My friend told me that they had just learned an equation to find the length of a function. I decided that it would be cool to try to find it myself. I got: [tex] L(x) = \int \sqrt(f'(x)^2 +1)dx [/tex] I got that by saying that the length of a line with a slope of a over a distance of h is: [tex] \sqrt(f'(x)^2 +1) [/tex] Am I right?
In general, when a function f is determined by a vectorfunction (so you have a parameter equation of the curve), the arc length is given by: [tex]\ell = \int_a^b {\left\| {\frac{{d\vec f}} {{dt}}} \right\|dt}[/tex] There are of course conditions such as df/dt has to exist, be continous, the arc has to be continous. Now when a function is given in the form "y = f(x)" you can choose x as parameter and the formula simplifies to: [tex]\ell = \int_a^b {\sqrt {1 + y'^2 } dx} [/tex] Which is probably what you meant