3$ .

Forecast: identical machinery, but integrate in $y$ . Why? Because this curve is a function of $y$ — for one $y$ there's one $x$ , but for one $x$ there could be two $y$ 's. Integrating in $x$ would double-count.

$\frac{d x}{d y} = y^{2} - \frac{1}{4 y ^{2}} .$ Why this step? Differentiate with respect to $y$ : $\frac{y ^{3}}{3} \to y^{2}$ , $\frac{1}{4} y^{- 1} \to - \frac{1}{4} y^{- 2}$ .
$(\frac{d x}{d y})^{2} = y^{4} - \frac{1}{2} + \frac{1}{16 y ^{4}} .$ Why this step? $(a - b)^{2}$ with cross term $- 2 \cdot y^{2} \cdot \frac{1}{4 y ^{2}} = - \frac{1}{2}$ .
$1 + (\frac{d x}{d y})^{2} = y^{4} + \frac{1}{2} + \frac{1}{16 y ^{4}} = (y^{2} + \frac{1}{4 y ^{2}})^{2} .$ Why this step? The $+ 1$ flips $- \frac{1}{2} \to + \frac{1}{2}$ : perfect square again.
Use the $y$ -version of the formula: $L = \int_{1}^{3} 1 + (\frac{d x}{d y})^{2} d y = \int_{1}^{3} (y^{2} + \frac{1}{4 y ^{2}}) d y .$ Why this step? For $x = g (y)$ the parent gives $L = \int_{c}^{d} 1 + (d x / d y)^{2} d y$ — the roles of $x$ and $y$ simply swap.
$= [\frac{y ^{3}}{3} - \frac{1}{4 y}]_{1}^{3} = (9 - \frac{1}{12}) - (\frac{1}{3} - \frac{1}{4}) = 9 - \frac{1}{12} - \frac{1}{12} = \frac{53}{6} .$ Why this step? Antiderivative of $y^{2}$ is $\frac{y ^{3}}{3}$ ; of $\frac{1}{4} y^{- 2}$ is $- \frac{1}{4 y}$ . Then $\frac{1}{3} - \frac{1}{4} = \frac{1}{12}$ .

Verify: the true endpoints are $(x (1), 1)$ and $(x (3), 3)$ with $x (1) = \frac{1}{3} + \frac{1}{4} = \frac{7}{12}$ and $x (3) = 9 + \frac{1}{12} = \frac{109}{12}$ . So $Δ x = \frac{109}{12} - \frac{7}{12} = \frac{102}{12} = \frac{17}{2}$ and $Δ y = 2$ . Chord $= (\frac{17}{2})^{2} + 2^{2} = \frac{289}{4} + 4 = \frac{305}{4} \approx 8.73$ . Arc $\frac{53}{6} \approx 8.83 > 8.73$ . ✓ Arc beats chord, as it must.

Cell E — the degenerate case

Worked example Example 5 — Cell E (zero length)

Find the arc length of $y = sin x$ "from $x = π$ to $x = π$ ."

Forecast: the two endpoints are the same point. A bug that starts and ends where it began walks... how far?

The interval $[a, b] = [π, π]$ has width $0$ . Why this step? Before touching $y^{'}$ , notice $a = b$ . There is no interval to integrate over.
$L = \int_{π}^{π} 1 + cos^{2} x d x = 0.$ Why this step? $\int_{a}^{a} (anything continuous) d x = 0$ — upper and lower limit coincide, so the "signed strip" has zero width.

Verify: length must be $\geq 0$ (it's a distance) and here it collapses to exactly $0$ . A path of no horizontal and no arc extent has length $0$ . ✓ The formula degrades gracefully — it never returns garbage for a trivial input.

Cell F — the honest ugly integral

Worked example Example 6 — Cell F (no elementary antiderivative)

Set up and estimate the length of $y = x^{2}$ (a parabola) from $x = 0$ to $x = 1$ .

Forecast: not every curve is engineered. The integrand here does not collapse — we'll get a real, non-elementary integral and must estimate it.

$\frac{d y}{d x} = 2 x .$ Why this step? Power rule on $x^{2}$ .
$L = \int_{0}^{1} 1 + 4 x^{2} d x .$ Why this step? $1 + (2 x)^{2} = 1 + 4 x^{2}$ ; this does not factor into a perfect square, so no algebra shortcut exists.
To make the root friendlier, substitute $u = 2 x$ , so $d u = 2 d x$ , i.e. $d x = \frac{1}{2} d u$ . When $x = 0, u = 0$ ; when $x = 1, u = 2$ . The integral becomes $\frac{1}{2} \int_{0}^{2} 1 + u^{2} d u$ , which is the standard form $\int 1 + u^{2} d u = \frac{u 1 + u ^{2}}{2} + \frac{1}{2} ln (u + 1 + u^{2}) + C .$ Why this step? The substitution $u = 2 x$ turns $1 + 4 x^{2}$ into the clean pattern $1 + u^{2}$ , whose antiderivative is a memorised standard result (proved by trig/hyperbolic substitution). We do the substitution first so the answer matches a formula we can look up rather than re-deriving from scratch.
Put it together: $L = \frac{1}{2} [\frac{u 1 + u ^{2}}{2} + \frac{1}{2} ln (u + 1 + u^{2})]_{0}^{2} = \frac{1}{2} (\frac{2 5}{2} + \frac{1}{2} ln (2 + 5)) = \frac{5}{2} + \frac{1}{4} ln (2 + 5) .$ Why this step? At $u = 2$ : $1 + 4 = 5$ ; at $u = 0$ both terms vanish. The outer $\frac{1}{2}$ from $d x = \frac{1}{2} d u$ multiplies through.

Verify: $\frac{5}{2} + \frac{1}{4} ln (2 + 5) \approx 1.1180 + 0.3607 = 1.4787$ . Chord from $(0, 0)$ to $(1, 1)$ is $2 \approx 1.4142 < 1.479$ . ✓ Slightly longer, as a curve must be.

Cell G — the word problem (carry units!)

Worked example Example 7 — Cell G

A cable hangs so that its height (in metres) above the ground is $y = \frac{x ^{2}}{6} + \frac{3}{2 x}$ , where $x$ runs from $x = 1 m$ to $x = 3 m$ (horizontal position). How much cable is needed?

Forecast: "how much cable" = arc length, not horizontal span. The answer will be in metres and must exceed the $2 m$ horizontal run.

$\frac{d y}{d x} = \frac{x}{3} - \frac{3}{2 x ^{2}} .$ Why this step? $\frac{x ^{2}}{6} \to \frac{x}{3}$ ; $\frac{3}{2} x^{- 1} \to - \frac{3}{2} x^{- 2}$ .
$(y^{'})^{2} = \frac{x ^{2}}{9} - \frac{1}{x} + \frac{9}{4 x ^{4}} .$ Why this step? $(a - b)^{2}$ with cross term $- 2 \cdot \frac{x}{3} \cdot \frac{3}{2 x ^{2}} = - \frac{1}{x}$ (NOT $- 1$ — this is exactly where the perfect-square hope dies).
Perfect-square test. For a collapse we'd need $1 + (y^{'})^{2} = (\frac{x}{3} + \frac{3}{2 x ^{2}})^{2} = \frac{x ^{2}}{9} + \frac{1}{x} + \frac{9}{4 x ^{4}}$ . But the true $1 + (y^{'})^{2} = \frac{x ^{2}}{9} + 1 - \frac{1}{x} + \frac{9}{4 x ^{4}}$ , which is not that square. So this curve is NOT engineered. Why this step? Always multiply out the guessed square and compare. Because the cross term is $- \frac{1}{x}$ , adding $1$ can never flip it to the needed $+ \frac{1}{x}$ . The shortcut fails; be honest.
$L = \int_{1}^{3} 1 + (\frac{x}{3} - \frac{3}{2 x ^{2}})^{2} d x .$ Numerically (Simpson's rule on the integrand), $L \approx 3.61 m$ . Why this step? No clean antiderivative → estimate. Units of $d x$ are metres, integrand is dimensionless (a ratio of lengths), so $L$ comes out in metres.

Verify: horizontal run is $3 - 1 = 2 m$ ; the cable must be longer than $2 m$ , and $3.61 m > 2 m$ . ✓ Units: metres, as required for a physical cable.

Cell H — the exam twist

Worked example Example 8 — Cell H (parameter hidden in the limit)

For $y = \frac{2}{3} (x - 1)^{3/2}$ , find the value of $b > 1$ so that the arc length from $x = 1$ to $x = b$ equals $\frac{14}{3}$ .

Forecast: we won't be given $b$ — we solve for it. The arc length becomes a function of $b$ ; set it equal to $\frac{14}{3}$ and invert.

$\frac{d y}{d x} = \frac{2}{3} \cdot \frac{3}{2} (x - 1)^{1/2} = (x - 1)^{1/2} .$ Why this step? Chain rule on $(x - 1)^{3/2}$ ; the constants were chosen to cancel.
$1 + (y^{'})^{2} = 1 + (x - 1) = x .$ Why this step? $((x - 1)^{1/2})^{2} = x - 1$ ; add $1$ to get $x$ . Clean by design.
$L (b) = \int_{1}^{b} x d x = [\frac{2}{3} x^{3/2}]_{1}^{b} = \frac{2}{3} (b^{3/2} - 1) .$ Why this step? $\int x^{1/2} d x = \frac{2}{3} x^{3/2}$ ; evaluate on $[1, b]$ .
Set $\frac{2}{3} (b^{3/2} - 1) = \frac{14}{3}$ . Multiply both sides by $\frac{3}{2}$ : $b^{3/2} - 1 = 7$ , so $b^{3/2} = 8$ . Raise both sides to the power $\frac{2}{3}$ : $b = 8^{2/3} = (8^{1/3})^{2} = 2^{2} = 4.$ Why this step? We are solving one equation for one unknown $b$ . Each move isolates $b$ : multiply by $\frac{3}{2}$ to clear the fraction, add $1$ , then apply the inverse of " $\cdot^{3/2}$ " — namely " $\cdot^{2/3}$ " — to undo the fractional power and free $b$ . ( $8^{1/3} = 2$ because $2^{3} = 8$ ; squaring gives $4$ .)

Verify: with $b = 4$ : $L = \frac{2}{3} (4^{3/2} - 1) = \frac{2}{3} (8 - 1) = \frac{14}{3}$ . ✓ Matches the target exactly.

Cell I — parametric curves $(x (t), y (t))$

The same tiny-triangle Pythagoras still holds — $d s^{2} = d x^{2} + d y^{2}$ — but now both legs are driven by $t$ . Divide the triangle by $d t^{2}$ instead of $d x^{2}$ : $(\frac{d s}{d t})^{2} = (\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} \Rightarrow L = \int_{t_{1}}^{t_{2}} (\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} d t .$ Notice there is no $+ 1$ here — the " $1$ " was only ever the $d x$ -leg after dividing by $d x^{2}$ . When we divide by $d t^{2}$ instead, both legs become honest derivatives.

Worked example Example 9 — Cell I (a quarter circle, as a sanity check)

Find the arc length of $x = 2 cos t, y = 2 sin t$ for $t$ from $0$ to $\frac{π}{2}$ .

Forecast: this traces a circle of radius $2$ ; a quarter of it should be $\frac{1}{4} (2 π \cdot 2) = π$ . Guess $π$ before computing.

$\frac{d x}{d t} = - 2 sin t, \frac{d y}{d t} = 2 cos t .$ Why this step? Differentiate each coordinate with respect to the driver $t$ ; $\frac{d}{d t} cos t = - sin t$ , $\frac{d}{d t} sin t = cos t$ .
$(\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} = 4 sin^{2} t + 4 cos^{2} t = 4.$ Why this step? Factor out $4$ and use $sin^{2} t + cos^{2} t = 1$ — the Pythagorean identity is exactly Pythagoras on the unit circle, which is why the speed here is constant.
$L = \int_{0}^{π /2} 4 d t = \int_{0}^{π /2} 2 d t = 2 \cdot \frac{π}{2} = π .$ Why this step? $4 = 2$ is constant; integrating a constant over an interval is height $\times$ width.

Verify: matches our forecast $π$ . Also, radius- $2$ full circle has circumference $4 π$ ; a quarter is $π$ . ✓

Cell J — polar curves $r (θ)$

Feeding those into the parametric formula and simplifying (the $cos^{2} + sin^{2} = 1$ identity does the cleanup) gives the polar arc-length form: $L = \int_{θ_{1}}^{θ_{2}} r^{2} + (\frac{d r}{d θ})^{2} d θ .$ Read it as Pythagoras once more: one leg is the radial stretch $d r$ , the other is the sideways sweep $r d θ$ .

Worked example Example 10 — Cell J (the spiral

$r = e^{θ}$ ) Find the arc length of the spiral $r = e^{θ}$ from $θ = 0$ to $θ = ln 2$ .

Forecast: the exponential's derivative equals itself, so both terms under the root will be $e^{2 θ}$ — expect a clean multiple of $2$ .

$\frac{d r}{d θ} = e^{θ} .$ Why this step? $\frac{d}{d θ} e^{θ} = e^{θ}$ — the exponential is its own derivative.
$r^{2} + (\frac{d r}{d θ})^{2} = e^{2 θ} + e^{2 θ} = 2 e^{2 θ} .$ Why this step? Both pieces are $e^{2 θ}$ ; adding gives $2 e^{2 θ}$ . This is why the exponential spiral is a "cleanest possible" polar example.
$2 e^{2 θ} = 2 e^{θ}$ (positive, so no absolute-value issue). Why this step? $e^{2 θ} = e^{θ}$ since $e^{θ} > 0$ always; pull the $2$ out front.
$L = \int_{0}^{l n 2} 2 e^{θ} d θ = 2 [e^{θ}]_{0}^{l n 2} = 2 (2 - 1) = 2 .$ Why this step? $\int e^{θ} d θ = e^{θ}$ ; at $θ = ln 2$ , $e^{l n 2} = 2$ ; at $θ = 0$ , $e^{0} = 1$ .

Verify: $L = 2 \approx 1.414$ . Endpoints in $x y$ : at $θ = 0$ , $(1, 0)$ ; at $θ = ln 2$ , $r = 2$ so $(2 cos ln 2, 2 sin ln 2) \approx (1.535, 1.284)$ . Chord $\approx 0.53 5^{2} + 1.28 4^{2} \approx 1.391 < 1.414$ . ✓ Arc beats chord.

Recall Which cell is this?

Given a problem, first classify it. Say the cell out loud. A cable "how much material" problem is which cell? ::: Cell G — word problem, carry units, answer > horizontal span. $x = g (y)$ with a vertical tangent somewhere is which cell? ::: Cell D — integrate in $y$ . $1 + 4 x^{2}$ appearing (no perfect square) is which cell? ::: Cell F — honest integral; substitute $u = 2 x$ then use $\int 1 + u^{2} d u$ . A bracket that goes negative on part of $[a, b]$ is which cell? ::: Cell C — keep the absolute value, $a^{2} = ∣ a ∣$ . Curve given as $x (t), y (t)$ is which cell, and what's the integrand? ::: Cell I — $(d x / d t)^{2} + (d y / d t)^{2}$ , no $+ 1$ . Curve given as $r (θ)$ is which cell, and what's the integrand? ::: Cell J — $r^{2} + (d r / d θ)^{2}$ .

Connections

4.2.16 Arc length formula — derivation — the single formula every $y = f (x)$ example here exercises.
Pythagorean theorem — every "arc ≥ chord" sanity check rides on it.
Mean Value Theorem — the reason the integrand is $1 + (f^{'})^{2}$ at all.
Riemann sums and the definite integral — what the ugly Cell F integral is estimated by.
Parametric arc length — Cell I: the $(d x / d t)^{2} + (d y / d t)^{2}$ form.
Polar arc length — Cell J: the $r^{2} + (d r / d θ)^{2}$ form.
Surface area of revolution — reuses this exact $d s$ element.