Mathematics of Physics and Engineering

The Tangent Vector and Arc Length^29

Lc{c,d) between the points r(c) and r(d) along a rectifiable curve C is

Lc(c,d) = lim Ln,

Theorem 1.3.1 Assume that r '(t) exists for all t G (a, b) and the vector
function r'(t) is continuous. Then the curve C is rectifiable and

Lc(c,d)= f ||r'(i)||dt. (1.3.10)

Proof. It follows from the assumptions of the theorem and from relation

(1.3.2) that Arj — r'(ti) At; + Vi, where AU — ti+1 - ti and the vectors

Vi satisfy max 0 <j<n-i \\vi\\/Ati —> 0 as max 0 <i<„_i At* —> 0. Therefore,

||Ari|| = Wr'itJWAti + SiAU, (1.3.11)

n— 1 n — 1 TI—1

5] ||Ar-iH = Y, \\r\U)\\ A^ + £

e

'

Ai

• i=0 i=0 i=0

where the numbers e$ satisfy
maxo<i<n—I £i —^ 0, n —> oo. Then (1.3.10)
follows after passing to the limit. •

EXERCISE 1.3.6? (a) Verify (1.3.11). Hint use the triangle inequality to esti-

mate \\r'(U) Ati + Vi\\ — \\r'(ti)\\ AU. (b) Show that a piece-wise smooth curve

is rectifiable. Hint: apply the above theorem to each smooth piece separately, and

then add the results.

EXERCISE 1.3.7. c Interpreting the graph of the function y — f(x) as a

curve in M^3 , and assuming that f'(x) exists and is continuous, show that

the length of this curve from (c, /(c), 0) to (d,f(d),Q), as given by (1.3.10),

is fc yjl + \f'(x)\^2 dx; the derivation of this result in ordinary calculus is

similar to the derivation of (1.3.10).

Given a point r(c) on a rectifiable curve C, we define the arc length

function s — s(t), t > c, as

s(t) = Lc(c,t)

It follows that ds/dt = \\r'{t)\\ > 0. We call ds = ||r'(*)||dt the line

element of the curve C. If r(t) = x(t) i + y(t) j + z(t) k, where (£, j, k) is

a cartesian coordinate system at O, then

'ds\

dt

2 .1..dr 2 /,\2 /.\2 /,\2

~dl -1 s)

+

(S)

+

(I) • <-

12

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