High School Calculus/The Length of a Plane Curve

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 $(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2$. Within that right triangle, the height $\mathrm{\Delta }y$ is the slope $\left(\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\right)$ times $\mathrm{\Delta }x$. This secant slope is close to the slope of the curve. Thus $\mathrm{\Delta }y$ is approximately $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\mathrm{\Delta }x$.

$\mathrm{\Delta }s\approx \sqrt{(\mathrm{\Delta }x{)}^2+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2(\mathrm{\Delta }x{)}^2}=\sqrt{1+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2}\mathrm{\Delta }x$ (1)

Now add these pieces and make them smaller. The infinitesimal triangle has $(\mathrm{d}s{)}^2=(\mathrm{d}x{)}^2+(\mathrm{d}y{)}^2$. Think of $\mathrm{d}s$ as $\sqrt{1+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2}\mathrm{d}x$ and integrate:

length of curve = $\int \mathrm{d}s=\int \sqrt{1+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2}\mathrm{d}x$. (2)