\(\)The length of a curve is called the *arc length*.

## Function graphs

:

The arc length of function graphs is explained using two examples.

### \(f(x)\)

Let \(f\) be a function of variable \(x).

$$

y = f(x)

$$

We can approximate the length of a small segment \(\Delta s\) using the Pythagorean theorem.

$$

\Delta s=\sqrt{(\Delta x)^2 + (\Delta y)^2}

\nonumber

$$

Adding all the segments gives us the approximate length of the curve

$$

\sum\sqrt{(\Delta x)^2 + (\Delta y)^2}

\nonumber

$$

When we bring \(\Delta s\rightarrow 0\), the approximation becomes the accurate representation. To find the arc length \(s\), we sum all the segments.

The arc length of function graph follows as

$$

\shaded{

s=\int\sqrt{(dx)^2 + (dy)^2}

}

\label{eq:functionint}

$$

Note that the limits are conveniently omitted for now. The examples show how to add these.

**Example**

Consider function \(f(x)\) between \(x=-1\) and \(x=1\)

$$

y = f(x) = x^2

\label{eq:functiongraph}

$$

Express \(dy\) in terms of \(dx\) in the example equation \(\eqref{eq:functiongraph}\)

$$

\newcommand{dv}[1]{\tfrac{d}{d #1}}

\begin{align*}

y&=x^2 \\

\dv{x}y &=2x \\

dy &= 2x\;dx

\end{align*}

$$

Substitute \(dy\) in the integral \(\eqref{eq:functionint}\), and place the bounds \(x=-1\) to \(1\) to find the curve length

$$

\begin{align*}

s &=

\int_{-1}^{1}\sqrt{(dx)^2 + (2x\;dx)^2}\;dx \\

&=\int_{-1}^{1}\sqrt{1+4x^2}\;dx \\

\end{align*}

$$

Solve using wolframalpha returns approximately \(3.2671\).

### \(f(\theta)\)

Functions such as a circle on the (x,y) plane are more naturally described using polar coordinates. Consider the polar function of a circle between \(0\) and \(\pi\):

$$

r=1 \;\land\; \theta \in \left[0,\pi\right)

$$

Since the radius is 1, the value of \(\theta\) reflects the arc length \(\Delta s\) in radians. Bringing \(\Delta \theta\rightarrow 0\), we find the arc length by summing all the tiny segments:

$$

s=\int d\theta

\label{eq:polarint}

$$

The arc length is found by placing the bounds \(x=-1\) to \(1\) in integral \(\eqref{eq:polarint}\). length

$$

s

= \int_{0}^{\pi}d\theta

= \left[ \theta \right]_{0}^{\pi}

= \pi

\nonumber

$$

## Parametric curve

A parametric curve is a function with one-dimensional input and a multi-dimensional output.

Determining the length of a parametric curve is best described using an example:

Consider parametric curve \(f(t)\)

$$

f(t)

=\left\{

\begin{array}{l}

f_x(t)=t^3-3t\\

f_y(t)=3t^2

\end{array}

\right.

\nonumber

$$

abbreviated this using vector notation

$$

f(t) =

\left\langle\;

t^3-3t,\;

3t^2\;

\right\rangle

\label{eq:parmcurve}

$$

What is the length of the curve between \(-1.5\) to \(1.5\)?

We find the *arc length* similar to function graphs using the integral \(\int\sqrt{(dx)^2+(dy)^2}\) where \(dx\) and \(dy\) represent the tiny change in \(x\) and \(y\) values from the start to the end of the line.

With parametric curves, since \(x\) and \(y\) are given as functions of \(t\), we write \(dx\) and \(dy\) in terms of \(dt\) by taking the derivative of these two functions.

$$

\left\{

\begin{array}{ c l l }

x=t^3-3t & \Rightarrow

\frac{d}{dt}x = 3t^2-3 & \Rightarrow

dx=(3t^2-3)\;dt

\\

y=3t^2 & \Rightarrow

\frac{d}{dt}y=6t & \Rightarrow

dy=6t\;dt

\end{array}

\right.

\nonumber

$$

Putting these into the integral

$$

\begin{align}

\int\sqrt{(dx)^2+(dx)^2}

&= \int\sqrt{((3t^2-3)dt)^2 + (6t\ dt)^2} \;dt \nonumber \\

&= \int\sqrt{(3t^2-3)^2 + (6t)^2} \;dt \nonumber \\

&= 3\int t^2+1 \;dt \label{eq:parametricfnc}

\end{align}

$$

Now everything is written in terms of \(t\). Place the bounds on the integral equation \(\eqref{eq:parametricfnc}\)

$$

\begin{align*}

3\int_{-2}^{2} t^2+1 \;dt

&= \left[ t^3+3t \right]_{-2}^{2} \\

&= (2^3-3(2)) – (3(-2)) \\

&= 28

\end{align*}

$$