# Arc Length

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Arc Length – HMC Calculus Tutorial
Arc Length

Suppose $f$ is continuously differentiable on the interval $[a,b]$.

Suppose $f$ is continuously differentiable on the interval $[a,b]$.

Let’s derive a formula for the length $L$ of the curve on the interval, called the arc length over $[a,b]$.

Mean Value Theorem
Mean Value Theorem

Let $f$ be differentiable on $(a,b)$ and continuous on $[a,b]$. Then there exists at least one point $c$ in $(a,b)$ for which $f'(c)=\frac{f(b)-f(a)}{b-a}{\small\textrm{.}}$

Riemann Sum
Riemann Sum

Let $f$ be continuous and non-negative on $[a,b]$ and let $a=x_0< x_1< \ldots< x_n=b$ be a partition of $[a,b]$.

For each $k$, $1 \leq k \leq n$, let $x_k^*$ be any point in the interval $[x_{k-1}, x_k]$. Then the sum $\sum_{i=1}^n f(x_i^*)\, \Delta x_i$ is called a Riemann sum for $f$ on $[a,b]$.

Definite Integral
Definite Integral

Let $f$ be defined on $[a,b]$ and let $a = x_0 < x_1 < ~…~ < x_n = b$ be a partition of $[a,b]$. For each $k$, $1 \leq k \leq n$, let $x_k^*$ be any point in $[x_{k-1}, x_k]$. Then the definite integral of $f$ over $[a,b]$ is defined as $\int^b_a f(x)\, dx=\lim_{\max \Delta x_i\to 0} \left(\sum_{i=1}^n f\left(x_i^*\right)\Delta x_i\right).$

We’ll start by subdividing the interval $[a,b]$ into $n$ subintervals $[x_0, x_1], [x_1, x_2],~…~, [x_{n-1},x_n]$
where $a=x_0 < x_1< … < x_{n-1} < x_n=b$.

Introduce the line segments between
$(x_0, f(x_0)) {\small\textrm{ and }} (x_1, f(x_1)),$
$(x_1, f(x_1)) {\small\textrm{ and }} (x_2, f(x_2)), … ,$
$(x_{n-1}, f(x_{n-1})) {\small\textrm{ and }} (x_n, f(x_n))$.

The resulting polygonal path approximates the curve given by $y=f(x)$, and its length approximates the arc length of $f(x)$ over $[a,b]$.

Let’s find the length of the polygonal path by adding up the lengths of the individual line segments. The $k$th line segment is the hypotenuse of a triangle with base $\Delta x_k$ and height $f(x_k)-f(x_{k-1})$, and so has length

$L_k=\sqrt{\left(\Delta x_k \right)^2+\left[f(x_k)-f(x_{k-1})\right]^2} {\small\textrm{.}}$ By the Mean Value Theorem, there exists $x_k^*\in [x_{k-1},x_k]$ such that $\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}=f'(x_k^*)$ so $f(x_k)-f(x_{k-1})=f'(x_k^*)(x_k-x_{k-1})=f'(x_k^*)\Delta x_k.$ Thus, $L_k=\sqrt{(\Delta x_k)^2+[f'(x_k^*)\Delta x_k]^2}=\sqrt{1+[f'(x_k^*)]^2}\, \Delta x_k.$ Finally, the length of the entire polygonal path is $\sum^n_{k=1} L_k=\sum^n_{k=1} \sqrt{1+[f'(x_k^*)]^2}\, \Delta x_k$ which has the form of a Riemann sum. Increasing the number of subintervals such that $\max \Delta x_k \to 0$, $\, \sum^n_{k=1} L_k \to L$. That is, $L=\lim_{\max \Delta x_k\to 0}\sum^n_{k=1} \sqrt{1+[f'(x_k^*)]^2}\, \Delta x_k=\int^b_a \sqrt{1+[f'(x)]^2}\, dx$ by the definition of the definite integral as a limit of Riemann sums. Thus, we have proved the following:

#### Arc Length

Let $f(x)$ be continuously differentiable on $[a,b]$. Then the arc length $L$ of $f(x)$ over $[a,b]$ is given by $L=\int^b_a \sqrt{1+[f'(x)]^2}\, dx.$ Similarly, if $x=g(y)$ with $g$ continuously differentiable on $[c,d]$, then the arc length $L$ of $g(y)$ over $[c,d]$ is given by $L=\int^d_c \sqrt{1+[g'(y)]^2}\, dy.$ These integrals often can only be computed using numerical methods.

#### Example

We can compute the arc length of the graph of $f(x)=x^{3/2}$ over $[0,1]$ as follows: \begin{eqnarray*} L=\int^1_0 \sqrt{1+[f'(x)]^2}\, dx &=& \int^1_0 \sqrt{1+[3x^{1/2}/2]^2}\, dx\\ &=& \int^1_0 \sqrt{1+9x/4}\, dx\\ &=& \left.\frac{8}{27}(1+9x/4)^{3/2}\right|^1_0\\ &=& (1+9/4)^{3/2}-(1)^{3/2}\\ &=& (13/4)^{3/2}-1\\ &\approx & 1.44. \end{eqnarray*}

### Key Concepts

Let $f(x)$ be continuously differentiable on $[a,b]$. Then the arc length $L$ of $f(x)$ over $[a,b]$ is given by $L = \int^b_a \sqrt{1+[f'(x)]^2}\, dx$ Similarly, if $x = g(y)$ with $g$ continuously differentiable on $[c,d]$, then the arc length $L$ of $g(y)$ over $[c,d]$ is given by $L = \int^d_c \sqrt{1+[g'(y)]^2}\, dy$