Mathematics of Physics and Engineering

Functions, Sets, Gradient 125

EXERCISE 3.I.3.^5 Show that if V/ = 0 in a domain G, then f is constant

in G. Hint: For point P g G and a unit vector u, define the point Bt, t £ R,

so that PBt = tu. Using the definition of open set show that Bt is in G for

sufficiently small \t\. Apply the mean-value theorem to g(t) = f(Bt) to show that

f(Bt) = f(P)- Given another point Pi in G, connect P and Pi with a continuous

curve in G, and apply the above argument several times as you move along the

curve from P to Pi.

Let us now introduce cartesian coordinates in the space A. Then every

point P will have coordinates (x, y) or (x, y, z) and we write f(P) = f(x, y)

or f(P) = f(x,y,z). Recall that the partial derivative fx is computed

by differentiating / with respect to x and treating all other variables as

constants:

fx{x 0 ,yo)= lim - - - (3.1.7)
Ai-»0 X
EXERCISE 3.1.4.C Verify that, in cartesian coordinates, fx = D\f.
An alternative notation for fx is df/dx. The partial derivatives fy, fz
are defined similar to (3.1.7) and coincide with the directional derivatives
in the j and K directions, respectively. The higher-order partial derivatives
are defined as partial derivatives of the corresponding lower-order partial
derivatives:

/** - dx , Jxy - dy , Jvx - Qx , etc.

EXERCISE 3.1.5? For two scalar fields f,g, verify the following properties of

the gradient, generalizing the product and quotient rules from one-variable

calculus:

V(/ 5 ) = fVg + gVf, V(f/g) = (gVf - fVg)/g^2.

The next two results should be familiar from a course in multi-variable

calculus:

• If the function f = f(x,y,z) has partial derivatives fx,fy,fz in some
  neighborhood of the point A and the derivatives are continuous at A, then
  the function is differentiate at A and

V/(A) = fx(A) i + fy(A) j + fz(A) «• (3-1-8)