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I understand the arc length formula is derived from adding the distances between a series of points on the curve, and using the mean value theorem to get:

$ L = \int_a^b \sqrt{ 1 + (f'(x))^2 } dx $

But is there an intuition here I'm missing? Something about taking the integral of the derivative seems like it should mean something..

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Divide the interval into $n$ equal parts, $a = x_0 \lt x_1 \lt \cdots \lt x_n = b$, with $x_{i+1} = x_i + \Delta x_i$.

Suppose you want to approximate the curve between $(x_i,f(x_i))$ and $(x_i+\Delta x,f(x_i+\Delta x))$. You could simply approximate it with the straight line between the two points, whose length is $$\sqrt{\left( f(x_i+\Delta x) - f(x_i)\right)^2 + (\Delta x)^2}.$$ In the picture below, the black line is the graph $y=f(x)$, and the green line is the line that joints $(x_i,f(x_i))$ on the bottom left and $(x_1+\Delta x,f(x_1+\Delta x))$ on the top right. enter image description here Then you would have that the arc length is approximated by the sum of the lengths $$\text{Arc Lenght} \approx \sum_{i=1}^n \sqrt{(f(x_i+\Delta x) - f(x_i)))^2 + (\Delta x)^2}$$ and take the limit as $n\to \infty$. Unfortunately, the expression in the sum is not of the form necessary to view it as a Riemann sum, so you cannot turn that limit into a limit of Riemann sums, and from there to an integral.

So we take a slightly different approach. Instead of approximating the length of the curve from $(x_i,f(x_i))$ to $(x_i+\Delta x, f(x_i+\Delta x))$ with the straight line between the two points, we will approximate it with the tangent line to the graph of $f$ at $x_i$, from $(x_i,f(x_i))$ to the point $x_i+\Delta x$. This is the blue line in the picture above.

If $\Delta x$ is small, then we know the tangent line is a very good approximation for the curve on $[x_i,x_i+\Delta x]$, so the line will be a good approximation to the length of the curve.

Now, the tangent line to $y=f(x)$ through the point $x_i$ is given by $$y = f(x_i) + f'(x_i)(x-x_i).$$ At $x_i+\Delta x$, the line goes through $f(x_i) + f'(x_i)\Delta x$. So this tangent line goes from $(x_i,f(x_i))$ to $(x_i+\Delta x ,f(x_i)+f'(x_i)\Delta x)$. The length of the line between those two points is \begin{align*} &\sqrt{\Bigl( (x_i+\Delta x) - x_i\Bigr)^2 + \Bigl((f(x_i)+f'(x_i)\Delta x) - f(x_i)\Bigr)^2}\\\ &\quad = \sqrt{ (\Delta x)^2 + \left(f'(x_i)\Delta x\right)^2} \\\ &\quad = \sqrt{\left(1 + \left(f'(x_i)\right)^2\right)\Delta x^2} = \left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x. \end{align*}

Adding all of these, we get an approximation to the arc length: $$\text{Arc Length} \approx \sum_{i=1}^n \left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x.$$ Now, these can be seen as Riemann sums. So if we take the limit as $n\to\infty$, the approximation gets better and better (because the tangent gets closer and closer to the curve, giving a better approximation). At the limit, we get the exact arc length, and the limit of the Riemann sums becomes the integral. So \begin{align*} \text{Arc Length} &= \lim_{n\to\infty}\sum_{i=1}^n\left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x\\\ &= \int_a^b \sqrt{1+(f'(x))^2}\,dx.{}{}{} \end{align*}

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I'd love to add a simple picture of both the naive attempt and the one using the tangent, but the "best" I can do locally is use Paint, and the results are awful; any recommendations? –  Arturo Magidin Mar 19 '11 at 5:20
    
I love Paint.NET. Picnik, if you're on the move. –  muntoo Mar 20 '11 at 4:44
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It might help instead to think of our function as a parametrized curve in the plane. In other words, consider the curve $(x(t), y(t))$ for $t\in[a,b]$. Then it's clear that the differential $ds$ which lies along the curve is given by $ds=\sqrt{dx^2+dy^2}$.

Now we integrate $ds$ with respect to $t$ and the integral becomes $\displaystyle \int_a^b{\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}}dt$.

Your formula follows by setting $x(t)=t$, and $y(t)=f(t)$.

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the infinitesimal pythagorean formula! –  yoyo Mar 19 '11 at 14:10
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