Calculus: Analytic Geometry and Calculus, with Vectors

7.1 Curves and lengths 411

that (x1, f (xi)) precedes (X2, f (X2)) when x1 < x2. Then C has finite length

;CI and

(7.171) If(b) - f(a)l < Cl I< lb - al + If(b) - f(a)I

To prove this, let P be a partition of the interval a < x < b, and, for

each k, let Pk = (xk, f(xk)). Let Axk = xk - xk-i and let

AYk = f(xk) - f(xk-1)
Then

(7.172) IPk1PkI = Oxk + AYx.

But

(7.173) DYk < L1xk^2 + AYk2 < l xkt2 + 21Axk IYkl +l Ykl2

(IAxkl +

Therefore,

(7.174) l'Ykl < IPk-iPkl < Ioxkl + IoYki.

But the numbers Oxk are all positive and the numbers DYk are all negative

(or all positive), and hence (why?) addition gives

(7.175) If(b) - f(a)I < IPk-lPxl < lb - at + If(b) -f(a)1.

k-1

Since this inequality holds for each partition P, the curve C has finite

length ICI. Moreover, the least upper bound of the central sums cannot

exceed the particular upper bound in (7.175), and no upper bound can

be less than If(b) - f (a) I. This proves Theorem 7.17.

We have been studying lengths of curves, that is,

lengths of ordered sets of points of special types. There

are other (and different) theories involving lengths of

unordered sets S in E3 and E2 and Ei. We need not be

authorities on these matters, but we should have at least

Figure 7.18

a vague idea that, in elementary mathematics, the part of the set S of

Figure 7.18 that lies "between" 11 and B has length L if there is a "natural

ordering" of the points of the part that yields a curve C having length L.

Problems 7.19

1 Tell how the length ICI of a curve C is defined.

2 Show that if a curve C having finite length runs from .4, through t12 to

-43i then .42 separates the curve into two parts each having finite length. It is

not quite so easy to show that "the whole is equal to the sum of its parts," but

it can be done. In particular, the length of a simple polygon is the sum of the

lengths of its straight segments.