📂Calculus

Length of a Curve

Length of a Plane Curve¹

Buildup

Suppose we have a smooth function $y=f(x)$ given as in figure (a) above, with $n+1$ points on it. The total length $s$ of the curve can be obtained by summing up the lengths $s_{k}$ of each arc divided by points. Moreover, the length of each arc can be approximated by the length between two points as shown in figure (b). As the number of points increases, the sum of these approximated lengths will get closer to the actual length $L$. Thus, we obtain the following equation.

$$L = \lim \limits_{n \to \infty} \sum \limits_{i=1}^{n} \left| P_{i-1}P_{i} \right|$$

At this time, the length of each component is as follows, by the Pythagorean theorem.

$$\left| P_{i-1}P_{i} \right| = \sqrt{(x_{i} - x_{i-1})^{2} + (y_{i} - y_{i-1})^{2}} = \left( \Delta x_{i} \right)^{2} + \left( \Delta y_{i} \right)^{2}$$

Furthermore, by the mean value theorem, we can know that there exists $x_{i}^{\ast} \in (x_{i-1}, x_{i})$ for which the following equation holds.

$$\begin{align*} f(x_{i}) - f(x_{i-1}) =&\ f^{\prime} (x_{i}^{\ast}) \left( x_{i} - x_{i-1} \right) \\ \Delta y_{i} =&\ f^{\prime} (x_{i}^{\ast}) \Delta x_{i} \end{align*}$$

Therefore, the length of each segment is as follows.

$$\begin{align*} \left| P_{i-1}P_{i} \right| =&\ \left( \Delta x_{i} \right)^{2} + \left( \Delta y_{i} \right)^{2} \\ =&\ \left( \Delta x_{i} \right)^{2} + \left[ f^{\prime} (x_{i}^{\ast}) \right]^{2} \left( \Delta x_{i} \right)^{2} \\ =&\ \sqrt{1 + \left[ f^{\prime} (x_{i}^{\ast}) \right]^{2}} \Delta x_{i} \end{align*}$$

Then, the length of the curve $L$ is as follows.

$$L = \lim \limits_{n \to \infty} \sum \limits_{i=1}^{n} \left| P_{i-1}P_{i} \right| = \lim \limits_{n \to \infty} \sum \limits_{i=1}^{n} \sqrt{1 + \left[ f^{\prime} (x_{i}^{\ast}) \right]^{2}} \Delta x_{i}$$

Since $\sqrt{1 + \left[ f^{\prime} (x) \right]^{2}}$ is continuous, the limit of the Riemann sum exists and it is integrable. Therefore, the length of the curve is defined as follows.

Definition

If $f^{\prime}$ is continuous at $[a,b]$(If $f$ is a smooth function), the length $L$ of the curve $y=f(x)$ is defined as follows.

$$L := \int_{a}^{b} \sqrt{1 + \left[ f^{\prime} (x) \right]^{2}} dx = \int_{a}^{b} \sqrt{1 + \left( \dfrac{d y}{d x} \right)^{2}} dx$$

From this, the arc length function, which represents the length of the curve from the starting point $P_{0}(a, f(a))$ to the point $Q(x,f(x))$, is naturally defined as follows.

$$s(x) = \int_{a}^{x} \sqrt{1 + \left[ f^{\prime} (t) \right]^{2}} dt$$

Therefore, $\dfrac{d s}{d x} = \sqrt{1 + [f^{\prime}(x)]^{2}} = \sqrt{1 + \left( \dfrac{d y}{d x} \right)^{2}}$ holds, and the length of the curve can be indicated as follows.

$$L = \int_{C} ds = \int_{a}^{b} \sqrt{1 + \left( \dfrac{d y}{d x} \right)^{2}} dx$$

Theorem

Suppose curve $C$ is represented by the parametric equation $x = f(t), y=g(t), \alpha \le t \le \beta$. If $f^{\prime}, g^{\prime}$ is continuous at $[\alpha, \beta]$(If $f, g$ is a smooth function), the length of curve $C$ is as follows.

$$L = \int_{\alpha}^{\beta} \sqrt{\left( \dfrac{d x}{d t} \right)^{2} + \left( \dfrac{d y}{d t} \right)^{2}} dt$$

Proof

Let’s assume $a = x(\alpha), b = x(\beta)$. By assumption, $\dfrac{d y}{d x} = \dfrac{\dfrac{d y}{d t}}{\dfrac{d x}{d t}}$ holds. Therefore,

$$L = \int_{a}^{b} \sqrt{1 + \left( \dfrac{d y}{d x} \right)^{2}} dx = \int_{\alpha}^{\beta} \sqrt{1 + \left(\dfrac{\dfrac{d y}{d t}}{\dfrac{d x}{d t}} \right)^{2}} \dfrac{d x}{d t} dt = \int_{\alpha}^{\beta} \sqrt{\left(\dfrac{d x}{d t}\right)^{2} + \left(\dfrac{d y}{d t}\right)^{2}}dt$$

■

Length of a Space Curve²

Similarly to the buildup above, when a curve in a 3-dimensional space is expressed as $\mathbf{r}(t) = \left( f(t), g(t), h(t) \right)$, the length of the curve is defined as follows.

$$\begin{align*} L =&\ \int_{a}^{b} \sqrt{\left[ f^{\prime}(t) \right]^{2} + \left[ g^{\prime}(t) \right]^{2} + \left[ h^{\prime}(t) \right]^{2}} dt \\ =&\ \int_{a}^{b} \sqrt{ \left( \dfrac{d x}{d t} \right)^{2} + \left( \dfrac{d y}{d t} \right)^{2} + \left( \dfrac{d z}{d t} \right)^{2} } dt \\ =&\ \int_{a}^{b} \left| \mathbf{r}^{\prime}(t) \right| dt \end{align*}$$

Likewise, the arc length function is as follows.

$$s(t) = \int_{a}^{t} \left| \mathbf{r}^{\prime}(u) \right| du = \int_{a}^{t} \sqrt{ \left( \dfrac{d x}{d u} \right)^{2} + \left( \dfrac{d y}{d u} \right)^{2} + \left( \dfrac{d z}{d u} \right)^{2} } du$$