Arc lengths

We have now learnt a great deal about various properties of curves. We can study their gradients, the areas under them and between them, we can even study properties of functions which have discontinuities and which extend all the way to \infty. We can look at the revolution of a curve about an axis and study the volume enclosed using cylinders of various forms and we have a good understanding of the link between finite sums of pieces which make up an area or volume and the limit that these pieces become infinitesimally small and thus how we end up with an integral from a Riemann sum.

There is one piece of the puzzle left, which is to know the length of a curve between two points. Let’s say we have some function f(x) and we want to know the length of the curve (ie. how long a piece of string would be that went along the curve and stretched from some point a to another point b on the curve). The most naive guess would be to lie the string straight between the points f(a) and f(b) and calculate the length like that. Looking at this figure

naivelength

 

we can see that this will give:

 

L=\sqrt{(b-a)^2+(f(b)-f(a))^2}

 

This is clearly a pretty bad approximation however.

Clearly a better approximation would be to split the line up into many parts and approximate each of these by a straight line. Already alarm bells should be ringing and you should be thinking of integration. You are going to add up lots of pieces and hopefully take the limit that each of these pieces will be very small. This sounds like the beginnings of an integral problem.

Let’s split the curve up into 3 parts, each with the same \Delta x but now \Delta y is a function of the position of the i^{th} point that we are looking at.

betterlength
We can write down a general expression for the i^{th} line approximation. The length of it will be:

 

L_i=\sqrt{\Delta x^2+\left(f(x_i+\Delta x)-f(x_i)\right)^2}

 

Let’s pull out a factor of \Delta x to make this look more like a term in a Riemann sum:

 

L_i=\Delta x\sqrt{1+\frac{\left(f(x_i+\Delta x)-f(x_i)\right)^2}{\Delta x^2}}

 

So the total length is going to be:

 

L\approx \sum_{i=0}^{n-1}\Delta x\sqrt{1+\frac{\left(f(x_i+\Delta x)-f(x_i)\right)^2}{\Delta x^2}}

 

Clearly, the more lines we draw along the curve, and the shorter they are, the better is going to be the approximation. We want to take the limit that these lines go to zero size. We can look at the term inside the square root and see that the expression \frac{\left(f(x_i+\Delta x)-f(x_i)\right)}{\Delta x} as the \Delta x goes smaller (ie. the number of pieces goes larger) will get closer and closer to our definition of the gradient of the slope, ie. the derivative of the function at x_i. That is:

 

L=\lim_{\Delta x\rightarrow 0}\sum_{i=0}^{n-1}\Delta x\sqrt{1+\frac{\left(f(x_i+\Delta x)-f(x_i)\right)^2}{\Delta x^2}}=\lim_{\Delta x\rightarrow 0}\sum_{i=0}^{n-1}\Delta x\sqrt{1+\left(\left.\frac{df(x)}{dx}\right|_{x=x_i}\right)^2}

 

=\int_a^b\sqrt{1+\left(\frac{df(x)}{dx}\right)^2}dx

 

where we first took the limit of the terms inside the square root, then we looked at the whole thing and took the limit that the number of pieces that we were studying went to infinity. This isn’t strictly the best way to do it, but it’s rather more intuitive to see that as we take the number of pieces smaller and smaller, the expression inside the square root looks more and more like the derivative, and then we have an infinite number of such pieces in our Riemann sum which add together to give the integral expression.

And that’s it! That’s the expression for the length of a curve from a to b. Now let’s look at some examples:
Let’s first work out the circumference of a circle using this method. A circle can be written as two curves: f(x)=\pm \sqrt{r^2-x^2} where the positive part is the half circle above the x axis, and the negative part is below the x axis. We can simply say then that the circumference of a circle is twice the length along the curve f(x)=\sqrt{r^2-x^2} from x=-r to x=r:

circ

 

We simply plug it into our equation for the length of a curve and see what we get. First we need the derivative
\frac{dy}{dx} which is \frac{-x}{\sqrt{r^2-x^2}}. Then the length of the curve is given by:

 

L=\int_{-r}^r\sqrt{1+\frac{x^2}{r^2-x^2}}dx=\int_{-r}^r\frac{r}{\sqrt{r^2-x^2}}dx

 

we use a substitution x=r\sin\theta to get:

 

L=\int_{-\frac{\pi}{2}}^\frac{\pi}{2} r d\theta=r\pi

 

So this is the length of the half circle, so the perimeter of a full circle is twice this =2\pi r as we know that it should be.
Let’s find an example which at first sight looks rather complicated. We will find the length of this arc between 0 and \frac{\pi}{4}:

 

y=\ln(\sec x)

 

\frac{dy}{dx}=\frac{\sec x\tan x}{\sec x}=\tan x

 

=\int_0^\frac{\pi}{4}\sqrt{1+f'(x)^2}dx=\int_0^\frac{\pi}{4}\sqrt{1+\tan^2 x}dx=\int_0^\frac{\pi}{4}\sqrt{\sec^2 x}dx

 

=\int_0^\frac{\pi}{4}\sec x dx=\left.\ln |\tan x+\sec x|\right|_{0}^\frac{\pi}{4}

 

=\ln(\sqrt{2}+1)

Note that we were able to equate \sqrt{\sec^2x} with \sec x only because we were interested in this function at 0 and \frac{\pi}{4} at which points it is positive.

Exercise: Calculate the arc length of the function y=\sqrt{x-x^2}+\arcsin\sqrt{x} between 0 and \pi.

It is important to note that Stewart introduces arc length using the formalism of parametric equations. We do not do this here and while it will not be examined, it is a useful technique to understand.

How clear is this post?