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Comparing equalities (8.73) and (8.74), we see that the graphof the function is well approximated by the plane (8.74) in a neighborhood of thepoint (x0 , y0 , z0 ). More precisely, the point (x, y, f (x, y)) of the graph of the function differs from the point (x, y, z(x, y)) of the#plane (8.74) by an amount that isinfinitesimal in comparison with the magnitude (x − x0 )2 + (y − y0 )2 of the displacement of its curvilinear coordinates (x, y) from the coordinates (x0 , y0 ) of thepoint (x0 , y0 , z0 ).By the uniqueness of the differential of a function, the plane (8.74) possessingthis property is unique and has the formz = f (x0 , y0 ) +∂f∂f(x0 , y0 )(x − x0 ) +(x0 , y0 )(y − y0 ).∂x∂y(8.75)This plane is called the tangent plane to the graph of the function z = f (x, y) at thepoint (x0 , y0 , f (x0 , y0 )).Thus, the differentiability of a function z = f (x, y) at the point (x0 , y0 )and the existence of a tangent plane to the graph of this function at the point(x0 , y0 , f (x0 , y0 )) are equivalent conditions.c.
The Normal VectorWriting Eq. (8.75) for the tangent plane in the canonical form∂f∂f(x0 , y0 )(x − x0 ) +(x0 , y0 )(y − y0 ) − z − f (x0 , y0 ) = 0,∂x∂ywe conclude that the vector∂f∂f(x0 , y0 ), (x0 , y0 ), −1∂x∂y(8.76)is the normal vector to the tangent plane. Its direction is considered to be the direction normal or orthogonal to the surface S (the graph of the function) at the point(x0 , y0 , f (x0 , y0 )).In particular, if (x0 , y0 ) is a critical point of the function f (x, y), then the normalvector to the graph at the point (x0 , y0 , f (x0 , y0 )) has the form (0, 0, −1) and consequently, the tangent plane to the graph of the function at such a point is horizontal(parallel to the xy-plane).The three graphs in Fig.
8.1 illustrate what has just been said.Figures 8.1a and c depict the location of the graph of a function with respect tothe tangent plane in a neighborhood of a local extremum (minimum and maximumrespectively), while Fig. 8.1b shows the graph in the neighborhood of a so-calledsaddle point.4728 Differential Calculus in Several VariablesFig. 8.1d. Tangent Planes and Tangent VectorsWe know that if a path Γ : I → R3 in R3 is given by differentiable functions x =x(t), y = y(t), z = z(t), then the vector (ẋ(0), ẏ(0), ż(0)) is the velocity vector attime t = 0. It is a direction vector of the tangent at the point x0 = x(0), y0 = y(0),z0 = z(0) to the curve in R3 that is the support of the path Γ .Now let us consider a path Γ : I → S on the graph of a function z = f (x, y)given in the form x = x(t), y = y(t), z = f (x(t), y(t)).
In this particular case wefind that∂f∂fẋ(0), ẏ(0), ż(0) = ẋ(0), ẏ(0), (x0 , y0 )ẋ(0) +(x0 , y0 )ẏ(0) ,∂x∂yfrom which it can be seen that this vector is orthogonal to the vector (8.76) normal to the graph S of the function at the point (x0 , y0 , f (x0 , y0 )). Thus we haveshown that if a vector (ξ, η, ζ ) is tangent to a curve on the surface S at the point(x0 , y0 , f (x0 , y0 )), then it is orthogonal to the vector (8.76) and (in this sense) liesin the plane (8.75) tangent to the surface S at the point in question. More preciselywe could say that the whole line x = x0 + ξ t, y = y0 + ηt, z = f (x0 , y0 ) + ζ t liesin the tangent plane (8.75).Let us now show that the converse is also true, that is, if a line x = x0 + ξ t,y = y0 + ηt, z = f (x0 , y0 ) + ζ t, or what is the same, the vector (ξ, η, ζ ), lies in theplane (8.75), then there is a path on S for which the vector (ξ, η, ζ ) is the velocityvector at the point (x0 , y0 , f (x0 , y0 )).The path can be taken, for example, to bex = x0 + ξ t,y = y0 + ηt,z = f (x0 + ξ t, y0 + ηt).In fact, for this path,ẋ(0) = ξ,ẏ(0) = η,ż(0) =∂f∂f(x0 , y0 )ξ +(x0 , y0 )η.∂x∂yIn view of the equality∂f∂f(x0 , y0 )ẋ(0) +(x0 , y0 )ẏ(0) − ż(0) = 0∂x∂y8.4 Real-valued Functions of Several Variables473Fig.
8.2and the hypothesis that∂f∂f(x0 , y0 )ξ +(x0 , y0 )η − ζ = 0.∂x∂yWe conclude thatẋ(0), ẏ(0), ż(0) = (ξ, η, ζ ).Hence the tangent plane to the surface S at the point (x0 , y0 , z0 ) is formed by thevectors that are tangents at the point (x0 , y0 , z0 ) to curves on the surface S passingthrough the point (see Fig.
8.2).This is a more geometric description of the tangent plane. In any case, one cansee from it that if the tangent to a curve is invariantly defined (with respect to thechoice of coordinates), then the tangent plane is also invariantly defined.We have been considering functions of two variables for the sake of visualizability, but everything that was said obviously carries over to the general case of afunctiony = f x1, . . . , xm(8.77)of m variables, where m ∈ N.At the point (x01 , . .
. , x0m , f (x01 , . . . , x0m )) the plane tangent to the graph of sucha function can be written in the formm ∂f 1x0 , . . . , x0m x i − x0i ;y = f x01 , . . . , x0m +i∂x(8.78)i=1the vector∂f∂f(x0 ), . . . , m (x0 ), −1∂x∂x 1is the normal vector to the plane (8.78). This plane itself, like the graph of the function (8.77), has dimension m, that is, any point is now given by a set (x 1 , .
. . , x m )of m coordinates.Thus, Eq. (8.78) defines a hyperplane in Rm+1 .Repeating verbatim the reasoning above, one can verify that the tangent plane(8.78) consists of vectors that are tangent to curves passing through the point4748 Differential Calculus in Several Variables(x01 , . . . , x0m , f (x01 , . .
. , x0m )) and lying on the m-dimensional surface S – the graphof the function (8.77).8.4.7 Problems and Exercises1. Let z = f (x, y) be a function of class C (1) (G; R).a) If ∂f∂y (x, y) ≡ 0 in G, can one assert that f is independent of y in G?b) Under what condition on the domain G does the preceding question have anaffirmative answer?2. a) Verify that for the function!f (x, y) =,xy xx 2 −y+y 2if x 2 + y 2 = 0,0,if x 2 + y 2 = 0,22the following relations hold:∂f∂ 2f(0, 0) = 1 = −1 =(0, 0).∂x∂y∂y∂xb) Prove that if the function f (x, y) has partial derivatives∂f∂xandin some∂2f∂x∂y (or ∂y∂x )∂2f∂2f∂y∂x (resp.
∂xθy )neighborhood U of the point (x0 , y0 ), and if the mixed derivativeexists in U and is continuous at (x0 , y0 ), then the mixed derivativealso exists at that point and the following equality holds:∂f∂y∂2f∂ 2f∂ 2f(x0 , y0 ) =(x0 , y0 ).∂x∂y∂y∂x3. Let x 1 , . . . , x m be Cartesian coordinates in Rm . The differential operatorΔ=m∂2i=1∂x i2,acting on functions f ∈ C (2) (G; R) according to the ruleΔf =m∂ 2f 1x , .
. . , xm ,2∂x ii=1is called the Laplacian.The equation Δf = 0 for the function f in the domain G ⊂ Rm is calledLaplace’s equation, and its solutions are called harmonic functions in the domain G.8.4 Real-valued Functions of Several Variables475a) Show that if x = (x 1 , . . .
, x m ) and45 m5 2,x, = 6xi ,i=1then for m > 2 the functionf (x) = ,x,−2−m2is harmonic in the domain Rm \0, where 0 = (0, . . . , 0).b) Verify that the function 11,x,2m· exp − 2 ,f x ,...,x ,t =√4a t(2a πt)mwhich is defined for t > 0 and x = (x 1 , . . . , x m ) ∈ Rm , satisfies the heat equation∂f= a 2 Δf,∂tthat is, verify thatthe function.∂f∂t= a2m∂2fi=1 ∂x i 2at each point of the domain of definition of4. Taylor’s formula in multi-index notation. The symbol α := (α1 , . .
. , αm ) consisting of nonnegative integers αi , i = 1, . . . , m, is called the multi-index α.The following notation is conventional:|α| := α1 + · · · + αm ,α! := α1 ! · · · αm !;finally, if a = (a1 , . . . , am ), thenαm.a α := a1α1 · · · ama) Verify that if k ∈ N, then(a1 + · · · + am )k =|α|=kk!αma α1 · · · am,α1 ! · · · αm ! 1or(a1 + · · · + am )k = k!aα ,α!|α|=kwhere thesummation extends over all sets α = (α1 , . .
. , αm ) of nonnegative integerssuch that mi=1 αi = k.4768 Differential Calculus in Several Variablesb) LetD α f (x) :=∂ |α| f(x).(∂x 1 )α1 · · · (∂x m )αmShow that if f ∈ C (k) (G; R), then the equality∂i1 ···ik f (x)hi1 · · · hik =i1 +···+im =k k!D α f (x)hα ,α!|α|=kwhere h = (h1 , . . . , hm ), holds at any point x ∈ G.c) Verify that in multi-index notation Taylor’s theorem with the Lagrange formof the remainder, for example, can be written asf (x + h) =n−1 11 αD f (x)hα +D α f (x + θ h)hα .α!α!|α|=n|α|=0d) Write Taylor’s formula in multi-index notation with the integral form of theremainder (Theorem 4).5. a) Let I m = {x = (x 1 , . . .
, x m ) ∈ Rm | |x i | ≤ ci , i = 1, . . . , m} be an m-dimensional closed interval and I a closed interval [a, b] ⊂ R. Show that if the functionf (x, y) = f (x 1 , . . . , x m , y) is defined and continuous on the set I m ×I , then for anypositive number ε > 0 there exists a number δ > 0 such that |f (x, y1 ) − f (x, y2 )| <ε if x ∈ I m , y1 , y2 ∈ I , and |y1 − y2 | < δ.b) Show that the functionbF (x) =f (x, y) dyais defined and continuous on the closed interval I m .c) Show that if f ∈ C(I m ; R), then the functionF(x, t) = f (tx)is defined and continuous on I m × I 1 , where I 1 = {t ∈ R | |t| ≤ 1}.d) Prove Hadamard’s lemma:If f ∈ C (1) (I m ; R) and f (0) = 0, there exist functions g1 , .
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