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 $$(\Delta x)^2 + (\Delta y)^2$$. 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)