Calculus: Analytic Geometry and Calculus, with Vectors

11.2 Increments, chain rule, and gradients 563

ticular direction of a vector. In any case, it should be recognized that

our function can be of interest to men as well as to boys. Everything
we do will depend upon the fundamental fact that there is an astonish-
ingly effective way of estimating the number Au, defined by

(11.21) Au = u(x + Ax, y + Ay, Z + AZ)- u(x,Y,z),

which represents the difference (or increment) between the values ofu
at two places. The basic trick is to divide and conquer the huge dis-

crepancy between the natures of the two terms in the right member of
(11.21) by subtracting and adding terms to obtain the telescopicsum in

(11.22) Au = u(x + Ax, y + Ay, z + Az) -u(x,y+Ay,z+Az)

+u(x,Y+AY,z+Az) -u(x,Y,z+AZ)

+ U (X' Y,Z +AZ) - u (x,Y,z)

Defining a function 0(t) by the formula

(11.221) 4(t) = u(t, Y + Ay, Z + AZ)

enables us to apply the mean-value theorem to the difference in the

first line, and similar functions apply to the other two. Thus we obtain

Au = u.,(x*, Y + AY, z + AZ) Ax + uy(x, Y*, Z + AZ) Ay + u=(x,Y,z*) Az,

where x* lies between x and x + Ax, y* lies between y and y + Ay, and

z* lies between z and z + Az. Our hypothesis that the derivatives are

continuous allows us to put this in the form

(11.222)

Au = [i& (x,y,z) + El] Ax + [uv(x,Y,z) + E2) Ay + [u,(x,Y,z) + E3) Az

where el, E2, e3 are quantities which approach zero as Ax, Ay, Az approach

zero. This formula can be put in the form

(11.223) Au =[+ei]Lix+ Cu + E2 Ay + CZ + E3J AZ.

Y

It is quite possible to rub out the epsilons, replace deltas by dees, write

the formula

(11.224) du = az dx + ay dy + az dz,

and then undertake to explain the antics. At least for. the present we

adopt different tactics.

Let three functions x, y, z be defined and differentiable over some interval

T of values of t, and let the point P(t) having coordinates x(t), y(t), z(t)

trace a curve C in R as t increases over T. Let

(11.225) w(t) = u(x(t), y(t), z(t)).