# Help with deriving arc length formula?

I'm having trouble deriving arc length of functions. I'm using a vector-oriented approach, if that's alright. I'm doing this because I don't understand the $1+(f')^2$ part of the standard formula, and I don't like using formulae I don't understand.

So far I have this: Let $d(\vec{u},\vec{v})$ be the Euclidean distance function between two vectors $\vec{u}$ and $\vec{v}$, defined by $$d(\vec{u},\vec{v})=\left \| \vec{v}-\vec{u} \right \|$$ Let $f(x)$ be any "reasonable" function. (By "reasonable", I mean continuously differentiable on the closed interval $[a,b]$, with no singularities, crazy fractal stuff, etc. I think the proof works as long as it's continuous, actually. I could be wrong.) To obtain the arc length of $f(x)$ on the interval $[a,b]$, we have to rectify the curve generated by $f(x)$. We add up the distances between infinitesimal parts of $f$: $$\int_{a}^{b}d(f(x),f(x+h))dx$$

We are working in 2D, so here's the explicit formula for $d(\vec{u},\vec{v})$. Let $\vec{u}=\left \langle x_1,y_1 \right \rangle$ and $\vec{v}=\left \langle x_2,y_2 \right \rangle$ $$\vec{v}-\vec{u}=\left \langle x_2-x_1,y_2-y_1 \right \rangle$$ $$d(\vec{u},\vec{v})=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ Now, let's substitute $\vec{u}=\left \langle a,f(a) \right \rangle$ and $\vec{v}=\left \langle a+h,f(a+h) \right \rangle$ where $a$ is some number on $f$ and $h$ is a small number (infinitesimal, you know?). $$d(\vec{u},\vec{v})=\lim_{h \to 0}\sqrt{((a+h)-(a))^2+(f(a+h)-f(a))^2}$$

That's where I hit a hitch. I'm not sure how to get rid of that h and replace it with a derivative or something more practical. Oh, and $((a+h)-a)^2=h^2$, right? $h$ is infinitesimal. Is squaring it an issue? Or does it become so small so as to be considered nothing?

If you multiply and divide by $h$ outside your square root, it becomes

\begin{align*} h \sqrt{\frac{\big((a + h) - a\big)^2}{h^2} + \frac{\big(f(a + h) - f(a)\big)^2}{h^2}} &= h \sqrt{1 + \left(\frac{f(a + h) - f(a)}{h}\right)^2} \end{align*}

Now if you partition your interval into $n$ little bits of length $h$ (replacing $h$ by the more suggest $\delta x$), and add up all these approximate lengths, you get

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

Now you should recognize this as a Riemann sum, and the term inside the square root will converge to $\sqrt{1 + f'(x)^2}$ if you have a "nice" function.

• "Multiply and divide by $h$ outside [the] square root"? I'm confused. – Aieou Dec 31 '15 at 8:47
• @Aieou Write $h/h$, and move the denominator inside the square root and leave the numerator as is. – user296602 Dec 31 '15 at 8:47
• Okay, let me see if I got this right. $\frac{h}{h}=1$, so obviously it's "free": multiplying by it does not actually change the equation. Then, as the next step: $\frac{h}{h}=\frac{h}{1}\frac{1}{h}$ Then we multiply the second half into the square root. This is the part that confuses me. Let's define $p$ as all of the stuff currently under the square root, for brevity's sake. You're telling me that $$\frac{1}{h}\sqrt{p}=\sqrt{\frac{p}{h^2}}$$ [comment continued] – Aieou Dec 31 '15 at 9:02
• After some testing, this does appear to be true, and everything works out from there. $\frac{h^2}{h^2}$ is definitely 1, and the term with $f$ in it is definitely the definition of the derivative. I guess what confuses me is how despite that multiplying by $\frac{h}{h}$ is definitely valid and doesn't really change anything, it somehow makes a 1 appear out of "nowhere" and turns that difference in a function into a derivative. Very strange. – Aieou Dec 31 '15 at 9:02
• @Aieou It's just a different phrasing of "factor an $h^2$ from each term inside the square root." I find it more natural to introduce it as I did in the answer, but they are equivalent ideas. – user296602 Dec 31 '15 at 9:03

An answer to a related question can be found here

From that you will see how to arrive at the well known result that the length of a parametric curve $c(t)= (c_1(t), \dots, c_n(t))$ is given by $$\ell(c) =\int_a^b ||c^\prime(t)|| \, dt$$

The special case you are considering in the heading is the case where the curve is given by a graph, i.e. $c(t) = (t, f(t))$. It is easy to see that in this case the previous formula reduces to

$$\ell(c)=\int_a^b\sqrt{1+f^\prime(t)^2}\,dt$$

• Wouldn't it be $c(t)=(t,f(t))$? – Aieou Dec 31 '15 at 9:18
• @Aieou yes, of course. Thank you! – Thomas Dec 31 '15 at 9:26