# Arc Length

The **arc length formula** uses the language of calculus to generalize and solve a classical problem in geometry: finding the length of any specific curve. Given a function $f$ that is defined and differentiable on the interval $[a, \, b]$, the length $L$ of the curve $y = f(x)$ in that interval is $L = \int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx.$

Note that the same formula can be applied to piecewise-differentiable functions by splitting up the interval of integration at the points of non-differentiability. This means that the formula holds for piecewise-linear functions; in fact, the formula originates from the ability of these piecewise-linear functions to model more complicated (or, more curvy) functions.

How would you calculate arc length for a curve? Before choosing an option, think about how each of the following possibilities might or might not yield a solution.

**A**. Create a bunch of Riemann rectangles, like with Riemann sums, and add up the lengths of the uppermost line segments, creating a Riemann staircase. Take the limit of all such totals.

**B**. Flatten out the curve by rotating each part of it to become horizontal. Then, take the length of that line.

**C**. Approximate the curve with a lot of line segments, and take the sum of their length. Take the limit of all such sums.

**D**. Draw the curve with a 1-unit think pen, so that a vertical 1-unit line segment bisected by the function is contained in the region. The length of the curve is then equal to the area of that region, which can be found with traditional calculus methods.

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## Explanation

There are two intuitive ways to understand the formula above. Both could be fully rigorized into a proof, but for that, refer to the next section.

The crux of the issue lies in the fact that a line segment of $y = mx + b$ from $x$ to $x + w$ has length $\sqrt{w^2 + (wm)^2} = w \sqrt{1 + m^2}$.

The following sections aim to suggest that $(1)$ it is possible to approximate a function with piecewise-linear functions, and $(2)$ no matter how a function is modeled by piecewise-linear functions, the arc length remains the same.

$(1)$ The first approach is to choose many points distributed across the function and draw tangent lines to the function at those points. By choosing enough points, it is always possible to make subsequent tangent lines intersect (i.e., not be parallel). The line segments between tangent line intersections then provides a rough estimate of the function itself and therefore of the function's length (and one that can be easily calculated!).

If the tangent line segments have slopes $m_1, \, m_2, \, \dots, \, m_n$ and widths $w_1, \, w_2, \, \dots, \, w_n$, then the length of the tangent line construction is $\sum_{i=1}^n \sqrt{1 + (m_i)^2} \cdot w_i.$ Recalling each $m_i$ is $\tfrac{dy}{dx}$ for the function and each $w_i$ is a difference of $x$'s, this formula looks remarkably similar to the above formula (just with a summation instead of an integral). In fact, as $n$ grows very large, they become equal.

Intuitively, this makes sense. Tangent lines serve to model a function well near their point of tangency, so their length should model a function's arc length well too. The main difficulty in formalizing this method is that, in order to stay "near" the function, subsequent tangent lines should intersect in between their points of tangency (this also allows for a more natural translation between $w_i$'s and $x_i$'s). This issue is resolved by choosing more points along the curve wherever the tangent lines are not "near" the function.

$(2)$ The second approach is to choose points regularly distributed across the function and connect them with line segments. If the interval has length $W = b - a$, then $n+1$ points can be chosen according to the rule $x_k = a + k \cdot \left( \frac{b-a}{n} \right)$ and $y_k = f(x_k)$. Then, the line connecting $(x_k, \, y_k)$ and $(x_{k+1}, \, y_{k+1})$ is both an approximation of the function itself and, as $n$ grows large, an approximation of a tangent line to the function at $x = x_k$.

The $k^\text{th}$ line segment has length $\sqrt{1 + \left( \frac{y_{k+1} - y_k}{x_{k+1} - x_k} \right)^2} \cdot (x_{k+1} - x_k)$. Summing over all segments, $\sum_{k = 1}^n \sqrt{1 + \left( \frac{y_{k+1} - y_k}{x_{k+1} - x_k} \right)^2} \cdot (x_{k+1} - x_k),$ yields an expression that looks very much like a Riemann sum (another way of expressing an integral). Recalling that $\frac{y_{k+1} - y_k}{x_{k+1} - x_k}$ is the approximation of the function's derivative (the slope of its tangent line), the result looks remarkably like the above formula. As $n$ grows large, the values are, in fact, equal.

Intuitively, this approximation method is very direct and geometric. It tries to approximate a function with a bunch of line segments based on the values of the function itself (and no additional information). It assumes that the length of the function can indeed be modeled well by many little line segments

Both these methods $(1)$ and $(2)$ take a function and attempt to decompose it into a connected series of line segments (a polygonal path). This presupposes that the length of a curve is equal to the limit of a sequence of lengths of polygonal paths. Happily, this is the leading definition of curve length, and curves whose lengths are not measurable by such a method are known as unrectifiable.

## Find the length of the curve $\displaystyle y=1+6x^{\frac{3}{2}},$ $0 \leq x \leq 1.$

Note $\begin{array}{c}\displaystyle{\frac{dy}{dx}}=9x^{\frac{1}{2}} & \Rightarrow & 1+\left(\frac{dy}{dx}\right)^2=1+81x, \end{array}$ and so the arc length formular gives $L=\int_{0}^{1} \sqrt{1+81x}\, dx.$ Substituting $u=1+81x$ and $du=81\,dx$, note that when $x=0,$ $u=1;$ and when $x=1,$ $u=82.$ Therefore, $L=\int_{1}^{82} u^{\frac{1}{2}} \left(\frac{1}{81}\, du\right)=\frac{2}{243}\left[u^{\frac{3}{2}}\right]_{1}^{82}=\frac{2}{243}(82\sqrt{82}-1). \ _\square$

Let $f(n)$ denote the arc length of the curve $y=\ln\left(x\right)$ in the interval $x\in\left [1,n\right ]$. Find $\lim_{n\to\infty}\big(n-f(n)\big).$ If your answer is of the form $\ln\left(\sqrt{a}-b\right)+\sqrt{c}-d$, where $a$, $b$, $c$, and $d$ are non-negative integers and $a$ and $c$ are not perfect squares, find $a+b+c+d$.

## Proof

The main proof of this result has elements in common with both of the above approaches, but at its core, it is an elaboration upon the second approach.

Suppose that a curve $C$ is defined by the equation $y=f(x),$ where the function $f$ is continuous in $a \leq x \leq b,$ and divide the interval $[a,b]$ into $n$ subintervals with endpoints $x_0, x_1, \ldots , x_n$ and equal width $\Delta x.$ Since $y_i=f(x_i),$ the point $P_i=(x_i,y_i)$ lies on $C$ and the line segments connecting adjacent two points of $P_0,P_1, \ldots, P_n,$ shown above, are an approximation to $C.$

The length $L$ of $C$ is approximately the sum of the lengths of these line segments. Therefore, it is possible to define the length $L$ of the curve $C$ with the equation $y=f(x),$ $a \leq x \leq b,$ as the limit of the sum of the lengths of these line segments:

$L=\lim_{n \to \infty} \sum_{i=1}^{n} \lvert{P_{i-1}P_i}\rvert, \qquad (1)$

where $\lvert{P_{i-1}P_i}\rvert$ denotes the distance between the two points $P_{i-1}$ and $P_i.$

Let $\Delta y_i=y_i-y_{i-1},$ then

$\lvert{P_{i-1}P_i}\rvert=\sqrt{(x_i-x_{i-1})^2+(y_i-y_{i-1})^2}=\sqrt{(\Delta x)^2+(\Delta y)^2} .$

By applying the mean value theorem to $f$ on the interval $[x_{i-1},x_i],$ note that there is a number ${x_i}^{*}$ between $x_{i-1}$ and $x_i$ such that

$f(x_i)-f(x_{i-1})=f'({x_i}^{*})(x_i-x_{i-1}),$

i.e. $\Delta y_i=f'({x_i}^{*}) \Delta x.$ Thus,

$\begin{aligned} \lvert{P_{i-1}P_i}\rvert &= \sqrt{(\Delta x)^2+(\Delta y)^2} \\ &= \sqrt{(\Delta x)^2+(f'({x_i}^{*}) \Delta x)^2} \\ &= \sqrt{1+(f'({x_i}^{*}))^2} ~\Delta x. \end{aligned}$

Therefore, by definition $(1),$

$L=\lim_{n \to \infty} \sum_{i=1}^{n} \lvert{P_{i-1}P_i}\rvert = L=\lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{1+(f'({x_i}^{*}))^2} ~ \Delta x.$

By the definition of definite integrals, this expression is equal to

$\int_{a}^{b} \sqrt{1+(f'(x))^2}\, dx.$

Hence, by using Leibniz notation for derivatives, the arc length formula becomes

$L=\int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx.$

## Find the length of an arc of curve

$y=\frac{a}{2}\left(e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right)$

between $x=0$ and $x=a$.

Note that $y=a\cosh\left(\dfrac{x}{a}\right) \Rightarrow \dfrac{dy}{dx} = \sinh\left(\dfrac{x}{a}\right)$ (using chain rule) . Now,

$\sqrt{1+\left(\dfrac{dy}{dx}\right)^2} = \sqrt{1+\sinh^2\left(\dfrac{x}{a}\right)} = \left| \cosh\left(\dfrac{x}{a}\right) \right|.$

Now length of the arc from $x=0$ to $x=a$ is given by the integral: $\displaystyle\int_{0}^{a} \cosh\left(\dfrac{x}{a}\right) \ dx.$

Substitute $y=\dfrac{x}{a} \Rightarrow dx=a \ dy$ and changing limits, the integral becomes $a\big[\sinh (y)\big]_{0}^{1} = a\sinh(1) = \dfrac{a\big(e^2-1\big)}{2e}.\ _\square$

## Parametric Form

(See full wiki here.)

The arc length $L$ of the graph of a parametric function $(x,y) = \left(x(t),y(t)\right)$ from $t=a$ to $t=b$ is given by

$L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right )^2 }\, dt.$

This follows readily from the earlier discussion by making the substitution $dx = \frac{dx}{dt} \, dt$. Formally, the same method of proof applies.

## Polar Form

(See full wiki here.)

The arc length $L$ of the graph of a polar function $r(\theta)$ from $t = a$ to $t = b$ is given by

$L = \int_a^b \sqrt{r^2 + \left( \frac{dr}{d\theta} \right)^2} \, d\theta.$

This formula follows immediately from the parametric form, upon noting that the parameter is $\theta$ and $x'(\theta) = \frac{d}{d\theta} \left( r(\theta) \cos \theta \right) = r'(\theta) \cos \theta - r(\theta) \sin \theta, \\ y'(\theta) = \frac{d}{d\theta} \left( r(\theta) \sin \theta \right) = r'(\theta) \sin \theta + r(\theta) \cos \theta.$

## See Also

- Gabriel's Horn
- Parametric Arc Length
- Polar Arc Length
- Volume by Rotation (Disc Method and Shell Method)