# High School Calculus/The Length of a Plane Curve

The graph of $y = x^{\frac {3}{2}}$ is a curve in the x-y plane. How long is that curve? A definite integral needs endpoints and we specify x = 0 and x = 4. The first problem is to know what "length function" to integrate.
Here is the unofficial reasoning that gives the length of the curve. A straight piece has $(\Delta x)^2 + (\Delta y)^2$. Within that right triangle, the height $\Delta y$ is the slope $\left (\frac {\Delta y}{\Delta x}\right)$ times $\Delta x$. This secant slope is close to the slope of the curve. Thus $\Delta y$ is approximately $\left (\frac {\operatorname {d}y}{\operatorname {d}x}\right)\Delta x$.
$\Delta s \approx \sqrt{(\Delta x)^2 + \left (\frac {\operatorname {d}y}{\operatorname {d}x}\right)^2(\Delta x)^2} = \sqrt{1 + \left (\frac {\operatorname {d}y}{\operatorname {d}x}\right)^2}\Delta x$ (1)
Now add these pieces and make them smaller. The infinitesimal triangle has $(\operatorname {d}s)^2 = (\operatorname {d}x)^2 + (\operatorname {d}y)^2$. Think of $\operatorname {d}s$ as $\sqrt{1 + \left(\frac {\operatorname {d}y}{\operatorname {d}x}\right)^2}\operatorname {d}x$ and integrate:
length of curve = $\int \operatorname {d}s = \int \sqrt {1 + \left(\frac {\operatorname {d}y}{\operatorname {d}x}\right)^2}\operatorname {d}x$. (2)