Calculus: Analytic Geometry and Calculus, with Vectors

11.2 Increments, chain rule, and gradients 567

(isothermal surfaces or equipotential surfaces, for example) upon which

u has a particular constant value. If Po(xo,yo,zo) and some sphere with

center at Po lie in our region R, then our hypotheses (including the

hypothesis that Vu ; 0 at Po) imply that there isa surface S consisting

of points P(x,y,z) for which u(x,y,z) = u(xo,yo,zo). Let C be a curve

which lies on S and passes through Po and has thevector v for its forward

tangent at Po. Supposing that a point moving along this curve has

coordinates x(t), y(t), z(t), we have

(11.27) ' u(x(t), Y(t), z(t)) = u(x0,Yo, zo)

and hence du/dt = 0. Therefore, (Vu).v = 0. Thus Vu is orthogonal
to each line which passes through Po and which is tangent to a curve on
S. As Figure 11.271 may suggest, this is just
what we mean when we say that Vu is a normal
to the surface S at the point Po. Therefore,
we can find the normal to the surface S having
the equation u(x,y)z) = c, at a point Po on V '0 Yu(zyz)=c
S, by finding the gradient of u at Po. To find
the plane tangent to S at Po is easy; it is the
plane through Po orthogonal to the gradient. Figure 11.271
A review of matters relating to equations of
lines and tangent planes may be in order. If the gradient of u at a point
Po(xo,yo,zo) is

Eli+Bj-}-Ck,

then the equations of the line through Po upon which this gradient lies are

(11.28) x - x- Y - Yo_Z - zo

A B C

and the equation of the plane through Po normal to the gradient is

(11.281) A(x - xo) + B(y - yo) + C(z - zn) = 0.

The equations (11.28) are correct because they say that the scalar com-

ponents of the vector from Po(xo,yo,zo) to P(x,y,z) are proportional to the

scalar components of the gradient. The equation (11.281) is correct

because it says that the vector from Po(xo,yo,zo) to P(x,y,z) is orthogonal

to the gradient.

Problems 11.29

1 This problem requires us to think about some ways in which the formulas

(11.21) to (11.25) can be used to solve problems. Suppose f is a given function

having continuous partial derivatives. Suppose we want f(x,y,z) but we cal-

culate f(xo,yo,zo) because it is easier to calculate f(xo,yo,zo) or because x, y, z are

unknown and xo, yo, zo are the numbers we got when we measured them. How