Arc length formula — derivation: Question bank

The whole formula we are stress-testing: $L = \int_{a}^{b} 1 + (\frac{d y}{d x})^{2} d x, d s = d x^{2} + d y^{2} .$ Keep the little right triangle in mind the whole way: base $d x$ , height $d y$ , hypotenuse $d s$ .

True or false — justify

Arc length can be shorter than the straight line joining the two endpoints.

False. A straight line is the shortest path between two points, so the curve — a sum of chords each at least as long as its own straight piece — is always

\geq

the endpoint chord (see Pythagorean theorem).

If a curve is horizontal (

y

constant) from

x = a

x = b

, its arc length is exactly

b - a

True. Then

d y / d x = 0

, so the integrand is

1 + 0 = 1

and

L = \int_{a}^{b} 1 d x = b - a

— a flat curve is just its own base.

Doubling every

y

-value of a curve doubles its arc length.

False. Stretching vertically changes the slope

d y / d x

nonlinearly inside a square root, so the length changes but not by a clean factor of 2.

Shifting a curve up by a constant (

y \to y + c

) leaves its arc length unchanged.

True. A vertical shift doesn't change

d y / d x

(the derivative of a constant is 0), so the integrand and the length are identical.

The arc length integrand

1 + (y^{'})^{2}

is always

\geq 1

True.

(y^{'})^{2} \geq 0

, so the thing under the root is at least 1, meaning arc always accumulates at least as fast as horizontal distance.

For a curve

x = g (y)

you can always just use

\int_{a}^{b} 1 + (d y / d x)^{2} d x

anyway.

False. If

x = g (y)

has vertical tangents (like a sideways parabola),

d y / d x

blows up; you must integrate in

y

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

Arc length always increases when you widen the interval

[a, b]

True. The integrand

1 + (y^{'})^{2} \geq 1 > 0

, so integrating over a larger interval adds a strictly positive amount of length.

The Mean Value Theorem step requires

f

to be differentiable everywhere on

[a, b]

Mostly true and worth respecting: Mean Value Theorem needs

f

continuous on the closed interval and differentiable on the open interior; a corner (non-differentiable point) breaks the clean single formula and forces splitting the integral.

Spot the error

L = \int_{a}^{b} y d x

, because we integrate to get totals."

\int y d x

computes the area under the curve, not the distance along it. Length must carry the

1 + (y^{'})^{2}

from the hypotenuse;

y

alone measures height, not travel.

L = \int_{a}^{b} ∣ f^{'} (x) ∣ d x

— slope tells you how far you go."

This counts only vertical travel

Δ y

and ignores the horizontal leg

Δ x

. The "

1

" under the root is exactly the missing

d x

term; drop it and you shorten every slanted step.

(\frac{x}{4} + \frac{1}{x})^{2} = \frac{x}{4} + \frac{1}{x}

for all

x

Only valid when the inside is

\geq 0

. In general

a^{2} = ∣ a ∣

; you must check the sign on the actual interval before dropping the absolute value.

(Δ x)^{2} + (Δ y)^{2} = Δ x + Δ y

You cannot split a square root over a sum;

a^{2} + b^{2} \neq = a + b

. That false step would ignore Pythagoras and always overestimate the hypotenuse.

"We pulled out

Δ x_{i}

(Δ x_{i})^{2} = Δ x_{i}

— no absolute value needed, so it's a lucky simplification."

It's not luck: we chose the labels so

x_{i - 1} < x_{i}

, making

Δ x_{i} > 0

, hence

(Δ x_{i})^{2} = Δ x_{i}

genuinely. The positivity is by construction, not by accident.

"The Riemann sum

\sum 1 + (f^{'} (x_{i}^{*}))^{2} Δ x_{i}

needs the

x_{i}^{*}

chosen at the left endpoint."

No — the Mean Value Theorem hands us a specific interior point

x_{i}^{*}

, and because any tag point works in the limit when

f^{'}

is continuous, the sum still converges to the same integral.

"A curve with a sharp corner has infinite arc length there."

False. A corner makes

f^{'}

undefined at one point but the length stays finite — you just split the integral at the corner and add the two finite pieces.

Why questions

Why does a "

+ 1

" appear inside the square root, and not a "

+ 2

" or nothing?

Because

d s^{2} = d x^{2} + d y^{2}

; dividing by

d x^{2}

gives

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

, and that lone

1

is precisely the horizontal-leg contribution

d x^{2} / d x^{2}

Why do we approximate with straight chords instead of, say, tiny circular arcs?

A straight segment is the only shape whose length we already know exactly (Pythagoras). We tile with the tool we have, then shrink the tiles so the approximation error vanishes.

Why does the Mean Value Theorem show up in an integration derivation?

It converts the measurable secant slope

Δ y /Δ x

into an actual derivative value

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

, so the sum becomes a Riemann sum for

1 + (f^{'})^{2}

rather than a mess of differences.

Why must

f^{'}

be continuous for the limit to be a definite integral?

Riemann integrability of

1 + (f^{'})^{2}

needs the integrand to be nice (continuous

f^{'}

is enough); a wildly discontinuous slope could make the limiting sum fail to converge to an integral.

Why is the same

d s

reused for surfaces of revolution?

Because a surface of revolution is built by spinning arc elements: each strip has length

d s

and circumference

2 π (radius)

, so

d A = 2 π r d s

reuses the very element we derived here.

Why can the same curve be written as

y = f (x)

x = g (y)

and give the same length?

Length is a geometric property of the path, independent of which variable you sweep; the two integrals are just different parametrizations of the identical

d s

, as parametric form makes explicit.

Why does the formula fail (or need care) at a vertical tangent?

There

d y / d x \to \infty

, so

1 + (y^{'})^{2}

diverges as an integrand in

x

; switch to integrating in

y

, where the slope

d x / d y

is finite.

Edge cases

What is the arc length over a zero-width interval,

a = b

Zero.

\int_{a}^{a} (\dots) d x = 0

— no interval, no path, no length.

What happens to the integrand at a point where

f^{'} (x) = 0

(a flat spot / local max)?

It equals

1 + 0 = 1

there; the curve is momentarily horizontal, so it accumulates length at the pure horizontal rate — perfectly finite and well behaved.

A curve has a vertical tangent (slope

\to \infty

) somewhere in

[a, b]

. Can we still find its length?

Yes, but not by integrating in

x

there. Reparametrize as

x = g (y)

or use a parameter $t$, so the infinite slope becomes a finite, integrable expression.

Does the arc length formula work for a curve given in polar coordinates like

r = 1 + cos θ

Not directly — you need the polar version

L = \int r^{2} + (d r / d θ)^{2} d θ

, derived from the same

d s^{2} = d x^{2} + d y^{2}

after converting.

f^{'}

has a jump discontinuity (a corner) at

x = c \in (a, b)

, how do we compute

L

Split:

L = \int_{a}^{c} + \int_{c}^{b}

. Each piece is smooth and finite; add them. The single corner contributes no length by itself.

What if the curve doubles back (not a function of

x

), like a full circle?

The

y = f (x)

formula fails the vertical-line test. Use parametric arc length, where

x

and

y

each depend on a parameter

t

, so the path can loop freely.

Recall One-line summary of every trap

Length lives in the hypotenuse $d s = d x^{2} + d y^{2}$ : never $y$ (that's area), never $∣ y^{'} ∣$ alone (that's vertical-only), never dropping the $∣ \cdot ∣$ (sign check), and never in $x$ where the tangent goes vertical (switch variables).

Connections

Arc length formula — derivation — the parent this bank interrogates.
Pythagorean theorem — why $d s$ has that shape.
Mean Value Theorem — the secant-to-derivative bridge tested above.
Riemann sums and the definite integral — why the limiting sum is an integral.
Parametric arc length, Polar arc length, Surface area of revolution — where the edge cases lead next.