Maths 14th Chapter Partial Derivatives PDF

14 Partial Derivatives Overview The volume of a right circular cylinder is a function V = pr 2h of its radius and its height, so it is a function V(r, h) of two variables r and h. The speed of sound...

14 Partial Derivatives Overview The volume of a right circular cylinder is a function V = pr 2h of its radius and its height, so it is a function V(r, h) of two variables r and h. The speed of sound through seawater is primarily a function of salinity S and temperature T. The monthly pay- ment on a home mortgage is a function of the principal borrowed P, the interest rate i, and the term t of the loan. These are examples of functions that depend on more than one inde- pendent variable. In this chapter we extend the ideas of single-variable differential calculus to functions of several variables. Their derivatives are more varied and interesting because of the dif- ferent ways the variables can interact. The applications of these derivatives are also more varied than for single-variable calculus, and in the next chapter we will see that the same is true for integrals involving several variables. 14.1 Functions of Several Variables Real-valued functions of several independent real variables are defined analogously to functions of a single variable. Points in the domain are now ordered pairs (triples, quadru- ples, n-tuples) of real numbers, and values in the range are real numbers. DeFinitionS Suppose D is a set of n-tuples of real numbers (x1, x2,... , xn). A real-valued function ƒ on D is a rule that assigns a unique (single) real number w = ƒ(x1, x2,... , xn) to each element in D. The set D is the function’s domain. The set of w-values taken on by ƒ is the function’s range. The symbol w is the dependent variable of ƒ, and ƒ is said to be a function of the n independent variables x1 to xn. We also call the xj ’s the function’s input variables and call w the function’s output variable. If ƒ is a function of two independent variables, we usually call the independent vari- ables x and y and the dependent variable z, and we picture the domain of ƒ as a region in the xy-plane (Figure 14.1). If ƒ is a function of three independent variables, we call the independent variables x, y, and z and the dependent variable w, and we picture the domain as a region in space. In applications, we tend to use letters that remind us of what the variables stand for. To say that the volume of a right circular cylinder is a function of its radius and height, we might write V = ƒ(r, h). To be more specific, we might replace the notation ƒ(r, h) by the formula 792 M14_HASS8986_14_SE_C14_792-882.indd 792 01/02/17 4:19 PM  14.1 Functions of Several Variables 793 y f (x, y) f(a, b) x z D 0 (a, b) 0 f (x, y) FIGURE 14.1 An arrow diagram for the function z = ƒ(x, y). that calculates the value of V from the values of r and h, and write V = pr 2h. In either case, r and h would be the independent variables and V the dependent variable of the function. As usual, we evaluate functions defined by formulas by substituting the values of the independent variables in the formula and calculating the corresponding value of the depen- dent variable. For example, the value of ƒ(x, y, z) = 2x2 + y 2 + z2 at the point (3, 0, 4) is ƒ(3, 0, 4) = 2(3)2 + (0)2 + (4)2 = 225 = 5. Domains and Ranges In defining a function of more than one variable, we follow the usual practice of excluding inputs that lead to complex numbers or division by zero. If ƒ(x, y) = 2y - x2, then y cannot be less than x2. If ƒ(x, y) = 1>(xy), then xy cannot be zero. The domain of a func- tion is assumed to be the largest set for which the defining rule generates real numbers, unless the domain is otherwise specified explicitly. The range consists of the set of output values for the dependent variable. EXAMPLE 1 (a) These are functions of two variables. Note the restrictions that may apply to their domains in order to obtain a real value for the dependent variable z. Function Domain Range z = 2y - x2 y Ú x2 3 0, q) 1 z = xy xy ≠ 0 (-q, 0) ∪ (0, q) z = sin xy Entire plane 3 -1, 14 (b) These are functions of three variables with restrictions on some of their domains. Function Domain Range w = 2x 2 + y 2 + z 2 Entire space 3 0, q) 1 w = 2 (x, y, z) ≠ (0, 0, 0) (0, q) x + y 2 + z2 w = xy ln z Half-space z 7 0 (-q, q) Functions of Two Variables Regions in the plane can have interior points and boundary points just like intervals on the real line. Closed intervals 3 a, b4 include their boundary points, open intervals (a, b) don’t include their boundary points, and intervals such as 3 a, b) are neither open nor closed. M14_HASS8986_14_SE_C14_792-882.indd 793 14/12/16 5:13 PM  794 Chapter 14 Partial Derivatives DEFINITIONS A point (x0, y0) in a region (set) R in the xy-plane is an interior point of R if it is the center of a disk of positive radius that lies entirely in R (Figure 14.2). A point (x0 , y0) is a boundary point of R if every disk centered at (x0 , y0) contains points that lie outside of R as well as points that lie in R. (The R (x0 , y0) boundary point itself need not belong to R.) The interior points of a region, as a set, make up the interior of the region. The region’s boundary points make up its boundary. A region is open if it con- (a) Interior point sists entirely of interior points. A region is closed if it contains all its boundary points (Figure 14.3). y y y (x0 , y0) x x x R 0 0 0 (b) Boundary point FIGURE 14.2 Interior points and bound- {(x, y) 0 x 2 + y 2 < 1} {(x, y) 0 x 2 + y 2 = 1} {(x, y) 0 x 2 + y 2 ≤ 1} ary points of a plane region R. An interior Open unit disk. Boundary of unit Closed unit disk. Every point an disk. (The unit Contains all point is necessarily a point of R. A bound- interior point. circle.) boundary points. ary point of R need not belong to R. FIGURE 14.3 Interior points and boundary points of the unit disk in the plane. As with a half-open interval of real numbers 3 a, b), some regions in the plane are neither open nor closed. If you start with the open disk in Figure 14.3 and add to it some, but not all, of its boundary points, the resulting set is neither open nor closed. The bound- ary points that are there keep the set from being open. The absence of the remaining boundary points keeps the set from being closed. Two interesting examples are the empty set and the entire plane. The empty set has no interior points and no boundary points. This implies that the empty set is open (because it does not contain points that are not interior points), and at the same time it is closed (because there are no boundary points that it fails to contain). The entire xy-plane is also both open and closed: open because every point in the plane is an interior point, and closed because it has no boundary points. The empty set and the entire plane are the only subsets of the plane that are both open and closed. Other sets may be open, or closed, or neither. y DEFINITIONS A region in the plane is bounded if it lies inside a disk of finite Interior points, where y − x 2 > 0 radius. A region is unbounded if it is not bounded. Examples of bounded sets in the plane include line segments, triangles, interiors of triangles, rectangles, circles, and disks. Examples of unbounded sets in the plane include Outside, The parabola lines, coordinate axes, the graphs of functions defined on infinite intervals, quadrants, y − x2 < 0 1 y − x2 = 0 half-planes, and the plane itself. is the boundary. x −1 0 1 EXAMPLE 2 Describe the domain of the function ƒ(x, y) = 2y - x2. FIGURE 14.4 The domain of ƒ(x, y) in Solution Since ƒ is defined only where y - x2 Ú 0, the domain is the closed, Example 2 consists of the shaded region unbounded region shown in Figure 14.4. The parabola y = x2 is the boundary of the and its bounding parabola. domain. The points above the parabola make up the domain’s interior. M14_HASS8986_14_SE_C14_792-882.indd 794 14/12/16 5:13 PM  14.1 Functions of Several Variables 795 Graphs, Level Curves, and Contours of Functions of Two Variables There are two standard ways to picture the values of a function ƒ(x, y). One is to draw and label curves in the domain on which ƒ has a constant value. The other is to sketch the sur- face z = ƒ(x, y) in space. DEFINITIONS The set of points in the plane where a function ƒ(x, y) has a con- stant value ƒ(x, y) = c is called a level curve of ƒ. The set of all points (x, y, ƒ(x, y)) in space, for (x, y) in the domain of ƒ, is called the graph of ƒ. The graph of ƒ is also called the surface z = ƒ(x, y). z The surface EXAMPLE 3 Graph ƒ(x, y) = 100 - x2 - y 2 and plot the level curves ƒ(x, y) = 0, 100 z = f (x, y) ƒ(x, y) = 51, and ƒ(x, y) = 75 in the domain of ƒ in the plane. f(x, y) = 75 = 100 − x 2 − y 2 is the graph of f. Solution The domain of ƒ is the entire xy-plane, and the range of ƒ is the set of real numbers less than or equal to 100. The graph is the paraboloid z = 100 - x2 - y 2, the f(x, y) = 51 positive portion of which is shown in Figure 14.5. (a typical The level curve ƒ(x, y) = 0 is the set of points in the xy-plane at which level curve in the function’s ƒ(x, y) = 100 - x2 - y 2 = 0, or x2 + y 2 = 100, domain) which is the circle of radius 10 centered at the origin. Similarly, the level curves 10 y 10 ƒ(x, y) = 51 and ƒ(x, y) = 75 (Figure 14.5) are the circles f(x, y) = 0 x ƒ(x, y) = 100 - x2 - y 2 = 51, or x2 + y 2 = 49 FIGURE 14.5 The graph and selected ƒ(x, y) = 100 - x2 - y 2 = 75, or x2 + y 2 = 25. level curves of the function ƒ(x, y) in The level curve ƒ(x, y) = 100 consists of the origin alone. (It is still a level curve.) Example 3. The level curves lie in the If x2 + y 2 7 100, then the values of ƒ(x, y) are negative. For example, the circle xy-plane,which is the domain of the x + y 2 = 144, which is the circle centered at the origin with radius 12, gives the constant 2 function ƒ(x, y). value ƒ(x, y) = -44 and is a level curve of ƒ. The curve in space in which the plane z = c cuts a surface z = ƒ(x, y) is made up of The contour curve f(x, y) = 100 − x 2 − y 2 = 75 the points that represent the function value ƒ(x, y) = c. It is called the contour curve is the circle x 2 + y 2 = 25 in the plane z = 75. ƒ(x, y) = c to distinguish it from the level curve ƒ(x, y) = c in the domain of ƒ. Figure 14.6 shows the contour curve ƒ(x, y) = 75 on the surface z = 100 - x2 - y 2 z z = 100 − x 2 − y 2 defined by the function ƒ(x, y) = 100 - x2 - y 2. The contour curve lies directly above the circle x2 + y 2 = 25, which is the level curve ƒ(x, y) = 75 in the function’s domain. Plane z = 75 100 The distinction between level curves and contour curves is often overlooked, and it is common to call both types of curves by the same name, relying on context to make it clear 75 which type of curve is meant. On most maps, for example, the curves that represent con- stant elevation (height above sea level) are called contours, not level curves (Figure 14.7). Functions of Three Variables 0 In the plane, the points where a function of two independent variables has a constant value y ƒ(x, y) = c make a curve in the function’s domain. In space, the points where a function of three independent variables has a constant value ƒ(x, y, z) = c make a surface in the x function’s domain. The level curve f(x, y) = 100 − x2 − y2 = 75 is the circle x 2 + y 2 = 25 in the xy-plane. FIGURE 14.6 A plane z = c paral- DEFINITION The set of points (x, y, z) in space where a function of three inde- lel to the xy-plane intersecting a surface pendent variables has a constant value ƒ(x, y, z) = c is called a level surface of ƒ. z = ƒ(x, y) produces a contour curve. M14_HASS8986_14_SE_C14_792-882.indd 795 14/12/16 5:13 PM  796 Chapter 14 Partial Derivatives "x 2 + y 2 + z 2 = 1 FIGURE 14.7 Contours on Mt. Washington in New Hampshire. (From Appalachian Mountain Club. "x 2 + y 2 + z 2 = 2 Copyright by Appalachian Mountain Club.) z "x 2 + y 2 + z 2 = 3 Since the graphs of functions of three variables consist of points (x, y, z, ƒ(x, y, z)) lying in a four-dimensional space, we cannot sketch them effectively in our three- dimensional frame of reference. We can see how the function behaves, however, by look- ing at its three-dimensional level surfaces. 1 EXAMPLE 4 Describe the level surfaces of the function 2 y 3 ƒ(x, y, z) = 2x2 + y 2 + z2. x Solution The value of ƒ is the distance from the origin to the point (x, y, z). Each level surface 2x2 + y 2 + z2 = c, c 7 0, is a sphere of radius c centered at the origin. Figure FIGURE 14.8 The level surfaces of 14.8 shows a cutaway view of three of these spheres. The level surface 2x2 + y 2 + z2 = 0 ƒ(x, y, z) = 2x2 + y 2 + z2 are concentric consists of the origin alone. spheres (Example 4). We are not graphing the function here; we are looking at level surfaces in the function’s z (x0, y0, z0) domain. The level surfaces show how the function’s values change as we move through its domain. If we remain on a sphere of radius c centered at the origin, the function maintains a constant value, namely c. If we move from a point on one sphere to a point on another, y the function’s value changes. It increases if we move away from the origin and decreases if we move toward the origin. The way the values change depends on the direction we take. x The dependence of change on direction is important. We return to it in Section 14.5. The definitions of interior, boundary, open, closed, bounded, and unbounded for regions in space are similar to those for regions in the plane. To accommodate the extra (a) Interior point dimension, we use solid balls of positive radius instead of disks. (x0, y0, z0) z DEFINITIONS A point (x0, y0, z0) in a region R in space is an interior point of R if it is the center of a solid ball that lies entirely in R (Figure 14.9a). A point y (x0, y0, z0) is a boundary point of R if every solid ball centered at (x0, y0, z0) con- x tains points that lie outside of R as well as points that lie inside R (Figure 14.9b). The interior of R is the set of interior points of R. The boundary of R is the set of boundary points of R. A region is open if it consists entirely of interior points. A region is closed if (b) Boundary point it contains its entire boundary. FIGURE 14.9 Interior points and bound- ary points of a region in space. As with Examples of open sets in space include the interior of a sphere, the open half-space regions in the plane, a boundary point need z 7 0, the first octant (where x, y, and z are all positive), and space itself. Examples of not belong to the space region R. closed sets in space include lines, planes, and the closed half-space z Ú 0. A solid sphere M14_HASS8986_14_SE_C14_792-882.indd 796 29/12/16 8:47 AM  14.1 Functions of Several Variables 797 with part of its boundary removed or a solid cube with a missing face, edge, or corner point is neither open nor closed. Functions of more than three independent variables are also important. For example, the temperature on a surface in space may depend not only on the location of the point P(x, y, z) on the surface but also on the time t when it is visited, so we would write T = ƒ(x, y, z, t). Computer Graphing Three-dimensional graphing software makes it possible to graph functions of two vari- ables. We can often get information more quickly from a graph than from a formula, since the surfaces reveal increasing and decreasing behavior, and high points or low points. EXAMPLE 5 The temperature w beneath the Earth’s surface is a function of the depth x beneath the surface and the time t of the year. If we measure x in feet and t as the w number of days elapsed from the expected date of the yearly highest surface temperature, we can model the variation in temperature with the function w = cos (1.7 * 10-2t - 0.2x)e-0.2x. 15 (The temperature at 0 ft is scaled to vary from +1 to -1, so that the variation at x feet can 25 x be interpreted as a fraction of the variation at the surface.) t Figure 14.10 shows a graph of the function. At a depth of 15 ft, the variation (change in vertical amplitude in the figure) is about 5% of the surface variation. At 25 ft, there is almost no variation during the year. FIGURE 14.10 This graph shows the The graph also shows that the temperature 15 ft below the surface is about half a year seasonal variation of the temperature out of phase with the surface temperature. When the temperature is lowest on the surface below ground as a fraction of surface (late January, say), it is at its highest 15 ft below. Fifteen feet below the ground, the seasons temperature (Example 5). are reversed. Figure 14.11 shows computer-generated graphs of a number of functions of two vari- ables together with their level curves. z z z y x y y x x y y y x x x (a) z = sin x + 2 sin y (b) z = (4x2 + y 2)e-x 2 - y2 (c) z = xye-y 2 FIGURE 14.11 Computer-generated graphs and level curves of typical functions of two variables. M14_HASS8986_14_SE_C14_792-882.indd 797 14/12/16 5:13 PM  798 Chapter 14 Partial Derivatives EXERCISES 14.1 Domain, Range, and Level Curves 17. ƒ(x, y) = y - x 18. ƒ(x, y) = 2y - x In Exercises 1–4, find the specific function values. 19. ƒ(x, y) = 4x2 + 9y 2 20. ƒ(x, y) = x2 - y 2 1. ƒ(x, y) = x2 + xy 3 21. ƒ(x, y) = xy 22. ƒ(x, y) = y>x2 a. ƒ(0, 0) b. ƒ(- 1, 1) 1 c. ƒ(2, 3) d. ƒ(- 3, - 2) 23. ƒ(x, y) = 24. ƒ(x, y) = 29 - x2 - y 2 216 - x2 - y 2 2. ƒ(x, y) = sin (xy) 25. ƒ(x, y) = ln ( x2 + y 2 ) 26. ƒ(x, y) = e-(x + y2) 2 p p a. ƒa2, b b. ƒa- 3, b y 6 12 27. ƒ(x, y) = sin-1 ( y - x) 28. ƒ(x, y) = tan-1 a x b 1 p c. ƒap, b d. ƒa- , -7b 29. ƒ(x, y) = ln ( x2 + y 2 - 1 ) 30. ƒ(x, y) = ln ( 9 - x2 - y 2 ) 4 2 x - y Matching Surfaces with Level Curves 3. ƒ(x, y, z) = Exercises 31–36 show level curves for six functions. The graphs of y 2 + z2 these functions are given on the next page (items a–f ), as are their 1 1 equations (items g–l). Match each set of level curves with the appro- a. ƒ(3, - 1, 2) b. ƒa1, , - b 2 4 priate graph and appropriate equation. 1 31. 32. c. ƒa0, - , 0b d. ƒ(2, 2, 100) 3 y y 4. ƒ(x, y, z) = 249 - x2 - y 2 - z2 a. ƒ(0, 0, 0) b. ƒ(2, - 3, 6) 4 5 6 c. ƒ(- 1, 2, 3) d. ƒa , , b 22 22 22 x x In Exercises 5–12, find and sketch the domain for each function. 5. ƒ(x, y) = 2y - x - 2 6. ƒ(x, y) = ln ( x2 + y 2 - 4 ) (x - 1)( y + 2) 7. ƒ(x, y) = 33. 34. ( y - x)( y - x3) y y sin (xy) 8. ƒ(x, y) = 2 x + y 2 - 25 9. ƒ(x, y) = cos-1 ( y - x2) 10. ƒ(x, y) = ln (xy + x - y - 1) x x 11. ƒ(x, y) = 2 ( x2 - 4 )( y 2 - 9 ) 1 12. ƒ(x, y) = ln ( 4 - x2 - y 2 ) In Exercises 13–16, find and sketch the level curves ƒ(x, y) = c on the same set of coordinate axes for the given values of c. We refer to these level curves as a contour map. 35. 36. y y 13. ƒ(x, y) = x + y - 1, c = -3, - 2, - 1, 0, 1, 2, 3 14. ƒ(x, y) = x2 + y 2, c = 0, 1, 4, 9, 16, 25 15. ƒ(x, y) = xy, c = -9, - 4, - 1, 0, 1, 4, 9 16. ƒ(x, y) = 225 - x2 - y 2 , c = 0, 1, 2, 3, 4 x x In Exercises 17–30, (a) find the function’s domain, (b) find the func- tion’s range, (c) describe the function’s level curves, (d) find the boundary of the function’s domain, (e) determine if the domain is an open region, a closed region, or neither, and (f) decide if the domain is bounded or unbounded. M14_HASS8986_14_SE_C14_792-882.indd 798 14/12/16 5:14 PM  14.1 Functions of Several Variables 799 a. z f. z y x x y b. z x y xy 2 g. z = - h. z = y 2 - y 4 - x2 x + y2 2 2 + y2 >4 c. z i. z = (cos x)(cos y) e- 2x 1 j. z = e - y cos x k. z = 4x2 + y 2 xy (x2 - y 2) l. z = x2 + y 2 Functions of Two Variables Display the values of the functions in Exercises 37–48 in two ways: (a) by sketching the surface z = ƒ(x, y) and (b) by drawing an assort- ment of level curves in the function’s domain. Label each level curve with its function value. x y 37. ƒ(x, y) = y 2 38. ƒ(x, y) = 2x 2 2 39. ƒ(x, y) = x + y 40. ƒ(x, y) = 2x2 + y 2 2 41. ƒ(x, y) = x - y 42. ƒ(x, y) = 4 - x2 - y 2 d. z 43. ƒ(x, y) = 4x2 + y 2 44. ƒ(x, y) = 6 - 2x - 3y 45. ƒ(x, y) = 1 - 0 y 0 46. ƒ(x, y) = 1 - 0 x 0 - 0 y 0 2 2 47. ƒ(x, y) = 2x + y + 4 48. ƒ(x, y) = 2x2 + y 2 - 4 Finding Level Curves In Exercises 49–52, find an equation for and sketch the graph of the level curve of the function ƒ(x, y) that passes through the given point. y 49. ƒ(x, y) = 16 - x2 - y 2, 1 2 22, 22 2 2 x 50. ƒ(x, y) = 2x - 1, (1, 0) 51. ƒ(x, y) = 2x + y 2 - 3 , (3, - 1) 2y - x 52. ƒ(x, y) = , (-1, 1) x + y + 1 e. z Sketching Level Surfaces In Exercises 53–60, sketch a typical level surface for the function. 53. ƒ(x, y, z) = x2 + y 2 + z2 54. ƒ(x, y, z) = ln ( x2 + y 2 + z2 ) 55. ƒ(x, y, z) = x + z 56. ƒ(x, y, z) = z y 57. ƒ(x, y, z) = x2 + y 2 58. ƒ(x, y, z) = y 2 + z2 59. ƒ(x, y, z) = z - x2 - y 2 x 60. ƒ(x, y, z) = ( x2 >25 ) + ( y 2 >16 ) + ( z2 >9 ) M14_HASS8986_14_SE_C14_792-882.indd 799 14/12/16 5:14 PM  800 Chapter 14 Partial Derivatives Finding Level Surfaces 2 70. ƒ(x, y) = (sin x)(cos y) e2x + y2>8 , 0 … x … 5p, In Exercises 61–64, find an equation for the level surface of the func- 0 … y … 5p, P(4p, 4p) tion through the given point. 71. ƒ(x, y) = sin (x + 2 cos y), -2p … x … 2p, 61. ƒ(x, y, z) = 2x - y - ln z, (3, - 1, 1) -2p … y … 2p, P(p, p) 62. ƒ(x, y, z) = ln ( x2 + y + z2 ) , (- 1, 2, 1) 0.1 72. ƒ(x, y) = e(x - y) sin (x2 + y 2), 0 … x … 2p, 63. g(x, y, z) = 2x + y + z , 1 1, -1, 22 2 2 2 2 -2p … y … p, P(p, -p) x - y + z Use a CAS to plot the implicitly defined level surfaces in Exercises 64. g(x, y, z) = , (1, 0, - 2) 2x + y - z 73–76. In Exercises 65–68, find and sketch the domain of ƒ. Then find an 73. 4 ln (x2 + y 2 + z2) = 1 74. x2 + z2 = 1 equation for the level curve or surface of the function passing through 75. x + y 2 - 3z2 = 1 the given point. x 65. ƒ(x, y) = a a y b , (1, 2) q x n 76. sin a b - (cos y) 2x2 + z2 = 2 2 n=0 Parametrized Surfaces Just as you describe curves in the plane 66. g(x, y, z) = a (x + y)n q parametrically with a pair of equations x = ƒ(t), y = g(t) defined on , (ln 4, ln 9, 2) n=0 n!zn some parameter interval I, you can sometimes describe surfaces in y space with a triple of equations x = ƒ(u, y), y = g(u, y), z = h(u, y) Lx 21 - u 2 du 67. ƒ(x, y) = , (0, 1) defined on some parameter rectangle a … u … b, c … y … d. Many computer algebra systems permit you to plot such surfaces in para- y z metric mode. (Parametrized surfaces are discussed in detail in Section Lx 1 + t L0 24 - u 2 dt du 68. g(x, y, z) = 2 + , 1 0, 1, 23 2 16.5.) Use a CAS to plot the surfaces in Exercises 77–80. Also plot several level curves in the xy-plane. COMPUTER EXPLORATIONS 77. x = u cos y, y = u sin y, z = u, 0 … u … 2, Use a CAS to perform the following steps for each of the functions in 0 … y … 2p Exercises 69–72. 78. x = u cos y, y = u sin y, z = y, 0 … u … 2, a. Plot the surface over the given rectangle. 0 … y … 2p b. Plot several level curves in the rectangle. 79. x = (2 + cos u) cos y, y = (2 + cos u) sin y, z = sin u, c. Plot the level curve of ƒ through the given point. 0 … u … 2p, 0 … y … 2p y 80. x = 2 cos u cos y, y = 2 cos u sin y, z = 2 sin u, 69. ƒ(x, y) = x sin + y sin 2x, 0 … x … 5p, 0 … y … 5p, 2 0 … u … 2p, 0 … y … p P(3p, 3p) 14.2 Limits and Continuity in Higher Dimensions In this section we develop limits and continuity for multivariable functions. The theory is similar to that developed for single-variable functions, but since we now have more than one independent variable, there is additional complexity that requires some new ideas. Limits for Functions of Two Variables If the values of ƒ(x, y) lie arbitrarily close to a fixed real number L for all points (x, y) suf- ficiently close to a point (x0 , y0), we say that ƒ approaches the limit L as (x, y) approaches (x0 , y0). This is similar to the informal definition for the limit of a function of a single vari- able. Notice, however, that when (x0 , y0) lies in the interior of ƒ’s domain, (x, y) can approach (x0 , y0) from any direction, not just from the left or the right. For the limit to exist, the same limiting value must be obtained whatever direction of approach is taken. We illustrate this issue in several examples following the definition. M14_HASS8986_14_SE_C14_792-882.indd 800 14/12/16 5:14 PM  14.2 Limits and Continuity in Higher Dimensions 801 DEFINITION We say that a function ƒ(x, y) approaches the limit L as (x, y) ap- proaches (x0 , y0), and write lim ƒ(x, y) = L (x, y) S (x0, y0) if, for every number e 7 0, there exists a corresponding number d 7 0 such that for all (x, y) in the domain of ƒ, 0 ƒ(x, y) - L 0 6 e whenever 0 6 2(x - x0)2 + ( y - y0)2 6 d. The definition of limit says that the distance between ƒ(x, y) and L becomes arbitrarily small whenever the distance from (x, y) to (x0, y0) is made sufficiently small (but not 0). The definition applies to interior points (x0, y0) as well as boundary points of the domain of ƒ, although a boundary point need not lie within the domain. The points (x, y) that approach (x0 , y0) are always taken to be in the domain of ƒ. See Figure 14.12. y f (x, y) D d (x 0 , y0 ) z z 0 0 L−ε L L+ε FIGURE 14.12 In the limit definition, d is the radius of a disk centered at (x0, y0). For all points (x, y) within this disk, the function values ƒ(x, y) lie inside the corresponding interval (L - e, L + e). As for functions of a single variable, it can be shown that lim x = x0 (x, y) S (x0, y0) lim y = y0 (x, y) S (x0, y0) lim k = k (any number k). (x, y) S (x0, y0) For example, in the first limit statement above, ƒ(x, y) = x and L = x0. Using the defini- tion of limit, suppose that e 7 0 is chosen. If we let d equal this e, we see that if 0 6 2(x - x0)2 + ( y - y0)2 6 d = e, then 2(x - x0)2 6 e (x - x0)2 … (x - x0)2 + (y - y0)2 0 x - x0 0 6 e 2a2 = 0 a 0 0 ƒ(x, y) - x0 0 6 e. x = ƒ(x, y) That is, 0 ƒ(x, y) - x0 0 6 e whenever 0 6 2(x - x0)2 + (y - y0)2 6 d. So a d has been found satisfying the requirement of the definition, and therefore we have proved that lim ƒ(x, y) = lim x = x0. (x, y) S (x0, y0) (x, y) S (x0, y0) M14_HASS8986_14_SE_C14_792-882.indd 801 14/12/16 5:14 PM  802 Chapter 14 Partial Derivatives As with single-variable functions, the limit of the sum of two functions is the sum of their limits (when they both exist), with similar results for the limits of the differences, constant multiples, products, quotients, powers, and roots. These facts are summarized in Theorem 1. THEOREM 1—Properties of Limits of Functions of Two Variables The following rules hold if L, M, and k are real numbers and lim ƒ(x, y) = L and lim g(x, y) = M. (x, y) S (x0 , y0) (x, y) S (x0 , y0) 1. Sum Rule: lim (ƒ(x, y) + g(x, y)) = L + M (x, y) S (x0 , y0) 2. Difference Rule: lim (ƒ(x, y) - g(x, y)) = L - M (x, y) S (x0 , y0) 3. Constant Multiple Rule: lim kƒ(x, y) = kL (any number k) (x, y) S (x0 , y0) 4. Product Rule: lim (ƒ(x, y) # g(x, y)) = L # M (x, y) S (x0 , y0) ƒ(x, y) L 5. Quotient Rule: lim = , M≠0 (x, y) S (x0 , y0) g(x, y) M 6. Power Rule: lim 3 ƒ(x, y)4 n = L n, n a positive integer (x, y) S (x0 , y0) n n 7. Root Rule: lim 2ƒ(x, y) = 2L = L 1>n, (x, y) S (x0 , y0) n a positive integer, and if n is even, we assume that L 7 0. Although we will not prove Theorem 1 here, we give an informal discussion of why it is true. If (x, y) is sufficiently close to (x0 , y0), then ƒ(x, y) is close to L and g(x, y) is close to M (from the informal interpretation of limits). It is then reasonable that ƒ(x, y) + g(x, y) is close to L + M; ƒ(x, y) - g(x, y) is close to L - M; kƒ(x, y) is close to kL; ƒ(x, y)g(x, y) is close to LM; and ƒ(x, y) > g(x, y) is close to L > M if M ≠ 0. When we apply Theorem 1 to polynomials and rational functions, we obtain the use- ful result that the limits of these functions as (x, y) S (x0 , y0) can be calculated by evaluat- ing the functions at (x0 , y0). The only requirement is that the rational functions be defined at (x0 , y0). EXAMPLE 1 In this example, we can combine the three simple results following the limit definition with the results in Theorem 1 to calculate the limits. We simply substi- tute the x- and y-values of the point being approached into the functional expression to find the limiting value. x - xy + 3 0 - (0)(1) + 3 (a) lim 2 3 = = -3 (x, y) S (0,1) x y + 5xy - y (0) (1) + 5(0)(1) - (1)3 2 (b) lim 2x2 + y 2 = 2(3)2 + (-4)2 = 225 = 5 (x, y) S (3, -4) x2 - xy EXAMPLE 2 Find lim. (x, y) S (0, 0) 2x - 2y M14_HASS8986_14_SE_C14_792-882.indd 802 14/12/16 5:14 PM  14.2 Limits and Continuity in Higher Dimensions 803 Solution Since the denominator 2x - 2y approaches 0 as (x, y) S (0, 0), we cannot use the Quotient Rule from Theorem 1. If we multiply numerator and denominator by 2x + 2y, however, we produce an equivalent fraction whose limit we can find: x2 - xy 1 x2 - xy 21 2x + 2y 2 Multiply by a form lim = lim (x, y) S (0,0) 2x - 2y (x, y) S (0,0) 1 2x - 2y 21 2x + 2y 2 equal to 1. x 1 x - y 21 2x + 2y 2 = lim x - y Algebra (x, y) S (0,0) Cancel the nonzero = lim x 1 2x + 2y 2 factor (x - y). (x, y) S (0,0) = 0 1 20 + 20 2 = 0 Known limit values We can cancel the factor (x - y) because the path y = x (where we would have x - y = 0) is not in the domain of the function x2 - xy ƒ(x, y) =. 2x - 2y z 4xy 2 EXAMPLE 3 Find lim if it exists. (x, y) S (0,0) x + y 2 2 Solution We first observe that along the line x = 0, the function always has value 0 when y ≠ 0. Likewise, along the line y = 0, the function has value 0 provided x ≠ 0. So if the limit does exist as (x, y) approaches (0, 0), the value of the limit must be 0 (see 1 1 y Figure 14.13). To see if this is true, we apply the definition of limit. Let e 7 0 be given, but arbitrary. We want to find a d 7 0 such that x 4xy 2 FIGURE 14.13 The surface graph shows ` - 0` 6 e whenever 0 6 2x2 + y 2 6 d the limit of the function in Example 3 must x2 + y 2 be 0, if it exists. or 4 0 x 0 y2 6 e whenever 0 6 2x2 + y 2 6 d. x2 + y 2 Since y 2 … x2 + y 2 we have that 4 0 x 0 y2 y2 2 2 … 4 0 x 0 = 4 2x2 … 4 2x2 + y 2. … 1 x + y x + y2 2 So if we choose d = e >4 and let 0 6 2x2 + y 2 6 d, we get 4xy 2 e ` - 0 ` … 4 2x2 + y 2 6 4d = 4a b = e. 2 x + y 2 4 It follows from the definition that 4xy 2 lim = 0. (x, y) S (0,0) x2 + y 2 M14_HASS8986_14_SE_C14_792-882.indd 803 14/12/16 5:14 PM  804 Chapter 14 Partial Derivatives y EXAMPLE 4 If ƒ(x, y) = x , does lim ƒ(x, y) exist? (x, y) S (0, 0) Solution The domain of ƒ does not include the y-axis, so we do not consider any points (x, y) where x = 0 in the approach toward the origin (0, 0). Along the x-axis, the value of the function is ƒ(x, 0) = 0 for all x ≠ 0. So if the limit does exist as (x, y) S (0, 0), the value of the limit must be L = 0. On the other hand, along the line y = x, the value of the function is ƒ(x, x) = x>x = 1 for all x ≠ 0. That is, the function ƒ approaches the value 1 along the line y = x. This means that for every disk of radius d centered at (0, 0), the disk will contain points (x, 0) on the x-axis where the value of the function is 0, and also points (x, x) along the line y = x where the value of the function is 1. So no matter how small we choose d as the radius of the disk in Figure 14.12, there will be points within the disk for which the function values differ by 1. Therefore, the limit cannot exist because we can take e to be any number less than 1 in the limit definition and deny that L = 0 or 1, or any other real number. The limit does not exist because we have different limiting values along different paths approaching the point (0, 0). Continuity As with functions of a single variable, continuity is defined in terms of limits. z DEFINITION A function ƒ(x, y) is continuous at the point (x0 , y0) if 1. ƒ is defined at (x0 , y0), 2. lim ƒ(x, y) exists, (x, y) S (x0 , y0) x 3. lim ƒ(x, y) = ƒ(x0 , y0). (x, y) S (x0 , y0) −y A function is continuous if it is continuous at every point of its domain. (a) As with the definition of limit, the definition of continuity applies at boundary points as well as interior points of the domain of ƒ. The only requirement is that each point (x, y) y near (x0 , y0) be in the domain of ƒ. −0.8 0 0.8 A consequence of Theorem 1 is that algebraic combinations of continuous functions −1 1 are continuous at every point at which all the functions involved are defined. This means −0.8 0.8 that sums, differences, constant multiples, products, quotients, and powers of continuous 0 x functions are continuous where defined. In particular, polynomials and rational functions of two variables are continuous at every point at which they are defined. 0.8 −0.8 1 −1 0.8 0 −0.8 EXAMPLE 5 Show that (b) 2xy FIGURE 14.14 (a) The graph of , (x, y) ≠ (0, 0) ƒ(x, y) = x + y 2 2 2xy 0, (x, y) = (0, 0) , (x, y) ≠ (0, 0) ƒ(x, y) = x2 + y 2 0, (x, y) = (0, 0). is continuous at every point except the origin (Figure 14.14). The function is continuous at every point Solution The function ƒ is continuous at every point (x, y) except (0, 0) because its val- except the origin. (b) The values of ƒ ues at points other than (0, 0) are given by a rational function of x and y, and therefore at are different constants along each line those points the limiting value is simply obtained by substituting the values of x and y into y = mx, x ≠ 0 (Example 5). that rational expression. M14_HASS8986_14_SE_C14_792-882.indd 804 14/12/16 5:14 PM  14.2 Limits and Continuity in Higher Dimensions 805 At (0, 0), the value of ƒ is defined, but ƒ has no limit as (x, y) S (0, 0). The reason is that different paths of approach to the origin can lead to different results, as we now see. For every value of m, the function ƒ has a constant value on the “punctured” line y = mx, x ≠ 0, because 2xy 2x(mx) 2mx2 2m ƒ(x, y) ` = 2 2 ` = 2 2 = 2 =. y = mx x + y y = mx x + (mx) x + m2x2 1 + m2 Therefore, ƒ has this number as its limit as (x, y) approaches (0, 0) along the line: 2m lim ƒ(x, y) = lim c ƒ(x, y) ` d =. (x, y) S (0,0) (x, y) S (0,0) y = mx 1 + m2 along y = mx This limit changes with each value of the slope m. There is therefore no single number we may call the limit of ƒ as (x, y) approaches the origin. The limit fails to exist, and the func- tion is not continuous at the origin. Examples 4 and 5 illustrate an important point about limits of functions of two or more variables. For a limit to exist at a point, the limit must be the same along every approach path. This result is analogous to the single-variable case where both the left- and right-sided limits had to have the same value. For functions of two or more variables, if we ever find paths with different limits, we know the function has no limit at the point they approach. z 1 Two-Path Test for Nonexistence of a Limit If a function ƒ(x, y) has different limits along two different paths in the domain of ƒ as (x, y) approaches (x0 , y0), then lim(x, y)S(x0, y0) ƒ(x, y) does not exist. −1 EXAMPLE 6 Show that the function 1 y 2x2y 1 ƒ(x, y) = x + y2 4 x (Figure 14.15) has no limit as (x, y) approaches (0, 0). −1 Solution The limit cannot be found by direct substitution, which gives the indeterminate form 0 > 0. We examine the values of ƒ along parabolic curves that end at (0, 0). Along the (a) curve y = kx2, x ≠ 0, the function has the constant value y 2x2y 2x2 ( kx2 ) 2kx4 2k k = 10 ƒ(x, y) ` = ` = = 4 =. k= 3 y = kx2 x + y y = kx2 x + ( kx ) 4 2 4 2 2 x + k 2x4 1 + k 2 k= 1 Therefore, 2k lim ƒ(x, y) = lim c ƒ(x, y) ` d =. (x, y) S (0,0) (x, y) S (0,0) y = kx2 1 + k2 x along y = kx2 k = −0.1 This limit varies with the path of approach. If (x, y) approaches (0, 0) along the parabola y = x2, for instance, k = 1 and the limit is 1. If (x, y) approaches (0, 0) along the x-axis, k = −1 k = 0 and the limit is 0. By the two-path test, ƒ has no limit as (x, y) approaches (0, 0). It can be shown that the function in Example 6 has limit 0 along every straight line (b) path y = mx (Exercise 57). This implies the following observation: FIGURE 14.15 (a) The graph of ƒ(x, y) = 2x2y> ( x4 + y 2 ). (b) Along each Having the same limit along all straight lines approaching (x0 , y0) does not imply path y = kx2 the value of ƒ is constant, but that a limit exists at (x0 , y0). varies with k (Example 6). M14_HASS8986_14_SE_C14_792-882.indd 805 14/12/16 5:14 PM  806 Chapter 14 Partial Derivatives Whenever it is correctly defined, the composition of continuous functions is also con- tinuous. The only requirement is that each function be continuous where it is applied. The proof, omitted here, is similar to that for functions of a single variable (Theorem 9 in Section 2.5). Continuity of Compositions If ƒ is continuous at (x0 , y0) and g is a single-variable function continuous at ƒ(x0 , y0), then the composition h = g ∘ f defined by h(x, y) = g(ƒ(x, y)) is con- tinuous at (x0, y0). For example, the composite functions xy ex - y, cos 2 , ln (1 + x2y 2) x + 1 are continuous at every point (x, y). Functions of More Than Two Variables The definitions of limit and continuity for functions of two variables and the conclusions about limits and continuity for sums, products, quotients, powers, and compositions all extend to functions of three or more variables. Functions like y sin z ln (x + y + z) and x - 1 are continuous throughout their domains, and limits like ex + z e1 - 1 1 lim = = , P S (1,0,-1) z2 + cos 2xy (-1) + cos 0 2 2 where P denotes the point (x, y, z), may be found by direct substitution. Extreme Values of Continuous Functions on Closed, Bounded Sets The Extreme Value Theorem (Theorem 1, Section 4.1) states that a function of a single variable that is continuous at every point of a closed, bounded interval 3 a, b4 takes on an absolute maximum value and an absolute minimum value at least once in 3 a, b4. The same holds true of a function z = ƒ(x, y) that is continuous on a closed, bounded set R in the plane (like a line segment, a disk, or a filled-in triangle). The function takes on an abso- lute maximum value at some point in R and an absolute minimum value at some point in R. The function may take on a maximum or minimum value more than once over R. Similar results hold for functions of three or more variables. A continuous function w = ƒ(x, y, z) must take on absolute maximum and minimum values on any closed, bounded set (such as a solid ball or cube, spherical shell, or rectangular solid) on which it is defined. We will learn how to find these extreme values in Section 14.7. EXERCISES 14.2 Limits with Two Variables 1 1 2 Find the limits in Exercises 1–12. 3. lim 2x2 + y 2 - 1 4. lim ax + yb (x, y) S (3,4) (x, y) S (2, -3) 3x2 - y 2 + 5 x 1. lim 2. lim x2 + y 3 (x, y) S (0,0) x 2 + y 2 + 2 (x, y) S (0,4) 2y 5. lim sec x tan y 6. lim cos (x, y) S (0,p>4) (x, y) S (0,0) x + y + 1 M14_HASS8986_14_SE_C14_792-882.indd 806 14/12/16 5:14 PM  14.2 Limits and Continuity in Higher Dimensions 807 7. lim ex - y 8. lim ln 0 1 + x2 y 2 0 x2 + y 2 1 (x, y) S (0,ln 2) (x, y) S (1,1) 34. a. g(x, y) = b. g(x, y) = x2 - 3x + 2 x2 - y ey sin x 3 9. lim x 10. lim cos 2 xy (x, y) S (0,0) (x, y) S (1>27, p3) Continuity for Three Variables x sin y cos y + 1 11. lim 12. lim At what points (x, y, z) in space are the functions in Exercises 35–40 (x, y) S (1, p>6) x 2 + 1 (x, y) S (p>2,0) y - sin x continuous? 35. a. ƒ(x, y, z) = x2 + y 2 - 2z2 Limits of Quotients b. ƒ(x, y, z) = 2x2 + y 2 - 1 Find the limits in Exercises 13–24 by rewriting the fractions first. 36. a. ƒ(x, y, z) = ln xyz b. ƒ(x, y, z) = ex + y cos z x2 - 2xy + y 2 x2 - y 2 13. lim - 14. lim 1 1 (x, y) S (1,1) x y (x, y) S (1,1) x - y 37. a. h(x, y, z) = xy sin z b. h(x, y, z) = x≠y x≠y x 2 + z2 - 1 xy - y - 2x + 2 1 1 15. lim 38. a. h(x, y, z) = b. h(x, y, z) = (x, y) S (1,1) x - 1 0y0 + 0z0 0 xy 0 + 0 z 0 x≠1 y + 4 39. a. h(x, y, z) = ln ( z - x2 - y 2 - 1 ) 16. lim (x, y) S (2, -4) x2y - xy + 4x2 - 4x 1 x≠-4, x≠ x2 b. h(x, y, z) = z - 2x2 + y 2 x - y + 2 2x - 2 2y 17. lim 40. a. h(x, y, z) = 24 - x2 - y 2 - z2 (x, y) S (0,0) 2x - 2y x≠y 1 b. h(x, y, z) = x + y - 4 22x - y - 2 4 - 2x2 + y 2 + z2 - 9 18. lim 19. lim (x, y) S (2,2) 2x + y - 2 (x, y) S (2,0) 2x - y - 4 x + y≠ 4 2x - y≠ 4 No Limit Exists at the Origin 2x - 2y + 1 By considering different paths of approach, show that the functions in 20. lim Exercises 41–48 have no limit as (x, y) S (0, 0). (x, y) S (4,3) x - y - 1 x≠y + 1 x x4 41. ƒ(x, y) = - 42. ƒ(x, y) = sin ( x + y 2 2 ) 1 - cos (xy) 2x2 + y 2 x + y2 4 21. lim 22. lim xy (x, y) S (0,0) x2 + y 2 (x, y) S (0,0) z z 3 3 x + y x - y 23. lim 24. lim (x, y) S (1,-1) x + y (x, y) S (2,2) x4 - y 4 x Limits with Three Variables Find the limits in Exercises 25–30. y x