High School Calculus/The Length of a Plane Curve

Length of a Plane Curve

The graph of ${\displaystyle y=x^{\frac {3}{2}}}$  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 ${\displaystyle (\Delta x)^{2}+(\Delta y)^{2}}$ . Within that right triangle, the height ${\displaystyle \Delta y}$  is the slope ${\displaystyle \left({\frac {\Delta y}{\Delta x}}\right)}$  times ${\displaystyle \Delta x}$ . This secant slope is close to the slope of the curve. Thus ${\displaystyle \Delta y}$  is approximately ${\displaystyle \left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)\Delta x}$ .

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

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

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