Scalar functions of several variables (Sect. 14.1).

Transcription

1 Scalar functions of several variables (Sect. 14.1). Functions of several variables. On open, closed sets. Functions of two variables: Graph of the function. Level curves, contour curves. Functions of three variables. Level surfaces. z f(,) Scalar functions of several variables. Definition A scalar function of n variables is a function f : D R n R R, where n N, the set D is called the domain of the function, and the set R is called the range of the function. Remark: Comparison between f : R 2 R with r : R R 2. A scalar function of two variables is a function f : R 2 R (, ) f (, ). A vector function on the plane is a function r : R R 2 t r(t) = (t), (t).

2 Functions of several variables. An eample of a scalar-valued function of two variables, T : R 2 R is the temperature T of a plane surface, sa a table. Each point (, ) on the table is associated with a number, its temperature T (, ). An eample of a scalar-valued function of three variables, T : R 3 R is the temperature T of this room. Each point (,, z) in the room is associated with a number, its temperature T (,, z). Another eample of a scalar function of three variables is the speed of the air in the room. An eample of a vector-valued function of three variables, v : R 3 R 3, is the velocit of the air in the room. Scalar functions of several variables. Find the maimum domain D and range R sets where the function f : D R 2 R R given b f (, ) = is defined. Solution: The function f (, ) = is defined for all points (, ) R 2, therefore, D = R 2. The values of the function f are non-negative, that is, f (, ) = for all (, ) D. Therefore, the range space is R = [0, ).

3 Scalar functions of several variables. Find the maimum domain D and range R sets where the function f : D R 2 R R given b f (, ) = is defined. Solution: The function f (, ) = is defined for points (, ) R 2 such that 0. Therefore, D={(,) : > } D = {(, ) R 2 : }. The values of the function f are non-negative, that is, f (, ) = 0 for all (, ) D. Therefore, the range space is R = [0, ). = Scalar functions of several variables. Find the maimum domain D and range R sets where the function f : D R 2 R R given b f (, ) = 1/ is defined. Solution: The function f (, ) = 1/ is defined for points (, ) R 2 such that > 0. Therefore, D={(,) : > } D = {(, ) R 2 : > }. The values of the function f are positive, that is, f (, ) = 1/ > 0 for all (, ) D. Therefore, the range space is R = (0, ). =

4 Scalar functions of several variables (Sect. 14.1). Functions of several variables. On open, closed sets. Functions of two variables: Graph of the function. Level curves, contour curves. Functions of three variables. Level surfaces. On open and closed sets in R n. We first generalize from R 3 to R n the definition of a ball of radius r centered at ˆP. Definition A set B r (ˆP) R n, with n N and r > 0, is a ball of radius r centered at ˆP = (ˆ 1,, ˆ n ) iff B r (ˆP) = {( 1,, n ) R n : ( 1 ˆ 1 ) ( n ˆ n ) 2 < r 2 }. Remark: A ball B r (ˆP) contains the points inside a sphere of radius r centered at ˆP without the points of the sphere.

5 On open and closed sets in R n. Definition A point P S R n, with n N, is called an interior point iff there is a ball B r (P) S. A point P S R n, with n N, is called a boundar point iff ever ball B r (P) contains points in S and points outside S. The boundar of a set S is the set of all boundar points of S. R 2 Boundar point S Interior point Boundar On open and closed sets in R n. Definition A set S R n, with n N, is called open iff ever point in S is an interior point. The set S is called closed iff S contains its boundar. A set S is called bounded iff S is contained in ball, otherwise S is called unbounded open and bounded bounded closed and unbounded closed and bounded

6 On open and closed sets in R n. Find and describe the maimum domain of the function f (, ) = ln( 2 ). Solution: The maimum domain of f is the set D = {(, ) R 2 : > 2 }. D D is an open, unbounded set. Scalar functions of several variables (Sect. 14.1). Functions of several variables. On open, closed sets. Functions of two variables: Graph of the function. Level curves, contour curves. Functions of three variables. Level surfaces.

7 The graph of a function of two variables is a surface in R 3. Definition The graph of a function f : D R 2 R is the set of all points (,, z) in R 3 of the form (,, f (, )). The graph of a function f is also called the surface z = f (, ). Draw the graph of f (, ) = Solution: The graph of f is the surface z = This is a paraboloid along the z ais. z f(,) = Level curves, contour curves. Definition The level curves of a function f : D R 2 R R are the curves in the domain D R 2 of f solutions of the equation f (, ) = k, where k R is a constant in the range of f. The contour curves of function f are the curves in R 3 given b the intersection of the graph of f with horizontal planes z = k, where k R is a constant in the range of f. Curves of constant f in D R 2 are called level curves. Curves of constant f in R 3 are called contour curves.

8 Level curves, contour curves. Find and draw few level curves and contour curves for the function f (, ) = Solution: z f(,) = The level curves are solutions of the equation = k with k 0. These curves are circles of radius k in D = R 2. contour curves The contour curves are the circles {(,, z) : = k, z = k}. D level curves Level curves, contour curves. Find the maimum domain, range of, and graph the function 1 f (, ) = Solution: Since the denominator never vanishes, hence D = R 2. 1 Since 0 < , the + 2 range of f is R = (0, 1]. z = f (, ) 1 The contour curves are circles on horizontal planes in (0, 1].

9 Level curves, contour curves. N Given the topographic map in the figure, which wa do ou choose to the summit? W S E 1000 z Solution: From the east Scalar functions of several variables (Sect. 14.1). Functions of several variables. On open, closed sets. Functions of two variables: Graph of the function. Level curves, contour curves. Functions of three variables. Level surfaces.

10 Scalar functions of three variables. Definition The graph of a scalar function of three variables, f : D R 3 R R, is the set of points in R 4 of the form (,, z, f (,, z)) for ever (,, z) D. Remark: The graph a function f : D R 3 R requires four space dimensions. We cannot picture such graph. Definition The level surfaces of a function f : D R 3 R R are the surfaces in the domain D R 3 of f solutions of the equation f (,, z) = k, where k R is a constant in the range of f. Scalar functions of three variables. Draw one level surface of the function f : D R 3 R R 1 f (,, z) = z 2. Solution: The domain of f is D = R 3 and its range is R = (0, ). Writing k = 1/R 2, the level surfaces f (,, z) = k are spheres z 2 = R 2. z R

11 Limits and continuit for f : R n R (Sect. 14.2). The limit of functions f : R n R. : Computing a limit b the definition. Properties of limits of functions. s: Computing limits of simple functions. Continuous functions f : R n R. Computing limits of non-continuous functions: Two-path test for the non-eistence of limits. The sandwich test for the eistence of limits. The limit of functions of several variables. Definition The function f : D R n R, with n N, has the number L R as limit at the point ˆP R n, denoted as lim P ˆP f (P) = L, iff the following holds: For ever number ɛ > 0 there eists a number δ > 0 such that if P ˆP < δ then f (P) L < ɛ. Remarks: In Cartesian coordinates P = ( 1,, n ), ˆP = (ˆ 1,, ˆ n ). Then, P ˆP is the distance between points in R n, P ˆP = P ˆP = ( 1 ˆ 1 ) ( n ˆ n ) 2. f (P) L R is the absolute value of real numbers.

12 The limit of functions f : R 2 R. The function with values f (, ) has the number L as limit at the point P 0 = ( 0, 0 ) iff holds: For all points P = (, ) near P 0 = ( 0, 0 ) the value of f (, ) differs little from L. z We denote it as follows: lim f (, ) = L (,) ( 0, 0 ) L f(,) (, ) 0 0 Limits and continuit for f : R n R (Sect. 14.2). The limit of functions f : R n R. : Computing a limit b the definition. Properties of limits of functions. s: Computing limits of simple functions. Continuous functions f : R n R. Computing limits of non-continuous functions: Two-path test for the non-eistence of limits. The sandwich test for the eistence of limits.

13 Computing limits b definition usuall is not eas. Use the definition of limit to compute lim (,) (0,0) Solution: The function above is not defined at (0, 0). First: Guess what the limit L is. Along the line = 0 the function above vanishes for all 0. So, if L eists, it must be L = 0. Fi an number ɛ > 0. Given that ɛ, find a number δ > 0 such that ( 0) 2 + ( 0) 2 < δ < ɛ. Use the definition of limit to compute lim (,) (0,0) Solution: Given an ɛ > 0, find a number δ > 0 such that < δ < ɛ Recall: , that is, Then = = Choose δ = ɛ/2. If < δ, then We conclude that L = < 2δ = ɛ

14 Limits and continuit for f : R n R (Sect. 14.2). The limit of functions f : R n R. : Computing a limit b the definition. Properties of limits of functions. s: Computing limits of simple functions. Continuous functions f : R n R. Computing limits of non-continuous functions: Two-path test for the non-eistence of limits. The sandwich test for the eistence of limits. Properties of limits of functions. Theorem If f, g : D R n R, with n N, and lim P ˆP f (P) = L, lim P ˆP g(p) = M, then the following statements hold: 1. lim P ˆP f (P) ± g(p) = L ± M; 2. If k R, then lim P ˆP kf (P) = kl; 3. lim P ˆP f (P) g(p) = LM; ( f (P) ) 4. If M 0, then lim = L P ˆP g(p) M. 5. If k Z and s N, then lim P ˆP [ f (P) ] r/s = L r/s. Remark: The Theorem above implies that: If f : D R n R is a rational function f = R/S, (quotient of two polnomials), with S(ˆP) 0, then lim P ˆP f (P) = f (ˆP).

15 Limits and continuit for f : R n R (Sect. 14.2). The limit of functions f : R n R. : Computing a limit b the definition. Properties of limits of functions. s: Computing limits of simple functions. Continuous functions f : R n R. Computing limits of non-continuous functions: Two-path test for the non-eistence of limits. The sandwich test for the eistence of limits. Limits of R/S at ˆP where S(ˆP) 0 are simple to find. Compute lim. (,) (1,2) Solution: The function above is a rational function in and and its denominator does not vanish at (1, 2). Therefore lim = 1 + 2(2) 1, (,) (1,2) 1 2 that is, lim = 4. (,) (1,2)

16 Limits and continuit for f : R n R (Sect. 14.2). The limit of functions f : R n R. : Computing a limit b the definition. Properties of limits of functions. s: Computing limits of simple functions. Continuous functions f : R n R. Computing limits of non-continuous functions: Two-path test for the non-eistence of limits. The sandwich test for the eistence of limits. Continuous functions f : R n R. Definition A function f : D R n R, with n N, is called continuous at ˆP D iff holds lim P ˆP f (P) = f (ˆP). Remarks: The definition above sas: (a) f (ˆP) is defined; (b) lim P ˆP f (P) = L eists; (c) L = f (ˆP). A function f : D R n R is continuous iff f is continuous at ever point in D. Continuous functions have graphs without holes or jumps.

17 Continuous functions f : R 2 R. Polnomial functions are continuous in R n. For eample, P 2 (, ) = a 0 + b 1 + b 2 + c c 2 + c 3 2. Rational functions f = R/S are continuous on their domain. For eample, f (, ) = , with ±. Composition of continuous functions are continuous. For eample, f (, ) = cos( ). Continuous functions f : R 2 R. Compute lim (,) ( π,0) cos( ). Solution: The function f (, ) = cos( ) is continuous for all (, ) R 2. Therefore, lim (,) ( cos( ) = cos(π + 0), π,0) that is, lim (,) ( cos( ) = 1. π,0)

18 Limits and continuit for f : R n R (Sect. 14.2). The limit of functions f : R n R. : Computing a limit b the definition. Properties of limits of functions. s: Computing limits of simple functions. Continuous functions f : R n R. Computing limits of non-continuous functions: Two-path test for the non-eistence of limits. The sandwich test for the eistence of limits. Two-path test for the non-eistence of limits. Theorem If a function f : D R n R, with n N, has two different limits along to different paths as P approaches ˆP, then lim P ˆP f (P) does not eist. Remark: Consider the case f : D R 2 R: If f (, ) L 1 along a path C 1 as (, ) ( 0, 0 ), f (, ) L 2 along a path C 2 as (, ) ( 0, 0 ), L 1 L 2, then lim f (, ) does not eist. (,) ( 0, 0 ) When side limits do not agree, the limit does not eist.

19 Two-path test for the non-eistence of limits. When side limits do not agree, the limit does not eist. f (, ) = z 1 f ( ) = C 2 C 1 C 2 C 1 C 3 Two-path test for the non-eistence of limits. Compute lim (,) (0,0) Solution: f (, ) = (3 2 )/( ) is not continuous at (0, 0). We tr to show that the limit above does not eist. If path C 1 is the -ais, ( = 0), then, f (, 0) = 3 2 = 3, lim 2 (,0) (0,0) If path C 2 is the -ais, ( = 0), then, f (, 0) = 3. f (0, ) = 0, lim f (0, ) = 0. (0,) (0,0) Therefore, lim (,) (0,0) does not eist

20 Two-path test for the non-eistence of limits. Remark: In the eample above one could compute the limit for arbitrar lines, that is, C m given b = m, with m a constant. That is, 3 2 f (, m) = 2 + 2m 2 2 = m 2. The limits along these paths are: lim f (, m) = 3 (,m) (0,0) 1 + 2m 2 which are different for each value of m. This agrees what we concluded: lim (,) (0,0) does not eist The sandwich test for the eistence of limits. Theorem If functions f, g, h : D R n R, with n N, satisf: (a) g(p) f (P) h(p) for all P near ˆP D; (b) lim P ˆP g(p) = L = lim P ˆP h(p); then lim P ˆP f (P) = L. h, g : R R h L g P = 0

21 The sandwich test for the eistence of limits. Compute lim (,) (0,0) Solution: f (, ) = 2 2 is not continuous at (0, 0). + 2 The Two-Path Theorem does not prove non-eistence of the limit. Consider paths C m given b = m, with m R. Then f (, m) = 2 m 2 + m 2 2 = m 1 + m 2, which implies lim f (, m) = 0, m R. (,m) (0,0) We cannot conclude that the limit does not eist. We cannot conclude that the limit eists. The sandwich test for the eistence of limits. Compute lim (,) (0,0) Solution: Notice: 2 1, for all (, ) (0, 0) So,, for all (, ) (0, 0). Hence, Since lim 0 = 0, the Sandwich Theorem with g =, h =, implies 2 lim (,) (0,0) = 0.

22 Partial derivatives and differentiabilit (Sect. 14.3). Partial derivatives of f : D R 2 R. Higher-order partial derivatives. The Mied Derivative Theorem. s of implicit partial differentiation. Partial derivatives of f : D R n R. Net class: Partial derivatives and continuit. Differentiable functions f : D R 2 R. Differentiabilit and continuit. A primer on differential equations. Partial derivatives of f : D R 2 R. Definition Given a function f : D R 2 R, the partial derivative of f (, ) with respect to at a point (, ) D is given b 1 [ ] f (, ) = lim f ( + h, ) f (, ). h 0 h The partial derivative of f (, ) with respect to at a point (, ) D is given b 1 [ ] f (, ) = lim f (, + h) f (, ). h 0 h Remark: To compute f (, ) derivate f (, ) keeping constant. To compute f (, ) derivate f (, ) keeping constant.

23 Computing f (, ) at ( 0, 0 ). Evaluate the function f at = 0. The result is a single variable function f (, 0 ). Compute the derivative of f (, 0 ) and evaluate it at = 0. The result is f ( 0, 0 ). Find f (1, 3) for f (, ) = /4. Solution: f (, 3) = 2 + 9/4; f (, 3) = 2; f (1, 3) = 2. To compute f (, ) derivate f (, ) keeping constant. Computing f (, ) at ( 0, 0 ). Evaluate the function f at = 0. The result is a single variable function f ( 0, ). Compute the derivative of f ( 0, ) and evaluate it at = 0. The result is f ( 0, 0 ). Find f (1, 3) for f (, ) = /4. Solution: f (1, ) = /4; f (1, ) = /2; f (1, 3) = 3/2. To compute f (, ) derivate f (, ) keeping constant.

24 Geometrical meaning of partial derivatives. f ( 0, 0 ) is the slope of the line tangent to the graph of f (, ) containing the point ( 0, 0, f ( 0, 0 ) ) and belonging to a plane parallel to the z-plane. z f (, ) f (, ) 0 f (, ) 0 0 f (, ) 0 0 f (, ) ( 0, ) f ( 0, 0 ) is the slope of the line tangent to the graph of f (, ) containing the point ( 0, 0, f ( 0, 0 ) ) and belonging to a plane parallel to the z-plane. Partial derivatives can be computed on an point in D. Find the partial derivatives of f (, ) = Solution: f (, ) = 2( + 2) (2 ) ( + 2) 2 f (, ) = 5 ( + 2) 2. f (, ) = ( 1)( + 2) (2 )(2) 5 ( + 2) 2 f (, ) = ( + 2) 2.

25 The derivative of a function is a new function. Recall: The derivative of a function f : R R is itself a function. The derivative of function f () = 2 at an arbitrar point is the function f () = 2. = f ( ) = f ( ) The same statement is true for partial derivatives. The partial derivatives of a function are new functions. Definition Given a function f : D R 2 R R, the functions partial derivatives of f (, ) are denoted b f (, ) and f (, ), and the are given b the epressions 1 f (, ) = lim [f ( + h, ) f (, )], h 0 h 1 f (, ) = lim [f (, + h) f (, )]. h 0 h Notation: Partial derivatives of f (, ) are denoted in several was: f (, ), f (, ), f (, ). f (, ), f (, ), f (, )

26 The partial derivatives of a paraboloid are planes Find the functions partial derivatives of f (, ) = Solution: f (, ) = 2 f (, ) = 2. f (, ) = f (, ) = 2. The partial derivatives of a paraboloid are planes. z f(,) z f (,) f (,) z The partial derivatives of a function are new functions. Find the partial derivatives of f (, ) = 2 ln(). Solution: f (, ) = 2 ln(), f (, ) = 2. Find the partial derivatives of f (, ) = Solution: f (, ) = 2, f (, ) = 2.

27 Partial derivatives and differentiabilit (Sect. 14.3). Partial derivatives of f : D R 2 R. Higher-order partial derivatives. The Mied Derivative Theorem. s of implicit partial differentiation. Partial derivatives of f : D R n R. Higher-order partial derivatives. Remark: Higher derivatives of a function are partial derivatives of its partial derivatives. The second partial derivatives of f (, ) are the following: 1 f (, ) = lim h 0 h [f ( + h, ) f (, )], 1 f (, ) = lim h 0 h [f (, + h) f (, )], 1 f (, ) = lim h 0 h [f (, + h) f (, )], 1 f (, ) = lim h 0 h [f ( + h, ) f (, )]. Notation: f, 2 f 2, f, and also f, 2 f, f.

28 Higher-order partial derivatives. Find all second order derivatives of the function f (, ) = 3 e Solution: f (, ) = 3 2 e 2, f (, ) = 2 3 e f (, ) = 6e 2, f (, ) = 4 3 e 2. f = 6 2 e 2, f = 6 2 e 2. Partial derivatives and differentiabilit (Sect. 14.3). Partial derivatives of f : D R 2 R. Higher-order partial derivatives. The Mied Derivative Theorem. s of implicit partial differentiation. Partial derivatives of f : D R n R.

29 Higher-order partial derivatives sometimes commute. Theorem If the partial derivatives f, f, f and f of a function f : D R 2 R eist and all are continuous functions, then holds f = f. Find f and f for f (, ) = cos(). Solution: f = sin(), f = sin() cos(). f = sin(), f = sin() cos(). Partial derivatives and differentiabilit (Sect. 14.3). Partial derivatives of f : D R 2 R. Higher-order partial derivatives. The Mied Derivative Theorem. s of implicit partial differentiation. Partial derivatives of f : D R n R.

30 s of implicit partial differentiation. Remark: Implicit differentiation rules for partial derivatives are similar to those for functions of one variable. Find z(, ) of the function z defined implicitl b the equation z + e 2z/ + cos(z) = 0. Solution: z + ( z) + 2 ( z)e 2z/ ( z) sin(z) = 0. Compute z as a function of, and z(, ), as follows, ( z) [ + 2 e2z/ sin(z) ] = z, z that is, ( z) = [ + 2 e2z/ sin(z) ]. s of implicit partial differentiation. Remark: Implicit differentiation rules for partial derivatives are similar to those for functions of one variable. Find z(, ) of the function z defined implicitl b the equation z + e 2z/ + cos(z) = 0. Solution: z + ( z) + ( 2 ( z) 2 2 z ) e 2z/ ( z) sin(z) = 0. Compute z as a function of, and z(, ), as follows, that is, ( z) = ( z) [ + 2 e2z/ sin(z) ] = z z e2z/, [ z + 2 z e 2z/ ] [ e2z/ sin(z) ].

31 Partial derivatives and differentiabilit (Sect. 14.3). Partial derivatives of f : D R 2 R. Higher-order partial derivatives. The Mied Derivative Theorem. s of implicit partial differentiation. Partial derivatives of f : D R n R. Partial derivatives of f : D R n R. Definition Given a function f : D R n R, with n N, the partial derivative of f ( 1,, n ) with respect to i, with i = 1,, n, at a point ( 1,, n ) D is given b f i 1 [ = lim f (1,, i + h,, n ) f ( 1,, n ) ]. h 0 h Remark: To compute f i derivate f with respect to i keeping all other variables j constant. Notation: f i, f i, f i, i f, i f.

32 Partial derivatives of f : D R n R. Compute all first partial derivatives of the function 1 φ(,, z) = z. 2 Solution: φ = 1 2 ( z 2 ) 3/2 φ = ( z 2 ) 3/2. Analogousl, the other partial derivatives are given b φ = ( z 2 ) 3/2, φ z z = ( z 2 ) 3/2. Partial derivatives of f : D R n R. 1 Verif that φ(,, z) = satisfies the Laplace z2 equation : φ + φ + φ zz = 0. Solution: Recall: φ = / ( z 2 ) 3/2. Then, 1 φ = ( z 2 ) 3/ ( z 2 ) 5/2. Denote r = z 2, then φ = r 3 r 5 Analogousl, φ = , and φ r 3 r 5 zz = 1 + 3z2. Then, r 3 r 5 φ + φ + φ zz = 3 r 3 + 3( z 2 ) r 5 = 3 r 3 + 3r 2 r 5. We conclude that φ + φ + φ zz = 0.