Additivity for Piecewise Smooth Curves

Why Piecewise Curves Need a Length Rule

A curve can be smooth on several intervals yet fail to have a single tangent direction where those intervals meet. A path made of two straight segments, for example, can turn sharply at a corner. The arc-length formula still applies, but it must be used on the smooth pieces separately.

The key fact is that dividing a path at a join does not change its total length: the length of the whole path is the sum of the lengths of its pieces. This lets us calculate paths with corners without pretending that they are smooth at the corners.

Definition: A continuous curve $\mathbf{r}:[a,b]\to\mathbb{R}^n$ is piecewise smooth if there is a finite partition $a=t_0<t_1<\cdots<t_m=b$ such that the restriction of $\mathbf{r}$ to each interval $[t_{i-1},t_i]$ is continuously differentiable, with one-sided derivatives allowed at the interval endpoints. The points $t_1,\ldots,t_{m-1}$ are the joins. The curve need not be differentiable at a join.

On each smooth piece, the parametric arc-length formula gives the length as the integral of speed. Here speed is $\|\mathbf{r}'(t)\|$, the magnitude of the velocity vector. A zero speed at some parameter value does not prevent the formula from applying.

Theorem: If a continuous piecewise smooth curve $\mathbf{r}:[a,b]\to\mathbb{R}^n$ has a partition $a=t_0<t_1<\cdots<t_m=b$ on whose intervals it is continuously differentiable, then its length is $$ L=\sum_{i=1}^{m}L_i =\sum_{i=1}^{m}\int_{t_{i-1}}^{t_i}\|\mathbf{r}'(t)\|\,dt, $$ where $L_i$ is the length of the restriction of the curve to $[t_{i-1},t_i]$. The formula allows the derivative, and therefore the tangent direction, to change at the joins.

Why the Lengths Add

The additivity has a geometric justification as well as an integral one. For a continuous curve, its length can be defined as the supremum of the lengths of polygonal paths formed by choosing finitely many points on the curve in parameter order and joining consecutive points with straight segments.

Let $L_i$ denote the length of the $i$th piece. Any polygonal approximation to the whole curve can be refined by inserting every join $t_i$. Inserting a point cannot shorten the polygonal path: if a segment from $A$ to $C$ is split at $B$, the triangle inequality gives $\|C-A\|\leq \|B-A\|+\|C-B\|$. After inserting the joins, the part of the polygonal path on each interval has length at most $L_i$. Thus every polygonal approximation to the full curve has length at most $\sum_i L_i$, so the full curve has length at most that sum.

For the reverse inequality, choose a polygonal approximation on each piece whose length is within $\varepsilon/m$ of $L_i$, where $\varepsilon>0$. Combining these finitely many approximations gives a polygonal approximation to the full curve with length at least $\sum_i L_i-\varepsilon$. Since this holds for every positive $\varepsilon$, the full curve has length at least $\sum_i L_i$. The two inequalities establish additivity.

On each smooth piece, the parametric arc-length formula identifies $L_i$ with the integral of speed over that piece. Adding those integrals gives the theorem’s formula. Equivalently, this is the ordinary additivity of definite integrals applied separately on the intervals where the derivative is defined.

Locate the joins.
Partition the parameter interval so each piece is smooth on its own interval.

Find the speed on each piece.
Differentiate that piece’s parametrization and take the magnitude of its velocity.

Integrate and add.
Calculate each piece’s length over its own parameter interval, then sum the lengths.

Worked Examples

Worked Example: A Path with a Right-Angle Corner

A path travels from $(0,0)$ to $(3,0)$, then from $(3,0)$ to $(3,4)$. Find its length using separate parametrizations for the two segments.

For the first segment, let $\mathbf{r}_1(t)=(t,0)$ for $0\leq t\leq 3$. Its velocity and speed are

\mathbf{r}_1'(t)=(1,0), \qquad \|\mathbf{r}_1'(t)\|=\sqrt{1^2+0^2}=1.

Therefore its length is

L_1=\int_0^3 1\,dt=3.

For the second segment, let $\mathbf{r}_2(u)=(3,u)$ for $0\leq u\leq 4$. Its velocity and speed are

\mathbf{r}_2'(u)=(0,1), \qquad \|\mathbf{r}_2'(u)\|=\sqrt{0^2+1^2}=1.

Its length is

L_2=\int_0^4 1\,du=4.

The end of the first segment, $(3,0)$, is the start of the second, so the pieces join continuously. By additivity, the path length is $L_1+L_2=3+4=7$. As a check, the answer is also the sum of the segment distances: $3+4=7$.

Worked Example: A Graph with Different Slopes on Either Side of a Corner

Consider the graph defined by $f(x)=2x$ for $-1\leq x\leq 0$ and $f(x)=3x$ for $0\leq x\leq 2$. The two formulas agree at $x=0$, so the graph is continuous there, but its slopes differ. Find its length.

For a graph $y=f(x)$ on an interval where $f$ is continuously differentiable, its length is $\int \sqrt{1+(f'(x))^2}\,dx$. On the first interval, $f'(x)=2$, so

L_1=\int_{-1}^{0}\sqrt{1+2^2}\,dx =\int_{-1}^{0}\sqrt{5}\,dx =\sqrt{5}\bigl(0-(-1)\bigr) =\sqrt{5}.

On the second interval, $f'(x)=3$, so

L_2=\int_{0}^{2}\sqrt{1+3^2}\,dx =\int_{0}^{2}\sqrt{10}\,dx =\sqrt{10}(2-0) =2\sqrt{10}.

Thus the graph’s total length is $L_1+L_2=\sqrt{5}+2\sqrt{10}$. To check the piece lengths directly, the first runs from $(-1,-2)$ to $(0,0)$, whose distance is $\sqrt{1^2+2^2}=\sqrt{5}$. The second runs from $(0,0)$ to $(2,6)$, whose distance is $\sqrt{2^2+6^2}=\sqrt{40}=2\sqrt{10}$. These agree with the integrals.

Worked Example: Joining a Circular Arc to a Line Segment

A path follows the quarter-circle parametrized by $\mathbf{r}_1(t)=(2\cos t,2\sin t)$ for $0\leq t\leq \pi/2$, then follows the vertical segment parametrized by $\mathbf{r}_2(u)=(0,2+u)$ for $0\leq u\leq 3$. Find the total length.

The first piece starts at $(2,0)$ and ends at $(0,2)$. Its velocity and speed are

\mathbf{r}_1'(t)=(-2\sin t,2\cos t), \qquad \|\mathbf{r}_1'(t)\| =\sqrt{4\sin^2t+4\cos^2t} =2.

Its length is therefore

L_1=\int_0^{\pi/2}2\,dt =2\left(\frac{\pi}{2}-0\right) =\pi.

The second piece starts at $(0,2)$ and ends at $(0,5)$. Its velocity has constant magnitude:

\mathbf{r}_2'(u)=(0,1), \qquad \|\mathbf{r}_2'(u)\|=1, \qquad L_2=\int_0^3 1\,du=3.

The endpoints match at $(0,2)$, so the pieces form a continuous path. Its length is $\pi+3$. As a check, the circular piece has radius $2$ and angle $\pi/2$, giving arc length $2(\pi/2)=\pi$; the segment has length $3$. Their sum agrees with the integral calculation.

What Additivity Does—and Does Not—Mean

Additivity concerns the length traveled along the parametrized path, not merely the distance between its endpoints. A path can turn, loop, or even retrace part of itself. Each traversal contributes its own length. For example, traveling along a segment and then returning along the same segment gives twice the segment’s length, even though the starting and ending points coincide.

Likewise, a corner does not add an extra length at the join. The join is a point, and the two pieces already account for all the distance traveled as the curve approaches and leaves it. The sum is the lengths of the pieces, with no separate “corner correction.”

Key takeaway: A finite number of corners does not prevent a curve from having a well-defined length. Split at the joins, apply the arc-length formula on each smooth piece, and add the resulting lengths.

A Common Mistake: Using One Slope Across a Corner

Consider $f(x)=2x$ on $[-1,0]$ and $f(x)=3x$ on $[0,2]$. A mistaken calculation might use the right-hand slope $f'(x)=3$ throughout $[-1,2]$, giving

\int_{-1}^{2}\sqrt{1+3^2}\,dx =3\sqrt{10}.

This is wrong because the slope is $2$, not $3$, on $[-1,0]$. The derivative is not one constant formula on the full interval. The correction is to split at $x=0$ and calculate $\sqrt{5}+2\sqrt{10}$, as in the worked example. More generally, do not apply a smooth-curve formula across an interval as though the derivative had a single expression when its formula changes at an interior join.

Check Your Understanding

Use the additivity rule and the appropriate arc-length formula to answer these questions.

A path consists of two smooth pieces that meet at one point. State the formula for the length of the whole path in terms of the lengths of the pieces.
Why is it permissible for the derivative of a piecewise smooth curve to fail to exist at a join?
A graph has slope $1$ on $[-2,0]$ and slope $2$ on $[0,1]$, and the pieces meet continuously. Set up the two length integrals.
If a path traverses the same segment twice, should that segment’s length be counted once or twice? Explain.
What inequality justifies that inserting a join into a polygonal approximation cannot decrease its length?