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)