There are many applications for the arc length of functions. I came across it when I needed to calculate the length of a density function in an interval [a,b]. The formula reads

\displaystyle s=\int\limits_{a}^{b}\sqrt{1+\big[f'(x)\big]^{2}}\,dx

Do you grasp how this formula arises? No? Let me show you!

I imagined how I zoom into a function f(x)ย  until single points become visisble. Actually, this is not possible from a mathematical point of view – but let’s try for simplicity’s sake ๐Ÿ™‚ The following sketch shows what I mean:

Grid of infinitesimal points

Grid of infinitesimal points

As you see, the horizontal distance is denoted by dx while the vertical distance is denoted by dy. dy can be expressed in terms of dx and the derivative f'(x):

\displaystyle dy=f'(x)\cdot dx

At the same time, Pythagoras’ theorem holds and allows us to calculate the infinitesimal distance ds. I also applied some manipulations to the expression.

\displaystyle ds=\sqrt{\big(dx\big)^{2}+\big(dy\big)^{2}}=\sqrt{\big(dx\big)^{2}+\big(f'(x)\big)^{2}\cdot \big(dx\big)^{2}}=\sqrt{1+\big(f'(x)\big)^{2}} \, dx

Adding these endlessly, small sections of the line together is done via a sum. Just write the integral sign in front of the formula.

\displaystyle s=\int\limits_{a}^{b} ds =\int\limits_{a}^{b} \sqrt{1+\big[f'(x)\big]^{2}}\, dx

Done. ๐Ÿ˜‰