# Math Insight

### Length of curves

The basic point here is a formula obtained by using the ideas of calculus: the length of the graph of $y=f(x)$ from $x=a$ to $x=b$ is $$\hbox{ arc length }=\int_a^b\;\sqrt{1+\left({dy\over dx}\right)^2}\;dx$$ Or, if the curve is parametrized in the form $$x=f(t)\;\;\;\;\;y=g(t)$$ with the parameter $t$ going from $a$ to $b$, then $$\hbox{ arc length }=\int_a^b\; \sqrt{\left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2}\;dt$$

This formula comes from approximating the curve by straight lines connecting successive points on the curve, using the Pythagorean Theorem to compute the lengths of these segments in terms of the change in $x$ and the change in $y$. In one way of writing, which also provides a good heuristic for remembering the formula, if a small change in $x$ is $dx$ and a small change in $y$ is $dy$, then the length of the hypotenuse of the right triangle with base $dx$ and altitude $dy$ is (by the Pythagorean theorem) $$\hbox{ hypotenuse }=\sqrt{dx^2+dy^2}= \sqrt{1+\left({dy\over dx}\right)^2}\;dx$$

Unfortunately, by the nature of this formula, most of the integrals which come up are difficult or impossible to ‘do’. But if one of these really mattered, we could still estimate it by numerical integration.

#### Exercises

1. Find the length of the curve $y=\sqrt{1-x^2}$ from $x=0$ to $x=1$.
2. Find the length of the curve $y={ 1 \over 4 }(e^{2x}+e^{-2x})$ from $x=0$ to $x=1$.
3. Set up (but do not evaluate) the integral to find the length of the piece of the parabola $y=x^2$ from $x=3$ to $x=4$.