On-line Math 21

6.4  Arclength¹

The idea that would occur to me, if I had to measure the length of a curve, would be to drive a bunch of nails in the blackboard along the curve, and loop a string through those nails (keeping it taut and on the ``right'' side of the nails). Then I'd unwrap it and hold it alongside a ruler. If I wanted to be more accurate, I'd go get more nails.

Let's see how that works without really driving nails into the blackboard. If the curve is y = f(x) , x Î [a,b] , then we can, as usual, break the interval from a to b up into a number ( n ) of subintervals, a = x₀ < x₁ < ... < x_n-1 < x_n = b . The nails would then be placed at the points {(x₀,f(x₀)),... ,(x_n,f(x_n))} . The length of a line segment from (x_i-1,f(x_i-1)) to (x_i,f(x_i)) is

Ds = Ö   (xi-xi-1)2+(f(xi)-f(xi-1))2   = Ö   Dx2+Dy2   ,

and so the length of the string will be:

n å i = 1  Ö   (xi-xi-1)2+(f(xi)-f(xi-1))2   = n å i = 1  Ö   Dx2+Dy2   .

This is the first of these applications of integration formulas that doesn't seem to be approaching an integral. There is no dx . But we can make it appear by:

length: @ n å i = 1  Ö   Dx2+Dy2   = n å i = 1  Ö   dx2+dy2   = n å i = 1    æ  ú Ö 1+ æ ç è dy dx ö ÷ ø 2      dx

Now we should see the standard idea of an application of integration coming out of this: take the number of terms to ¥, or the distance between the subintervals to 0, and you'd get:

length: = lim n® ¥  n å i = 1    æ  ú Ö 1+ æ ç è dy dx ö ÷ ø 2      dx = ó õ b a    æ  ú Ö 1+ æ ç è dy dx ö ÷ ø 2      dx.

The problems have to be very carefully chosen or else the integrals will be impossible. That is a hint, since the problems in the text are carefully chosen so that the integrals will not be impossible. If your solution is impossible to integrate, perhaps you set the problem up incorrectly. If the set-up is right, you might be missing something in doing the integral.

Example 1 Find the arc-length of the curve y = x^3/2 , x Î [0,4] .

Solution

length = ó õ 4 0    æ  ú Ö 1+( 3 2 x1/2)2   dx = ó õ 4 0    æ  ú Ö 1+ 9x 4   dx = 2 3 · 4 9 æ ç è 1+ 9x 4 ö ÷ ø 3/2   ê ê ê 4 0  = 8 27 ( 103/2-1) .

Example 2 Find the arc-length of the curve

y =

x2 2

-

ln(x) 4

, x Î (2,4] .

Solution

length = ó õ 4 2    æ  ú Ö 1+ æ ç è x- 1 4x ö ÷ ø 2     dx = ó õ 4 2    æ  ú Ö 1+ æ ç è x2- 1 2 + 1 16x2 ö ÷ ø   dx = ó õ 4 2    æ  ú Ö x2+ 1 2 + 1 16x2   dx = ó õ 4 2    æ  ú Ö æ ç è x+ 1 4x ö ÷ ø 2     dx = ó õ 4 2  æ ç è x+ 1 4x ö ÷ ø dx = x2 2 + 1 4 lnx ê ê ê 4 2  = æ ç è 8+ 1 4 ln4 ö ÷ ø - æ ç è 2+ 1 4 ln2 ö ÷ ø .

This is a standard trick, where the +1 was exactly what was needed to change the one perfect square to another, by changing the cross-term from -1/2 to 1/2 .

Example 3 Find the arc-length of the parabola y = x² , x Î [-2,2]

Footnotes:

¹This section starts out all wrong in the books. They always talk about defining the arclength of a curve by some formula. We are not really defining the idea of the length of a curve. It seems rather condescending to say that we don't already know what the length of a curve is. That's just a mathematician's way of talking. What they mean to say is that we need to come up with a way of measuring the length of a curve.

Copyright (c) 2000 by David L. Johnson.