THE IMPLICIT FUNCTION THEOREM

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1 THE IMPLICIT FUNCTION THEOREM ALEXANDRU ALEMAN 1. Motivation and statement We want to understand a general situation wic occurs in almost any area wic uses matematics. Suppose we are given number of equations wic involve a iger number of unknown variables, say (1.1) f j (x 1,..., x n, y 1,... y m ) = 0, 1 j m. Common sense says tat we migt not be able to solve tese equations uniquely, but we migt be able to express te variables y k in terms of n variables x j, i.e. we could write y k = g k (x 1,..., x n ), 1 k m, were g k are functions of n variables. In order to develop a meaningful discussion we must assume tat (1.1) is satisfied at least at one point, tat is, tere exist x 0 1,..., x 0 n, y 0 1,..., y 0 m suc tat (1.2) f j (x 0 1,..., x 0 n, y 0 1,... y 0 m) = 0, 1 j m. In tis case we require tat te functions y k = g k (x 1,..., x n ), 1 k m, satisfy g k (x 0 1,..., x 0 n) = y 0 k, 1 k m. Let us refer to tis question as Problem (1.1). As is to expect, tis is a tricky problem; Just imagine we want to decide weter te system x 2 + y 2 + z 2 = 1 x 3 + y 3 + z 3 = 2, as solutions of te form x, y(x), z(x). Some more tractable examples are listed below. Examples 1.1. Linear equations. 1) Let us start wit linear equations. a) Te equation x + y = 5, 1

2 2 ALEXANDRU ALEMAN as no unique solution, but eac unknown x, y can be expressed in terms of te oter x = 5 y, y = 5 x. b) Tis is not always te case as te following linear system sows: x + y + z = 2 x + y + 2z = 3 wic as solutions x = 1 y, z = 1, i.e. x, y cannot be expressed in terms of z. On te oter and, we can say tat y, z can be expressed in terms of x, or x, z can be expressed in terms of y. Tere is a general result in Linear Algebra wic says tat given n linear equations wit n+m unknowns ten tere exists a number k n suc tat all unknowns can be expressed in terms of k of tem. As we ave seen above tese k unknowns are not arbitrary. c) In te linear case it is quite easy to decide wic variables can be used to express te oters. Here is an example wic illustrates tis. Consider te linear system a 1 x + b 1 y + c 1 z + d 1 u = 1 a 2 x + b 2 y + c 2 z + d 2 u = 2, wit unknowns x, y, z, u, and constant coefficients a 1, a 2, b 1, b 2, c 1, c 2, d 1, d 2. Suppose tat te determinant of te matrix ( ) c1 d (1.3) 1 c 2 d 2 is non-zero. Ten writing te system as c 1 z + d 1 u = 1 a 1 x b 1 y c 2 z + d 2 u = 2 a 2 x b 2 y, it follows by Cramer s rule tat z, u can be expressed in terms of x, y. It is important to note tat in all linear cases, te remaining unknowns are expressed as functions of te given ones. For nonlinear expressions tis migt fail. 2) Te equation of a circle. Consider te equation Its general solution is given by x 2 + y 2 = 1, x, y [ 1, 1] x = ± 1 y 2, or y = ± 1 x 2. Tus we need two functions to express x in terms of y, or vice versa. Tis problem can be removed using te assumption (1.3). For example,

3 if x 0 = THE IMPLICIT FUNCTION THEOREM 3 3 4, y 0 = 1 2, ten for x is close to x0, te function y = + 1 x 2, satisfies te equation as well as te condition y(x 0 ) = y 0. However, if y 0 = 1 ten tere are always two solutions to Problem (1.1). Tese examples reveal tat a solution of Problem (1.1) migt require: To restrict te domains of definition of te functions g k we are looking for. For example, we could consider tem as defined only in a neigborood of te point (x 0 1,..., x 0 n) considered in (1.2) (see Example 2). Some additional conditions regarding te independence of te equations (1.1) wit respect to te y-variables as illustrated in Example 1c). Te second type of condition may appear mysterious, but it is just te nonlinear version of te condition in Example 1c), and it can be easily explained using derivatives. Before we do tat let us give a precise formulation of Problem (1.1) for continuously differentiable functions. Assume tat te functions f 1,..., f m ave continuous partial derivatives in a neigborood V of p 0 = (x 0 1,..., x 0 n, y 0 1,..., y 0 m) R n+m and satisfy: f j (x 0 1,..., x 0 n, y 0 1,... y 0 m) = 0, 1 j m. Does tere exists a neigborood U of x 0 = (x 0 1,..., x 0 n) R n and uniquely determined functions, g 1,..., g m : U R wit continuous partial derivatives in U, suc tat and g k (x 0 1,..., x 0 n) = y 0 k, 1 k m, f j (x, g 1 (x),..., g m (x)) = 0, x = (x 1,... x n ) U, 1 j m? Let us look at te simplest case wen n = m = 1. Suppose tat f : R 2 R as continuous partial derivatives, and tat Problem (1.1) can be solved in te terms described above. In oter words, f satisfies f(x 0, y 0 ) = 0, for some (x 0, y 0 ) R 2, and tere exists an interval I containing x 0 and a function g : I R wit a continuous derivative on I suc tat g(x 0 ) = y 0 and f(x, g(x)) = 0, x I.

4 4 ALEXANDRU ALEMAN By differentiating tis equality w.r.t. x and using te cain rule we obtain (x, g(x)) + x y (x, g(x))g (x) = 0, x I. Ten obviously (x, g(x)) = 0 wenever (x, g(x)) = 0 x y Tis is a very restrictive condition wic we definitely want to avoid! In fact, Example 2) sows tat wen x 0 = 1, y 0 = 0, tere are two functions g as above and none of tem is differentiable at x 0. Tus te assumption we sall use ere is tat at least near (x 0, y 0 ) we ave (x, y) 0. y Note tat in tis case we ave g (x) = (x, g(x))/ (x, g(x)), x y in particular, g is continuous at tose points At te cost of working wit smaller sets we may assume as well tat y (x0, y 0 ) 0, since te continuity of te partial derivate implies tat tis olds in a neigborood of (x 0, y 0 ). and g (x) = (x, g(x))/ (x, g(x)), x y (x, g(x)) 0, y A similar situation occurs in iger dimensions as well. Suppose tat f 1, f 2 : R 3 R as continuous partial derivatives, and tat Problem (1.1) can be solved, i.e. f j (x 0, y 0, z 0 ) = 0, j = 1, 2, for some (x 0, y 0, z 0 ) R 3, and tere exists an interval I containing x 0 and two functions g 1, g 2 : I R wit continuous derivatives on I, suc tat g 1 (x 0 ) = y 0, g 2 (x 0 ) = z 0, and f j (x, g 1 (x), g 2 (x)) = 0, x I, j = 1, 2. By differentiating tese equalities w.r.t. x and using again te cain rule we obtain j x (x, g 1(x), g 2 (x))+ j y (x, g 1(x), g 2 (x))g 1(x)+ j z (x, g 1(x), g 2 (x))g 2(x) = 0,

5 THE IMPLICIT FUNCTION THEOREM 5 wen x I, j = 1, 2. Tis can be rewritten as ) ( ) 1 1 y z + ( 1 x 2 x (x,g 1 (x),g 2 (x)) 2 y 2 z (x,g1 (x),g 2 (x)) ( ) g 1 (x) g 2(x) = 0. If te 2 2 matrix above is not invertible, tis creates similar complications as in te scalar case. For tis reason, we sall assume it is invertible near te point (x 0, y 0, z 0 ). Again, by continuity it suffices to assume tat tis matrix computed at x 0 is invertible, in wic case it will be invertible in a (possibly smaller) neigborood of tis point. Since a matrix is invertible if and only if it as a nonzero determinant, we can reformulate tis in terms of te Jacobian determinants (see CC:128, p. 732). Our condition is equivalent to (f 1, f 2 ) (y, z) (x 0,y 0,z 0 ) 0. Also note tat in tis case we can apply Cramer s rule to find an implicit formula for te derivatives g 1, g 2. Recall tat te Cramer rule says tat if a b c d 0, ten te unique solution of te linear system is given by x = u v a c ax + by = u cx + dy = v c d a b b, y = d a c In our case tis gives (f 1,f 2 ) g 1(x) (x,y) = g 2(x) (x,g1 (x),g 2 (x)), = (f 1,f 2 ) (y,z) u v b d (f 1,f 2 ) (z,x) (f 1,f 2 ) (y,z) We are now ready to state te Implicit Function Teorem. (x,g1 (x),g 2 (x)). Teorem 1.1. Assume tat te functions f 1,..., f m ave continuous partial derivatives in a neigborood V of p 0 = (x 0 1,..., x 0 n, y 0 1,..., y 0 m) R n+m and satisfy: (1) f j (x 0 1,..., x 0 n, y 0 1,... y 0 m) = 0, 1 j m,

6 6 ALEXANDRU ALEMAN (2) (f 1,..., f m ) (y 1,..., y m ) 0. p0 Ten tere exists a neigborood U of x 0 = (x 0 1,..., x 0 n) R n and uniquely determined functions, g 1,..., g m : U R wit continuous partial derivatives in U, suc tat (3) g k (x 0 1,..., x 0 n) = y 0 k, 1 k m, (4) (x 1,..., x n, g 1 (x),..., g m (x)) V and (5) f j (x, g 1 (x),..., g m (x)) = 0, x = (x 1,... x n ) U, 1 j m. Moreover, for 1 j, 1 k m (6) g k (x) = x j (f 1,...,f m) (y 1,...,y k 1,x j,y k,...,y m) (f 1,...,f m) (y 1,...,y m) (x,g1 (x),...,g m(x)). 2. Exercises Exercise 2.1. Sow tat Problem (1.1) is solvable for te system x 2 + y 2 + z 2 = 1 x 3 + y 3 + z 3 = 2, and any point (x 0, y 0, z 0 ) were te equations are satisfied, and y 0, z 0 0, y 0 z 0. Solution We only need to verify ypotesis (3) in Teorem 1.1 for te functions f 1 (x, y, z) = x 2 + y 2 + z 2, f 2 (x, y, z) = x 3 + y 3 + z 3, tat is, (f 1, f 2 ) (y, z) 0. (x 0,y 0,z 0 ) Since 1 y = 2y, 1 z = 2z, 2 y = 2 3y2, z = 3z2, te determinant on te left is 2y 2z 3y 2 3z 2 = 6y 0 (z 0 ) 2 6(y 0 ) 2 z 0 = 6y 0 z 0 (z 0 y 0 ). (x 0,y 0,z 0 ) Under te conditions on y 0, z 0, tis determinant does not vanis and te solution is complete. A more callenging related problem is te following:

7 THE IMPLICIT FUNCTION THEOREM 7 Exercise 2.2. Let S be te set of solutions of te system x 2 + y 2 + z 2 = 1 x 3 + y 3 + z 3 = 2, Sow tat S {(x, y, z) : 0 < y < z}, is a smoot curve, in t sense tat tere exists an interval I R and a continuously differentiable function u : I R 3 wit u(i) = S {(x, y, z) : 0 < y < z}. Exercise 2.3. a) Construct a continuously differentiable bijective function g : R 2 R 2 wose inverse is not differentiable on R 2. b) Let V R 3 be open, and let f : V R 3 ave continuous partial derivatives in V. Sow tat if te Jacobian matrix Df(p 0 ) of f at p 0 V is non-singular ten tere exists a neigborood U of f(p 0 ) and a function : U V wit continuous partial derivatives in U, suc tat f((y)) = y. Solution a) Te typical function on R wit tis property is (x) = x 3. Indeed, is strictly increasing on R wit (R) = R, i.e. is bijective. On te oter and, 1 = 3 x wic is not differentiable at te origin. On R 2 we can consider te function ( ) x 3 1 (x, y) =, y wit inverse ( 3 ) x 1 1 (x, y) =, y wic is not differentiable at te origin because it does not ave a partial derivative in x at tat point. b) Te equality f(y) = x g(x, y) = f(x) y = 0 leads to system of te form (1.1) wit 3 equations and 6 unknowns. Since g(p 0, f(p 0 )) = 0, and g (x 1, x 2, x 3 ) 0, (p0,f(p 0 ) we can apply Teorem 1.1 wic gives a unique function defined near f(p 0 ) wit continuous partial derivatives, suc tat g((y), y) = 0 f((y)) = y.

8 8 ALEXANDRU ALEMAN Question 2.1. Does tere exist a neigborood V 1 V of p 0 suc tat te function f from b) is injective on V 1 and f(v 1 ) is open? 3. Proof of te Implicit Function Teorem Let us note from te start te following: If we can sow tat tere exists a neigborood U of x 0 = (x 0 1,..., x 0 n) R n and uniquely determined functions, g 1,..., g m : U R wose partial derivatives exist in U, suc tat (3),(4),(5) old, ten g k, 1 k m ave continuous partial derivatives and (6) olds. Tis s just a repetition of te argument in te last example preceding te statement of te teorem. Suppose we ave found differentiable functions g 1,..., g m : U R as above. From (5) and te cain rule we obtain for fixed 1 j n k y 1 ((x, g 1 (x),..., g m (x)) g 1 x j k y m ((x, g 1 (x),..., g m (x)) g m x j = k x j ((x, g 1 (x),..., g m (x)). Tis can be seen as a linear system wit matrix ( k y l ((x, g 1 (x),..., g m (x))) 1 k m and rigt and side ( 1 x j (((x, g 1 (x),..., g m (x),..., m x j ((x, g 1 (x),..., g m (x). By Cramer s rule its solution is just (6). Now if (6) olds and eac g k is continuous, ten te partial derivatives of g k are obviously continuous. Tus we need to prove te following claim: (*) Tere exists a neigborood U of x 0 = (x 0 1,..., x 0 n) R n and uniquely determined functions, g 1,..., g m : U R wose partial derivatives exist in U, suc tat (3),(4),(5) old. We sall prove te claim by induction. We start wit te case wen m = 1. Tus we want to solve Problem (1.1) for one equation f(x 1,..., x n, y) = 0 assuming tat f(x 0 1,..., x 0 n, y 0 ) = 0, and y (x0 1,..., x 0 n, y 0 ) 0.

9 THE IMPLICIT FUNCTION THEOREM 9 Also, we can assume witout loss tat y (x0, y 0 ) > 0, oterwise we can cange from f to f. Since V is open and te function y y (x0, y) is assumed to be continuous, tere exists a cube Q = [x 0 1 a, x a]... [x 0 n a, x 0 n + a] [y 0 a, y 0 + a] V, suc tat (x, y) > 0, (x, y) Q. y In particular, since f(x 0, y) vanises at y 0 and is strictly increasing on [y 0, y 0 + a], it satisfies f(x 0, y 0 a) < 0, f(x 0, y 0 + a) > 0. At tis point we use uniform continuity. Recall tat a continuous realvalued function on a compact subset K of R n is uniformly continuous, i.e. given ε > 0 tere exists δ > 0 suc tat if u, v K and u v < δ ten (u) (v) < ε. Tis means tat for every ε > 0 tere exists δ > 0 wic can be cosen in (0, a], suc tat wenever x x 0 < δ, f(x, y) f(x 0, y) < ε, y [y 0 a, y 0 + a]. If ε is sufficiently small we conclude tat for x x 0 < δ we ave f(x, y 0 a) < 0, f(x, y 0 + a) > 0. Since for eac x as above, y f(x, y) is strictly increasing on [y 0 a, y 0 + a] tere exists a unique y = g(x) [y 0 a, y 0 + a] wit f(x, g(x)) = 0. Clearly, g(x 0 ) = 0 wic gives (3), (4) and (5)wit a function g defined on U = {x : x x = < ε}, wit values in [y 0 a, y 0 + a]. It remains to sow tat g as partial derivatives in U. Tis is te ard part of te proof! Fortunately it isn t very long. If you did not already note it, te only information available about g is tat f(x, g(x)) = 0, so tat all furter information about tis function must emerge from it.

10 10 ALEXANDRU ALEMAN a) g is continuous on U. Indeed, if we assume te contrary, tere exists x U and a sequence (x m ) in U wit lim m xm = x, g(x) lim m g(xm ) = y [y 0 a, y 0 + a]. But ten f(x, ) is strictly increasing on [y 0 a, y 0 a] and satisfies f(x, y) = f(x, g(x)) = 0. tis is a contradiction wic sows te continuity of g. b) g as partial derivatives in U. If {e 1,..., e n } denotes te canonical basis in R n, we want to sow tat for eac x U, te limits g(x + e j ) g(x) lim, 1 j n 0 exist. Recall tat f(x, g(x)) = 0, and write for fixed x and 0 wit sufficiently small (we need tat x te j, x + te j, t [0, ] U, (3.4) 0 = f(x + e j, g(x + e j )) f(x, g(x)) = f(x + e j, g(x + e j )) f(x, g(x + e j )) Apply te mean value teorem to te function u(t) = f(x + te j, g(x + e j )) to conclude tat f(x + e j, g(x + e j )) f(x, g(x + e j )) + f(x, g(x + e j)) f(x, g(x)) t [0, ], u() u(0) = = u (s) = x (x + se j, g(x + e j ), were s [0, 1]. For te second term on te rigt we consider te function v(t) = f(x + e j, tg(x + e j ) + (1 t)g(x)) t [0, 1]. Te function is well defined and by te mean-value teorem it satisfies wic, by te cain rule implies v(1) v(0) = v (r), r [0, 1], f(x, g(x+e j )) f(x, g(x)) = y (x, rg(x+e j)+(1 r)g(x))(g(x+e j ) g(x)), for some r [0, 1], and tus f(x, g(x + e j )) f(x, g(x)) = y (x, rg(x+e j)+(1 r)g(x)) g(x + e j) g(x).

11 THE IMPLICIT FUNCTION THEOREM 11 If we replace tese in (3.4) we obtain (3.5) (3.6) 0 = f(x + e j, g(x + e j )) f(x, g(x)) = x (x + se j, g(x + e j ) + y (x, rg(x + e j) + (1 r)g(x)) g(x + e j) g(x), for some s [0, ], r [0, 1], wic may depend of (and x). Now use te continuity of g, to conclude tat uniformly on [0, ], and, x y lim 0 x (x + se j, g(x + e j )) = (x, g(x)), x lim 0 y (x, rg(x+e j)+(1 r)g(x)) g(x + e j) g(x) uniformly on [0, 1]. Tus must exist for all x U. lim 0 g(x + e j ) g(x), = (x, g(x)) 0, x U, y Tus we ave verified te claim (*) for m = 1. As pointed out at te beginning, we prove te general case by induction. Assume (*) olds for some positive integer m. We want to prove tat te claim olds for m + 1 as well. Assume tat f 1,..., f m+1 ave continuous partial derivatives in a neigborood V of p 0 = (x 0 1,..., x 0 n, y 0 1,..., y 0 m+1) R n+m+1 and satisfy: Note tat f j (x 0 1,..., x 0 n, y1, 0... ym+1) 0 = 0, 1 j m + 1, (f 1,..., f m+1 ) (y 1,..., y m+1 ) 0. p0 (f 1,..., f l 1, f l+1,... f m+1 ) (y 1,..., y k 1, y k+1,..., y m+1 ) is obtained from (f 1,...,f m+1 ) (y 1,...,y m+1 by deleting te l-t row and te k-t column, i.e. it is a minor of order m of tis determinant. Ten at ) least

12 12 ALEXANDRU ALEMAN one of tese minors must be non-zero at p 0, oterwise (f 1,..., f m+1 ) (y 1,..., y m+1 ) = 0. p0 Since te functions and variables can be permuted, tere is no arm if we assume tat (f 2,..., f m+1 ) (y 2,..., y m+1 ) 0. p0 Ten we can treat y 1 as an x-variable, and look at te last m equations, in order to arrive at te conclusion tat f 2,..., f m+1 ave continuous partial derivatives in a neigborood V of p 0 = (x 0 1,..., x 0 n, y 0 1,..., y 0 m+1) R n+1+m and satisfy: f j (x 0 1,..., x 0 n, y1, 0... ym+1) 0 = 0, 2 j m + 1, (f 2,..., f m+1 ) (y 2,..., y m+1 ) 0. p0 But ten te induction ypotesis implies tat tere exists a neigborood U of x 0 = (x 0 1,..., x 0 n, y 0 1) R n+1 and uniquely determined functions, g 2,..., g m+1 : U R wose partial derivatives exist in U, suc tat g k (x 0 1,..., x 0 n, y 1 ) = y 0 k, 2 k m + 1, (x 1,..., x n, y 1, g 2 (x, y 1 ),..., g m+1 (x, y 1 )) V, (x, y 1 ) = (x 1,... x n, y 1 ) U and for 2 j m + 1, f j (x 1,..., x n, y 1, g 2 (x, y 1 ),..., g m+1 (x, y 1 )) = 0, (x, y 1 ) = (x 1,... x n, y 1 ) U. Recall from te beginning of te proof tat in tis case te partial derivatives of g 2,..., g m+1 are continuous in U and replace tis in te first equation to obtain f 2 (x 1,..., x n, y 1, g 2 (x, y 1 ),..., g m+1 (x, y 1 )) = 0, (x, y 1 ) = (x 1,... x n, y 1 ) U, tat is, we arrived at te case wen m = 1. Since tis was solved above, it follows tat tere exists a neigborood U of x 0 and a function g 1 : U R wose partial derivatives exist in U, suc tat g(x 0 ) = y1, 0 (x 0, g 1 (x)) U, and f k (x 1,..., x n, g 1 (x) 1, g 2 (x, g 1 (x)),..., g m+1 (x, g 1 )(x)) = 0, x = (x 1,... x n ) U. By te cain rule, te partial derivatives of, g k (x, g 1 (x)) exist in U and we are done.

13 THE IMPLICIT FUNCTION THEOREM An Application Let us return to te situation considered in Exercise 2.3. We ave seen tat te inverse of a bijective differentiable function is not necessarily differentiable. It is of great importance to find (simple) additional conditions wic prevent tis. Wit elp of te Implicit Function Teorem tis becomes an easy task. However, te result below is actually equivalent to Teorem 1.1. Teorem 4.1. Let U R n be open and let f : U R n be injective and ave continuous partial derivatives on U. on U. If f = (f 1,..., f n ) satisfies (f 1,..., f n ) (x) 0, x U, (x 1,..., x n ) ten f(u) is open and f 1 : f(u) U as continuous partial derivatives on f(u). Proof. Let y 0 f(u), say y 0 = f(x 0 ), and consider again te function g : R n U R n defined by g(x, y) = f(x) y, as in te solution of Exercise 2.3. If we write g = (g 1,..., g n ), te equality g(x, y) = 0 is equivalent to g j (x 1,..., x n, y 1,..., y m ) = 0, 1 j n. We also ave tat g(x 0, y 0 ) = 0, and (g 1,..., g n ) (x 1,..., x n ) 0. (x 0,y 0 ) By Teorem 1.1 it follows tat tere exists a neigborood W of y 0 and a function : W U wit continuous partial derivatives on W, suc tat g(y, (y)) = 0, y W. Equivalently, f((y)) = y, y W. In particular, since y 0 = f(u) was arbitrary, it follows tat f(u) is open (any point in f(u) as a neigborood W contained in f(u)). We obviously ave also f(f 1 (y)) = y, y W, f 1 (y) = (y), for all y W, because f is injective. But ten f 1 must ave continuous partial derivatives in W, ence in all of f(u).

14 14 ALEXANDRU ALEMAN 5. Preparation for te written exam- set 1 Exercise 1. Consider te function f : R 4 R defined by f(x, y, z, u) = u 3 x 3x 2 y 3zy z 4. Sow tat if (x 0.y 0, z 0, u 0 ) {(x, y, z, u) : f(x, y, z, u) = 0, xu 0}, tere exists a neigborood U R 3 of (x 0, y 0, z 0 ) and a function g : U R wit continuous partial derivatives suc tat f(x, y, z, g(x, y, z)) = 0, (x, y, z) U. Exercise 2. Find te maximum and minimum values of 2xy + e z on te set {(x, y, z) : x 2 + 4y 2 + e z = 1, 1 z 0}. Exercise 3. Sow tat te curve C given by r(t) = 2 cos t sin ti + 2 cos 2 tj + 2 sin tk, 0 t π lies on a spere centered at te origin. Find C zds. Exercise 4. Sow tat te field F (x, y, z) = yzi + xzj + xyk is conservative and find a potential (find φ wit φ = F ). Find te field lines and te equipotential surfaces of F. Exercise 5. Let r = xi+yj+zk, and r = r 2. Sow tat if f : R R is twice differentiable on R ten 2 f(r) = 4rf (r) + 6f (r).

15 THE IMPLICIT FUNCTION THEOREM 15 Recall tat 2 is te Laplacian, 2 u = 2 u + 2 u + 2 u. x 2 y 2 z 2 Exercise 6. Use a line integral to compute te area enclosed by te curve 4x 2/3 + 9y 2/3 = 16. Exercise 7. Use a surface integral to compute te volume of te ellipsoid {(x, y, z) : 4x 2 + 9y 2 + z 2 1, }. Exercise 8. Consider te curve C given by r(t) = (1 + sin t)i + (1 + cos t)j+(3 cos t sin t)k in R 3. Sow it lies in te plane {(x, y, z) : x+ y + z = 5}, and it encloses a surface wose projection on te xy-plane is a disc. Use Stokes s teorem to evaluate were F (x, y, z) = ye x i + e x j + z 2 k. C F dr, 6. Hints 1. Note tat on te given set we ave u function teorem. 0. Apply te implicit 2. Note tat te gradient of te Lagrange function as te form (2y, 2x, e z ) λ(2x, 2y, e z ) so te critical point occurs wen λ = 1, and x = y = 0, wic forces z = 0. Tus te extrema can only occur on te boundary wen z = 0 or z = 1. If z = 0, x = y = 0 and 2xy + e z = 1. If z = 1, x2 + y 2 = 1 e 1 and we can consider te maximum or minimum of

16 16 ALEXANDRU ALEMAN 2xy + e 1 wen x 2 + y 2 = 1 e 1. Anoter application of te Lagrange metod gives tat tese are ± (1 e 1 )/2 3. We ave Terefore ds = 2dt and r 2 = 4 cos 2 t sin 2 t + +4 cos 4 t + 4 sin 2 t = 4. C π zds = 4 sin tdt = If φ(x, y, z) = xyz ten φ = F. Te equipotential surfaces are ten given by {(x, y, z) : φ(x, y, z) = c} = {(x, y, z) : xyz = c}. Te field lines are given by te differential equations so tat dx yz = dy xz = dz xy, xdx = ydy = zdz, x 2 = y 2 + C 1, y 2 = z 2 + C We ave f(r) = f(x 2 + y 2 + z 2 ) = u(x, y, z), ence by te cain and product rule u x = f 2 u (r)2x, x = f (r)4x 2 + 2f (r). 2 Since tese relations are symmetric in x, y, z te result follows. 6. If C denotes te given curve ten te area enclosed by it can be computed wit Green s teorem and it is given by xdy. To compute C tis integral we can use te parametrization r = 2 cos 3 ti+2 sin 3 tj, 0 t 2π and obtain 2π xdy = 12 cos 4 t sin 2 tdt. C 0

17 THE IMPLICIT FUNCTION THEOREM By te divergence teorem and symmetry, te volume is given by div(xi + yj + zk)dv = 2 (xi + yj + zk) ˆNdS, were S = {(x, y, z) : 4x 2 + 9y 2 + z 2 = 1,, z 0}. Te surface as a projection on te xy-plane. S 8. It is easy to see tat te curve lies in te given plane. Note tat t (1+sin t, 1+cos t) parametrizes te te circle of radius one centered at (1, 1) in te xy-plane, so tat te function G(x, y, z) = (x, y, 5 x y) maps tis circle onto C. Te surface enclosed is te image by G of te disc D of radius one centered at (1, 1) in te xy-plane. Stokes s teorem gives F dr = curlf ˆNdS, C S were S is described by z = 5 x y, x, y D. Te outer unit normal is te same as te one of te plane x + y + z = 5, tat is ˆN = (1, 1, 1), and Finally, ds = 3. curlf = e x k, curlf ˆN = e x. 7. Preparation for te written exam- set 2 Exercise 1. Consider te function f : R 2 R defined by Compute f(x, y) = e x + x 2 y + y + y 3. x g(x, y) = Sow tat tere exists a solution u of te differential equation y u (x) = ex + 2xu(x) 1 + x 2 + 3u 2 (x).

18 18 ALEXANDRU ALEMAN defined on an interval ( a, a), a > 0 wit u(0) = 0. Exercise 2. Find te minimum value of x 2 +y 2 +z 2 wen x+y+z = 1. Exercise 3. If te curve C is given by r(t) = t 2 cos ti+t 2 sin tj+tk, 0 t π, evaluate C (x2 + y 2 )ds. Exercise 4. Sow tat te field F (x, y, z) = 2x z i + 2y z j + (3 x2 + y 2 z 2 ) k, is conservative and find a potential (find φ wit φ = F ). Find te field lines and te equipotential surfaces of F. Exercise 5. Let r = xi + yj + zk, and r = r 2. Let f, g : R R be differentiable on R. Sow tat te field F (x, y, z) = f(r)i + g(r)j cannot be conservative unless f, g are constant on [0, ). Exercise 6. Use a line integral to compute te area enclosed by te curve r = (e cos t + sin t)i + cos tj, 0 t 2π. Exercise 7. Use a surface integral to compute te volume of te cone {(x, y, z) : x 2 + 4y 2 = z, 0 z 1}. Exercise 8. For F (x, y, z) = yi + x 2 j + zk, use Stokes s teorem to compute te circulation of tis field around te oriented boundary of {(x, y, z) : z = 16 (x 1) 2 y 2, z 0} were te normal points outwards.

19 THE IMPLICIT FUNCTION THEOREM Hints 1. We ave x g(x, y) = y = ex + 2xy 1 + x 2 + 3y 2, in particular, we see tat > 0 on y R2. Since f(0, 0) = 1, we can apply te implicit function teorem to find a continuously differentiable function u defined on a neigborood of 0 suc tat f(x, u(x)) = 1, u(0) = 0 and u x (x) = y (x, u(x)). 2. Te gradient of te Lagrange function is (2x, 2y, 2z, 0) λ(1, 1, 1, x+ y + z 1) and it vanises wen 2x = 2y = 2z = λ. Ten λ = 2 3, and x 2 + y 2 + z 2 = 1 3. Since for x = 1, y = z = 0, we ave x z 2 = 1 tis must be te minimum. 3. We ave ds = (2t cos t t 2 sin t) 2 + (2t sin t + t 2 cos t) 2 + t 2 = 5t 2 + t 4, and on C x 2 + y 2 = t 4. Terefore (x 2 + y 2 )ds = π C 0 Integrate by parts. t 4 5t 2 + t 2 dt = 1 2 π 2 0 x xdx. 4. Te potential is given by φ(x, y, z) = 3z + x2 + y 2, z and terefore te equipotential surfaces are given by {(x, y, z) : 3z 2 + x 2 + y 2 = Cz}, C R.

20 20 ALEXANDRU ALEMAN Te field lines are given by te differential equations zdx 2x = zdy 2y = z 2 dz 3z x 2 y 2. From te first equation we get x = Cy, and te second becomes i.e. zdz 3z (C 2 + 1)y 2 = dy y, z 3 + C2 + 1 ln 3z (C 2 + 1)y 2 = ln y We ave F (x, y, z) = f(x 2 + y 2 + z 2 )i + g(x 2 + y 2 + z 2 )j + 0k = F 1 i + F 2 j + F 3 k. If F is conservative we must ave F 1 y = frac F 2 x, frac F 1 z = frac F 3 x, frac F 2 z = frac F 3 y. Tis gives 2yf (x 2 +y 2 +z 2 ) = 2xg (x 2 +y 2 +z 2 ), 2zf (x 2 +y 2 +z 2 ) = 0, 2zg (x 2 +y 2 +z 2 ) = 0, i.e. f (x 2 +y 2 +z 2 ) = g (x 2 +y 2 +z 2 ) = 0, wenever z 0. Since every t (0, ) can be written as t = x 2 + y 2 + z 2 for some x, y, z R wit t z 0 (for example z =, x = 0, y = t,) it follows tat f, g are 2 2 constant on (0, ) and te desired result follows from te continuity of f, g. 6. By Green s teorem, te area is given by C xdy = 2π 0 (e cos t + sin t)( sin t)dt = 0 + 2π 0 1 cos(2t) dt = π. 2

21 THE IMPLICIT FUNCTION THEOREM I ll take tis out of te collection since te calculations are too tideous. Sorry! 8. Tis is te surface of a cone and its boundary is te circle C = {(x 1) 2 + y 2 = 16} in te xy-plane. Stokes s teorem gives F dr = curlf ˆNdS, C S for any surface S wit tis boundary. Te most convenient one is te disc D = {(x 1) 2 + y 2 16} in te xy-plane. Terefore taking also into account te orientation, F dr = (2x + 1)k kdxdy = π. C D