13.5 DIRECTIONAL DERIVATIVES and the GRADIENT VECTOR

Transcription

1 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus DIRECTIONAL DERIVATIVES and te GRADIENT VECTOR Directional Derivatives In Section 13.3 te partial derivatives f x and f y were defined as f x (x,y) = f y (x,y) = f(x +,y) " f(x,y) f(x,y +) " f(x,y) (if te it exists and is finite) and (if te it exists and is finite). Tese partial derivative measured te instantaneous rate of cange of z = f(x,y) as we moved in te increasing x direction (wile olding y constant) and in te increasing y direction (wile olding x constant). Sometimes, owever, we are interested in te rate of cange of z = f(x,y) as we move in some oter direction, and tat leads to te idea of a directional derivative to measure te instantaneous rate of cange of z = f(x,y) as we move in any direction. Definition: Te directional derivative of z = f(x,y) in te direction of a unit vector u = a, b is D u f(x,y) = f(x +a,y +b) " f(x,y) (if te it exists and is finite). Figures 1a and 1b illustrate te slope of te line tangent to te curve were te plane above te vector u intersects te surface z = f(x,y). Using tis definition can be tedious and algebraically messy, but it is wort doing once. Fortunately we will soon see a muc more efficient metod.

2 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus 2 Example 1: Use tis definition to calculate D u f( 1, 2) for f(x,y) = 1 + xy and u = 0.6, 0.8 Solution: D u f( 1, 2) = f(1 +0.6,2 +0.8) " f(1,2) = [1 +(1 +0.6)(2 +0.8)] " 3 [ (0.6) +0.8] " 3 = 2(0.6) +0.8 = 2 Practice 1: Calculate D u f( 2, 1) for f(x,y) = x 2 + 2x + 3y +1 and u = 5 13, For more complicated functions it becomes extremely difficult to use te definition to calculate directional derivatives. Fortunately tere is a muc easier way given by te next teorem. Teorem: If ten f(x,y) is differentiable, f as a directional derivative in te direction of every unit vector u = a, b, and D u f(x,y) = f x (x,y) a + f y (x,y) b. A proof of tis result uses a multivariable "Cain Rule" tat we will discuss in Section Example 2: Verify tat tis teorem gives te same answer for D u f(x,y) as our use of te directional derivative definition in Example 1. Solution: f x (x,y)=y and f y (x,y)=x so f x ( 1, 2)= 2 and f y ( 1, 2)= 1. u = 0.6, 0.8 so D u f( 1, 2) = (2)(0.6) + (1)(0.8) = 2, te same result as in Example 1. Practice 2: Verify tat tis teorem gives te same answer for D u f(x,y) as your use of te directional derivative definition in Practice 1. Example 3: Calculate te directional derivatives of z = f(x,y) = x + 5x 2 y 3 at te point (2,1) in te directions of te unit vectors (a) u = 0.6, 0.8, (b) u = 0.6, 0.8, (c) u = 0.8, 0.6, (d) u = i = 1, 0, and (e) u =j = 0, 1. Solution: f x (x,y) = xy 3 and f y (x,y) = 15x 2 y 2 so f x (2,1) = 21 and f y (2,1) = 60. Ten, by te previous Teorem, (a) for u = 0.6, 0.8, D u f(2,1) = f x (2,1) a + f y (2,1) b = (21)(0.6) + (60)(0.8) = 60.6.

3 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus 3 (b) for u = 0.6, 0.8, D u f(2,1) = (21)( 0.6) + (60)( 0.8) = (c) for u = 0.8, 0.6, D u f(2,1) = (21)(0.8) + (60)(0.6) = (d) for u = i = 1, 0, D u f(2,1) = (21)(1) + (60)(0) = 21 ( = f x (2,1) ). (e) for u = j = 0, 1, D u f(2,1) = (21)(0) + (60)(1) = 60 ( = f y (2,1) ). Te Gradient Vector You migt ave noticed tat te pattern D u f(x,y) = f x (x,y) a + f y (x,y) b for calculating te directional derivative can be viewed as te dot product of te vector f x (x,y), f y (x,y) wit te unit direction vector u = a, b. Tis vector f x (x,y), f y (x,y) sows up in a variety of contexts and is called te gradient of f. Definition of te Gradient Vector: Te gradient vector of f(x,y) is f(x,y) = f x (x,y), f y (x,y) = f x i + f y j. Te symbol " f " is read as "grad f" or "del f." Example 4: Calculate f(x,y) and f(0,1) for (a) f(x,y) = x + 5x 2 y 3 and (b) f(x,y) = y 2 + sin(x). Solution: (a) For f(x,y) = x + 5x 2 y 3, f x (x,y) = xy 3 and f y (x,y) = 15x 2 y 2 f(x,y) = xy 3, 15x 2 y 2 and f(0,1) = 1, 0. (b) For f(x,y) = y 2 + sin(x), f x (x,y) = cos(x) and f y (x,y) = 2y so f(x,y) = cos(x), 2y and f(0,1) = 1, 2. so Practice 3: Calculate f(x,y) and f(1,2) for (a) f (x, y) = e xy + 2x 3 y + y 2 and (b) f(x,y) = cos(2x+3y). Te directional derivative can now be written simply as te dot product of te gradient and te unit direction vector. D u f(x,y) = f(x,y). u Te gradient vector f(x,y) is useful for muc more tan a compact notation for directional derivatives. f(x,y) as a number of special features tat make it useful for investigating te beavior of surfaces.

4 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus 4 Tree very important properties of te gradient vector f(x,y) : (1) At a point (x,y), te maximum value of te directional derivative D u f(x,y) is f(x,y). (2) At a point (x,y), te maximum value of D u f(x,y) occurs wen u as te same direction as te gradient vector f(x,y). (At eac point (x,y), te gradient vector f(x,y) "points" in te direction of maximum increase for f(x,y).) (3) At a point (x,y), te gradient vector f(x,y) is normal (perpendicular) to te level curve tat goes troug te point (x,y). One of te beauties of matematics is tat sometimes a result like te powerful and non-obvious properties of te gradient can be proven in rater simple ways. Proof of (1) and (2): D u f(x,y) = f(x,y). u = f(x,y) u cos( θ ) θ is te angle between te vectors f(x,y) and u = f(x,y) cos( θ ) Te maximum value of cos( θ ) is 1 (wen θ = 0), so te maximum value of D u f(x,y) is f(x,y) and tis maximum occurs wen θ = 0, wen f(x,y) and u ave te same direction. Proof of (3): f is constant along te level curve at te point (x,y) so D u f(x,y) = 0 wen u is te direction of te level curve. But D u f(x,y) = f(x,y). u so f(x,y) is perpendicular to u, te direction of te level curve. Example 5: Find (a) te maximum rate of cange of f(x,y) = xe y at te point (2,0,) and (b) te direction in wic tis maximum rate of cange occurs. Solution: f x (x,y) = e y and f y (x,y) = xe y so f x (2,0) = e 0 = 1 and f y (2,0) = 2e 0 = 2. (a) Te maximum value of te rate of cange of f is D u f(x,y) = f(1,2) = 1, 2 = 5. (b) Tis maximum value occurs wen u is in te direction of f(1,2): u = 1/ 5, 2/ 5. Practice 4: Find (a) te maximum rate of cange of f (x, y) = 2x + 3y at te point (5, 2) and (b) te direction in wic tis maximum rate of cange occurs. Fig. 2 sows several level curves for a function z = f(x,y) and te

5 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus 5 gradient vector at several locations. (Note: te lengts of tese gradient vectors are exaggerated.) Practice 5: Sketc te gradient vector f(x,y) for te function f in Fig. 2 at A, B and C. A ball placed at (x,y) will begin to roll in te direction u = f(x,y). Cbing to a (local) maximum Property (2) is te foundation for using te gradient vector f(x,y) in iterative metods for finding local maximums of functions of several variables: (i) At any point (x,y) we take a sort "step" in te direction of f(x,y) tis takes us "upill" along te steepest route at tat point. (ii) Repeat step (i) until a (local) maximum is reaced. Property (3) provides an easy way to geometrically determine te direction of te gradient from te level curves of a surface. Fig. 3 sows level curves for a function z = f(x,y) and te upill gradient pats for several starting points. Practice 6: Sketc te upill gradient pat for te function f in Fig. 3 at starting points A, B and C. Beyond z = f(x,y) So far all of te examples ave dealt wit functions of two variables, z = f(x,y), but tat was just for convenience. Te definitions and ideas of gradient vectors and directional derivatives and teir properties extend in a natural way to functions of tree (or more) variables. Extensions to w = f(x,y,z) Definition: f(x,y,z) = f x (x,y,z), f y (x,y,z), f z (x,y,z) = f x i + f y j + f z k. Teorem: For a differentiable function f(x,y,z) and a unit direction vector u = a, b, c, D u f(x,y,z) = f(x,y,z). u Features: (1) Te maximum value of te directional derivative D u f(x,y,z) is f(x,y,z). (2) Te maximum value of D u f(x,y,z) occurs wen u as te same direction as f(x,y,z). (2) Te gradient vector f(x,y,z) is normal (perpendicular) to te level surface troug te point (x,y,z).

6 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus 6 Tese same ideas also extend very naturally to functions of more tan tree variables. For example, if x, y and z (all in meters) give te location in a room ten w=f(x,y,z) could be te temperature ( o C) at tat location. Ten instead of a level curve (in 2D), te points (x,y,z) were w=70 would be a level surface in 3D. D u f(x,y,z) would be te instantaneous rate of cange of temperature at location (x,y,z) in te direction u, and te units of D u f(x,y,z) would be o C/m. Te gradient vector f(x,y,z) would still give te maximum value of te directional derivative and would point in te direction of maximum rate of temperature increase. A eat-seeking flying bug in te room would follow a pat in te direction of te gradient at eac point. PROBLEMS In problems 1 4, find te directional derivative of f at te given point in te direction indicated by te given angle θ. (Note: θ is te angle te direction vector u makes wit te positive x axis, so te components of u are cos(θ), sin(θ). ) 1. f(x, y) = x 2 y 2 + 2x 4 y at (1, 2) wit θ = π/3 2. f(x, y) = (x 2 y) 3 at (3, 1) wit θ = 3π/4 3. f(x, y) = y x at (1, 2) wit θ = π/2 4. f(x, y) = sin( x + 2y ) at (4, 2) wit θ = 2π/3 In problems 5 8, (a) find te gradient of f, (b) evaluate te gradient at te given point P, and (c) find te rate of cange of f at P in te direction of te given vector u. 5. f(x, y) = x 3 4x 2 y + y 2 at P = (0, 1) wit u = 3/5, 4/5. 6. f(x, y) = e x. sin( y ) at P = (1, π/4) wit u = 1/ 5, 2/ f(x, y, z) = xy 2 z 3 at P = (1, 2, 1) wit u = 1/ 3, 1/ 3, 1/ f(x, y, z) = xy + yz 2 + xz 3 at P = (2, 0, 3) wit u = 2/3, 1/3, 2/3. In problems 9 14, find te directional derivative of te given function at te given point in te direction of te given vector v. 9. f(x, y) = x y at (5, 1) wit v = 12, f(x, y) = x/y at (6, 2) wit v = 1, g(x, y) = x. e xy at ( 3, 0) wit v = 2i + 3j. 12. g(x, y) = e x cos( y ) at (1, π/6) wit v = i j. 13. f(x, y, z) = xyz at (2, 4, 2) wit v = 4, 2, f(x, y, z) = z 3 x 2 y at (1, 6, 2) wit v = 3, 4, 12.

7 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus 7 In problems 15 20, find te maximum rate of cange of f at te given point and te direction in wic it occurs. 15. f(x, y) = x. e y + 3y at te point ( 1, 0) 16. f(x, y) = ln( x 2 + y 2 ) at te point ( 1, 2) 17. f(x, y) = x 2 + 2y at te point ( 4, 10) 18. f(x, y, z) = x + y/z at te point ( 4, 3, 1) 19. f(x, y) = cos( 3x + 2y ) at te point ( π/6, π/8) 20. f(x, y, z) = x y + y z at te point ( 4, 2, 1) 21. At eac dot in Fig. 4 sketc te gradient vector. 22. At eac dot in Fig. 5 sketc te gradient vector. 23. At eac dot in Fig. 6 sketc te upill gradient pat. 24. At eac dot in Fig. 7 sketc te upill gradient pat. 25. Sow tat a differentiable function f decreases most rapidly at (x,y) in te direction opposite to te gradient vector, tat is, in te direction f(x,y). 26. Use te result of Problem 23 to find te direction in wic te function f(x,y) = x 4 y x 2 y 3 decreases fastest at te point (2, 3). 27. Te temperature T in a metal ball is inversely proportional to te distance from te center of te ball, wic we take to be te origin. Te temperature at te point (1, 2, 2) is 120 o. (a) Find te rate of cange of T at (1, 2, 2) in te direction toward te point (2, 1, 3). (b) Sow tat at any given point in te ball te direction of greatest increase in temperature is given by a vector tat points toward te origin. 28. Te temperature at a point (x, y, z) is given by T(x,y,z) = 200. e ( x2 3y 2 9z 2 ) were T is measured in o C and x, y, and z in meters. (a) Find te rate of cange of temperature at te point P(2, 1, 2) in te direction toward te point (3, 3, 3). (b) In wic direction does te temperature increase fastest at P?

8 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus 8 (c) Find te maximum rate of increase at P. 29. Suppose tat over a certain region of space te electrical potential V is given by V(x,y,z) = 5x 2 3xy + xyz. (a) Find te rate of cange of te potential at P(3, 4, 5) in te direction of te vector v = i + j k. (b) In wic direction does V cange most rapidly at P? (c) Wat is te maximum rate of cange at P? 30. Suppose tat you are cbing a ill wose sape is given by te equation z = x y 2 and you are standing at te point wit coordinates (60, 100, 764). (a) In wic direction sould you proceed initially in order to reac te top of te ill fastest? (b) If you cb in tat direction, at wat angle above te orizontal will you be cbing initially? 31. Let F be a function of two variables tat as continuous partial derivatives and consider te points A(1,3), B(3,3), C(1,7), and D(6,15). Te directional derivative of A in te direction of te vector AB is 3, and te direction derivative at A in te direction of AC is 26. Find te direction derivative of f at A in te direction of te vector AD. Practice Answers Practice 1: f(x,y) = x 2 + 2x + 3y +1 and u = 5 13,12 13 " f 2 + 5,2 +12 D u f(2,1) = $ # % & f (2,1) &" ( # = '( $ " % # 13 $ % + 3 " # $ % +1 ) +,12 * + &" = $ # % + " $ # % + " $ # % +1 ) ( + ' *,12 = Practice 2: f x (x,y)= 2x+2 and f y (x,y)= 3 so f x ( 2, 1)= 6 and f y ( 2, 1)= 3. u = 5 13,12 13 so = 66 13! D u f( 2, 1) = (6) 5 $! # & + (3) 12 $ # & = 66, te same result as in Example 2 but muc easier. " 13% " 13% 13 Practice 3: (a) For f (x, y) = e xy + 2x 3 y + y 2, f x (x,y) = y! e xy + 6x 2 y and f y (x,y) = x! e xy + 2x 3 + 2y so f(x,y) = y! e xy + 6x 2 y, x! e xy + 2x 3 + 2y and f(1,2) = 2e 2 +12, e (b) For f(x,y) = cos(2x+3y), f x (x,y) = -2sin(2x+3y) and f y (x,y) = -3sin(2x+3y) f( 1, 2) = -2sin(2x+3y), -3sin(2x+3y) and f( 1, 2) = -2sin(8), -3sin(8). so Practice 4: f x (x,y) = 1 2x + 3y and f y (x,y) = 3 2 2x + 3y so f x ( 5, 2) = 1 4 and f y ( 5, 2) = 3 8.

9 13.5 Directional Derivatives and te Gradient Vector Contemporary Calculus 9 (a) Te maximum value of te rate of cange of f is D u f(5,2) = f(5,2) = 1 4, 3 8 = 13 8! (b) Tis maximum value occurs wen u is in te direction of f(5,2): u =!f(5,2)!f(5,2) = 2 13, Practice 5: See Fig. 9. Note tat eac gradient vector is perpendicular to te level curve and points upill. Practice 6: See Fig. 8. Note tat upill gradient pat is always perpendicular to te level curves. Selected Answers (a) f(x,y) = 3x 2 8xy, 4x 2 + 2y (b) 0, 2 (c) 8/5 7. (a) f(x,y,z) = y 2 z 3, 2xyz 3, 3xy 2 z 2 (b) 4, 4, 12 (c) 20/ / / / , 1/ 5, 2/ / 6, 4/ 17, 1/ (13/2), 3/ 13, 2/ 13 21, See Fig. 10. Note tat eac gradient vector is perpendicular to te level curve and points upill. 23. See Fig. 11. Note tat upill gradient pat is always perpendicular to te level curves. 27. (a) 40/( 3 3 ) 29. (a) 32/ 3 (b) u= 38, 6, 12 /(2 406 ) (c) /13