Geometric Bounds for Arc Length

Why Bound a Curve’s Length?

A length integral can be difficult to evaluate, and even a correct-looking calculation can contain an error. Geometry supplies two useful checks: a curve must be at least as long as the straight chord joining its endpoints, and it cannot be longer than the total distance traveled horizontally plus the total distance traveled vertically.

These bounds apply to curves that turn, reverse direction, or pass through several smooth pieces. The upper bound uses the accumulated coordinate changes, not merely the difference between the endpoint coordinates. That distinction matters whenever a curve changes direction.

Definitions and the Bounds

Let a plane curve be parametrized by $\mathbf{r}(t)=(x(t),y(t))$, for $a\le t\le b$. Assume it is piecewise $C^1$: there is a finite partition of $[a,b]$ such that $x$ and $y$ have continuous first derivatives on each piece. Its length is the sum of the speed integrals over those pieces, equivalently

L=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2}\,dt.

The endpoint chord is the straight-line distance between $\mathbf{r}(a)$ and $\mathbf{r}(b)$. The accumulated horizontal variation $V_x$ and vertical variation $V_y$ are the total distances traveled in the two coordinate directions:

V_x=\int_a^b |x'(t)|\,dt, \qquad V_y=\int_a^b |y'(t)|\,dt.

For a piecewise $C^1$ curve, these integrals are understood piece by piece. The absolute values ensure that motion in opposite directions adds to the total distance rather than canceling.

Theorem: For any piecewise $C^1$ plane curve $\mathbf{r}(t)=(x(t),y(t))$ on $[a,b]$, $$ \sqrt{(x(b)-x(a))^2+(y(b)-y(a))^2} \le L \le V_x+V_y. $$

The left side is a lower bound because the shortest route between two points is their straight-line segment. The right side is an upper bound because each small movement along the curve has length no greater than the sum of its horizontal and vertical movements.

Why the Inequalities Hold

Write $\mathbf{r}'(t)=(x'(t),y'(t))$ on each smooth piece. The displacement from the starting point to the endpoint is the integral of the velocity, taken piece by piece:

\mathbf{r}(b)-\mathbf{r}(a)=\int_a^b \mathbf{r}'(t)\,dt.

If the displacement is nonzero, let $\mathbf{e}$ be the unit vector in its direction. Taking the dot product with $\mathbf{e}$ gives its length. Since the dot product of a unit vector with any vector is no greater than the length of that vector,

\begin{aligned} |\mathbf{r}(b)-\mathbf{r}(a)| &=\mathbf{e}\cdot\bigl(\mathbf{r}(b)-\mathbf{r}(a)\bigr)\\ &=\int_a^b \mathbf{e}\cdot\mathbf{r}'(t)\,dt\\ &\le \int_a^b |\mathbf{r}'(t)|\,dt=L. \end{aligned}

If the displacement is zero, the lower bound simply says $0\le L$, which is true. This proves the chord bound, including for piecewise smooth curves.

For the upper bound, apply the pointwise inequality $\sqrt{u^2+v^2}\le |u|+|v|$. It follows by squaring both nonnegative sides: the square of the right side is $u^2+v^2+2|uv|$, which is at least $u^2+v^2$. Therefore, on each smooth piece,

\sqrt{(x'(t))^2+(y'(t))^2} \le |x'(t)|+|y'(t)|.

Integrating and adding the results over the pieces proves $L\le V_x+V_y$.

Equality cases: The chord bound is an equality when the curve travels along the chord in one direction without backtracking. The upper bound is an equality when $x'(t)y'(t)=0$ almost everywhere: at each instant, at most one coordinate changes.

Using the Bounds for a Graph

For a graph $y=f(x)$ on $[a,b]$, use the parametrization $\mathbf{r}(x)=(x,f(x))$. If $f$ is piecewise $C^1$, its length is $\int_a^b\sqrt{1+(f'(x))^2}\,dx$. The endpoint chord and coordinate variations give

\sqrt{(b-a)^2+(f(b)-f(a))^2} \le L \le (b-a)+\int_a^b |f'(x)|\,dx.

The horizontal variation is $b-a$ because $x$ increases across the interval. The vertical variation is the integral of $|f'|$, which accounts for any rises and falls. If the graph is traversed in the opposite direction, the length and bounds are unchanged; the horizontal variation is then $|b-a|$.

Worked Example: Checking the Length of a Parabola Segment

Find the length of $y=x^2/2$ from $x=0$ to $x=2$, then check it against the geometric bounds.

Here $f'(x)=x$, so the length integral is

L=\int_0^2\sqrt{1+x^2}\,dx.

An antiderivative is $F(x)=\frac12\bigl(x\sqrt{1+x^2}+\ln(x+\sqrt{1+x^2})\bigr)$. To check it, differentiate:

F'(x) =\frac12\left(\frac{1+2x^2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right) =\frac{1+x^2}{\sqrt{1+x^2}} =\sqrt{1+x^2}.

At $x=2$, $F(2)=\sqrt5+\frac12\ln(2+\sqrt5)$; at $x=0$, $F(0)=0$. Thus

L=\sqrt5+\frac12\ln(2+\sqrt5).

The endpoints are $(0,0)$ and $(2,2)$, so the chord has length $\sqrt{2^2+2^2}=2\sqrt2$. The horizontal variation is $2$, and the vertical variation is $\int_0^2 x\,dx=2$. Hence $2\sqrt2\le L\le4$. For a numerical check, rounded only at the end to three decimal places, $L\approx2.958$, while $2\sqrt2\approx2.828$. The computed length lies between the bounds.

Bounds for Parametric Curves

The same theorem works directly with a parametrization, without converting the curve to a graph. This is useful when a curve turns back in one coordinate or when one coordinate is not a single-valued function of the other. The chord depends only on the endpoints; the upper bound reflects the full coordinate travel.

Worked Example: A Quarter-Circle Arc

Consider $\mathbf{r}(t)=(\cos t,\sin t)$, for $0\le t\le\pi/2$. Its derivative is $(-\sin t,\cos t)$, so its speed is

\sqrt{(-\sin t)^2+(\cos t)^2} =\sqrt{\sin^2t+\cos^2t}=1.

Therefore $L=\int_0^{\pi/2}1\,dt=\pi/2$. The endpoints are $(1,0)$ and $(0,1)$, giving chord length $\sqrt{(0-1)^2+(1-0)^2}=\sqrt2$.

For the upper bound, $x'(t)=-\sin t$ and $y'(t)=\cos t$. On this interval, $\sin t$ and $\cos t$ are nonnegative, so

V_x=\int_0^{\pi/2}\sin t\,dt=1, \qquad V_y=\int_0^{\pi/2}\cos t\,dt=1.

Thus the bounds are $\sqrt2\le L\le2$. Indeed, $L=\pi/2$, and to three decimal places $1.414<1.571<2.000$.

Accumulated Variation Matters

Endpoint differences can hide motion that reverses direction. The chord uses net displacement, but the upper bound must count all the horizontal and vertical travel. A piecewise smooth example makes this distinction clear.

Worked Example: A Two-Segment Zigzag

Let the curve travel in a straight line from $(0,0)$ to $(1,1)$, then in a straight line from $(1,1)$ to $(2,0)$. The first segment has displacement $(1,1)$, so its length is $\sqrt{1^2+1^2}=\sqrt2$. The second has displacement $(1,-1)$, so its length is also $\sqrt2$. By additivity,

L=\sqrt2+\sqrt2=2\sqrt2.

The endpoints are $(0,0)$ and $(2,0)$, so the chord length is $2$. The horizontal coordinate increases by $1$ on each segment, giving $V_x=1+1=2$. The vertical coordinate rises by $1$ and then falls by $1$, giving $V_y=1+1=2$. The upper bound is $V_x+V_y=4$, and the check is

2\le 2\sqrt2\le4.

The endpoint vertical difference is zero, but the curve has vertical variation $2$. Using only the endpoint difference would miss the upward and downward travel.

A Common Mistake

A frequent error is to use the endpoint chord as an upper bound, or to replace total vertical variation by the absolute difference in endpoint heights. For the zigzag above, the incorrect calculation would be

L\le |2-0|+|0-0|=2.

But the actual length is $2\sqrt2$, which is greater than $2$. The calculation has used net changes in both coordinates, not the accumulated coordinate travel. The correct upper-bound calculation is $V_x+V_y=2+2=4$; the endpoint chord $2$ is a lower bound.

Practical check: A computed length must be at least the endpoint chord and no greater than the sum of the total horizontal and vertical variations. If either inequality fails, check the endpoints, the speed, and whether direction changes were included in the variation.

How to Apply the Bounds

For a length calculation, first identify the curve’s endpoints and compute their straight-line distance. Then find the total variation of each coordinate, using absolute values of the coordinate derivatives on every smooth piece. Compare the resulting lower and upper bounds with the proposed length.

These inequalities do not usually determine the exact length: the bounds can be separated, as in the examples. Their value is that they provide independent geometric checks. A length outside the interval is certainly incorrect; a length inside it passes this check, though it may still contain an error.

Check Your Understanding

Use the endpoint chord and accumulated coordinate variations to answer the following.

For a piecewise $C^1$ curve, write the chord lower bound and the variation upper bound in terms of $x(t)$ and $y(t)$.
A curve starts at $(1,2)$ and ends at $(4,6)$. What is its endpoint chord length?
Why does the upper bound use $\int |y'(t)|\,dt$ rather than $|y(b)-y(a)|$?
For $\mathbf{r}(t)=(t,2t)$, $0\le t\le1$, compute the chord, length, and sum of coordinate variations.
Can a length calculation that falls between the two bounds still be wrong? Explain.