# Foundations



The length of a curve is called the arc length.

## Arc length

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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*}

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