Page 70 - Vector Analysis

66 CHAPTER 2. Differentiation of Functions of Several Variables

Let U Ď Rn be an open set, a P U and f : U Ñ R be a real-valued function. Suppose

that f P C 1(U ; R) and (∇f )(a) ‰ 0. Then B f (a) ‰ 0 for some 1 ď k ď n. W.L.O.G.,

B xk

we can assume that B f (a) ‰ 0. By the implicit function theorem, there exists an open

B xn

neighborhood V Ď Rn´1 of (a1, ¨ ¨ ¨ , an´1) and an open neighborhood W Ď R of an, as well as

a C 1-function φ : V Ñ R such that in a neighborhood of a the level set ␣x P U ˇ f (x) = f (a)(
ˇ

can be represented by xn = φ(x1, ¨ ¨ ¨ , xn´1); that is,

() @ (x1, ¨ ¨ ¨ , xn´1) P V .
f x1, ¨ ¨ ¨ , xn´1, φ(x1, ¨ ¨ ¨ , xn´1) = f (a)

Moreover,

()
´ fxj (x1, ¨ ¨ ¨ , xn´1, φ(x1, ¨ ¨ ¨ , xn´1))
φxj (x1, ¨ ¨ ¨ , xn´1) = fxn x1, ¨ ¨ ¨ , xn´1, φ(x1, ¨ ¨ ¨ , xn´1) @ (x1, ¨ ¨ ¨ , xn´1) P V .

Consider the collection of vectors tvjunj=´11 given by

vj = B ˇ( x1, ¨ ¨ ¨ , xn´1, φ(x1, ¨ ¨ ¨ ) (x1, ¨ ¨ ¨ , xn´1) P V .
B xj ˇ , xn´1)
ˇx=a

Then vj1 s are tangent vectors of the level surface. If tejujn=1 is the standard basis of Rn, then

vj = ej + ( ¨ ¨ ¨ , 0, φxj (a1, ¨ ¨ ¨ , ) = ej ´ ( ¨¨ , 0, fxj (a) ) .
0, an´1) 0, ¨
fxn (a)

Therefore, the gradient vector (∇f )(a) is perpendicular to vj for all 1 ď j ď n ´ 1 which
conclude the following

Proposition 2.81. Let U Ď Rn be open and f P C 1(U ; R); that is, f : U Ñ R is contin-

uously differentiable. Then if (∇f )(x0) ‰ 0, the vector (∇f )(x0) is the unit normal to
}(∇f )(x0)}Rn
␣x ˇ f (x0)(
the level set P U ˇ f (x) = at x0.

Example 2.82. Find the normal to S = ␣(x, y , z ) ˇ x2 + y2 + z2 = 3( at (1, 1, 1) P S.
ˇ

Solution: Take f (x, y, z) = x2 + y2 + z2 ´ 3. Then (∇f )(x, y, z) = (2x, 2y, 2z); thus

(∇f )(1, 1, 1) = (2, 2, 2) is normal to S at (1, 1, 1).

Example 2.83. Consider the surface

S = ␣(x, y, z) P R3 ˇ x2 ´ y2 + xyz = 1(.
ˇ

Find the tangent plane of S at (1, 0, 1).