The Arc Length of a Piecewise Smooth Curve

# The Arc Length of a Piecewise Smooth Curve

Recall from the Integrals of Complex Functions Along Piecewise Smooth Curves page that if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$ is continuous, $\gamma : [a, b] \to \mathbb{C}$ is a piecewise smooth curve with $\gamma ([a, b]) \subset A$ where $a = a_0 < a_1 < ... < a_n = b$ is a partition on $[a, b]$] for which [[$\gamma'$ exists on each open subinterval $(a_k, a_{k+1})$ and is continuous on each closed subinterval $[a_k, a_{k+1}]$ for all $k \in \{0, 1, ..., n-1\}$, then the integral of $f$ along the curve $\gamma$ is:

(1)
\begin{align} \quad \int_{\gamma} f(z) \: dz = \int_a^b f(\gamma(t)) \cdot \gamma'(t) \: dt = \sum_{k=0}^{n-1} \int_{a_k}^{a_{k+1}} f(\gamma(t)) \cdot \gamma'(t) \: dt \end{align}

We will now define the arclength of a piecewise smooth curve.

 Definition: Let $\gamma :[a, b] \to \mathbb{C}$ be a piecewise smooth curve where $\gamma(t) = x(t) + iy(t)$. The Arc Length of $\gamma$ denoted $l (\gamma)$ is defined as $\displaystyle{l(\gamma) = \int_a^b \mid \gamma'(t) \mid \: dt = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2} \: dt}$.

For example, consider the piecewise smooth curve $\gamma : [0, 2\pi] \to \mathbb{C}$ that is the unit circle parameterized as $\gamma(t) = e^{it}$. Then $\gamma'(t) = ie^{it}$ and the arc length of $\gamma$ is:

(2)
\begin{align} \quad l(\gamma) &= \int_0^{2\pi} \mid \gamma'(t) \mid \: dt \\ &= \int_0^{2 \pi} \mid ie^{it} \mid \: dt \\ &= \int_0^{2 \pi} 1 \: dt \\ &= \left [ t \right ]_{0}^{2\pi} \\ &= 2\pi \end{align}

Of course this is to be expected since we already know that the unit circle has circumference $2\pi$. In general, it should be noted that computing the arc length of a piecewise smooth curve may be quite difficult since evaluating integrals involving square roots can be rather cumbersome.

Furthermore, it should be noted that the domain for which the curve $\gamma$ is traced is also important in determining the arc length of a curve. For example, if $\sigma : [0, 4\pi] \to \mathbb{C}$ is defined by $\sigma (t) = e^{it}$ then graphically, $\gamma$ and $\sigma$ are identical. However, $\sigma$ traces the unit circle in the counterclockwise direction TWICE, and so the arc length of $\sigma$ will be $2 \cdot 2\pi = 4\pi$ in accordance.

 Theorem 2: Let $A \subseteq \mathbb{C}$ be open, and let $f : A \to \mathbb{C}$ be a continuous function on $A$. Let $\gamma : [a, b] \to \mathbb{C}$ be a piecewise smooth curve such that $\gamma ([a, b]) \subset A$. If there exists an $M > 0$ such that $\mid f(z) \mid \leq M$ for all points $z$ on $\gamma$ then $\displaystyle{\biggr \lvert \int_{\gamma} f(z) \: dz \biggr \rvert \leq M l(\gamma)}$.
• Proof: Suppose that there exists an $M > 0$ such that for all $z$ on $\gamma$ we have that $\mid f(z) \mid \leq M$. Then:
(3)
\begin{align} \quad \biggr \lvert \int_{\gamma} f(z) \: dz \biggr \rvert &= \biggr \lvert \int_a^b f(\gamma(t)) \cdot \gamma'(t) \: dt \biggr \rvert \\ & \leq \int_a^b \mid f(\gamma(t)) \mid \mid \gamma'(t) \mid \: dt \\ & \leq \int_a^b M \mid \gamma'(t) \mid \: dt \\ & \leq M \int_a^b \mid \gamma'(t) \mid \: dt \\ & \leq M l(\gamma) \quad \blacksquare \end{align}