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Unformatted text preview: 5E-15(pp 0922-0931) 1/18/06 2:46 PM Page 922 CHAPTER 15 We can visualize a function of two variables by its graph (a surface) or by its level curves (a contour map). P artial Derivatives 5E-15(pp 0922-0931) 1/18/06 2:46 PM Page 923 So far we have dealt with the calculus of functions of a single variable. But, in the real world, physical quantities often depend on two or more variables, so in this chapter we turn our attention to functions of several variables and extend the basic ideas of differential calculus to such functions. |||| 15.1 Functions of Several Variables In this section we study functions of two or more variables from four points of view: ■ ■ ■ ■ verbally (by a description in words) numerically (by a table of values) algebraically (by an explicit formula) visually (by a graph or level curves) Functions of Two Variables The temperature T at a point on the surface of the Earth at any given time depends on the longitude x and latitude y of the point. We can think of T as being a function of the two variables x and y, or as a function of the pair x, y . We indicate this functional dependence by writing T f x, y . The volume V of a circular cylinder depends on its radius r and height h. In fact, we r 2h. We say that V is a function of r and h, and we write V r, h r 2h. know that V Definition A function f of two variables is a rule that assigns to each ordered pair of real numbers x, y in a set D a unique real number denoted by f x, y . The set D is the domain of f and its range is the set of values that f takes on, that is, f x, y x, y D. We often write z f x, y to make explicit the value taken on by f at the general point x, y . The variables x and y are independent variables and z is the dependent variable. [Compare this with the notation y f x for functions of a single variable.] A function of two variables is just a function whose domain is a subset of 2 and whose range is a subset of . One way of visualizing such a function is by means of an arrow diagram (see Figure 1), where the domain D is represented as a subset of the xy-plane. y (x, y) f 0 x f (a, b) 0 f(x, y) z D (a, b) FIGURE 1 923 5E-15(pp 0922-0931) 924 ❙❙❙❙ 1/18/06 2:46 PM Page 924 CHAPTER 15 PARTIAL DERIVATIVES x+y+1=0 If a function f is given by a formula and no domain is specified, then the domain of f is understood to be the set of all pairs x, y for which the given expression is a welldefined real number. y x=1 EXAMPLE 1 Find the domains of the following functions and evaluate f 3, 2 . _1 0 x sx (a) f x, y _1 y x 1 x ln y 2 (b) f x, y 1 x SOLUTION FIGURE 2 Domain of f(x, y)= (a) œ„„„„„„„ x+y+1 x-1 s3 f 3, 2 2 3 1 s6 2 1 The expression for f makes sense if the denominator is not 0 and the quantity under the square root sign is nonnegative. So the domain of f is D y x=¥ 0 x x, y x y 1 0, x 1 The inequality x y 1 0, or y x 1, describes the points that lie on or above the line y x 1, while x 1 means that the points on the line x 1 must be excluded from the domain. (See Figure 2.) (b) f 3, 2 3 ln 2 2 3 3 ln 1 0 Since ln y 2 x is defined only when y 2 x 0, that is, x y 2, the domain of f is D x, y x y 2 . This is the set of points to the left of the parabola x y 2. (See Figure 3.) FIGURE 3 Domain of f (x, y)=x ln(¥-x) Not all functions are given by explicit formulas. The function in the next example is described verbally and by numerical estimates of its values. EXAMPLE 2 In regions with severe winter weather, the wind-chill index is often used to describe the apparent severity of the cold. This index W is a subjective temperature that depends on the actual temperature T and the wind speed v. So W is a function of T and v, and we can write W f T, v . Table 1 records values of W compiled by the National Weather Service of the U.S. and the Meteorological Service of Canada. Wind speed (km / h) TABLE 1 Wind-chill index as a function of air temperature and wind speed T v 5 10 15 20 25 30 40 50 60 70 80 |||| THE NEW WIND-CHILL INDEX A new wind-chill index was introduced in November of 2001 and is more accurate than the old index at measuring how cold it feels when it’s windy. The new index is based on a model of how fast a human face loses heat. It was developed through clinical trials in which volunteers were exposed to a variety of temperatures and wind speeds in a refrigerated wind tunnel. Actual temperature (°C) 5 4 3 2 1 1 0 1 1 2 2 3 0 2 3 4 5 6 6 7 8 9 9 10 5 7 9 11 12 12 13 14 15 16 16 17 10 13 15 17 18 19 20 21 22 23 23 24 15 19 21 23 24 25 26 27 29 30 30 31 20 24 27 29 30 32 33 34 35 36 37 38 25 30 33 35 37 38 39 41 42 43 44 45 30 36 39 41 43 44 46 48 49 50 51 52 35 41 45 48 49 51 52 54 56 57 58 60 40 47 51 54 56 57 59 61 63 64 65 67 5E-15(pp 0922-0931) 1/18/06 2:46 PM Page 925 SECTION 15.1 FUNCTIONS OF SEVERAL VARIABLES ❙❙❙❙ 925 For instance, the table shows that if the temperature is 5 C and the wind speed is 50 km h, then subjectively it would feel as cold as a temperature of about 15 C with no wind. So f TABLE 2 . Year P L K 1899 1900 1901 1902 1903 1904 1905 1906 1907 1908 1909 1910 1911 1912 1913 1914 1915 1916 1917 1918 1919 1920 1921 1922 100 101 112 122 124 122 143 152 151 126 155 159 153 177 184 169 189 225 227 223 218 231 179 240 100 105 110 117 122 121 125 134 140 123 143 147 148 155 156 152 156 183 198 201 196 194 146 161 100 107 114 122 131 138 149 163 176 185 198 208 216 226 236 244 266 298 335 366 387 407 417 431 5, 50 EXAMPLE 3 In 1928 Charles Cobb and Paul Douglas published a study in which they modeled the growth of the American economy during the period 1899–1922. They considered a simplified view of the economy in which production output is determined by the amount of labor involved and the amount of capital invested. While there are many other factors affecting economic performance, their model proved to be remarkably accurate. The function they used to model production was of the form bL K 1 P L, K 1 where P is the total production (the monetary value of all goods produced in a year), L is the amount of labor (the total number of person-hours worked in a year), and K is the amount of capital invested (the monetary worth of all machinery, equipment, and buildings). In Section 15.3 we will show how the form of Equation 1 follows from certain economic assumptions. Cobb and Douglas used economic data published by the government to obtain Table 2. They took the year 1899 as a baseline and P, L, and K for 1899 were each assigned the value 100. The values for other years were expressed as percentages of the 1899 figures. Cobb and Douglas used the method of least squares to fit the data of Table 2 to the function 1.01L0.75K 0.25 P L, K 2 (See Exercise 71 for the details.) If we use the model given by the function in Equation 2 to compute the production in the years 1910 and 1920, we get the values P 147, 208 1.01 147 0.75 208 0.25 161.9 P 194, 407 1.01 194 0.75 407 0.25 235.8 which are quite close to the actual values, 159 and 231. The production function (1) has subsequently been used in many settings, ranging from individual firms to global economic questions. It has become known as the Cobb-Douglas production function. Its domain is L, K L 0, K 0 because L and K represent labor and capital and are therefore never negative. y ≈+¥=9 EXAMPLE 4 Find the domain and range of t x, y _3 15 3 x x2 s9 y2 SOLUTION The domain of t is D x, y 9 x2 y2 0 x, y x 2 y2 9 which is the disk with center 0, 0 and radius 3. (See Figure 4.) The range of t is FIGURE 4 9-≈-¥ Domain of g(x, y)=œ„„„„„„„„„ zz s9 x2 y 2, x, y D 5E-15(pp 0922-0931) 926 ❙❙❙❙ 1/18/06 2:46 PM Page 926 CHAPTER 15 PARTIAL DERIVATIVES Since z is a positive square root, z 9 x2 0. Also y2 9 s9 ? x2 y2 3 So the range is z0 z 3 0, 3 Graphs Another way of visualizing the behavior of a function of two variables is to consider its graph. Definition If f is a function of two variables with domain D, then the graph of f is the set of all points x, y, z in z 3 such that z f x, y and x, y is in D. { x, y, f(x, y)} S Just as the graph of a function f of one variable is a curve C with equation y f x , so the graph of a function f of two variables is a surface S with equation z f x, y . We can visualize the graph S of f as lying directly above or below its domain D in the xy-plane (see Figure 5). f(x, y) 0 D x (x, y, 0) y EXAMPLE 5 Sketch the graph of the function f x, y 6 3x 2y. SOLUTION The graph of f has the equation z 6 3x 2y, or 3x 2y z 6, which represents a plane. To graph the plane we first find the intercepts. Putting y z 0 in the equation, we get x 2 as the x-intercept. Similarly, the y-intercept is 3 and the z-intercept is 6. This helps us sketch the portion of the graph that lies in the first octant (Figure 6). FIGURE 5 z (0, 0, 6) (0, 3, 0) (2, 0, 0) FIGURE 6 y x The function in Example 5 is a special case of the function f x, y ax by c which is called a linear function. The graph of such a function has the equation z a x b y c, or ax b y z c 0, so it is a plane. In much the same way that linear functions of one variable are important in single-variable calculus, we will see that linear functions of two variables play a central role in multivariable calculus. EXAMPLE 6 Sketch the graph of t x, y s9 x2 y 2. 5E-15(pp 0922-0931) 1/18/06 2:46 PM Page 927 S ECTION 15.1 FUNCTIONS OF SEVERAL VARIABLES z (3, 0, 0) 927 SOLUTION The graph has equation z s9 x 2 y 2. We square both sides of this equation to obtain z 2 9 x 2 y 2, or x 2 y 2 z 2 9, which we recognize as an equation of the sphere with center the origin and radius 3. But, since z 0, the graph of t is just the top half of this sphere (see Figure 7). (0, 0, 3) 0 ❙❙❙❙ (0, 3, 0) y EXAMPLE 7 Use a computer to draw the graph of the Cobb-Douglas production function x P L, K FIGURE 7 Graph of g(x, y)=œ„„„„„„„„„ 9-≈-¥ 1.01L0.75K 0.25. SOLUTION Figure 8 shows the graph of P for values of the labor L and capital K that lie between 0 and 300. The computer has drawn the surface by plotting vertical traces. We see from these traces that the value of the production P increases as either L or K increases, as is to be expected. 300 200 P 100 0 300 FIGURE 8 200 100 K 00 100 200 300 L EXAMPLE 8 Find the domain and range and sketch the graph of h x, y 4x 2 y 2. SOLUTION Notice that h x, y is defined for all possible ordered pairs of real numbers x, y , so the domain is 2, the entire x y-plane. The range of h is the set 0, of all nonnegative real numbers. [Notice that x 2 0 and y 2 0, so h x, y 0 for all x and y.] The graph of h has the equation z 4 x 2 y 2, which is the elliptic paraboloid that we sketched in Example 4 in Section 13.6. Horizontal traces are ellipses and vertical traces are parabolas (see Figure 9). z FIGURE 9 Graph of h (x, y)=4≈+¥ x y Computer programs are readily available for graphing functions of two variables. In most such programs, traces in the vertical planes x k and y k are drawn for equally spaced values of k and parts of the graph are eliminated using hidden line removal. 5E-15(pp 0922-0931) 928 ❙❙❙❙ 1/18/06 2:46 PM Page 928 CHAPTER 15 PARTIAL DERIVATIVES Figure 10 shows computer-generated graphs of several functions. Notice that we get an especially good picture of a function when rotation is used to give views from different vantage points. In parts (a) and (b) the graph of f is very flat and close to the xy-plane 2 2 except near the origin; this is because e x y is very small when x or y is large. z z x y x (b) f(x, y)=(≈+3¥)e _≈_¥ (a) f(x, y)=(≈+3¥)e _≈_¥ z z x y x (c) f(x, y)=sin x+sin y y (d) f(x, y)= sin x sin y xy FIGURE 10 Level Curves So far we have two methods for visualizing functions: arrow diagrams and graphs. A third method, borrowed from mapmakers, is a contour map on which points of constant elevation are joined to form contour curves, or level curves. Definition The level curves of a function f of two variables are the curves with equations f x, y k, where k is a constant (in the range of f ). A level curve f x, y k is the set of all points in the domain of f at which f takes on a given value k. In other words, it shows where the graph of f has height k. You can see from Figure 11 the relation between level curves and horizontal traces. The level curves f x, y k are just the traces of the graph of f in the horizontal plane z k projected down to the xy-plane. So if you draw the level curves of a function and visual- 5E-15(pp 0922-0931) 1/18/06 2:46 PM Page 929 S ECTION 15.1 FUNCTIONS OF SEVERAL VARIABLES ❙❙❙❙ 929 z 40 00 45 45 00 50 00 LONESOME MTN. 0 A 55 00 B y 50 x 00 k=45 450 f(x, y)=20 FIGURE 11 0 k=40 k=35 k=30 k=25 k=20 e Lon som ee e Cr k FIGURE 12 Visual 15.1A animates Figure 11 by showing level curves being lifted up to graphs of functions. FIGURE 13 World mean sea-level temperatures in January in degrees Celsius ize them being lifted up to the surface at the indicated height, then you can mentally piece together a picture of the graph. The surface is steep where the level curves are close together. It is somewhat flatter where they are farther apart. One common example of level curves occurs in topographic maps of mountainous regions, such as the map in Figure 12. The level curves are curves of constant elevation above sea level. If you walk along one of these contour lines you neither ascend nor descend. Another common example is the temperature function introduced in the opening paragraph of this section. Here the level curves are called isothermals and join locations with the same temperature. Figure 13 shows a weather map of the world indicating the average January temperatures. The isothermals are the curves that separate the colored bands. 5E-15(pp 0922-0931) 930 ❙❙❙❙ 1/18/06 2:46 PM CHAPTER 15 PARTIAL DERIVATIVES y EXAMPLE 9 A contour map for a function f is shown in Figure 14. Use it to estimate the 50 5 values of f 1, 3 and f 4, 5 . SOLUTION The point (1, 3) lies partway between the level curves with z-values 70 and 80. We estimate that 4 3 2 80 70 60 1 0 Page 930 50 2 3 1 f 1, 3 80 70 60 4 73 f 4, 5 56 Similarly, we estimate that x 5 EXAMPLE 10 Sketch the level curves of the function f x, y F IGURE 14 k y 6 3x 2y for the values 6, 0, 6, 12. SOLUTION The level curves are 6 0 x _6 k= 0 k= 6 k= 12 k= FIGURE 15 Contour map of f(x, y)=6-3x-2y 3x 2y k or 3x 2y k 6 0 This is a family of lines with slope 3 . The four particular level curves with 2 k 6, 0, 6, and 12 are 3x 2y 12 0, 3x 2y 6 0, 3x 2y 0, and 3x 2y 6 0. They are sketched in Figure 15. The level curves are equally spaced parallel lines because the graph of f is a plane (see Figure 6). EXAMPLE 11 Sketch the level curves of the function t x, y s9 x2 y2 for k 0, 1, 2, 3 SOLUTION The level curves are s9 x2 y2 k or x2 y2 k2 9 This is a family of concentric circles with center 0, 0 and radius s9 k 2. The cases k 0, 1, 2, 3 are shown in Figure 16. Try to visualize these level curves lifted up to form a surface and compare with the graph of t (a hemisphere) in Figure 7. (See TEC Visual 15.1A.) y k=3 k=2 k=1 k=0 (3, 0) 0 x FIGURE 16 Contour map of g(x, y)=œ„„„„„„„„„ 9-≈-¥ EXAMPLE 12 Sketch some level curves of the function h x, y 4x 2 SOLUTION The level curves are 4x 2 y2 k or x2 k4 y2 k 1 y 2. 5E-15(pp 0922-0931) 1/18/06 2:46 PM Page 931 S ECTION 15.1 FUNCTIONS OF SEVERAL VARIABLES ❙❙❙❙ 931 which, for k 0, describe a family of ellipses with semiaxes sk 2 and sk. Figure 17(a) shows a contour map of h drawn by a computer with level curves corresponding to k 0.25, 0.5, 0.75, . . . , 4. Figure 17(b) shows these level curves lifted up to the graph of h (an elliptic paraboloid) where they become horizontal traces. We see from Figure 17 how the graph of h is put together from the level curves. y z Visual 15.1B demonstrates the connection between surfaces and their contour maps. x x FIGURE 17 y The graph of h (x, y)=4≈+¥ is formed by lifting the level curves. (a) Contour map (b) Horizontal traces are raised level curves EXAMPLE 13 Plot level curves for the Cobb-Douglas production function of Example 3. SOLUTION In Figure 18 we use a computer to draw a contour plot for the Cobb-Douglas production function 1.01P 0.75K 0.25 P L, K K 300 200 220 180 100 140 100 FIGURE 18 100 200 300 L Level curves are labeled with the value of the production P. For instance, the level curve labeled 140 shows all values of the labor L and capital investment K that result in a production of P 140. We see that, for a fixed value of P, as L increases K decreases, and vice versa. For some purposes, a contour map is more useful than a graph. That is certainly true in Example 13. (Compare Figure 18 with Figure 8.) It is also true in estimating function values, as in Example 9. 5E-15(pp 0932-0941) 932 ❙❙❙❙ 1/18/06 4:05 PM Page 932 CHAPTER 15 PARTIAL DERIVATIVES Figure 19 shows some computer-generated level curves together with the corresponding computer-generated graphs. Notice that the level curves in part (c) crowd together near the origin. That corresponds to the fact that the graph in part (d) is very steep near the origin. z y z x x y (a) Level curves of f(x, y)=_xye_≈_¥ (b) Two views of f(x, y)=_xye_≈_¥ z y x y x FIGURE 19 (c) Level curves of f(x, y)= _3y ≈+¥+1 (d) f(x, y)= _3y ≈+¥+1 Functions of Three or More Variables A function of three variables, f , is a rule that assigns to each ordered triple x, y, z in a 3 domain D a unique real number denoted by f x, y, z . For instance, the temperature T at a point on the surface of the Earth depends on the longitude x and latitude y of the point and on the time t, so we could write T f x, y, t . EXAMPLE 14 Find the domain of f if f x, y, z ln z x y sin z y SOLUTION The expression for f x, y, z is defined as long as z y 0, so the domain of f is D x, y, z 3 z y This is a half-space consisting of all points that lie above the plane z y. 5E-15(pp 0932-0941) 1/18/06 4:05 PM Page 933 SECTION 15.1 FUNCTIONS OF SEVERAL VARIABLES ❙❙❙❙ 933 It’s very difficult to visualize a function f of three variables by its graph, since that would lie in a four-dimensional space. However, we do gain some insight into f by examining its level surfaces, which are the surfaces with equations f x, y, z k, where k is a constant. If the point x, y, z moves along a level surface, the value of f x, y, z remains fixed. z ≈+¥+z@=9 EXAMPLE 15 Find the level surfaces of the function ≈+¥+z@=4 f x, y, z x2 y2 z2 SOLUTION The level surfaces are x 2 y 2 z 2 k, where k 0. These form a family of concentric spheres with radius sk. (See Figure 20.) Thus, as x, y, z varies over any sphere with center O, the value of f x, y, z remains fixed. y x ≈+¥+z@=1 FIGURE 20 Functions of any number of variables can also be considered. A function of n variables is a rule that assigns a number z f x 1, x 2 , . . . , x n to an n-tuple x 1, x 2 , . . . , x n of real numbers. We denote by n the set of all such n-tuples. For example, if a company uses n different ingredients in making a food product, ci is the cost per unit of the ith ingredient, and x i units of the ith ingredient are used, then the total cost C of the ingredients is a function of the n variables x 1, x 2 , . . . , x n : 3 C f x 1, x 2 , . . . , x n c1 x 1 c2 x 2 cn x n The function f is a real-valued function whose domain is a subset of n. Sometimes we will use vector notation to write such functions more compactly: If x x 1, x 2 , . . . , x n , we often write f x in place of f x 1, x 2 , . . . , x n . With this notation we can rewrite the function defined in Equation 3 as fx cx where c c1, c2 , . . . , cn and c x denotes the dot product of the vectors c and x in Vn . In view of the one-to-one correspondence between points x 1, x 2 , . . . , x n in n and their position vectors x x 1, x 2 , . . . , x n in Vn , we have three ways of looking at a function f defined on a subset of n : 1. As a function of n real variables x 1, x 2 , . . . , x n 2. As a function of a single point variable x 1, x 2 , . . . , x n 3. As a function of a single vector variable x x 1, x 2 , . . . , x n We will see that all three points of view are useful. |||| 15.1 Exercises 1. In Example 2 we considered the function W f T, v , where W is the wind-chill index, T is the actual temperature, and v is the wind speed. A numerical representation is given in Table 1. (a) What is the value of f 15, 40 ? What is its meaning? (b) Describe in words the meaning of the question “For what value of v is f 20, v 30 ?” Then answer the question. (c) Describe in words the meaning of the question “For what value of T is f T, 20 49 ?” Then answer the question. (d) What is the meaning of the function W f 5, v ? Describe the behavior of this function. (e) What is the meaning of the function W f T, 50 ? Describe the behavior of this function. 5E-15(pp 0932-0941) 934 ❙❙❙❙ 1/18/06 4:06 PM Page 934 CHAPTER 15 PARTIAL DERIVATIVES 2. The temperature-humidity index I (or humidex, for short) is the TABLE 3 Apparent temperature as a function of temperature and humidity Actual temperature (°F) Relative humidity (%) (a) (b) (c) (d) h 20 30 40 50 60 70 80 77 78 79 81 82 82 84 86 88 90 87 90 93 96 100 93 96 101 107 114 99 104 110 120 132 15 20 30 40 50 2 2 2 2 2 2 15 4 4 5 5 5 5 5 20 5 7 8 8 9 9 9 30 9 13 16 17 18 19 19 40 14 21 25 28 31 33 33 19 29 36 40 45 48 50 60 144 2 50 124 100 10 24 37 47 54 62 67 69 106 95 5 10 93 90 t √ 83 85 Duration (hours) Wind speed (knots) perceived air temperature when the actual temperature is T and the relative humidity is h, so we can write I f T, h . The following table of values of I is an excerpt from a table compiled by the National Oceanic and Atmospheric Administration. T 6. Let f x, y (a) (b) (c) (d) ln x y 1 . Evaluate f 1, 1 . Evaluate f e, 1 . Find and sketch the domain of f . Find the range of f . x 2e 3 xy. (a) Evaluate f 2, 0 . (b) Find the domain of f . (c) Find the range of f . 7. Let f x, y What is the value of f 95, 70 ? What is its meaning? For what value of h is f 90, h 100? For what value of T is f T, 50 88? What are the meanings of the functions I f 80, h and I f 100, h ? Compare the behavior of these two functions of h. 8. Find and sketch the domain of the function f x, y 3. Verify for the Cobb-Douglas production function P L, K 4. The wind-chill index W discussed in Example 2 has been modeled by the following function: W T, v 13.12 0.6215T 11.37v 0.16 2 Check to see how closely this model agrees with the values in Table 1 for a few values of T and v. 5. The wave heights h in the open sea depend on the speed v of the wind and the length of time t that the wind has been blowing at that speed. Values of the function h f v, t are recorded in feet in the following table. (a) What is the value of f 40, 15 ? What is its meaning? (b) What is the meaning of the function h f 30, t ? Describe the behavior of this function. (c) What is the meaning of the function h f v, 30 ? Describe the behavior of this function. 2 esz x y . (a) Evaluate f 2, 1, 6 . (b) Find the domain of f . (c) Find the range of f . ln 25 x 2 (a) Evaluate t 2, 2, 4 . (b) Find the domain of t. (c) Find the range of t. 10. Let t x, y, z 11–20 |||| sx 13. f x, y ln 9 15. f x, y 2 x2 x sy 2 18. f x, y sx 2 x x2 s1 20. f x, y, z ■ 3y 3y x 2 y2 1 x2 ln 4 y2 4x 2 ln 16 ■ x x sy 4 x ln y sy 1 19. f x, y, z 14. f x, y 9y 2 sx 5y y 17. f x, y ■ z2 . 12. f x, y y 3x 16. f x, y ■ y2 Find and sketch the domain of the function. 11. f x, y 0.3965Tv 0.16 y 2. What is the range of f ? x 9. Let f x, y, z 1.01L 0.75K 0.25 discussed in Example 3 that the production will be doubled if both the amount of labor and the amount of capital are doubled. Is this also true for the general production function P L, K bL K 1 ? s1 ■ y2 z2 4y 2 ■ x2 z2 ■ ■ ■ ■ ■ ■ 5E-15(pp 0932-0941) 1/18/06 4:06 PM Page 935 S ECTION 15.1 FUNCTIONS OF SEVERAL VARIABLES 21–29 |||| 3 23. f x, y 1 25. f x, y 1 27. f x, y graph is a cone. The other is for a function t whose graph is a paraboloid. Which is which, and why? 22. f x, y 4x 2 x 24. f x, y y 2 y2 28. f x, y s16 x 29. f x, y sx 2 y cos x 26. f x, y x 3 x2 y I y2 ■ ■ ■ y II 1 y2 ■ 935 32. Two contour maps are shown. One is for a function f whose Sketch the graph of the function. 21. f x, y ❙❙❙❙ 2 16y 2 x ■ ■ ■ ■ ■ ■ ■ x ■ 30. Match the function with its graph (labeled I–VI). Give reasons for your choices. (a) f x, y x (c) f x, y (e) f x, y 1 x z I (b) f x, y xy (d) f x, y y 1 x2 y2 y2 2 x 33. Locate the points A and B in the map of Lonesome Mountain y sin( x (f) f x, y (Figure 12). How would you describe the terrain near A? Near B ? 22 y ) 34. Make a rough sketch of a contour map for the function whose z II graph is shown. z y x z III y x z IV y x x z V z VI y x y 35–36 |||| A contour map of a function is shown. Use it to make a rough sketch of the graph of f . y 35. x y y 36. 14 13 12 11 x _8 y _6 mate the values of f 3, 3 and f 3, about the shape of the graph? _4 x 31. A contour map for a function f is shown. Use it to esti- 2 . What can you say x 8 y ■ ■ 37–44 ■ |||| ■ ■ ■ ■ ■ ■ ■ ■ ■ Draw a contour map of the function showing several level curves. 0 37. f x, y 1 30 20 10 x x2 ln x 40. f x, y ey x y y 41. f x, y 38. f x, y 42. f x, y y sec x 44. f x, y y x2 xy 39. f x, y 70 60 50 40 1 sx 43. f x, y ■ ■ x ■ y ■ 2 ■ ■ ■ ■ ■ y2 ■ y2 ■ ■ 5E-15(pp 0932-0941) ❙❙❙❙ 936 1/18/06 4:07 PM Page 936 CHAPTER 15 PARTIAL DERIVATIVES 64. (a) t x, y 45–46 |||| Sketch both a contour map and a graph of the function and compare them. x2 45. f x, y 46. f x, y ■ 9y 2 ■ ■ 9x 2 s36 ■ ■ 4y 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ T x, y at the point x, y . The level curves of T are called isothermals because at all points on an isothermal the temperature is the same. Sketch some isothermals if the temperature function is given by 100 1 x 2 2y 2 65. f x, y ■ xy-plane, then the level curves of V are called equipotential curves because at all points on such a curve the electric potential is the same. Sketch some equipotential curves if V x, y c sr 2 x 2 y 2, where c is a positive constant. xy xy 51. f x, y 52. f x, y ■ ■ 2 3 ■ 50. f x, y x 3 yx ■ ■ ■ sin ye ■ ■ ■ ■ ■ x2 57. z 4y 2 sin x sin y ■ 59–62 ■ |||| ■ ■ ■ x 2 y 2e x3 58. z 1 55. z ■ 54. z y2 ■ ■ x 60. f x, y, z x2 3y 2 61. f x, y, z x2 y2 62. f x, y, z x2 ■ 63–64 ■ |||| ■ ■ and 3xy 2 1 4 ■ y2 ■ ■ ■ ■ ■ ■ ■ Describe how the graph of t is obtained from the graph of f . 63. (a) t x, y (c) t x, y f x, y 2 f x, y ■ ■ ■ ■ e cx 2 ■ ■ ■ ■ ■ (b) t x, y (d) t x, y 2 f x, y 2 f x, y ■ xy ■ ■ x2 ■ y2 ■ ■ y2 . How does the shape of the graph depend sx 2 lns x 2 y2 e sx f x, y y2 f x, y 2 f x, y y2 sin(s x 2 y2) 1 sx 2 y2 ; 71. (a) Show that, by taking logarithms, the general CobbDouglas function P z2 ■ ■ 68. f x, y ln ■ ■ In general, if t is a function of one variable, how is the graph t(s x 2 y 2 ) obtained from the graph of t? of f x, y 5z 2 ■ ■ y y2 f x, y y2 ■ ■ f x, y 5z 3y ■ 10 xy ■ x2 y2 sin 2x ■ 4y 2 ■ x x2 f x, y on c? Describe the level surfaces of the function. 59. f x, y, z ■ ; 70. Graph the functions 56. z sin s x 2 ■ ; 69. Use a computer to investigate the family of functions x 53–58 |||| Match the function (a) with its graph (labeled A–F on page 937) and (b) with its contour map (labeled I–VI). Give reasons for your choices. 53. z ■ x2 y2 x ye ■ 67. f x, y (dog saddle) ■ ■ |||| Use a computer to graph the function using various domains and viewpoints. Comment on the limiting behavior of the function. What happens as both x and y become large? What happens as x, y approaches the origin? (monkey saddle) 3 2 ; 67–68 |||| Use a computer to graph the function using various domains and viewpoints. Get a printout of one that, in your opinion, gives a good view. If your software also produces level curves, then plot some contour lines of the same function and compare with the graph. y3 ■ x4 3x ; 49–52 x3 ■ f x, y 4 |||| Use a computer to graph the function using various domains and viewpoints. Get a printout that gives a good view of the “peaks and valleys.” Would you say the function has a maximum value? Can you identify any points on the graph that you might consider to be “local maximum points”? What about “local minimum points”? 66. f x, y 48. If V x, y is the electric potential at a point x, y in the 49. f x, y (b) t x, y 2, y 3, y ; 65–66 ■ 47. A thin metal plate, located in the x y-plane, has temperature T x, y fx fx (c) t x, y ■ P K bL K 1 ln b can be expressed as ln L K (b) If we let x ln L K and y ln P K , the equation in part (a) becomes the linear equation y x ln b. Use Table 2 (in Example 3) to make a table of values of ln L K and ln P K for the years 1899–1922. Then use a graphing calculator or computer to find the least squares regression line through the points ln L K , ln P K . (c) Deduce that the Cobb-Douglas production function is P 1.01L 0.75K 0.25. ■ 5E-15(pp 0932-0941) 1/18/06 4:07 PM Page 937 ❙❙❙❙ S ECTION 15.1 FUNCTIONS OF SEVERAL VARIABLES z A z B z C y y y x x D E z x z z F y x x I x y II y III y y x x IV x V y VI y y x 0 y x x 937 5E-15(pp 0932-0941) ❙❙❙❙ 938 1/18/06 4:07 PM Page 938 CHAPTER 15 PARTIAL DERIVATIVES |||| 15.2 Limits and Continuity Let’s compare the behavior of the functions f x, y sin x 2 y 2 x2 y2 and x2 x2 t x, y y2 y2 as x and y both approach 0 [and therefore the point x, y approaches the origin]. TABLE 1 Values of f x, y y 1.0 0.5 0.2 1.0 0.455 0.759 0.5 0.759 0.2 T ABLE 2 Values of t x, y y 1.0 0.5 0.2 0 0.2 0.5 1.0 1.0 0.000 0.600 0.923 1.000 0.923 0.600 0.000 0.759 0.5 0.600 0.000 0.724 1.000 0.724 0.000 0.600 0.986 0.829 0.2 0.923 0.724 0.000 1.000 0.000 0.724 0.923 1.000 0.990 0.841 0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.986 0.829 0.2 0.923 0.724 0.000 1.000 0.000 0.724 0.923 0.986 0.990 0.986 0.959 0.759 0.5 0.600 0.000 0.724 1.000 0.724 0.000 0.600 0.829 0.841 0.829 0.759 0.455 1.0 0.000 0.600 0.923 1.000 0.923 0.600 0.000 0 0.2 0.5 1.0 0.829 0.841 0.829 0.759 0.455 0.959 0.986 0.990 0.986 0.959 0.829 0.986 0.999 1.000 0.999 0 0.841 0.990 1.000 0.2 0.829 0.986 0.999 0.5 0.759 0.959 1.0 0.455 0.759 x x Tables 1 and 2 show values of f x, y and t x, y , correct to three decimal places, for points x, y near the origin. (Notice that neither function is defined at the origin.) It appears that as x, y approaches (0, 0), the values of f x, y are approaching 1 whereas the values of t x, y aren’t approaching any number. It turns out that these guesses based on numerical evidence are correct, and we write lim x, y l 0, 0 sin x 2 y 2 x2 y2 1 and lim x, y l 0, 0 x2 x2 y2 y2 does not exist In general, we use the notation lim x, y l a, b f x, y L to indicate that the values of f x, y approach the number L as the point x, y approaches the point a, b along any path that stays within the domain of f . In other words, we can make the values of f x, y as close to L as we like by taking the point x, y sufficiently close to the point a, b , but not equal to a, b . A more precise definition follows. 1 Definition Let f be a function of two variables whose domain D includes points arbitrarily close to a, b . Then we say that the limit of f x, y as x, y approaches a, b is L and we write lim x, y l a, b if for every number f x, y L f x, y L 0 there is a corresponding number whenever x, y D and 0 sx 0 such that a 2 y b 2 5E-15(pp 0932-0941) 1/18/06 4:08 PM Page 939 SECTION 15.2 LIMITS AND CONTINUITY ❙❙❙❙ 939 Other notations for the limit in Definition 1 are lim f x, y L xla ylb and f x, y l L as x, y l a, b Notice that f x, y L is the distance between the numbers f x, y and L, and 2 2 is the distance between the point x, y and the point a, b . Thus, yb sx a Definition 1 says that the distance between f x, y and L can be made arbitrarily small by making the distance from x, y to a, b sufficiently small (but not 0). Figure 1 illustrates Definition 1 by means of an arrow diagram. If any small interval L ,L is given around L , then we can find a disk D with center a, b and radius 0 such that f maps all the points in D [except possibly a, b ] into the interval L ,L . y (x, y) ∂ D f (a, b) 0 x ( 0 ) L-∑ L L+∑ z FIGURE 1 z L+∑ L L-∑ S 0 x (a, b) D∂ y FIGURE 2 y b 0 FIGU R E 3 a Another illustration of Definition 1 is given in Figure 2 where the surface S is the graph of f . If 0 is given, we can find 0 such that if x, y is restricted to lie in the disk D and x, y a, b , then the corresponding part of S lies between the horizontal planes and z L . zL For functions of a single variable, when we let x approach a, there are only two possible directions of approach, from the left or from the right. We recall from Chapter 2 that if lim x l a f x lim x l a f x , then lim x l a f x does not exist. For functions of two variables the situation is not as simple because we can let x, y approach a, b from an infinite number of directions in any manner whatsoever (see Figure 3) as long as x, y stays within the domain of f . Definition 1 says that the distance between f x, y and L can be made arbitrarily small by making the distance from x, y to a, b sufficiently small (but not 0). The definition refers only to the distance between x, y and a, b . It does not refer to the direction of approach. Therefore, if the limit exists, then f x, y must approach the same limit no matter how x, y approaches a, b . Thus, if we can find two different paths of approach along which the function f x, y has different limits, then it follows that lim x, y l a, b f x, y does not exist. x If f x, y l L 1 as x, y l a, b along a path C1 and f x, y l L 2 as x, y l a, b along a path C2 , where L 1 L 2 , then lim x, y l a, b f x, y does not exist. EXAMPLE 1 Show that x, y l 0, 0 SOLUTION Let f x, y Then y lim 0 gives f x, 0 x2 x2 x2 y2 does not exist. y2 y 2 x 2 y 2 . First let’s approach 0, 0 along the x-axis. x 2 x 2 1 for all x 0, so f x, y l 1 as x, y l 0, 0 along the x-axis 5E-15(pp 0932-0941) 940 ❙❙❙❙ 1/18/06 4:09 PM Page 940 CHAPTER 15 PARTIAL DERIVATIVES y We now approach along the y-axis by putting x all y 0, so f=_1 f x, y l x f=1 SOLUTION If y xy x 2 If x 0, then f 0, y 0y 1 f= 2 lim x, y l 0, 0 f x, y exist? 0. Therefore x, y l 0, 0 along the x-axis 0, so as x, y l 0, 0 along the y-axis Although we have obtained identical limits along the axes, that does not show that the given limit is 0. Let’s now approach 0, 0 along another line, say y x. For all x 0, f x, x x f=0 y 2 , does as 2 f x, y l 0 y f=0 x, y l 0, 0 along the y-axis 0 x2 0, then f x, 0 f x, y l 0 y=x as 1 for (See Figure 4.) Since f has two different limits along two different lines, the given limit does not exist. (This confirms the conjecture we made on the basis of numerical evidence at the beginning of this section.) EXAMPLE 2 If f x, y FIGURE 4 1 y2 y2 0. Then f 0, y f x, y l 1 2 Therefore as x2 x 2 x 1 2 2 x, y l 0, 0 along y x (See Figure 5.) Since we have obtained different limits along different paths, the given limit does not exist. FIGURE 5 Figure 6 sheds some light on Example 2. The ridge that occurs above the line y x cor1 responds to the fact that f x, y 2 for all points x, y on that line except the origin. z y In Visual 15.2 a rotating line on the surface in Figure 6 shows different limits at the origin from different directions. x FIGURE 6 f(x, y)= xy ≈+¥ EXAMPLE 3 If f x, y xy 2 x2 y4 , does lim x, y l 0, 0 f x, y exist? SOLUTION With the solution of Example 2 in mind, let’s try to save time by letting x, y l 0, 0 along any nonvertical line through the origin. Then y is the slope, and f x, y So x mx 2 x mx f x, m x f x, y l 0 2 as 4 m 2x 3 x m 4x 4 2 x, y l 0, 0 along y m x, where m m 2x 1 m 4x 2 mx 5E-15(pp 0932-0941) 1/18/06 4:09 PM Page 941 SECTION 15.2 LIMITS AND CONTINUITY |||| Figure 7 shows the graph of the function in Example 3. Notice the ridge above the parabola x y 2. 0.5 _0.5 2 0 x 2 _2 0y _2 941 Thus, f has the same limiting value along every nonvertical line through the origin. But that does not show that the given limit is 0, for if we now let x, y l 0, 0 along the parabola x y 2, we have y2 y2 y4 1 f x, y f y 2, y 22 4 y y 2y 4 2 f x, y l 1 2 so z0 ❙❙❙❙ as y2 x, y l 0, 0 along x Since different paths lead to different limiting values, the given limit does not exist. Now let’s look at limits that do exist. Just as for functions of one variable, the calculation of limits for functions of two variables can be greatly simplified by the use of properties of limits. The Limit Laws listed in Section 2.3 can be extended to functions of two variables. The limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on. In particular, the following equations are true. FIGURE 7 lim 2 x, y l a, b x a lim y x, y l a, b b lim x, y l a, b c c The Squeeze Theorem also holds. EXAMPLE 4 Find lim x, y l 0, 0 3x 2y if it exists. x2 y2 SOLUTION As in Example 3, we could show that the limit along any line through the origin is 0. This doesn’t prove that the given limit is 0, but the limits along the parabolas y x 2 and x y 2 also turn out to be 0, so we begin to suspect that the limit does exist and is equal to 0. Let 0. We want to find 0 such that 3x 2 y x y2 0 2 3x 2 y x2 y2 that is, But x 2 x2 y 2 since y 2 3x 2 y x2 y2 |||| Another way to do Example 4 is to use the Squeeze Theorem instead of Definition 1. From (2) it follows that 3y 0 0, so x 2 x 2 0 3 sx 2 sx 2 y2 y2 1 and therefore 3 sy 2 sx 2 sx 2 0 0 y2 3y 3 and let 0 Thus, if we choose lim whenever 3x 2 y x2 y2 3 x, y l 0, 0 whenever 3 sx 2 y2 y2 y2 , then 3 3 3 Hence, by Definition 1, and so the first inequality in (3) shows that the given limit is 0. lim x, y l 0, 0 3x 2y x2 y2 0 Continuity Recall that evaluating limits of continuous functions of a single variable is easy. It can be accomplished by direct substitution because the defining property of a continuous function 5E-15(pp 0942-0951) 942 ❙❙❙❙ 1/18/06 2:48 PM Page 942 CHAPTER 15 PARTIAL DERIVATIVES is limx l a f x f a . Continuous functions of two variables are also defined by the direct substitution property. 4 Definition A function f of two variables is called continuous at a, b if lim f x, y x, y l a, b f a, b We say f is continuous on D if f is continuous at every point a, b in D. The intuitive meaning of continuity is that if the point x, y changes by a small amount, then the value of f x, y changes by a small amount. This means that a surface that is the graph of a continuous function has no hole or break. Using the properties of limits, you can see that sums, differences, products, and quotients of continuous functions are continuous on their domains. Let’s use this fact to give examples of continuous functions. A polynomial function of two variables (or polynomial, for short) is a sum of terms of the form cx m y n, where c is a constant and m and n are nonnegative integers. A rational function is a ratio of polynomials. For instance, x4 f x, y 5x 3 y 2 6 xy 4 7y 6 is a polynomial, whereas 2 xy 1 x2 y2 t x, y is a rational function. The limits in (2) show that the functions f x, y x, t x, y y, and h x, y c are continuous. Since any polynomial can be built up out of the simple functions f , t, and h by multiplication and addition, it follows that all polynomials are continuous on 2. Likewise, any rational function is continuous on its domain because it is a quotient of continuous functions. EXAMPLE 5 Evaluate lim x, y l 1, 2 x2y3 x 3y 2 2y . 3x x 2 y 3 x 3 y 2 3x 2y is a polynomial, it is continuous everywhere, so we can find the limit by direct substitution: SOLUTION Since f x, y lim x, y l 1, 2 x2y3 x 3y 2 3x 2y EXAMPLE 6 Where is the function f x, y 12 23 x2 x2 13 22 31 22 y2 continuous? y2 SOLUTION The function f is discontinuous at 0, 0 because it is not defined there. Since f is a rational function, it is continuous on its domain, which is the set D x, y x, y 0, 0 . EXAMPLE 7 Let t x, y x2 x2 0 y2 y2 if x, y 0, 0 if x, y 0, 0 11 5E-15(pp 0942-0951) 1/18/06 2:49 PM Page 943 S ECTION 15.2 LIMITS AND CONTINUITY ❙❙❙❙ 943 Here t is defined at 0, 0 but t is still discontinuous at 0 because lim x, y l 0, 0 t x, y does not exist (see Example 1). |||| Figure 8 shows the graph of the continuous function in Example 8. EXAMPLE 8 Let z 3x 2y x2 y2 0 f x, y y x We know f is continuous for x, y Also, from Example 4, we have lim x, y l 0, 0 FIGURE 8 if x, y 0, 0 if x, y 0, 0 0, 0 since it is equal to a rational function there. f x, y lim x, y l 0, 0 3x 2y x y2 0 2 f 0, 0 2 Therefore, f is continuous at 0, 0 , and so it is continuous on . Just as for functions of one variable, composition is another way of combining two continuous functions to get a third. In fact, it can be shown that if f is a continuous function of two variables and t is a continuous function of a single variable that is defined on the range of f , then the composite function h t f defined by h x, y t f x, y is also a continuous function. 2 z0 EXAMPLE 9 Where is the function h x, y _2 _2 _2 y0 0 2 2 x arctan y x continuous? SOLUTION The function f x, y y x is a rational function and therefore continuous except on the line x 0. The function t t arctan t is continuous everywhere. So the composite function t f x, y arctan y x h x, y FIGURE 9 The function h(x, y)=arctan (y/x) is discontinuous where x=0. is continuous except where x h above the y-axis. 0. The graph in Figure 9 shows the break in the graph of Functions of Three or More Variables Everything that we have done in this section can be extended to functions of three or more variables. The notation f x, y, z lim x, y, z l a, b, c L means that the values of f x, y, z approach the number L as the point x, y, z approaches the point a, b, c along any path in the domain of f. Because the distance between two points x, y, z and a, b, c in 3 is given by s x a 2 y b2 z c 2, we can write the precise definition as follows: For every number 0 there is a corresponding number 0 such that f x, y, z L whenever 0 sx a 2 and x, y, z is in the domain of f. The function f is continuous at a, b, c if lim x, y, z l a, b, c f x, y, z f a, b, c y b 2 z c 2 5E-15(pp 0942-0951) 944 ❙❙❙❙ 1/18/06 2:49 PM Page 944 CHAPTER 15 PARTIAL DERIVATIVES For instance, the function 1 f x, y, z x2 y2 z2 1 is a rational function of three variables and so is continuous at every point in 3 except where x 2 y 2 z 2 1. In other words, it is discontinuous on the sphere with center the origin and radius 1. If we use the vector notation introduced at the end of Section 15.1, then we can write the definitions of a limit for functions of two or three variables in a single compact form as follows. If f is defined on a subset D of n, then lim x l a f x L means that for every number 0 there is a corresponding number 0 such that 5 fx L whenever x D and 0 x a Notice that if n 1, then x x and a a, and (5) is just the definition of a limit for functions of a single variable. For the case n 2, we have x x, y , a a, b , and x a y b 2, so (5) becomes Definition 1. If n 3, then sx a2 x x, y, z , a a, b, c , and (5) becomes the definition of a limit of a function of three variables. In each case the definition of continuity can be written as lim f x fa xla |||| 15.2 Exercises 1. Suppose that lim f x, y 6. What can you say about the value of f 3, 1 ? What if f is continuous? x, y l 3, 1 7. 2. Explain why each function is continuous or discontinuous. 9. (a) The outdoor temperature as a function of longitude, latitude, and time (b) Elevation (height above sea level) as a function of longitude, latitude, and time (c) The cost of a taxi ride as a function of distance traveled and time 3–4 11. 13. Use a table of numerical values of f x, y for x, y near the origin to make a conjecture about the value of the limit of f x, y as x, y l 0, 0 . Then explain why your guess is correct. x 2y 3 ■ ■ ■ x 3y 2 2 xy ■ ■ 5 ■ ■ ■ ■ ■ ■ 5–20 |||| Find the limit, if it exists, or show that the limit does not exist. 5. 6. lim x, y l 5, lim x, y l 6, 3 2 x5 4 x 3y 15. 17. 2x y x 2 2y 2 4. f x, y ■ 18. 19. y2 x y cos y 3x 2 y 2 lim x, y l 0, 0 10. xy lim sx x, y l 0, 0 2 y 12. 2 2x y x y2 lim 2y ■ 14. 4 x, y l 0, 0 x lim sx 2 x, y l 0, 0 2 lim x 2 sin 2 y 2x 2 y 2 lim 6x 3 y 2x 4 y 4 lim x4 x2 lim x 2 sin2 y x 2 2y 2 x, y l 0, 0 x, y l 0, 0 x, y l 0, 0 y y2 xy 1 sin 16. y4 y2 xy4 lim 2 y8 z2 1 e lim x 2 2y 2 x2 y2 lim xy x2 yz 2 y2 xy x2 yz y2 x xz 2 z4 lim x, y l 0, 0 3z 2 z2 x, y, z l 0, 0, 0 x, y, z l 0, 0, 0 x, y, z l 0, 0, 0 ■ x, y l 0, 0 2 lim x, y, z l 3, 0, 1 5x y 2 20. x y cos x x2 x, y l 0, 0 8. 2 |||| 3. f x, y x2 lim ■ ■ ■ zx z2 ■ ■ ■ ■ ■ ■ ■ 5E-15(pp 0942-0951) 1/18/06 2:50 PM Page 945 S ECTION 15.3 PARTIAL DERIVATIVES x2y3 2x2 y2 1 ; 21–22 |||| Use a computer graph of the function to explain why the limit does not exist. 21. 22. 2x 2 lim 3x y 4y 2 3x 5y 2 xy3 lim x x, y l 0, 0 ■ 2 y ■ 6 ■ ■ ■ ■ ■ ■ ■ ■ Find h x, y continuous. t2 24. t t ■ ; 25–26 f x, y 2x 3y st st 1 , 1 f x, y x2 ■ ■ ■ ■ ■ 6 37. ■ ■ ■ ■ 38. ■ e1 27–36 xy ■ ■ ■ 26. f x, y ■ ■ ■ 30. F x, y sin x y e x y2 ■ ■ 1 x2 1 ■ 28. F x, y arctan( x ex 2 y ln x 2 32. G x, y sin 1 33. f x, y, z 34. f x, y, z |||| 15.3 0, 0 ■ ■ ■ ■ ■ ■ ■ ■ ■ y3 y2 lim x2 y 2 ln x 2 x, y l 0, 0 ■ ■ x 1 x y x2 x2 sx y ■ ■ ■ ■ ■ ■ lim x, y, z l 0, 0, 0 x yz y2 x2 z2 ; 40. At the beginning of this section we considered the function y 2 sin x 2 y 2 x2 y2 and guessed that f x, y l 1 as x, y l 0, 0 on the basis of numerical evidence. Use polar coordinates to confirm the value of the limit. Then graph the function. 4 y2 sy y2 ■ f x, y y2 2 ■ ■ y 2 ■ y2 39. Use spherical coordinates to find y2 ■ x, y l 0, 0 sy ) sx 31. G x, y ■ x3 x2 Determine the set of points at which the function is continuous. 29. F x, y if x, y lim |||| 27. F x, y 0, 0 ■ |||| 25. f x, y if x, y y2 |||| Use polar coordinates to find the limit. [If r, are polar coordinates of the point x, y with r 0, note that r l 0 as x, y l 0, 0 .] y ■ 0, 0 ■ Graph the function and observe where it is discontinuous. Then use the formula to explain what you have observed. ■ if x, y 37–38 st, ■ 0, 0 t f x, y and the set on which h is |||| 23. t t xy xy x2 0 36. f x, y ■ 23–24 if x, y 945 2 x, y l 0, 0 ■ 35. f x, y ❙❙❙❙ 41. Show that the function f given by f x on z2 42. If c n . [Hint : Consider x a 2 x x is continuous a x a .] Vn , show that the function f given by f x continuous on n. z c x is Partial Derivatives On a hot day, extreme humidity makes us think the temperature is higher than it really is, whereas in very dry air we perceive the temperature to be lower than the thermometer indicates. The National Weather Service has devised the heat index (also called the temperature-humidity index, or humidex, in some countries) to describe the combined effects of temperature and humidity. The heat index I is the perceived air temperature when the actual temperature is T and the relative humidity is H. So I is a function of T and H and we can write I f T, H . The following table of values of I is an excerpt from a table compiled by the National Weather Service. 5E-15(pp 0942-0951) 946 ❙❙❙❙ 1/18/06 2:50 PM Page 946 CHAPTER 15 PARTIAL DERIVATIVES TABLE 1 Relative humidity (%) Heat index I as a function of temperature and humidity H 50 55 60 65 70 75 80 85 90 90 96 98 100 103 106 109 112 115 119 92 100 103 105 108 112 115 119 123 128 94 104 107 111 114 118 122 127 132 137 96 109 113 116 121 125 130 135 141 146 98 114 118 123 127 133 138 144 150 157 100 119 124 129 135 141 147 154 161 168 T Actual temperature (°F) If we concentrate on the highlighted column of the table, which corresponds to a relative humidity of H 70%, we are considering the heat index as a function of the single f T, 70 . Then t T describes how the variable T for a fixed value of H. Let’s write t T heat index I increases as the actual temperature T increases when the relative humidity is 70%. The derivative of t when T 96 F is the rate of change of I with respect to T when T 96 F : t 96 lim t 96 hl0 h h t 96 f 96 lim hl0 h, 70 h We can approximate it using the values in Table 1 by taking h t 98 t 96 t 96 f 98, 70 2 2 and 133 2 t 94 t 96 f 96, 70 t 96 2: 125 2 f 94, 70 2 f 96, 70 f 96, 70 118 2 125 2 4 3.5 Averaging these values, we can say that the derivative t 96 is approximately 3.75. This means that, when the actual temperature is 96 F and the relative humidity is 70%, the apparent temperature (heat index) rises by about 3.75 F for every degree that the actual temperature rises! Now let’s look at the highlighted row in Table 1, which corresponds to a fixed temperf 96, H , ature of T 96 F. The numbers in this row are values of the function G H which describes how the heat index increases as the relative humidity H increases when the actual temperature is T 96 F. The derivative of this function when H 70% is the rate of change of I with respect to H when H 70%: G 70 By taking h G 70 G 70 lim G 70 hl0 5 and h h G 70 lim f 96, 70 hl0 h h f 96, 70 5, we approximate G 70 using the tabular values: G 75 G 70 f 96, 75 5 G 65 G 70 5 f 96, 70 130 5 f 96, 65 f 96, 70 5 125 5 121 125 5 1 0.8 5E-15(pp 0942-0951) 1/18/06 2:50 PM Page 947 SECTION 15.3 PARTIAL DERIVATIVES ❙❙❙❙ 947 By averaging these values we get the estimate G 70 0.9. This says that, when the temperature is 96 F and the relative humidity is 70%, the heat index rises about 0.9 F for every percent that the relative humidity rises. In general, if f is a function of two variables x and y, suppose we let only x vary while keeping y fixed, say y b, where b is a constant. Then we are really considering a function of a single variable x, namely, t x f x, b . If t has a derivative at a, then we call it the partial derivative of f with respect to x at a, b and denote it by fx a, b . Thus 1 fx a, b ta where tx f x, b By the definition of a derivative, we have ta lim ta hl0 h h ta and so Equation 1 becomes 2 fx a, b lim fa hl0 h, b h f a, b Similarly, the partial derivative of f with respect to y at a, b , denoted by fy a, b , is obtained by keeping x fixed x a and finding the ordinary derivative at b of the function G y f a, y : 3 fy a, b lim f a, b hl0 h h f a, b With this notation for partial derivatives, we can write the rates of change of the heat index I with respect to the actual temperature T and relative humidity H when T 96 F and H 70% as follows: f T 96, 70 3.75 fH 96, 70 0.9 If we now let the point a, b vary in Equations 2 and 3, fx and fy become functions of two variables. 4 If f is a function of two variables, its partial derivatives are the functions fx and fy defined by fx x, y fy x, y lim fx hl0 lim hl0 f x, y h, y h f x, y h h f x, y 5E-15(pp 0942-0951) 948 ❙❙❙❙ 1/18/06 2:51 PM Page 948 CHAPTER 15 PARTIAL DERIVATIVES There are many alternative notations for partial derivatives. For instance, instead of fx we can write f1 or D1 f (to indicate differentiation with respect to the first variable) or f x. But here f x can’t be interpreted as a ratio of differentials. Notations for Partial Derivatives If z f x, y , we write fx x, y fx f x x fy x, y fy f y y f x, y z x f1 D1 f Dx f f x, y z y f2 D2 f Dy f To compute partial derivatives, all we have to do is remember from Equation 1 that the partial derivative with respect to x is just the ordinary derivative of the function t of a single variable that we get by keeping y fixed. Thus, we have the following rule. Rule for Finding Partial Derivatives of z f x, y 1. To find fx , regard y as a constant and differentiate f x, y with respect to x. 2. To find fy , regard x as a constant and differentiate f x, y with respect to y. EXAMPLE 1 If f x, y x3 x2y3 2y 2, find fx 2, 1 and fy 2, 1 . SOLUTION Holding y constant and differentiating with respect to x, we get fx x, y and so 3x 2 fx 2, 1 3 22 2 xy 3 2 2 13 16 Holding x constant and differentiating with respect to y, we get fy x, y 3x 2 y 2 fy 2, 1 3 2 2 12 4y 41 8 Interpretations of Partial Derivatives z T¡ S C¡ T™ P (a, b, c) 0 C™ y x (a, b, 0) FIGURE 1 The partial derivatives of f at (a, b) are the slopes of the tangents to C ¡ and C ™. To give a geometric interpretation of partial derivatives, we recall that the equation z f x, y represents a surface S (the graph of f ). If f a, b c, then the point P a, b, c lies on S. By fixing y b, we are restricting our attention to the curve C1 in which the vertical plane y b intersects S. (In other words, C1 is the trace of S in the plane y b.) Likewise, the vertical plane x a intersects S in a curve C2 . Both of the curves C1 and C2 pass through the point P. (See Figure 1.) f x, b , so the slope of its Notice that the curve C1 is the graph of the function t x fx a, b . The curve C2 is the graph of the function G y f a, y , tangent T1 at P is t a fy a, b . so the slope of its tangent T2 at P is G b Thus, the partial derivatives fx a, b and fy a, b can be interpreted geometrically as the slopes of the tangent lines at P a, b, c to the traces C1 and C2 of S in the planes y b and x a. 5E-15(pp 0942-0951) 1/18/06 2:51 PM Page 949 S ECTION 15.3 PARTIAL DERIVATIVES z z=4-≈-2¥ ❙❙❙❙ 949 As we have seen in the case of the heat index function, partial derivatives can also be interpreted as rates of change. If z f x, y , then z x represents the rate of change of z with respect to x when y is fixed. Similarly, z y represents the rate of change of z with respect to y when x is fixed. C¡ EXAMPLE 2 If f x, y y=1 4 x2 2y 2, find fx 1, 1 and fy 1, 1 and interpret these num- bers as slopes. SOLUTION We have (1, 1, 1) fx x, y x FIGURE 2 z z=4-≈-2¥ C™ x=1 (1, 1, 1) y 2x fy x, y 4y fx 1, 1 y (1, 1) 2 2 fy 1, 1 4 The graph of f is the paraboloid z 4 x 2 2y 2 and the vertical plane y 1 intersects it in the parabola z 2 x 2, y 1. (As in the preceding discussion, we label it C1 in Figure 2.) The slope of the tangent line to this parabola at the point 1, 1, 1 is fx 1, 1 2. Similarly, the curve C2 in which the plane x 1 intersects the paraboloid is the parabola z 3 2y 2, x 1, and the slope of the tangent line at 1, 1, 1 is fy 1, 1 4. (See Figure 3.) Figure 4 is a computer-drawn counterpart to Figure 2. Part (a) shows the plane y 1 intersecting the surface to form the curve C1 and part (b) shows C1 and T1 . [We have used the vector equations r t t, 1, 2 t 2 for C1 and r t 1 t, 1, 1 2 t for T1 .] Similarly, Figure 5 corresponds to Figure 3. 2 x (1, 1) FIGURE 3 4 4 3 3 z2 z2 1 1 0 0 0 y 1 1 FIGURE 4 2 0 x y 1 1 (a) 2 x (b) 4 4 3 3 z2 z2 1 1 0 FIGURE 5 0 0 0 0 y 1 1 2 x 0 0 0 y 1 1 2 x 5E-15(pp 0942-0951) 950 ❙❙❙❙ 1/18/06 2:51 PM Page 950 CHAPTER 15 PARTIAL DERIVATIVES EXAMPLE 3 If f x, y sin x 1 , calculate y f f and . x y SOLUTION Using the Chain Rule for functions of one variable, we have f x f y |||| Some computer algebra systems can plot surfaces defined by implicit equations in three variables. Figure 6 shows such a plot of the surface defined by the equation in Example 4. cos cos EXAMPLE 4 Find z x 1 x y x 1 y y 1 x 1 y cos x x 1 1 y x cos y 1 y 1 x y 1 y 2 x and z y if z is defined implicitly as a function of x and y by the equation x3 SOLUTION To find z y3 z3 6 xy z 1 x, we differentiate implicitly with respect to x, being careful to treat y as a constant: 3x 2 z x 3z 2 6y z 6 xy x2 z2 z x 2yz 2 xy 0 Solving this equation for z x, we obtain z x FIGURE 6 Similarly, implicit differentiation with respect to y gives In Visual 15.3 you can zoom and rotate the surfaces in Figure 6 and Exercises 41–44. y2 z2 z y 2xz 2 xy Functions of More than Two Variables Partial derivatives can also be defined for functions of three or more variables. For example, if f is a function of three variables x, y, and z, then its partial derivative with respect to x is defined as fx x, y, z lim hl0 fx h, y, z h f x, y, z and it is found by regarding y and z as constants and differentiating f x, y, z with respect w x can be interpreted as the rate of change of w with to x. If w f x, y, z , then fx respect to x when y and z are held fixed. But we can’t interpret it geometrically because the graph of f lies in four-dimensional space. In general, if u is a function of n variables, u f x 1, x 2 , . . . , x n , its partial derivative with respect to the ith variable x i is u xi lim hl0 f x1 , . . . , xi 1 , xi h, xi 1 , . . . , xn h f x1 , . . . , xi , . . . , xn 5E-15(pp 0942-0951) 1/18/06 2:52 PM Page 951 SECTION 15.3 PARTIAL DERIVATIVES ❙❙❙❙ 951 and we also write u xi f xi fx i fi Di f e x y ln z. EXAMPLE 5 Find fx , fy , and fz if f x, y, z SOLUTION Holding y and z constant and differentiating with respect to x, we have ye x y ln z fx Similarly, xe x y ln z fy and e xy z fz Higher Derivatives If f is a function of two variables, then its partial derivatives fx and fy are also functions of two variables, so we can consider their partial derivatives fx x , fx y , fy x , and fy y , which are called the second partial derivatives of f . If z f x, y , we use the following notation: fx x fxx fx y fxy x fyx fy y fyy y f22 f y f y f x f x x f21 2 f x y f12 fy x f11 2 2 2 z x2 2 z yx f yx 2 2 z xy f xy 2 f y 2 2 z y2 Thus, the notation fx y (or 2 f y x) means that we first differentiate with respect to x and then with respect to y, whereas in computing fyx the order is reversed. EXAMPLE 6 Find the second partial derivatives of x3 f x, y x2y3 2y 2 SOLUTION In Example 1 we found that fx x, y 3x 2 2 xy 3 6x 2y 3 fy x, y 3x 2 y 2 4y Therefore fxx fyx x x 3x 2 3x 2 y 2 2 xy 3 4y 6 xy 2 fxy fyy y y 3x 2 3x 2 y 2 2 xy 3 6 xy 2 4y 6x 2 y 4 5E-15(pp 0952-0961) 952 ❙❙❙❙ 1/18/06 3:07 PM Page 952 CHAPTER 15 PARTIAL DERIVATIVES 20 z0 _20 |||| Figure 7 shows the graph of the function f in Example 6 and the graphs of its first- and second-order partial derivatives for 2 x 2, 2 y 2. Notice that these graphs are consistent with our interpretations of fx and fy as slopes of tangent lines to traces of the graph of f . For instance, the graph of f decreases if we start at 0, 2 and move in the positive x-direction. This is reflected in the negative values of fx. You should compare the graphs of fy x and fyy with the graph of fy to see the relationships. _40 _2 _1 y 0 _2 _1 1 0x 22 1 f 40 40 z 20 z 20 0 _20 _2 _1 y 0 0 _2 _1 1 0x 22 1 _2 _1 y fx z0 _2 _1 y 0 1 _2 _1 1 0x 22 fxx 1 20 z0 _20 _20 40 20 z0 _2 _1 1 0x 22 fy 40 20 1 0 _2 _1 1 0x 22 _20 _40 _2 _1 y 0 _40 _2 _1 1 0x 22 1 _2 _1 y 0 fxy fyx fyy FIGURE 7 Notice that fx y fyx in Example 6. This is not just a coincidence. It turns out that the mixed partial derivatives fx y and fyx are equal for most functions that one meets in practice. The following theorem, which was discovered by the French mathematician Alexis Clairaut (1713–1765), gives conditions under which we can assert that fx y fyx . The proof is given in Appendix F. |||| Alexis Clairaut was a child prodigy in mathematics, having read l’Hospital’s textbook on calculus when he was ten and presented a paper on geometry to the French Academy of Sciences when he was 13. At the age of 18, Clairaut published Recherches sur les courbes à double courbure, which was the first systematic treatise on three-dimensional analytic geometry and included the calculus of space curves. Clairaut’s Theorem Suppose f is defined on a disk D that contains the point a, b . If the functions fx y and fyx are both continuous on D, then fx y a, b fyx a, b Partial derivatives of order 3 or higher can also be defined. For instance, 2 fx yy fx y y y f 3 yx 2 y f x 5E-15(pp 0952-0961) 1/18/06 3:08 PM Page 953 SECTION 15.3 PARTIAL DERIVATIVES and using Clairaut’s Theorem it can be shown that fx yy continuous. EXAMPLE 7 Calculate fxx y z if f x, y, z SOLUTION fx 953 fyyx if these functions are yz . sin 3x yz 3 cos 3x fxx fyx y ❙❙❙❙ 9 sin 3x fxx y 9 z cos 3x fxx y z 9 cos 3x yz yz yz 9y z sin 3x yz Partial Differential Equations Partial derivatives occur in partial differential equations that express certain physical laws. For instance, the partial differential equation 2 u x2 2 u y2 0 is called Laplace’s equation after Pierre Laplace (1749–1827). Solutions of this equation are called harmonic functions and play a role in problems of heat conduction, fluid flow, and electric potential. EXAMPLE 8 Show that the function u x, y e x sin y is a solution of Laplace’s equation. ux e x sin y uy u xx SOLUTION e x sin y u yy u xx u yy e x sin y e x cos y e x sin y e x sin y 0 Therefore, u satisfies Laplace’s equation. The wave equation 2 u t2 u(x, t) x FIGURE 8 2 a2 u x2 describes the motion of a waveform, which could be an ocean wave, a sound wave, a light wave, or a wave traveling along a vibrating string. For instance, if u x, t represents the displacement of a vibrating violin string at time t and at a distance x from one end of the string (as in Figure 8), then u x, t satisfies the wave equation. Here the constant a depends on the density of the string and on the tension in the string. EXAMPLE 9 Verify that the function u x, t sin x a t satisfies the wave equation. SOLUTION u xx sin x ut t a 2 sin x ux ut cos x at a cos x So u satisfies the wave equation. at at at a 2 u xx 5E-15(pp 0952-0961) 954 ❙❙❙❙ 1/18/06 3:08 PM Page 954 CHAPTER 15 PARTIAL DERIVATIVES The Cobb-Douglas Production Function In Example 3 in Section 15.1 we described the work of Cobb and Douglas in modeling the total production P of an economic system as a function of the amount of labor L and the capital investment K. Here we use partial derivatives to show how the particular form of their model follows from certain assumptions they made about the economy. If the production function is denoted by P P L, K , then the partial derivative P L is the rate at which production changes with respect to the amount of labor. Economists call it the marginal production with respect to labor or the marginal productivity of labor. Likewise, the partial derivative P K is the rate of change of production with respect to capital and is called the marginal productivity of capital. In these terms, the assumptions made by Cobb and Douglas can be stated as follows. (i) If either labor or capital vanishes, then so will production. (ii) The marginal productivity of labor is proportional to the amount of production per unit of labor. (iii) The marginal productivity of capital is proportional to the amount of production per unit of capital. Because the production per unit of labor is P L, assumption (ii) says that P L P L for some constant . If we keep K constant K tion becomes an ordinary differential equation: 5 dP dL K0 , then this partial differential equa- P L If we solve this separable differential equation by the methods of Section 10.3 (see also Exercise 75), we get 6 P L, K0 C1 K0 L Notice that we have written the constant C1 as a function of K0 because it could depend on the value of K0 . Similarly, assumption (iii) says that P K P K and we can solve this differential equation to get 7 P L 0, K C2 L 0 K Comparing Equations 6 and 7, we have 8 P L, K bL K 5E-15(pp 0952-0961) 1/18/06 3:08 PM Page 955 S ECTION 15.3 PARTIAL DERIVATIVES ❙❙❙❙ 955 where b is a constant that is independent of both L and K. Assumption (i) shows that 0 and 0. Notice from Equation 8 that if labor and capital are both increased by a factor m, then P mL, mK b mL mK m bL K m P L, K If 1, then P mL, mK mP L, K , which means that production is also increased by a factor of m. That is why Cobb and Douglas assumed that 1 and therefore bL K 1 P L, K This is the Cobb-Douglas production function that we discussed in Section 15.1. |||| 15.3 Exercises (b) In general, what can you say about the signs of W T and W v ? (c) What appears to be the value of the following limit? 1. The temperature T at a location in the Northern Hemisphere depends on the longitude x, latitude y, and time t, so we can write T f x, y, t . Let’s measure time in hours from the beginning of January. (a) What are the meanings of the partial derivatives T x, T y, and T t ? (b) Honolulu has longitude 158 W and latitude 21 N. Suppose that at 9:00 A.M. on January 1 the wind is blowing hot air to the northeast, so the air to the west and south is warm and the air to the north and east is cooler. Would you expect fx 158, 21, 9 , fy 158, 21, 9 , and ft 158, 21, 9 to be positive or negative? Explain. lim vl of the wind and the length of time t that the wind has been blowing at that speed. Values of the function h f v, t are recorded in feet in the following table. Duration (hours) Actual temperature (°C) Wind speed (km / h) T 30 40 50 t 5 10 15 20 30 40 50 60 70 10 18 20 21 22 23 24 26 27 29 30 30 20 30 33 34 35 36 37 25 37 39 41 42 43 44 2 2 2 2 2 2 2 15 4 4 5 5 5 5 5 20 5 7 8 8 9 9 9 30 9 13 16 17 18 19 19 40 14 21 25 28 31 33 33 19 29 36 40 45 48 50 60 24 37 47 54 62 67 69 23 15 10 Wind speed (knots) from Table 1 in Section 15.1. 20 v 50 3. The wind-chill index W is the perceived temperature when the actual temperature is T and the wind speed is v, so we can write W f T, v . The following table of values is an excerpt v v 4. The wave heights h in the open sea depend on the speed v 2. At the beginning of this section we discussed the function I f T, H , where I is the heat index, T is the temperature, and H is the relative humidity. Use Table 1 to estimate fT 92, 60 and fH 92, 60 . What are the practical interpretations of these values? W (a) Estimate the values of fT 15, 30 and fv 15, 30 . What are the practical interpretations of these values? (a) What are the meanings of the partial derivatives h v and h t ? (b) Estimate the values of fv 40, 15 and ft 40, 15 . What are the practical interpretations of these values? (c) What appears to be the value of the following limit? lim tl h t 5E-15(pp 0952-0961) 956 ❙❙❙❙ 1/18/06 3:09 PM Page 956 CHAPTER 15 PARTIAL DERIVATIVES 8. A contour map is given for a function f . Use it to estimate 5–6 |||| Determine the signs of the partial derivatives for the function f whose graph is shown. fx 2, 1 and fy 2, 1 . z y 3 _4 0 1 x 2 8 6 _2 10 12 14 16 4 2 y 1 5. (a) fx 1, 2 (b) fy 1, 2 6. (a) fx (b) fy 1, 2 (d) fyy 1, 2 (c) fxx ■ ■ 3 x 18 16 4 x 2 y 2, find fx 1, 2 and fy 1, 2 and interpret these numbers as slopes. Illustrate with either hand-drawn sketches or computer plots. 9. If f x, y 1, 2 1, 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 10. If f x, y s4 x 2 4y 2, find fx 1, 0 and fy 1, 0 and interpret these numbers as slopes. Illustrate with either handdrawn sketches or computer plots. 7. The following surfaces, labeled a, b, and c, are graphs of a function f and its partial derivatives fx and fy. Identify each surface and give reasons for your choices. ; 11–12 |||| Find fx and fy and graph f , fx , and fy with domains and viewpoints that enable you to see the relationships between them. x2 11. f x, y 8 ■ ■ y2 ■ x2y ■ 12. f x, y ■ ■ ■ xe ■ x2 y2 ■ ■ ■ ■ 4 13–34 z0 _4 _8 _3 _2 _1 13. f x, y a 0 y Find the first partial derivatives of the function. |||| 1 2 3 2 0 _2 x 3x 2y 4 14. f x, y x5 3x 3y 2 xe 3 y 15. z 16. z 17. f x, y cos r ln r 2 0 y 1 2 3 2 _2 x x y 2z3 27. w ln x 2y 29. u 0 34. u 4 sx 2 1 sin x 1 ■ ■ ■ arctan ( x s t ) y 35–38 _8 _3 _2 _1 c 0 y 1 2 3 2 0 _2 x |||| cos t 2 dt 28. w 3z x2 2 sr 2 s2 xy z 2 xn nxn ■ ■ ■ ■ Find the indicated partial derivatives. 35. f x, y sx 2 36. f x, y sin 2 x t2 xy2 t 2z 32. f x, y, z, t 2x2 ■ x 2e y z z0 _4 x y t2 26. f x, y, z 3y z x y z 2 tan y t 33. u st 2 s 2 22. f x, t xe t sin 31. f x, y, z, t 8 20. f s, t s2 xy 30. u 25. f x, y, z b 18. f x, y te w t 23. u _4 y y y ln x 24. f x, y sin 21. f r, s z0 _3 _2 _1 x x 19. w 4 3xy 4 y2; fx 3, 4 3y ; fy 6, 4 ■ ■ ■ ■ 5E-15(pp 0952-0961) 1/18/06 3:10 PM Page 957 S ECTION 15.3 PARTIAL DERIVATIVES 37. f x, y, z xy z; fz 3, 2, 1 38. f u, v, w w tan u v ; fv 2, 0, 3 ■ ■ ■ ■ ■ ■ fx 3, 2 , fx 3, 2.2 , and fx y 3, 2 . ■ ■ ■ ■ ■ y ■ 1.8 2.0 2.5 12.5 10.2 9.3 3.0 18.1 17.5 15.9 3.5 39–40 |||| Use the definition of partial derivatives as limits (4) to find fx x, y and fy x, y . x2 ■ ■ 2y 2 xy ■ ■ 40. f x, y ■ ■ ■ s3x ■ ■ y ■ ■ ■ 41–44 |||| Use implicit differentiation to find z x and z y. (You can see what these surfaces look like in TEC Visual 15.3.) 41. x 2 y2 43. x ■ z2 z 42. y z 3x yz ■ ■ ■ ■ ■ ■ x ■ 2y ■ 20.0 22.4 26.1 the following partial derivatives are positive or negative at the point P. (a) fx (b) fy (c) fxx (d) fxy (e) fyy 3z ■ 2.2 66. Level curves are shown for a function f . Determine whether z ln x 44. sin x y z arctan y z ■ 957 65. Use the table of values of f x, y to estimate the values of x 39. f x, y ❙❙❙❙ ■ y 45–46 Find z |||| x and z 45. (a) z fx 46. (a) z y. f xty fxy (c) z ■ ■ 47–52 (b) z ■ ■ ■ 10 8 f xy ■ ■ ■ y 4 x 2 3x y 49. z xx 51. u 3 4 2 P ■ 48. f x, y e s sin t ■ ■ ■ ■ 50. z y ■ ■ ■ ■ ln 3x 5y 67. Verify that the function u y tan 2 x 52. v ■ x y2 sx ■ ■ x sin x 2y 54. u x 4y 2 55. u ln sx 2 y2 56. u ■ ■ x ye y ■ ■ ■ ■ ■ ■ ■ 2 xy 5 ■ kt sin k x is a solution of the u xx. 2 68. Determine whether each of the following functions is a solution ■ |||| Verify that the conclusion of Clairaut’s Theorem holds, that is, u x y u yx . 53. u 22 e heat conduction equation u t 53–56 ■ 6 Find all the second partial derivatives. |||| 47. f x, y ■ fx (b) z ty ■ ■ of Laplace’s equation u xx u yy 0. (a) u x 2 y 2 (b) u x 2 y 2 (c) u x 3 3x y 2 (d) u ln s x 2 y 2 (e) u sin x cosh y cos x sinh y (f) u e x cos y e y cos x ■ 69. Verify that the function u 57–64 1 s x 2 y 2 z 2 is a solution of the three-dimensional Laplace equation u xx u yy u zz 0. Find the indicated partial derivative. |||| 3x y 4 57. f x, y 2 58. f x, t x 3 y 2; ct xe ; 59. f x, y, z fxxy, 2z ; 3y 23 60. f r, s, t wave equation u t t a 2u xx. (a) u sin k x sin a k t (b) u t a 2t 2 x 2 (c) u x at 6 x at 6 (d) u sin x a t ln x a t ftxx fttt , cos 4 x 70. Show that each of the following functions is a solution of the fyyy r ln r s t ; frss, fxy z, fy zz frst 3 u e r sin ; 61. u 71. If f and t are twice differentiable functions of a single vari- r2 able, show that the function 3 62. z usv w; 2z y ; u x, t uvw 3 x 63. w z w zyx 3 , x e a1 x1 show that y 72. If u 6 ■ u x y2 z3 x a y bz c; 64. u ■ ■ ■ ■ a2 x2 an x n 2 ■ ■ at tx at is a solution of the wave equation given in Exercise 70. w 2 fx ■ ■ ■ ■ ■ u 2 x1 , where a 2 1 2 u 2 x2 a2 2 a2 n 2 u 2 xn u 1, 5E-15(pp 0952-0961) 958 ❙❙❙❙ 1/18/06 3:11 PM Page 958 CHAPTER 15 PARTIAL DERIVATIVES xe y 73. Show that the function z ye x is a solution of the 82. If a, b, c are the sides of a triangle and A, B, C are the opposite equation 3 3 z x3 3 z y3 3 z x y2 x y x 2 angles, find A a, A the Law of Cosines. z are fx x, y believe it? bL K satisfies the equation L P L K P K 2 y 2 z 2 16 intersects the plane y 2 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point 1, 2, 2 . P L 86. In a study of frost penetration it was found that the temperature T at time t (measured in days) at a depth x (measured in feet) can be modeled by the function 76. The temperature at a point x, y on a flat metal plate is given by T x, y 60 1 x 2 y 2 , where T is measured in C and x, y in meters. Find the rate of change of temperature with respect to distance at the point 2, 1 in (a) the x-direction and (b) the y-direction. T x, t 77. The total resistance R produced by three conductors with resis- tances R1, R2, R3 connected in a parallel electrical circuit is given by the formula 1 R1 1 R 1 R2 ; 1 R3 R1. T0 x T1e sin t x where 2 365 and is a positive constant. (a) Find T x. What is its physical significance? (b) Find T t. What is its physical significance? (c) Show that T satisfies the heat equation Tt k Txx for a certain constant k. (d) If 0.2, T0 0, and T1 10, use a computer to graph T x, t . (e) What is the physical significance of the term x in the expression sin t x? 87. Use Clairaut’s Theorem to show that if the third-order partial derivatives of f are continuous, then 78. The gas law for a fixed mass m of an ideal gas at absolute tem- perature T , pressure P, and volume V is PV is the gas constant. Show that V T P V T P m RT , where R fx yy T P T V T mR 11.37v 0.16 90. If f x, y 0.3965 T v 0.16 K v2 K 3 sx 3 y 3, find fx 0, 0 . 91. Let f x, y ; 81. The kinetic energy of a body with mass m and velocity v is K 1 mv 2. Show that 2 2 2 x x 2 y 2 3 2e sin x y , find fx 1, 0 . [Hint: Instead of finding fx x, y first, note that it’s easier to use Equation 1 or Equation 2.] 89. If f x, y where T is the temperature C and v is the wind speed km h . When T 15 C and v 30 km h, by how much would you expect the apparent temperature to drop if the actual temperature decreases by 1 C ? What if the wind speed increases by 1 km h ? K m fyyx two variables have? (b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of three variables. 1 80. The wind-chill index is modeled by the function 0.6215 T fyx y 88. (a) How many nth-order partial derivatives does a function of 79. For the ideal gas of Exercise 78, show that 13.12 y. Should you 3x 85. The ellipsoid 4 x 2 C1 K0 L by solving the differential equation (See Equation 5.) W 4y and fy x, y 6 x x 2 2 y 2 intersects the plane x 1 in a parabola. Find parametric equations for the tangent line to this parabola at the point 1, 2, 4 . Use a computer to graph the paraboloid, the parabola, and the tangent line on the same screen. P dP dL Find R x ; 84. The paraboloid z 75. Show that the Cobb-Douglas production function satisfies P L, K0 c by implicit differentiation of 83. You are told that there is a function f whose partial derivatives y 74. Show that the Cobb-Douglas production function P b, A CAS (a) (b) (c) (d) (e) x 3y x2 0 xy 3 y2 if x, y 0, 0 if x, y 0, 0 Use a computer to graph f . Find fx x, y and fy x, y when x, y 0, 0 . Find fx 0, 0 and fy 0, 0 using Equations 2 and 3. Show that fxy 0, 0 1 and fyx 0, 0 1. Does the result of part (d) contradict Clairaut’s Theorem? Use graphs of fxy and fyx to illustrate your answer. 5E-15(pp 0952-0961) 1/18/06 3:11 PM Page 959 SECTION 15.4 TANGENT PLANES AND LINEAR APPROXIMATIONS |||| 15.4 ❙❙❙❙ 959 Tangent Planes and Linear Approximations One of the most important ideas in single-variable calculus is that as we zoom in toward a point on the graph of a differentiable function, the graph becomes indistinguishable from its tangent line and we can approximate the function by a linear function. (See Section 3.10.) Here we develop similar ideas in three dimensions. As we zoom in toward a point on a surface that is the graph of a differentiable function of two variables, the surface looks more and more like a plane (its tangent plane) and we can approximate the function by a linear function of two variables. We also extend the idea of a differential to functions of two or more variables. Tangent Planes z T¡ C¡ P T™ C™ 0 y x FIGURE 1 The tangent plane contains the tangent lines T¡ and T™. Suppose a surface S has equation z f x, y , where f has continuous first partial derivatives, and let P x 0 , y0 , z0 be a point on S. As in the preceding section, let C1 and C2 be the curves obtained by intersecting the vertical planes y y0 and x x 0 with the surface S. Then the point P lies on both C1 and C2. Let T1 and T2 be the tangent lines to the curves C1 and C2 at the point P. Then the tangent plane to the surface S at the point P is defined to be the plane that contains both tangent lines T1 and T2 . (See Figure 1.) We will see in Section 15.6 that if C is any other curve that lies on the surface S and passes through P, then its tangent line at P also lies in the tangent plane. Therefore, you can think of the tangent plane to S at P as consisting of all possible tangent lines at P to curves that lie on S and pass through P. The tangent plane at P is the plane that most closely approximates the surface S near the point P. We know from Equation 13.5.7 that any plane passing through the point P x 0 , y0 , z0 has an equation of the form Ax x0 By y0 By dividing this equation by C and letting a the form z 1 z0 ax Cz z0 A C and b x0 by 0 B C, we can write it in y0 If Equation 1 represents the tangent plane at P, then its intersection with the plane y must be the tangent line T1. Setting y y0 in Equation 1 gives z z0 ax x0 y y0 y0 and we recognize these as the equations (in point-slope form) of a line with slope a. But from Section 15.3 we know that the slope of the tangent T1 is fx x 0 , y0 . Therefore, a fx x 0 , y0 . Similarly, putting x x 0 in Equation 1, we get z z0 b y y0 , which must represent the tangent line T2 , so b fy x 0 , y0 . |||| Note the similarity between the equation of a tangent plane and the equation of a tangent line: y y0 f x 0 x x 0 2 Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface z f x, y at the point P x 0 , y0 , z0 is z z0 fx x 0 , y0 x x0 fy x 0 , y0 y y0 5E-15(pp 0952-0961) ❙❙❙❙ 960 1/18/06 3:11 PM Page 960 CHAPTER 15 PARTIAL DERIVATIVES EXAMPLE 1 Find the tangent plane to the elliptic paraboloid z 2x 2 y 2 at the point 1, 1, 3 . SOLUTION Let f x, y 2x 2 y 2. Then fx x, y 4x fy x, y 2y fx 1, 1 4 fy 1, 1 2 Then (2) gives the equation of the tangent plane at 1, 1, 3 as z 4x 1 z or 3 4x 2y 2y 3 Figure 2(a) shows the elliptic paraboloid and its tangent plane at (1, 1, 3) that we found in Example 1. In parts (b) and (c) we zoom in toward the point (1, 1, 3) by restricting the domain of the function f x, y 2 x 2 y 2. Notice that the more we zoom in, the flatter the graph appears and the more it resembles its tangent plane. Visual 15.4 shows an animation of Figures 2 and 3. 40 40 20 20 20 0 z0 z0 _20 _20 40 z 1 _20 _4 _2 y 0 2 44 2 _2 0 x _4 _2 0 _2 y 0 0 2 2 0 y x 1 2 (b) (a) 1 2 x (c) FIGURE 2 The elliptic paraboloid z=2≈+¥ appears to coincide with its tangent plane as we zoom in toward (1, 1, 3). In Figure 3 we corroborate this impression by zooming in toward the point (1, 1) on a contour map of the function f x, y 2 x 2 y 2. Notice that the more we zoom in, the more the level curves look like equally spaced parallel lines, which is characteristic of a plane. 1.5 1.2 1.05 FIGURE 3 Zooming in toward (1, 1) on a contour map of f(x, y)=2≈+¥ 0.5 1.5 0.8 1.2 0.95 1.05 5E-15(pp 0952-0961) 1/18/06 3:12 PM Page 961 S ECTION 15.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ❙❙❙❙ 961 Linear Approximations In Example 1 we found that an equation of the tangent plane to the graph of the function f x, y 2 x 2 y 2 at the point (1, 1, 3) is z 4 x 2y 3. Therefore, in view of the visual evidence in Figures 2 and 3, the linear function of two variables L x, y 4x 2y 3 is a good approximation to f x, y when x, y is near (1, 1). The function L is called the linearization of f at (1, 1) and the approximation f x, y 4x 2y 3 is called the linear approximation or tangent plane approximation of f at (1, 1). For instance, at the point (1.1, 0.95) the linear approximation gives f 1.1, 0.95 4 1.1 2 0.95 3 3.3 2 1.1 2 0.95 2 3.3225. But if which is quite close to the true value of f 1.1, 0.95 we take a point farther away from (1, 1), such as (2, 3), we no longer get a good approxi11 whereas f 2, 3 17. mation. In fact, L 2, 3 In general, we know from (2) that an equation of the tangent plane to the graph of a function f of two variables at the point a, b, f a, b is z f a, b fx a, b x a fy a, b y b The linear function whose graph is this tangent plane, namely 3 L x, y f a, b fx a, b x a fy a, b y b is called the linearization of f at a, b and the approximation 4 z f x, y f a, b fx a, b x xy x2 0 x xy if (x, y)≠(0, 0), ≈+¥ f(0, 0)=0 f(x, y)= fy a, b y b is called the linear approximation or the tangent plane approximation of f at a, b . We have defined tangent planes for surfaces z f x, y , where f has continuous first partial derivatives. What happens if fx and fy are not continuous? Figure 4 pictures such a function; its equation is y f x, y FIGURE 4 a y2 if x, y 0, 0 if x, y 0, 0 You can verify (see Exercise 42) that its partial derivatives exist at the origin and, in fact, fx 0, 0 0 and fy 0, 0 0, but fx and fy are not continuous. The linear approximation 1 0, but f x, y would be f x, y x. So a function of two 2 at all points on the line y variables can behave badly even though both of its partial derivatives exist. To rule out such behavior, we formulate the idea of a differentiable function of two variables. Recall that for a function of one variable, y f x , if x changes from a to a x, we defined the increment of y as y fa x fa 5E-15(pp 0962-0971) 962 ❙❙❙❙ 1/18/06 3:15 PM Page 962 CHAPTER 15 PARTIAL DERIVATIVES In Chapter 3 we showed that if f is differentiable at a, then y 5 |||| This is Equation 3.6.5. a fa x x where l 0 as xl0 Now consider a function of two variables, z f x, y , and suppose x changes from a to x and y changes from b to b y. Then the corresponding increment of z is z 6 fa x, b y f a, b Thus, the increment z represents the change in the value of f when x, y changes from a, b to a x, b y . By analogy with (5) we define the differentiability of a function of two variables as follows. f x, y , then f is differentiable at a, b if z can be 7 Definition If z expressed in the form z where 1 and 2 fx a, b l 0 as x, x fy a, b y x 1 y 2 y l 0, 0 . Definition 7 says that a differentiable function is one for which the linear approximation (4) is a good approximation when x, y is near a, b . In other words, the tangent plane approximates the graph of f well near the point of tangency. It’s sometimes hard to use Definition 7 directly to check the differentiability of a function, but the following theorem provides a convenient sufficient condition for differentiability. 8 Theorem If the partial derivatives fx and fy exist near a, b and are continuous at a, b , then f is differentiable at a, b . |||| Theorem 8 is proved in Appendix F. xe xy is differentiable at (1, 0) and find its linearization there. Then use it to approximate f 1.1, 0.1 . EXAMPLE 2 Show that f x, y SOLUTION The partial derivatives are fx x, y fx 1, 0 |||| Figure 5 shows the graphs of the function f and its linearization L in Example 2. e xy x ye xy fy x, y 1 x 2e xy fy 1, 0 1 Both fx and fy are continuous functions, so f is differentiable by Theorem 8. The linearization is 6 L x, y 4 f 1, 0 1 z 2 1x fx 1, 0 x 1 1 1y fy 1, 0 y x y 0.1 0 1 The corresponding linear approximation is 0 xe xy 1 x FIGURE 5 01 0y x 0.1 1.1 y _1 so f 1.1, Compare this with the actual value of f 1.1, 0.1 1.1e 0.11 0.98542. 5E-15(pp 0962-0971) 1/18/06 3:15 PM Page 963 S ECTION 15.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ❙❙❙❙ 963 EXAMPLE 3 At the beginning of Section 15.3 we discussed the heat index (perceived temperature) I as a function of the actual temperature T and the relative humidity H and gave the following table of values from the National Weather Service. Relative humidity (%) H 50 55 60 65 70 75 80 85 90 90 96 98 100 103 106 109 112 115 119 92 100 103 105 108 112 115 119 123 128 94 104 107 111 114 118 122 127 132 137 96 109 113 116 121 125 130 135 141 146 98 114 118 123 127 133 138 144 150 157 100 119 124 129 135 141 147 154 161 168 T Actual temperature (°F) Find a linear approximation for the heat index I f T, H when T is near 96 F and H is near 70%. Use it to estimate the heat index when the temperature is 97 F and the relative humidity is 72%. SOLUTION We read from the table that f 96, 70 125. In Section 15.3 we used the tabular values to estimate that fT 96, 70 3.75 and fH 96, 70 0.9. (See pages 946–947.) So the linear approximation is f T, H f 96, 70 125 fT 96, 70 T 3.75 T 96 96 0.9 H fH 96, 70 H 70 70 In particular, f 97, 72 Therefore, when T 125 97 F and H 3.75 1 0.9 2 130.55 72%, the heat index is I 131 F Differentials For a differentiable function of one variable, y f x , we define the differential dx to be an independent variable; that is, dx can be given the value of any real number. The differential of y is then defined as dy 9 (See Section 3.10.) Figure 6 shows the relationship between the increment y and the differential dy: y represents the change in height of the curve y f x and dy represents the change in height of the tangent line when x changes by an amount d x x. For a differentiable function of two variables, z f x, y , we define the differentials dx and dy to be independent variables; that is, they can be given any values. Then the differential dz, also called the total differential, is defined by y y=ƒ Îy dx=Îx 0 a tangent line y=f(a)+fª(a)(x-a) FIGURE 6 dy a+Îx f x dx x 10 dz fx x, y dx fy x, y dy z dx x z dy y (Compare with Equation 9.) Sometimes the notation d f is used in place of dz. 5E-15(pp 0962-0971) 964 ❙❙❙❙ 1/18/06 3:16 PM Page 964 CHAPTER 15 PARTIAL DERIVATIVES If we take dx tial of z is x x a and dy dz y fx a, b x y a b in Equation 10, then the differen- fy a, b y b So, in the notation of differentials, the linear approximation (4) can be written as f x, y dz f a, b Figure 7 is the three-dimensional counterpart of Figure 6 and shows the geometric interpretation of the differential dz and the increment z : dz represents the change in height of the tangent plane, whereas z represents the change in height of the surface z f x, y x, b y. when x, y changes from a, b to a z { a+Îx, b+Îy, f(a+Îx, b+Îy)} surface z=f(x, y) Îz dz { a, b, f(a, b)} f(a, b) 0 dx f(a, b) y = Îx x (a, b, 0) (a+Îx, b+Îy, 0) Îy=dy tangent plane z-f(a, b)=f x (a, b)(x-a)+f y (a, b)(y-b) FIGURE 7 EXAMPLE 4 (a) If z f x, y x 2 3xy y 2, find the differential dz. (b) If x changes from 2 to 2.05 and y changes from 3 to 2.96, compare the values of z and dz. SOLUTION (a) Definition 10 gives z dx x dz (b) Putting x 2, dx |||| In Example 4, dz is close to z because the tangent plane is a good approximation to the surface z x 2 3xy y 2 near 2, 3, 13 . (See Figure 8.) x dz z dy y 0.05, y 22 2x 3y d x 3, and d y 3 3 0.05 3x 2y d y y 23 32 0.04, we get 0.04 0.65 The increment of z is 60 z 40 f 2.05, 2.96 2.05 z 20 2 f 2, 3 3 2.05 2.96 2.96 2 22 32 3 32 0.6449 0 _20 5 4 FIGURE 8 3 x 2 1 0 0 42 y Notice that z dz but dz is easier to compute. EXAMPLE 5 The base radius and height of a right circular cone are measured as 10 cm and 25 cm, respectively, with a possible error in measurement of as much as 0.1 cm 5E-15(pp 0962-0971) 1/18/06 3:16 PM Page 965 S ECTION 15.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ❙❙❙❙ 965 in each. Use differentials to estimate the maximum error in the calculated volume of the cone. r 2h 3. So the SOLUTION The volume V of a cone with base radius r and height h is V differential of V is dV V dr r V dh h r2 dh 3 2 rh dr 3 Since each error is at most 0.1 cm, we have r 0.1, h 0.1. To find the largest error in the volume we take the largest error in the measurement of r and of h. Therefore, we take dr 0.1 and dh 0.1 along with r 10, h 25. This gives 500 3 dV 100 3 0.1 0.1 20 Thus, the maximum error in the calculated volume is about 20 cm3 63 cm3 . Functions of Three or More Variables Linear approximations, differentiability, and differentials can be defined in a similar manner for functions of more than two variables. A differentiable function is defined by an expression similar to the one in Definition 7. For such functions the linear approximation is f x, y, z f a, b, c fx a, b, c x a fy a, b, c y b fz a, b, c z c and the linearization L x, y, z is the right side of this expression. If w f x, y, z , then the increment of w is w fx x, y y, z z f x, y, z The differential dw is defined in terms of the differentials dx, dy, and dz of the independent variables by dw w x dx w y dy w z dz EXAMPLE 6 The dimensions of a rectangular box are measured to be 75 cm, 60 cm, and 40 cm, and each measurement is correct to within 0.2 cm. Use differentials to estimate the largest possible error when the volume of the box is calculated from these measurements. SOLUTION If the dimensions of the box are x, y, and z, its volume is V dV V dx x V dy y V dz z yz d x xz d y x yz and so x y dz We are given that x 0.2, y 0.2, and z 0.2. To find the largest error in the volume, we therefore use dx 0.2, dy 0.2, and dz 0.2 together with x 75, y 60, and z 40: V dV 60 40 0.2 75 40 0.2 75 60 0.2 1980 Thus, an error of only 0.2 cm in measuring each dimension could lead to an error of as much as 1980 cm3 in the calculated volume! This may seem like a large error, but it’s only about 1% of the volume of the box. 5E-15(pp 0962-0971) 3:17 PM Page 966 CHAPTER 15 PARTIAL DERIVATIVES |||| 15.4 Exercises 1–6 |||| Find an equation of the tangent plane to the given surface at the specified point. 4x 2 2. z 9x 2 3. z s4 4. z y ln x, 5. z y cos x 6. z ex 1. z ■ 2 ■ y 2 y 2 6x x2 y2 1, 2, 4 2y, 5, 3y 2y 2, 1, Duration (hours) 1, 1 1, 1, 1 ■ ■ ■ ■ ■ ■ ■ ■ ; 7–8 |||| Graph the surface and the tangent plane at the given point. (Choose the domain and viewpoint so that you get a good view of both the surface and the tangent plane.) Then zoom in until the surface and the tangent plane become indistinguishable. x2 7. z 2 8. z ■ CAS 3y 2, xy arctan x y , ■ ■ 1, 1, 5 1, 1, ■ ■ 4 ■ ■ ■ ■ ■ ■ ■ 9–10 |||| Draw the graph of f and its tangent plane at the given point. (Use your computer algebra system both to compute the partial derivatives and to graph the surface and its tangent plane.) Then zoom in until the surface and the tangent plane become indistinguishable. 9. f x, y ■ ■ sin 2 x 4x2 x4 s1 1 10. f x, y ■ x 2 y 2 15 e ■ cos 2 y , 4y2 , y4 ■ ■ ■ ■ ■ 13. f x, y e cos x y, 15. f x, y tan x 2y , 16. f x, y sin 2 x 3y , ■ ■ ■ 1 0, 0 ■ ■ ■ x y, 6, 3 14. f x, y sx e 4y, 3, 0 3, 2 ■ ■ ■ 50 5 7 8 8 9 9 9 30 9 13 16 17 18 19 19 40 14 21 25 28 31 33 33 50 19 29 36 40 45 48 50 24 37 47 54 62 67 69 from Table 1 in Section 15.1. Wind speed (km / h) 1, 0 ■ 40 Use the table to find a linear approximation to the wave height function when v is near 40 knots and t is near 20 hours. Then estimate the wave heights when the wind has been blowing for 24 hours at 43 knots. ■ ■ ■ Actual temperature (°C) 12. f x, y 1, 4 x 30 22. The wind-chill index W is the perceived temperature when the actual temperature is T and the wind speed is v, so we can write W f T, v . The following table of values is an excerpt |||| Explain why the function is differentiable at the given point. Then find the linearization L x, y of the function at that point. x sy, 20 20 11–16 11. f x, y 15 the heat index function when the temperature is near 94 F and the relative humidity is near 80%. Then estimate the heat index when the temperature is 95 F and the relative humidity is 78%. 2, 3, f 2, 3 ■ 10 21. Use the table in Example 3 to find a linear approximation to 1, 1, 1 ■ 5 60 2, 2, 2 ■ t v y, , of the wind and the length of time t that the wind has been blowing at that speed. Values of the function h f v, t are recorded in the following table. 1, 2, 18 1, 4, 0 ■ 20. The wave heights h in the open sea depend on the speed v Wind speed (knots) ❙❙❙❙ 966 1/18/06 v T 20 30 40 50 60 70 10 18 20 21 22 23 23 15 24 26 27 29 30 30 20 30 33 34 35 36 37 25 37 39 41 42 43 44 17. Find the linear approximation of the function f x, y s20 f 1.95, 1.08 . x2 7y 2 at 2, 1 and use it to approximate ; 18. Find the linear approximation of the function f x, y ln x 3y at 7, 2 and use it to approximate f 6.9, 2.06 . Illustrate by graphing f and the tangent plane. 19. Find the linear approximation of the function f x, y, z s x 2 y 2 z 2 at 3, 2, 6 and use it to 1.97 2 5.99 2. approximate the number s 3.02 2 Use the table to find a linear approximation to the wind-chill index function when T is near 15 C and v is near 50 km h. Then estimate the wind-chill index when the temperature is 17 C and the wind speed is 55 km h. 23–28 23. z 25. u |||| Find the differential of the function. 3 x ln y 2 24. v y cos x y t 26. u rs e sin 2t 5E-15(pp 0962-0971) 1/18/06 3:17 PM Page 967 ❙❙❙❙ S ECTION 15.5 THE CHAIN RULE 27. w ■ lns x 2 ■ ■ y2 ■ z2 ■ 28. w ■ ■ 37. If R is the total resistance of three resistors, connected in paral- x ye xz ■ ■ ■ ■ lel, with resistances R1, R2, R3, then ■ x 2 x y 3y 2 and x, y changes from 3, 2.96, 0.95 , compare the values of z and dz. 31. The length and width of a rectangle are measured as 30 cm and 33. Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height 12 cm if the tin is 0.04 cm thick. 34. Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1 cm thick and the metal in the sides is 0.05 cm thick. 35. A boundary stripe 3 in. wide is painted around a rectangle first decimal place and then multiplied together. Use differentials to estimate the maximum possible error in the computed product that might result from the rounding. 39– 40 |||| values of Show that the function is differentiable by finding 1 and 2 that satisfy Definition 7. x2 39. f x, y ■ ■ ■ y2 ■ 40. f x, y ■ ■ ■ 5y 2 ■ ■ ■ ■ tiable at a, b , then f is continuous at a, b . Hint: Show that lim x, y l 0, 0 fa x, b y f a, b 42. (a) The function xy f x, y 36. The pressure, volume, and temperature of a mole of an ideal |||| 15.5 ■ xy 41. Prove that if f is a function of two variables that is differen- whose dimensions are 100 ft by 200 ft. Use differentials to approximate the number of square feet of paint in the stripe. gas are related by the equation PV 8.31T , where P is measured in kilopascals, V in liters, and T in kelvins. Use differentials to find the approximate change in the pressure if the volume increases from 12 L to 12.3 L and the temperature decreases from 310 K to 305 K. 1 R3 38. Four positive numbers, each less than 50, are rounded to the 32. The dimensions of a closed rectangular box are measured as 80 cm, 60 cm, and 50 cm, respectively, with a possible error of 0.2 cm in each dimension. Use differentials to estimate the maximum error in calculating the surface area of the box. 1 R2 If the resistances are measured in ohms as R1 25 , R2 40 , and R3 50 , with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of R. 1 to 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle. 1 R1 1 R 5x 2 y 2 and x, y changes from 1, 2 to 1.05, 2.1 , compare the values of z and dz. 29. If z 30. If z 967 x2 0 y2 if x, y 0, 0 if x, y 0, 0 was graphed in Figure 4. Show that fx 0, 0 and fy 0, 0 both exist but f is not differentiable at 0, 0 . [Hint: Use the result of Exercise 41.] (b) Explain why fx and fy are not continuous at 0, 0 . The Chain Rule Recall that the Chain Rule for functions of a single variable gives the rule for differentiating a composite function: If y f x and x t t , where f and t are differentiable functions, then y is indirectly a differentiable function of t and 1 dy dt dy dx dx dt For functions of more than one variable, the Chain Rule has several versions, each of them giving a rule for differentiating a composite function. The first version (Theorem 2) deals with the case where z f x, y and each of the variables x and y is, in turn, a function of a variable t . This means that z is indirectly a function of t, z f t t , h t , and the Chain Rule gives a formula for differentiating z as a function of t. We assume that f is differentiable (Definition 15.4.7). Recall that this is the case when fx and fy are continuous (Theorem 15.4.8). 5E-15(pp 0962-0971) 968 ❙❙❙❙ 1/18/06 3:24 PM Page 968 CHAPTER 15 PARTIAL DERIVATIVES f x, y is a differentiable function of x 2 The Chain Rule (Case 1) Suppose that z and y, where x t t and y h t are both differentiable functions of t. Then z is a differentiable function of t and dz dt f dx x dt f dy y dt Proof A change of t in t produces changes of x in x and y in y. These, in turn, produce a change of z in z, and from Definition 15.4.7 we have f x z f y x y x 1 y 2 where 1 l 0 and 2 l 0 as x, y l 0, 0 . [If the functions 1 and 2 are not defined at 0, 0 , we can define them to be 0 there.] Dividing both sides of this equation by t, we have z t f x x t f y y t x t 1 2 y t If we now let t l 0, then x t t t t t l 0 because t is differentiable and therefore continuous. Similarly, y l 0. This, in turn, means that 1 l 0 and 2 l 0, so dz dt lim tl0 z t f lim x tl0 x t f lim y tl0 f dx x dt f dy y dt f dx x dt y t lim tl0 lim tl0 x t lim tl0 2 lim tl0 y t f dy y dt 0 dx dt 1 0 dy dt Since we often write z x in place of f x, we can rewrite the Chain Rule in the form |||| Notice the similarity to the definition of the differential: z z dx dy dz x y dz dt EXAMPLE 1 If z x2y z dx x dt 3xy 4, where x z dy y dt sin 2 t and y cos t, find dz dt when t SOLUTION The Chain Rule gives dz dt z dx x dt 2 xy z dy y dt 3y 4 2 cos 2 t x2 12 xy 3 sin t It’s not necessary to substitute the expressions for x and y in terms of t. We simply 0. 5E-15(pp 0962-0971) 1/18/06 3:19 PM Page 969 SECTION 15.5 THE CHAIN RULE ❙❙❙❙ 969 y observe that when t (0, 1) 0 we have x dz dt 0 sin 0 0 and y 3 2 cos 0 0 cos 0 0 1. Therefore, sin 0 6 t0 x FIGURE 1 The curve x=sin 2t, y=cos t The derivative in Example 1 can be interpreted as the rate of change of z with respect to t as the point x, y moves along the curve C with parametric equations x sin 2 t, y cos t. (See Figure 1.) In particular, when t 0, the point x, y is 0, 1 and dz dt 6 is the rate of increase as we move along the curve C through 0, 1 . If, for instance, z T x, y x 2 y 3xy 4 represents the temperature at the point x, y , then the composite function z T sin 2 t, cos t represents the temperature at points on C and the derivative dz dt represents the rate at which the temperature changes along C. EXAMPLE 2 The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by the equation PV 8.31T . Find the rate at which the pressure is changing when the temperature is 300 K and increasing at a rate of 0.1 K s and the volume is 100 L and increasing at a rate of 0.2 L s. SOLUTION If t represents the time elapsed in seconds, then at the given instant we have T 300, dT dt 0.1, V 100, dV dt 0.2. Since P 8.31 T V the Chain Rule gives dP dt P dT T dt P dV V dt 8.31 0.1 100 8.31 dT V dt 8.31 300 100 2 8.31T dV V 2 dt 0.2 0.04155 The pressure is decreasing at a rate of about 0.042 kPa s. We now consider the situation where z f x, y but each of x and y is a function of two variables s and t : x t s, t , y h s, t . Then z is indirectly a function of s and t and we wish to find z s and z t. Recall that in computing z t we hold s fixed and compute the ordinary derivative of z with respect to t. Therefore, we can apply Theorem 2 to obtain z t z x x t z y y t A similar argument holds for z s and so we have proved the following version of the Chain Rule. 3 The Chain Rule (Case 2) Suppose that z and y, where x t s, t and y z s z x x s f x, y is a differentiable function of x h s, t are differentiable functions of s and t. Then z y y s z t z x x t z y y t 5E-15(pp 0962-0971) 970 ❙❙❙❙ 1/18/06 3:20 PM Page 970 CHAPTER 15 PARTIAL DERIVATIVES EXAMPLE 3 If z e x sin y, where x st 2 and y s 2t, find z s and z t. SOLUTION Applying Case 2 of the Chain Rule, we get z s z x z y x s 2 z x 2 2ste st sin s 2t z y x x s x t s y y s t y t s y t e x sin y 2st e x cos y s 2 2 s 2e st cos s 2t Case 2 of the Chain Rule contains three types of variables: s and t are independent variables, x and y are called intermediate variables, and z is the dependent variable. Notice that Theorem 3 has one term for each intermediate variable and each of these terms resembles the one-dimensional Chain Rule in Equation 1. To remember the Chain Rule it’s helpful to draw the tree diagram in Figure 2. We draw branches from the dependent variable z to the intermediate variables x and y to indicate that z is a function of x and y. Then we draw branches from x and y to the independent variables s and t. On each branch we write the corresponding partial derivative. To find z s we find the product of the partial derivatives along each path from z to s and then add these products: z z x e x cos y 2st 2ste st cos s 2t z y x t e x sin y t 2 2 t 2e st sin s 2t z t y s t z s FIGURE 2 z x z y x s y s Similarly, we find z t by using the paths from z to t. Now we consider the general situation in which a dependent variable u is a function of n intermediate variables x 1 , . . . , x n , each of which is, in turn, a function of m independent variables t1 , . . . , tm . Notice that there are n terms, one for each intermediate variable. The proof is similar to that of Case 1. 4 The Chain Rule (General Version) Suppose that u is a differentiable function of the n variables x 1 , x 2 , . . . , x n and each x j is a differentiable function of the m variables t1 , t2 , . . . , tm . Then u is a function of t1 , t2 , . . . , tm and u ti for each i u x1 x1 ti u x2 x2 ti u xn xn ti 1, 2, . . . , m. w x u z y v FIGURE 3 u v u EXAMPLE 4 Write out the Chain Rule for the case where w y y u, v , z z u, v , and t t u, v . t v u v f x, y, z, t and x x u, v , SOLUTION We apply Theorem 4 with n 4 and m 2. Figure 3 shows the tree diagram. Although we haven’t written the derivatives on the branches, it’s understood that if a branch leads from y to u, then the partial derivative for that branch is y u. With the aid 5E-15(pp 0962-0971) 1/18/06 3:21 PM Page 971 SECTION 15.5 THE CHAIN RULE ❙❙❙❙ 971 of the tree diagram we can now write the required expressions: w w v x x 4y value of u s when r x r s FIGURE 4 r s t w z w t v z v t v y w v y rs 2e t, and z r 2s sin t, find the SOLUTION With the help of the tree diagram in Figure 4, we have y t z t u w y 2 z 3, where x rse t, y 2, s 1, t 0. EXAMPLE 5 If u u w y u x w u z u y w x x u w u s z t r s u x x s u y y s 4 x 3 y re t t When r 2, s 1, and t 2yz 3 2rse 64 2 f s2 EXAMPLE 6 If t s, t x4 0, we have x u s t 2, t 2 z s u z t 3y 2z 2 r 2 sin t 2, and z 2, y 16 4 0, so 00 192 s 2 and f is differentiable, show that t satisfies the equation t SOLUTION Let x s2 t 2 and y t s t2 t t s 0 s 2. Then t s, t f x, y and the Chain Rule gives t s f x x s f y y s f 2s x t t f x x t f y y t f x t t 2 st f y 2s f 2t y 2t Therefore t t s s f x 2 st f y 2 st f x 2 st EXAMPLE 7 If z and y 0 f x, y has continuous second-order partial derivatives and x 2rs, find (a) z r and (b) 2z r 2. SOLUTION (a) The Chain Rule gives z r z x z y x r z 2r x y r z 2s y (b) Applying the Product Rule to the expression in part (a), we get 2 5 f y z r2 2r r 2 z x z x 2r 2s r z y z x 2s r z y r2 s2 5E-15(pp 0972-0981) ❙❙❙❙ 972 1/18/06 3:28 PM Page 972 CHAPTER 15 PARTIAL DERIVATIVES z x x But, using the Chain Rule again (see Figure 5), we have z x r y z x x x r 2 r sr FIGURE 5 z y r z y y r z 2s yx z y x y r 2 z 2r x2 s z x y x r y 2 2 z z 2s y2 2r xy Putting these expressions into Equation 5 and using the equality of the mixed secondorder derivatives, we obtain 2 z r2 2 2 z x z x 2 2 z x2 2r 2r 2 4r 2 2 z 2s 2 s 2r yx 2 z x2 8rs z xy 2 2s z y2 2 z xy 4s 2 z y2 Implicit Differentiation The Chain Rule can be used to give a more complete description of the process of implicit differentiation that was introduced in Sections 3.7 and 15.3. We suppose that an equation of the form F x, y 0 defines y implicitly as a differentiable function of x, that is, y f x , where F x, f x 0 for all x in the domain of f . If F is differentiable, we can apply Case 1 of the Chain Rule to differentiate both sides of the equation F x, y 0 with respect to x. Since both x and y are functions of x, we obtain F dx x dx But dx d x 1, so if F y 0 0 we solve for dy d x and obtain dy dx 6 F dy y dx F x F y Fx Fy To derive this equation we assumed that F x, y 0 defines y implicitly as a function of x. The Implicit Function Theorem, proved in advanced calculus, gives conditions under which this assumption is valid. It states that if F is defined on a disk containing a, b , where F a, b 0, Fy a, b 0, and Fx and Fy are continuous on the disk, then the equation F x, y 0 defines y as a function of x near the point a, b and the derivative of this function is given by Equation 6. EXAMPLE 8 Find y if x 3 y3 6 xy. 5E-15(pp 0972-0981) 1/18/06 3:28 PM Page 973 S ECTION 15.5 THE CHAIN RULE ❙❙❙❙ 973 SOLUTION The given equation can be written as x3 F x, y y3 6 xy 0 so Equation 6 gives |||| The solution to Example 8 should be compared to the one in Example 2 in Section 3.7. dy dx 3x 2 3y 2 Fx Fy x2 y2 6y 6x 2y 2x Now we suppose that z is given implicitly as a function z f x, y by an equation of the form F x, y, z 0. This means that F x, y, f x, y 0 for all x, y in the domain of f . If F and f are differentiable, then we can use the Chain Rule to differentiate the equation F x, y, z 0 as follows: F x But x x x F y y x 1 and F x x z x F z F z x 0 y 0 so this equation becomes z x 0 If F z 0, we solve for z x and obtain the first formula in Equations 7. The formula for z y is obtained in a similar manner. F x F z z x 7 F y F z z y Again, a version of the Implicit Function Theorem gives conditions under which our assumption is valid. If F is defined within a sphere containing a, b, c , where F a, b, c 0, Fz a, b, c 0, and Fx , Fy , and Fz are continuous inside the sphere, then the 0 defines z as a function of x and y near the point a, b, c and this equation F x, y, z function is differentiable, with partial derivatives given by (7). EXAMPLE 9 Find z z and if x 3 x y x3 SOLUTION Let F x, y, z |||| The solution to Example 9 should be compared to the one in Example 4 in Section 15.3. y3 y3 z3 z3 6 xy z 6 xy z 1. 1. Then, from Equations 7, we have z x Fx Fz 3x 2 3z 2 6yz 6 xy x2 z2 2yz 2 xy z y Fy Fz 3y 2 3z 2 6xz 6 xy y2 z2 2 xz 2 xy 5E-15(pp 0972-0981) ❙❙❙❙ 974 1/18/06 Exercises Use the Chain Rule to find dz dt or dw dt. |||| 2 1. z 2 xy xy , x 2. z sx 3. z sin x cos y, 4. z x ln x 2 y, yz , xy ■ t, ■ x x 2 xy t arctan 2 x ■ x ■ e tan y, x r e cos , sin r ■ 3s ■ ■ ■ ■ sr 2 u , y t. st 2 t s ■ x2 ■ u , p t ■ ■ f x, y , where f is differentiable, x t t , y t3 2, t 3 5, h 3 7, h 3 4, fx 2, 7 fy 2, 7 8, find dz dt when t 3. ■ ■ 26. Y w tan Y , r ht, 6, and ■ F u s, t , v s, t , where F, u, and v are differentiable, u 1, 0 2, us 1, 0 2, u t 1, 0 6, v 1, 0 3, vs 1, 0 5, vt 1, 0 4, Fu 2, 3 1, and Fv 2, 3 10. Find Ws 1, 0 and Wt 1, 0 . 31. x ■ 0, 0 3 6 4 8 1, 2 6 3 2 5 18. w f x, y, z , where x y r, s, t y t, u , z x t, u , y z t, u 19. v f p, q, r , where p p x, y, z , q 20. u ■ f s, t , ■ ■ where s ■ ■ q x, y, z , r s w, x, y, z , t ■ ■ ■ r x, y, z t w, x, y, z ■ z v, ■ ■ y, v; u 1 ■ ■ z r; w t, p t r; 1 ■ ■ ■ ■ ■ x 2y 28. y 5 xe y y ■ y 2 z z 2 30. sin x ■ ■ ■ ■ ■ ■ ■ ■ ■ y. cos x ln x ■ 2 sin x cos y ■ x and z 34. y z ■ ye x 1 cos y ■ 32. x y z 3x yz arctan y z ■ x 2y 3 ■ y z z ■ ■ 36. Wheat production in a given year, W, depends on the average Use a tree diagram to write out the Chain Rule for the given case. Assume all functions are differentiable. |||| x r, s, t , y 2x ■ Celsius. A bug crawls so that its position after t seconds is given by x s1 t, y 2 1 t, where x and y are measured 3 in centimeters. The temperature function satisfies Tx 2, 3 4 and Ty 2, 3 3. How fast is the temperature rising on the bug’s path after 3 seconds? t r, s f 2 r s, s 2 4 r . Use the table of values in Exercise 15 to calculate tr 1, 2 and ts 1, 2 . where x u Use Equations 7 to find z |||| ■ 2 y, v 35. The temperature at a point x, y is T x, y , measured in degrees 16. Suppose f is a differentiable function of x and y, and f x, y , 0 u v , u r s, v s Y when r 1, s 0, t t ■ 2 33. x fy 1 1 ■ 31–34 calculate tu 0, 0 and tv 0, 0 . 17. u 1, y 2, t y sin t ; Use Equation 6 to find dy d x. |||| 29. cos x 15. Suppose f is a differentiable function of x and y, and f e u sin v, e u cos v . Use the table of values to t u, v 17–20 x yz, x pr cos , y pr sin , u when p 2, r 3, 0 ■ 27. sxy ■ fx Y , s ■ 27–30 14. Let W s, t t u , r 0 x cos t, s 2 25. u ve w; u 1, w ln u 2 v 2 w 2 , u x 2xy; R R , when x y 1 x y 2 13. If z f s 2, r y u when x t xe y z , x 2 u v, y M M , when u 3, v u v s ln t y w 24. M t, ■ 23. R w st y y ss st, tan , ■ se 2 t, ■ s and z y s 2 t, s e t cos t z t 1 x y, u , x 2t ■ t, v 22. u 1 e sin t, s y se , xy ■ x y, 9. z 12. z z t ■ x y 3, x u v 2 w 3, z z , when u 2, v z , u cos t t, y ■ 2 x y, 11. z e, ■ 8. z 10. z 1 t x2 21. z y Use the Chain Rule to find z |||| 7. z y t st sin t, 2 1 21–26 |||| Use the Chain Rule to find the indicated partial derivatives. 3 2t e y t, x y y e, x 2 t, 2t x x 4 2 2y , xe , 6. w 7–12 2 yz 5. w ■ Page 974 CHAPTER 15 PARTIAL DERIVATIVES |||| 15.5 1–6 3:30 PM ■ ■ ■ temperature T and the annual rainfall R. Scientists estimate that the average temperature is rising at a rate of 0.15°C year and rainfall is decreasing at a rate of 0.1 cm year. They also 2 and estimate that, at current production levels, W T W R 8. (a) What is the significance of the signs of these partial derivatives? (b) Estimate the current rate of change of wheat production, dW dt. 5E-15(pp 0972-0981) 1/18/06 3:32 PM Page 975 ❙❙❙❙ C HAPTER 15 THE CHAIN RULE 37. The speed of sound traveling through ocean water with salinity 44. If u 35 parts per thousand has been modeled by the equation C 1449.2 0.055T 2 4.6T 0.00029T 3 D 14 15 12 10 10 5 8 10 20 30 40 2 u y e 45. If z fx 46. If z f x, y , where x ■ ■ ■ 2 z y 2 ■ z t z s ■ t, show that s 2 z y 2 u t 0. t and y s ■ e s sin t, show that u s 2s z x y , show that z x ■ T 16 20 2 u x 0.016D where C is the speed of sound (in meters per second), T is the temperature (in degrees Celsius), and D is the depth below the ocean surface (in meters). A scuba diver began a leisurely dive into the ocean water; the diver’s depth and surrounding water temperature over time are recorded in the following graphs. Estimate the rate of change (with respect to time) of the speed of sound through the ocean water experienced by the diver 20 minutes into the dive. What are the units? e s cos t and y f x, y , where x 975 ■ ■ ■ ■ ■ 47–52 |||| Assume that all the given functions have continuous second-order partial derivatives. 47. Show that any function of the form t (min) 10 20 30 z 40 t (min) fx at tx is a solution of the wave equation 38. The radius of a right circular cone is increasing at a rate of 2 2 z 1.8 in s while its height is decreasing at a rate of 2.5 in s. At what rate is the volume of the cone changing when the radius is 120 in. and the height is 140 in.? 39. The length , width w, and height h of a box change with time. At a certain instant the dimensions are 1 m and w h 2 m, and and w are increasing at a rate of 2 m s while h is decreasing at a rate of 3 m s. At that instant find the rates at which the following quantities are changing. (a) The volume (b) The surface area (c) The length of a diagonal at t [Hint: Let u 48. If u x a t, v f x, y , where x 2 z x2 a2 2 x a t.] e s cos t and y 2 u x2 e s sin t, show that 2 u y2 e f x, y , where x r 2 (Compare with Example 7.) 2 u s2 2s u t2 s 2, y 49. If z 2 z 2rs, find r s. 40. The voltage V in a simple electrical circuit is slowly decreasing 50. If z as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohm’s Law, V IR, to find how the current I is changing at the moment when R 400 , I 0.08 A, dV dt 0.01 V s, and dR dt 0.03 s. 51. If z f x, y , where x r cos , y , and (c) 2z r . r sin , find (a) z f x, y , where x (b) z r sin , show that 41. The pressure of 1 mole of an ideal gas is increasing at a rate 2 z r2 2 2 z t z x2 2 2 x t z xy z 2y z 2x x t2 y t2 (b) Find a similar formula for 2z ■ ■ ■ ■ ■ ■ z z r 1 r 2 t s, t and y 2 2 2 1 r2 f x, y , where x (a) Show that west on Highway 83. Each car is approaching the intersection of these highways. At a certain moment, car A is 0.3 km from the intersection and traveling at 90 km h while car B is 0.4 km from the intersection and traveling at 80 km h. How fast is the distance between the cars changing at that moment? |||| 2 z y2 52. Suppose z 42. Car A is traveling north on Highway 16 and car B is traveling 43–46 2 z x2 of 0.05 kPa s and the temperature is increasing at a rate of 0.15 K s. Use the equation in Example 2 to find the rate of change of the volume when the pressure is 20 kPa and the temperature is 320 K. r cos , y ■ x t h s, t . 2 z y2 y t r, y t 2 s t. ■ ■ ■ ■ ■ Assume that all the given functions are differentiable. 43. If z and z f x, y , where x r cos and (b) show that z x 2 z y 2 and y z r 2 r sin , (a) find z 1 r2 z 2 r 53. A function f is called homogeneous of degree n if it satisfies the equation f t x, t y t n f x, y for all t, where n is a positive integer and f has continuous second-order partial derivatives. (a) Verify that f x, y x 2 y 2 x y 2 5 y 3 is homogeneous of degree 3. 5E-15(pp 0972-0981) ❙❙❙❙ 976 1/18/06 3:32 PM Page 976 CHAPTER 15 PARTIAL DERIVATIVES 55. If f is homogeneous of degree n, show that (b) Show that if f is homogeneous of degree n, then f f x y n f x, y x y [Hint: Use the Chain Rule to differentiate f t x, t y with respect to t.] fx t x, t y 56. Suppose that the equation F x, y, z 0 implicitly defines each of the three variables x, y, and z as functions of the other two: z f x, y , y t x, z , x h y, z . If F is differentiable and Fx , Fy , and Fz are all nonzero, show that zxy 1 xyz 54. If f is homogeneous of degree n, show that 2 x2 2 f x2 |||| 15.6 2 xy f xy 2 f y2 y2 nn t n 1fx x, y 1 f x, y Directional Derivatives and the Gradient Vector The weather map in Figure 1 shows a contour map of the temperature function T x, y for the states of California and Nevada at 3:00 P.M. on a day in October. The level curves, or isothermals, join locations with the same temperature. The partial derivative Tx at a location such as Reno is the rate of change of temperature with respect to distance if we travel east from Reno; Ty is the rate of change of temperature if we travel north. But what if we want to know the rate of change of temperature when we travel southeast (toward Las Vegas), or in some other direction? In this section we introduce a type of derivative, called a directional derivative, that enables us to find the rate of change of a function of two or more variables in any direction. 60 50 Reno San Francisco 60 70 Las Vegas 70 0 F IGURE 1 50 100 150 200 (Distance in miles) 80 Los Angeles Directional Derivatives y Recall that if z f x, y , then the partial derivatives fx and fy are defined as fx x0, y0 u (x¸, y¸) 0 lim f x0 hl0 1 sin ¨ fy x0, y0 ¨ cos ¨ x F IGURE 2 A unit vector u=ka, b l=kcos ¨, sin ¨ l lim hl0 f x0, y0 h, y0 h f x0, y0 h h f x0, y0 and represent the rates of change of z in the x- and y-directions, that is, in the directions of the unit vectors i and j. Suppose that we now wish to find the rate of change of z at x 0 , y0 in the direction of an arbitrary unit vector u a, b . (See Figure 2.) To do this we consider the surface S with equation z f x, y (the graph of f ) and we let z0 f x 0 , y0 . Then the point 5E-15(pp 0972-0981) 1/18/06 3:33 PM Page 977 S ECTION 15.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR ❙❙❙❙ 977 P x 0 , y0 , z0 lies on S. The vertical plane that passes through P in the direction of u intersects S in a curve C. (See Figure 3.) The slope of the tangent line T to C at the point P is the rate of change of z in the direction of u. z T P(x¸, y¸, z¸) Q(x, y, z) Visual 15.6A animates Figure 3 by rotating u and therefore T. S C Pª(x¸, y¸, 0) ha y u h hb Qª(x, y, 0) F IGURE 3 x If Q x, y, z is another point on C and P , Q are the projections of P, Q on the xy-plane, B then the vector P Q is parallel to u and so B PQ hu x0 h a, y for some scalar h. Therefore, x and z z h z0 f x0 h a, h b y0 h b, so x h a, y0 h hb x0 h a, y y0 h b, f x 0 , y0 h If we take the limit as h l 0, we obtain the rate of change of z (with respect to distance) in the direction of u, which is called the directional derivative of f in the direction of u. 2 Definition The directional derivative of f at x 0 , y0 in the direction of a unit vector u a, b is Du f x 0, y0 lim hl0 f x0 h a, y0 hb f x 0, y0 h if this limit exists. By comparing Definition 2 with (1), we see that if u i 1, 0 , then Di f fx and if u j 0, 1 , then Dj f fy . In other words, the partial derivatives of f with respect to x and y are just special cases of the directional derivative. 5E-15(pp 0972-0981) 978 ❙❙❙❙ 1/18/06 3:33 PM Page 978 CHAPTER 15 PARTIAL DERIVATIVES EXAMPLE 1 Use the weather map in Figure 1 to estimate the value of the directional derivative of the temperature function at Reno in the southeasterly direction. SOLUTION The unit vector directed toward the southeast is u i j s2, but we won’t need to use this expression. We start by drawing a line through Reno toward the southeast. (See Figure 4.) 60 50 Reno San Francisco 60 70 Las Vegas 70 0 FIGURE 4 50 100 150 200 (Distance in miles) 80 Los Angeles We approximate the directional derivative Du T by the average rate of change of the temperature between the points where this line intersects the isothermals T 50 and T 60. The temperature at the point southeast of Reno is T 60 F and the temperature at the point northwest of Reno is T 50 F. The distance between these points looks to be about 75 miles. So the rate of change of the temperature in the southeasterly direction is 60 Du T 50 10 75 75 0.13 F mi When we compute the directional derivative of a function defined by a formula, we generally use the following theorem. 3 Theorem If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector u a, b and Du f x, y fx x, y a fy x, y b Proof If we define a function t of the single variable h by th f x0 h a, y0 hb then by the definition of a derivative we have 4 t0 lim hl0 th t0 h Du f x 0 , y0 lim hl0 f x0 h a, y0 hb h f x 0, y0 5E-15(pp 0972-0981) 1/18/06 3:34 PM Page 979 S ECTION 15.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR On the other hand, we can write t h Chain Rule (Theorem 15.5.2) gives f dx x dh th If we now put h f x, y , where x 0, then x x0, y t0 5 f dy y dh h a, y x0 fx x, y a y0 ❙❙❙❙ 979 h b, so the fy x, y b y0 , and fx x 0 , y0 a fy x 0 , y0 b Comparing Equations 4 and 5, we see that Du f x 0 , y0 fx x 0 , y0 a fy x 0 , y0 b If the unit vector u makes an angle with the positive x-axis (as in Figure 2), then we can write u and the formula in Theorem 3 becomes cos , sin Du f x, y 6 fx x, y cos fy x, y sin EXAMPLE 2 Find the directional derivative Du f x, y if |||| The directional derivative Du f 1, 2 in Example 2 represents the rate of change of z in the direction of u. This is the slope of the tangent line to the curve of intersection of the surface z x 3 3xy 4y 2 and the vertical plane through 1, 2, 0 in the direction of u shown in Figure 5. x3 f x, y and u is the unit vector given by angle 4y 2 3xy 6. What is Du f 1, 2 ? SOLUTION Formula 6 gives z Du f x, y fx x, y cos 3x 2 0 (1, 2, 0) x 1 2 y π 6 u 3y [3 s3 x 2 2 31 fy x, y sin 6 s3 2 3x 3x (8 6 1 2 8y 3 s3 )y Therefore Du f 1, 2 FIGURE 5 1 2 [3 s3 1 (8 3 s3 ) 2 13 3 s3 2 The Gradient Vector Notice from Theorem 3 that the directional derivative can be written as the dot product of two vectors: 7 Du f x, y fx x, y a fy x, y b fx x, y , fy x, y fx x, y , fy x, y a, b u The first vector in this dot product occurs not only in computing directional derivatives but in many other contexts as well. So we give it a special name (the gradient of f ) and a special notation (grad f or f , which is read “del f ”). 5E-15(pp 0972-0981) 980 ❙❙❙❙ 1/18/06 3:35 PM Page 980 CHAPTER 15 PARTIAL DERIVATIVES 8 Definition If f is a function of two variables x and y, then the gradient of f is the vector function f defined by f x, y EXAMPLE 3 If f x, y f j y e x y, then sin x f x, y fx , fy f 0, 1 and f i x fx x, y , fy x, y ye x y, xe x y cos x 2, 0 With this notation for the gradient vector, we can rewrite the expression (7) for the directional derivative as Du f x, y 9 f x, y u This expresses the directional derivative in the direction of u as the scalar projection of the gradient vector onto u. |||| The gradient vector f 2, 1 in Example 4 is shown in Figure 6 with initial point 2, 1 . Also shown is the vector v that gives the direction of the directional derivative. Both of these vectors are superimposed on a contour plot of the graph of f . EXAMPLE 4 Find the directional derivative of the function f x, y point 2, 1 in the direction of the vector v 2i 2 xy 3 i 4 y at the 5 j. SOLUTION We first compute the gradient vector at 2, f x, y x2y3 1: 3x 2 y 2 4j y f 2, 1 4i Note that v is not a unit vector, but since v is ±f(2, _1) v v v u x (2, _1) 8j s29, the unit vector in the direction of v 2 i s29 5 j s29 Therefore, by Equation 9, we have Du f 2, 1 f 2, 1 u 4i FIGURE 6 42 85 s29 8j 2 i s29 5 j s29 32 s29 Functions of Three Variables For functions of three variables we can define directional derivatives in a similar manner. Again Du f x, y, z can be interpreted as the rate of change of the function in the direction of a unit vector u. 5E-15(pp 0972-0981) 1/18/06 3:35 PM Page 981 SECTION 15.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR ❙❙❙❙ 981 10 Definition The directional derivative of f at x 0 , y0 , z0 in the direction of a unit vector u a, b, c is Du f x 0 , y0 , z0 lim f x0 h b, z0 h h a, y0 hl0 hc f x 0 , y0 , z0 if this limit exists. If we use vector notation, then we can write both definitions (2 and 10) of the directional derivative in the compact form Du f x 0 11 lim f x0 hu h hl0 f x0 where x 0 x 0 , y0 if n 2 and x 0 x 0 , y0 , z0 if n 3. This is reasonable since the vector equation of the line through x0 in the direction of the vector u is given by x x 0 t u (Equation 13.5.1) and so f x 0 h u represents the value of f at a point on this line. If f x, y, z is differentiable and u a, b, c , then the same method that was used to prove Theorem 3 can be used to show that 12 Du f x, y, z fx x, y, z a fy x, y, z b fz x, y, z c For a function f of three variables, the gradient vector, denoted by f x, y, z f or grad f , is fx x, y, z , fy x, y, z , fz x, y, z or, for short, f 13 fx , fy , fz f i x f j y f k z Then, just as with functions of two variables, Formula 12 for the directional derivative can be rewritten as Du f x, y, z 14 f x, y, z u EXAMPLE 5 If f x, y, z x sin y z, (a) find the gradient of f and (b) find the directional derivative of f at 1, 3, 0 in the direction of v i 2 j k. SOLUTION (a) The gradient of f is f x, y, z fx x, y, z , fy x, y, z , fz x, y, z sin y z, x z cos y z, x y cos y z 5E-15(pp 0982-0991) 982 ❙❙❙❙ 1/18/06 3:45 PM Page 982 CHAPTER 15 PARTIAL DERIVATIVES (b) At 1, 3, 0 we have v i 2 j k is 0, 0, 3 . The unit vector in the direction of f 1, 3, 0 1 i s6 u 2 j s6 1 k s6 Therefore, Equation 14 gives Du f 1, 3, 0 f 1, 3, 0 u 1 i s6 3k 2 j s6 1 s6 3 1 k s6 3 2 Maximizing the Directional Derivative Suppose we have a function f of two or three variables and we consider all possible directional derivatives of f at a given point. These give the rates of change of f in all possible directions. We can then ask the questions: In which of these directions does f change fastest and what is the maximum rate of change? The answers are provided by the following theorem. Visual 15.6B provides visual confirmation of Theorem 15. 15 Theorem Suppose f is a differentiable function of two or three variables. The maximum value of the directional derivative Du f x is f x and it occurs when u has the same direction as the gradient vector f x . Proof From Equation 9 or 14 we have Du f fu f u cos f cos where is the angle between f and u. The maximum value of cos is 1 and this occurs when 0. Therefore, the maximum value of Du f is f and it occurs when 0, that is, when u has the same direction as f . EXAMPLE 6 y (a) If f x, y xe y, find the rate of change of f at the point P 2, 0 in the direction from 1 P to Q( 2, 2). (b) In what direction does f have the maximum rate of change? What is this maximum rate of change? Q 2 1 ±f(2, 0) 0 1 P 3x FIGURE 7 SOLUTION (a) We first compute the gradient vector: f x, y f 2, 0 1, 2 l |||| At 2, 0 the function in Example 6 increases fastest in the direction of the gradient vector f 2, 0 1, 2 . Notice from Figure 7 that this vector appears to be perpendicular to the level curve through 2, 0 . Figure 8 shows the graph of f and the gradient vector. e y, xe y fx , fy The unit vector in the direction of PQ of f in the direction from P to Q is Du f 2, 0 1.5, 2 is u f 2, 0 1( 3 5 ) u 2( 1, 2 4 5 ) 1 34 55 , 34 55 , , so the rate of change 5E-15(pp 0982-0991) 1/18/06 3:45 PM Page 983 S ECTION 15.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 983 (b) According to Theorem 15, f increases fastest in the direction of the gradient vector f 2, 0 1, 2 . The maximum rate of change is 20 15 z 10 f 2, 0 5 0 ❙❙❙❙ 1, 2 s5 EXAMPLE 7 Suppose that the temperature at a point x, y, z in space is given by 0 FIGURE 8 1 x 2 30 1 y 2 T x, y, z 80 1 x 2 2 y 2 3 z 2 , where T is measured in degrees Celsius and x, y, z in meters. In which direction does the temperature increase fastest at the point 1, 1, 2 ? What is the maximum rate of increase? SOLUTION The gradient of T is T T i x T j y 160 x x 2y 2 2 1 1 T k z x 2 160 2y 2 At the point 1, 1, 3z 3z 2 i 22 1 xi 2 320 y x 2y 2 2 2y j j 5 8 3z 22 i 1 480 z x 2y 2 2 3z 2 2 k 3z k 2 the gradient vector is T 1, 1, 160 256 2 i 2j 6k 2j 6k By Theorem 15 the temperature increases fastest in the direction of the gradient vector 5 T 1, 1, 2 i 2 j 6 k or, equivalently, in the direction of i 2 j 6 k or 8 the unit vector i 2 j 6 k s41. The maximum rate of increase is the length of the gradient vector: T 1, 1, 2 5 8 i 2j 5 s41 8 6k Therefore, the maximum rate of increase of temperature is 5 s41 8 4 C m. Tangent Planes to Level Surfaces Suppose S is a surface with equation F x, y, z k, that is, it is a level surface of a function F of three variables, and let P x 0 , y0 , z0 be a point on S. Let C be any curve that lies on the surface S and passes through the point P. Recall from Section 14.1 that the curve C is described by a continuous vector function r t x t , y t , z t . Let t0 be the parameter value corresponding to P ; that is, r t0 x 0 , y0 , z0 . Since C lies on S, any point x t , y t , z t must satisfy the equation of S, that is, 16 F x t ,y t ,z t k If x, y, and z are differentiable functions of t and F is also differentiable, then we can use the Chain Rule to differentiate both sides of Equation 16 as follows: 17 F dx x dt But, since F Fx , Fy , Fz and r t terms of a dot product as F dy y dt F dz z dt 0 x t , y t , z t , Equation 17 can be written in F rt 0 5E-15(pp 0982-0991) 984 ❙❙❙❙ 1/18/06 3:45 PM Page 984 CHAPTER 15 PARTIAL DERIVATIVES z In particular, when t ±F(x¸, y¸, z¸) tangent plane P 0 S x FIGURE 9 r ª(t¸) C y x 0 , y0 , z0 , so t0 we have r t0 F x0, y0, z0 18 r t0 0 Equation 18 says that the gradient vector at P, F x0 , y0 , z0 , is perpendicular to the tangent vector r t0 to any curve C on S that passes through P. (See Figure 9.) If F x0 , y0 , z0 0, it is therefore natural to define the tangent plane to the level surface F x, y, z k at P x 0 , y0 , z0 as the plane that passes through P and has normal vector F x0 , y0 , z0 . Using the standard equation of a plane (Equation 13.5.7), we can write the equation of this tangent plane as 19 Fx x 0 , y0 , z0 x Fy x 0 , y0 , z0 y x0 Fz x 0 , y0 , z0 z y0 z0 0 The normal line to S at P is the line passing through P and perpendicular to the tangent plane. The direction of the normal line is therefore given by the gradient vector F x0 , y0 , z0 and so, by Equation 13.5.3, its symmetric equations are 20 x x0 Fx x0 , y0 , z0 z z0 Fz x0 , y0 , z0 y y0 Fy x0 , y0 , z0 In the special case in which the equation of a surface S is of the form z f x, y (that is, S is the graph of a function f of two variables), we can rewrite the equation as F x, y, z and regard S as a level surface (with k z f x, y 0 0) of F . Then Fx x 0 , y0 , z0 fx x 0 , y0 Fy x 0 , y0 , z0 fy x 0 , y0 Fz x 0 , y0 , z0 1 so Equation 19 becomes fx x 0 , y0 x x0 fy x 0 , y0 y z y0 z0 0 which is equivalent to Equation 15.4.2. Thus, our new, more general, definition of a tangent plane is consistent with the definition that was given for the special case of Section 15.4. EXAMPLE 8 Find the equations of the tangent plane and normal line at the point 2, 1, 3 to the ellipsoid x2 4 z2 9 y2 3 SOLUTION The ellipsoid is the level surface (with k F x, y, z x2 4 3) of the function y2 z2 9 5E-15(pp 0982-0991) 1/18/06 3:45 PM Page 985 ❙❙❙❙ S ECTION 15.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR |||| Figure 10 shows the ellipsoid, tangent plane, and normal line in Example 8. 985 Therefore, we have Fx x, y, z x 2 Fy x, y, z 2y 2, 1, Fz x, y, z 2 2z 9 4 Fx 2 0 z 2, 1, 3 1 Fy 3 Fz Then Equation 19 gives the equation of the tangent plane at 2 1x 2 2y 1 4 2 3 z 3 2, 1, 2, 1, 3 2 3 3 as 0 which simplifies to 3x 6 y 2 z 18 0. By Equation 20, symmetric equations of the normal line are 6 y 02 2 0 x 2 x 2 y 1 FIGURE 10 1 2 z 3 2 3 Significance of the Gradient Vector We now summarize the ways in which the gradient vector is significant. We first consider a function f of three variables and a point P x 0 , y0 , z0 in its domain. On the one hand, we know from Theorem 15 that the gradient vector f x0, y0, z0 gives the direction of fastest increase of f . On the other hand, we know that f x0 , y0 , z0 is orthogonal to the level surface S of f through P. (Refer to Figure 9.) These two properties are quite compatible intuitively because as we move away from P on the level surface S, the value of f does not change at all. So it seems reasonable that if we move in the perpendicular direction, we get the maximum increase. In like manner we consider a function f of two variables and a point P x 0 , y0 in its domain. Again the gradient vector f x0 , y0 gives the direction of fastest increase of f . Also, by considerations similar to our discussion of tangent planes, it can be shown that f x0 , y0 is perpendicular to the level curve f x, y k that passes through P. Again this is intuitively plausible because the values of f remain constant as we move along the curve. (See Figure 11.) y ±f(x¸, y¸) P(x¸, y¸) level curve f(x, y)=k 0 FIGURE 11 x curve of steepest ascent 300 200 100 FIGURE 12 If we consider a topographical map of a hill and let f x, y represent the height above sea level at a point with coordinates x, y , then a curve of steepest ascent can be drawn as in Figure 12 by making it perpendicular to all of the contour lines. This phenomenon can also be noticed in Figure 12 in Section 15.1, where Lonesome Creek follows a curve of steepest descent. 5E-15(pp 0982-0991) 986 ❙❙❙❙ 1/18/06 3:45 PM Page 986 CHAPTER 15 PARTIAL DERIVATIVES Computer algebra systems have commands that plot sample gradient vectors. Each gradient vector f a, b is plotted starting at the point a, b . Figure 13 shows such a plot (called a gradient vector field ) for the function f x, y x 2 y 2 superimposed on a contour map of f. As expected, the gradient vectors point “uphill” and are perpendicular to the level curves. y _9 _6 _3 0 369 x FIGURE 13 |||| 15.6 Exercises directional derivative of this snowfall function at Muskegon, Michigan, in the direction of Ludington. What are the units? 1. A contour map of barometric pressure (in millibars) is shown for 7:00 A.M. on September 12, 1960, when Hurricane Donna was raging. Estimate the value of the directional derivative of the pressure function at Raleigh, North Carolina, in the direction of the eye of the hurricane. What are the units of the directional derivative? 60 70 50 70 100 80 70 60 Green Bay Ludington 50 40 70 40 Milwaukee 976 980 988 Raleigh 0 984 992 25 Muskegon Grand Rapids 50 75 100 (Distance in miles) 3. A table of values for the wind-chill index W f T, v is given in Exercise 3 on page 955. Use the table to estimate the value i j s2. of Du f 20, 30 , where u 996 1000 4–6 |||| Find the directional derivative of f at the given point in the direction indicated by the angle . 1004 0 50 100 150 200 (Distance in miles) 4. f x, y inches) near Lake Michigan. Estimate the value of the y 4, 2, 1 , 5. f x, y 2. The contour map shows the average annual snowfall (in x 2y3 s5x 4y, 4, 1 , 6. f x, y ■ ■ x sin x y , ■ ■ 4 6 2, 0 , ■ ■ 3 ■ ■ ■ ■ ■ ■ 5E-15(pp 0982-0991) 1/18/06 3:45 PM Page 987 S ECTION 15.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 7–10 5xy 2 7. f x, y 8. f x, y 4 x 3y, y ln x, xe 10. f x, y, z P 1, sx 11–17 ■ P 1, 2 , 3, , u , 2 3 y z, P 1, 3, 1 , ■ ■ 28. Find the directions in which the directional derivative of 43 55 , P 3, 0, 2 , u ■ rapidly at x in the direction opposite to the gradient vector, that is, in the direction of f x. (b) Use the result of part (a) to find the direction in which the function f x, y x 4 y x 2 y 3 decreases fastest at the point 2, 3 . 5 12 13 13 u 2 yz 9. f x, y, z ■ , , ,, ■ function f x, y ■ ■ ■ ■ |||| 11. f x, y 1 12. f x, y ln x 2 13. t s, t 2t s e, 14. t r, e r sin , 2 x sy, 3, 4 , v y2 , 2, 1 , 2, 0 , v 15. f x, y, z sx 2 y2 16. f x, y, z x 17. t x, y, z ■ x ■ 3i 1, 2, 2, 4, 1, 1 , v 32 ■ , x2 y2 2x 4 y is i j. 30. Near a buoy, the depth of a lake at the point with coordinates x, y is z 200 0.02 x 2 0.001y 3, where x, y, and z are measured in meters. A fisherman in a small boat starts at the point 80, 60 and moves toward the buoy, which is located at 0, 0 . Is the water under the boat getting deeper or shallower when he departs? Explain. ■ the distance from the center of the ball, which we take to be the origin. The temperature at the point 1, 2, 2 is 120 . (a) Find the rate of change of T at 1, 2, 2 in the direction toward the point 2, 1, 3 . (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin. 2j v 6, 6, 3 1, 2, 3 1, 1, 2 , v ■ sin x y at the point (1, 0) has the value 1. 31. The temperature T in a metal ball is inversely proportional to j z 2, 3z 2y ■ 3 1, 2 3, v z, y v i 0, 4, x2 29. Find all points at which the direction of fastest change of the 236 777 u ■ f x, y 21 33 Find the directional derivative of the function at the given point in the direction of the vector v. ■ 987 27. (a) Show that a differentiable function f decreases most |||| (a) Find the gradient of f . (b) Evaluate the gradient at the point P. (c) Find the rate of change of f at P in the direction of the vector u. ■ ❙❙❙❙ ■ 2j ■ k ■ ■ ■ 32. The temperature at a point x, y, z is given by 18. Use the figure to estimate Du f 2, 2 . y T x, y, z ±f (2, 2) x 19. Find the directional derivative of f x, y 33. Suppose that over a certain region of space the electrical poten- sxy at P 2, 8 in tial V is given by V x, y, z 5x 2 3xy x yz. (a) Find the rate of change of the potential at P 3, 4, 5 in the direction of the vector v i j k. (b) In which direction does V change most rapidly at P ? (c) What is the maximum rate of change at P ? the direction of Q 5, 4 . x2 20. Find the directional derivative of f x, y, z y2 z 2 at P 2, 1, 3 in the direction of the origin. 21–26 |||| Find the maximum rate of change of f at the given point and the direction in which it occurs. 21. f x, y y 2 x, 22. f p, q qe 23. f x, y sin x y , 2 24. f x, y, z (1, 1, 1) ln x y z , tan x ■ ■ 0, 0 (1, 0) 34 23 26. f x, y, z ■ pe q, xyz, 25. f x, y, z ■ 2, 4 p 1, 3 3z , 2y ■ 2, ■ 5, 1, 1 ■ ■ x 2 3y 2 9z 2 where T is measured in C and x, y, z in meters. (a) Find the rate of change of temperature at the point P 2, 1, 2 in the direction toward the point 3, 3, 3 . (b) In which direction does the temperature increase fastest at P ? (c) Find the maximum rate of increase at P. (2, 2) u 0 200 e ■ ■ ■ ■ 34. Suppose you are climbing a hill whose shape is given by the equation z 1000 0.01x 2 0.02y 2, where x, y, and z are measured in meters, and you are standing at a point with coordinates 50, 80, 847 . The positive x-axis points east and the positive y-axis points north. (a) If you walk due south, will you start to ascend or descend? At what rate? (b) If you walk northwest, will you start to ascend or descend? At what rate? (c) In which direction is the slope largest? What is the rate of ascent in that direction? At what angle above the horizontal does the path in that direction begin? 5E-15(pp 0982-0991) ❙❙❙❙ 988 1/18/06 3:46 PM Page 988 CHAPTER 15 PARTIAL DERIVATIVES 35. Let f be a function of two variables that has continuous partial derivatives and consider the points A 1, 3 , B 3, 3 , C 1, 7 , and D 6, 15 . The directional derivative of f at A in l the direction of the vector AB is 3 and the directional derivative l at A in the direction of AC is 26. Find the directional derivative l of f at A in the direction of the vector AD . 36. For the given contour map draw the curves of steepest ascent starting at P and at Q. 44. yz ■ ln x ■ 0, 0, 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ; 45–46 |||| Use a computer to graph the surface, the tangent plane, and the normal line on the same screen. Choose the domain carefully so that you avoid extraneous vertical planes. Choose the viewpoint so that you get a good view of all three objects. 45. x y 46. x y z Q z, ■ ■ yz 6, ■ zx 3, 1, 1, 1 1, 2, 3 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ x 2 4 y 2, find the gradient vector f 2, 1 and use it to find the tangent line to the level curve f x, y 8 at the point 2, 1 . Sketch the level curve, the tangent line, and the gradient vector. 47. If f x, y 60 20 50 40 30 x y 2, find the gradient vector t 3, 1 and use it to find the tangent line to the level curve t x, y 2 at the point 3, 1 . Sketch the level curve, the tangent line, and the gradient vector. 48. If t x, y P 37. Show that the operation of taking the gradient of a function has the given property. Assume that u and v are differentiable func- tions of x and y and a, b are constants. (a) a u b v au bv (b) u v uv vu u (c) vu v un (d) 1 x2 a2 y2 b2 written as uv v nu n 49. Show that the equation of the tangent plane to the ellipsoid 1 at the point x 0 , y0 , z0 can be xx 0 a2 y y0 b2 z z0 c2 1 2 50. Find the equation of the tangent plane to the hyperboloid u 38. Sketch the gradient vector f 4, 6 for the function f whose level curves are shown. Explain how you chose the direction and length of this vector. x 2 a 2 y 2 b 2 z 2 c 2 1 at x 0 , y0 , z0 and express it in a form similar to the one in Exercise 49. 51. Show that the equation of the tangent plane to the elliptic paraboloid z c written as _5 (4, 6) _3 y 2 b 2 at the point x 0 , y0 , z0 can be 2 yy0 b2 z z0 c 52. Find the points on the ellipsoid x 2 2y 2 3z 2 1 where the tangent plane is parallel to the plane 3x y 3z 1. _1 4 x2 a2 2 xx 0 a2 y 6 z2 c 2 53. Find the points on the hyperboloid x 2 y 2 2 z 2 1 where the normal line is parallel to the line that joins the points 3, 1, 0 and 5, 3, 6 . 0 1 3 5 2 54. Show that the ellipsoid 3x 2 2y 2 z 2 9 and the sphere y z 8 x 6y 8z 24 0 are tangent to each x other at the point 1, 1, 2 . (This means that they have a common tangent plane at the point.) 2 0 39–44 2 4 6 x Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point. |||| 39. x 2 y2 40. x 41. x 2y 2 2 2y 3z 2 z2 2 21, 2, z 2 4, 2, 42. x z 4 arctan y z , 43. z 1 xe y cos z, 1 1, 0, 0 2 55. Show that every plane that is tangent to the cone x 2 y2 z2 passes through the origin. 56. Show that every normal line to the sphere x 2 1, 1 y2 z2 r2 passes through the center of the sphere. 1, 1, 0 yz 2 2, 1, 1 , 1, 1 57. Show that the sum of the x-, y-, and z -intercepts of any tangent plane to the surface s x sy sz sc is a constant. 58. Show that the product of the x-, y-, and z -intercepts of any tangent plane to the surface x y z c 3 is a constant. 5E-15(pp 0982-0991) 1/18/06 3:46 PM Page 989 SECTION 15.7 MAXIMUM AND MINIMUM VALUES 59. Find parametric equations for the tangent line to the curve of 2 intersection of the paraboloid z 4 x 2 y 2 z 2 9 at the point y and the ellipsoid x 1, 1, 2 . z 3 intersects the cylinder x 2 y 2 5 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point 1, 2, 1 . (b) Graph the cylinder, the plane, and the tangent line on the same screen. ; 61. (a) Two surfaces are called orthogonal at a point of inter- section if their normal lines are perpendicular at that point. Show that surfaces with equations F x, y, z 0 and G x, y, z 0 are orthogonal at a point P where F 0 and G 0 if and only if Fx Gx Fy Gy Fz Gz 0 at P. (b) Use part (a) to show that the surfaces z 2 x 2 y 2 and x 2 y 2 z 2 r 2 are orthogonal at every point of intersection. Can you see why this is true without using calculus? |||| 15.7 989 3 sx y is continuous and the partial derivatives fx and fy exist at the origin but the directional derivatives in all other directions do not exist. (b) Graph f near the origin and comment on how the graph confirms part (a). 62. (a) Show that the function f x, y 2 60. (a) The plane y ❙❙❙❙ ; 63. Suppose that the directional derivatives of f x, y are known at a given point in two nonparallel directions given by unit vectors u and v. Is it possible to find f at this point? If so, how would you do it? 64. Show that if z lim x l x0 fx f x, y is differentiable at x 0 f x0 x f x0 x0 x x0 x 0 , y0 , then 0 [Hint: Use Definition 15.4.7 directly.] Maximum and Minimum Values As we saw in Chapter 4, one of the main uses of ordinary derivatives is in finding maximum and minimum values. In this section we see how to use partial derivatives to locate maxima and minima of functions of two variables. In particular, in Example 6 we will see how to maximize the volume of a box without a lid if we have a fixed amount of cardboard to work with. z absolute maximum local maximum y x absolute minimum local minimum FIGURE 1 |||| Notice that the conclusion of Theorem 2 can be stated in the notation of the gradient vectors as f a, b 0. 1 Definition A function of two variables has a local maximum at a, b if f x, y f a, b when x, y is near a, b . [This means that f x, y f a, b for all points x, y in some disk with center a, b .] The number f a, b is called a local maximum value. If f x, y f a, b when x, y is near a, b , then f a, b is a local minimum value. If the inequalities in Definition 1 hold for all points x, y in the domain of f , then f has an absolute maximum (or absolute minimum) at a, b . The graph of a function with several maxima and minima is shown in Figure 1. You can think of the local maxima as mountain peaks and the local minima as valley bottoms. 2 Theorem If f has a local maximum or minimum at a, b and the first-order partial derivatives of f exist there, then fx a, b 0 and fy a, b 0. Proof Let t x f x, b . If f has a local maximum (or minimum) at a, b , then t has a local maximum (or minimum) at a, so t a 0 by Fermat’s Theorem (see Theorem 4.1.4). But t a fx a, b (see Equation 15.3.1) and so fx a, b 0. Similarly, by applying Fermat’s Theorem to the function G y f a, y , we obtain fy a, b 0. If we put fx a, b 0 and fy a, b 0 in the equation of a tangent plane (Equation 15.4.2), we get z z0 . Thus, the geometric interpretation of Theorem 2 is that if the graph of f has a tangent plane at a local maximum or minimum, then the tangent plane must be horizontal. 5E-15(pp 0982-0991) 990 ❙❙❙❙ 1/18/06 3:46 PM Page 990 CHAPTER 15 PARTIAL DERIVATIVES A point a, b is called a critical point (or stationary point) of f if fx a, b 0 and fy a, b 0, or if one of these partial derivatives does not exist. Theorem 2 says that if f has a local maximum or minimum at a, b , then a, b is a critical point of f . However, as in single-variable calculus, not all critical points give rise to maxima or minima. At a critical point, a function could have a local maximum or a local minimum or neither. z x2 EXAMPLE 1 Let f x, y y2 fx x, y 2x 2x 6y 14. Then 2 fy x, y 2y These partial derivatives are equal to 0 when x 1 and y is 1, 3 . By completing the square, we find that (1, 3, 4) f x, y 0 x 1 2 y 3, so the only critical point 3 2 Since x 1 2 0 and y 3 2 0, we have f x, y 4 for all values of x and y. Therefore, f 1, 3 4 is a local minimum, and in fact it is the absolute minimum of f . This can be confirmed geometrically from the graph of f , which is the elliptic paraboloid with vertex 1, 3, 4 shown in Figure 2. y x 4 6 FIGURE 2 z=≈+¥-2x-6y+14 EXAMPLE 2 Find the extreme values of f x, y y2 x 2. SOLUTION Since fx 2 x and fy 2y, the only critical point is 0, 0 . Notice that for points on the x-axis we have y 0, so f x, y x 2 0 (if x 0). However, for 2 points on the y-axis we have x 0, so f x, y y 0 (if y 0). Thus, every disk with center 0, 0 contains points where f takes positive values as well as points where f takes negative values. Therefore, f 0, 0 0 can’t be an extreme value for f , so f has no extreme value. z x FIGURE 3 y Example 2 illustrates the fact that a function need not have a maximum or minimum value at a critical point. Figure 3 shows how this is possible. The graph of f is the hyperbolic paraboloid z y 2 x 2, which has a horizontal tangent plane (z 0) at the origin. You can see that f 0, 0 0 is a maximum in the direction of the x-axis but a minimum in the direction of the y-axis. Near the origin the graph has the shape of a saddle and so 0, 0 is called a saddle point of f . We need to be able to determine whether or not a function has an extreme value at a critical point. The following test, which is proved at the end of this section, is analogous to the Second Derivative Test for functions of one variable. z=¥-≈ 3 Second Derivatives Test Suppose the second partial derivatives of f are continuous on a disk with center a, b , and suppose that fx a, b 0 and fy a, b 0 [that is, a, b is a critical point of f ]. Let D (a) If D (b) If D (c) If D NOTE 1 D a, b fxx a, b fyy a, b fx y a, b 2 0 and fxx a, b 0, then f a, b is a local minimum. 0 and fxx a, b 0, then f a, b is a local maximum. 0, then f a, b is not a local maximum or minimum. In case (c) the point a, b is called a saddle point of f and the graph of f crosses its tangent plane at a, b . ■ 5E-15(pp 0982-0991) 1/18/06 3:46 PM Page 991 ❙❙❙❙ SECTION 15.7 MAXIMUM AND MINIMUM VALUES 991 N OTE 2 If D 0, the test gives no information: f could have a local maximum or local minimum at a, b , or a, b could be a saddle point of f . NOTE 3 To remember the formula for D it’s helpful to write it as a determinant: ■ ■ fxx fx y fyx fyy D fxx fyy fx y 2 EXAMPLE 3 Find the local maximum and minimum values and saddle points of f x, y x4 y4 4 xy 1. SOLUTION We first locate the critical points: 4x 3 fx 4y 4y 3 fy 4x Setting these partial derivatives equal to 0, we obtain the equations x3 y 0 z x9 x x x8 1 x x4 x 0 x 3 from the first equation into the second To solve these equations we substitute y one. This gives 0 y3 and 1 x4 1 x x2 1 x2 1 x4 1 so there are three real roots: x 0, 1, 1. The three critical points are 0, 0 , 1, 1 , and 1, 1 . Next we calculate the second partial derivatives and D x, y : fxx 12 x 2 D x, y fx y fxx fyy 4 fx y 2 fyy 144 x 2 y 2 12 y 2 16 Since D 0, 0 16 0, it follows from case (c) of the Second Derivatives Test that the origin is a saddle point; that is, f has no local maximum or minimum at 0, 0 . Since D 1, 1 128 0 and fxx 1, 1 12 0, we see from case (a) of the test that f 1, 1 1 is a local minimum. Similarly, we have D 1, 1 128 0 and , so f 1, 1 is also a local minimum. fxx 1, 1 12 0 1 The graph of f is shown in Figure 4. y x FIGURE 4 z=x$+y$-4xy+1 |||| A contour map of the function f in Example 3 is shown in Figure 5. The level curves near 1, 1 and 1, 1 are oval in shape and indicate that as we move away from 1, 1 or 1, 1 in any direction the values of f are increasing. The level curves near 0, 0 , on the other hand, resemble hyperbolas. They reveal that as we move away from the origin (where the value of f is 1), the values of f decrease in some directions but increase in other directions. Thus, the contour map suggests the presence of the minima and saddle point that we found in Example 3. y _0.5 0 0.5 0.9 1 1.1 1.5 2 x 3 FIGURE 5 5E-15(pp 0992-1001) 992 ❙❙❙❙ 1/18/06 3:49 PM Page 992 CHAPTER 15 PARTIAL DERIVATIVES In Module 15.7 you can use contour maps to estimate the locations of critical points. EXAMPLE 4 Find and classify the critical points of the function 10 x 2 y f x, y 5x 2 4y 2 x4 2y 4 Also find the highest point on the graph of f . SOLUTION The first-order partial derivatives are fx 20 xy 4x 3 10 x 10 x 2 fy 8y 8y 3 So to find the critical points we need to solve the equations 2x2 4 2 x 10 y 5 5 5x 2 4y or 10 y 0 4y 3 0 5 2x2 From Equation 4 we see that either x 0 y2 In the first case (x 0), Equation 5 becomes 4y 1 have the critical point 0, 0 . In the second case 10 y 5 2 x 2 0 , we get x2 6 5y 4y 3 0, so y 0 and we 2.5 and, putting this in Equation 5, we have 25y solve the cubic equation 7 0 12.5 12.5 21y 4y 3 4y 0. So we have to 0 Using a graphing calculator or computer to graph the function 4y 3 ty _3 21y 12.5 2.7 as in Figure 6, we see that Equation 7 has three real roots. By zooming in, we can find the roots to four decimal places: FIGURE 6 y 2.5452 y 0.6468 y 1.8984 (Alternatively, we could have used Newton’s method or a rootfinder to locate these roots.) From Equation 6, the corresponding x-values are given by x s5y 2.5 If y 2.5452, then x has no corresponding real values. If y 0.6468, then x 0.8567. If y 1.8984, then x 2.6442. So we have a total of five critical points, which are analyzed in the following chart. All quantities are rounded to two decimal places. 5E-15(pp 0992-1001) 1/18/06 3:49 PM Page 993 S ECTION 15.7 MAXIMUM AND MINIMUM VALUES Critical point Value of f 0, 0 fxx D ❙❙❙❙ 993 Conclusion 0.00 10.00 80.00 local maximum 2.64, 1.90 8.50 55.93 2488.71 local maximum 0.86, 0.65 1.48 5.87 187.64 saddle point Figures 7 and 8 give two views of the graph of f and we see that the surface opens downward. [This can also be seen from the expression for f x, y : The dominant terms are x 2 2y 4 when x and y are large.] Comparing the values of f at its local maximum points, we see that the absolute maximum value of f is f 2.64, 1.90 8.50. In other words, the highest points on the graph of f are 2.64, 1.90, 8.50 . z z x Visual 15.7 shows several families of surfaces. The surface in Figures 7 and 8 is a member of one of these families. y x y FIGURE 7 FIGURE 8 y 2 7 3 |||| The five critical points of the function f in Example 4 are shown in red in the contour map of f in Figure 9. 1 _1.48 _3 _1 0 _2 _0 30 _0.8 x 3 _3 _1 F IGURE 9 EXAMPLE 5 Find the shortest distance from the point 1, 0, x 2y z 2 to the plane 4. SOLUTION The distance from any point x, y, z to the point 1, 0, d sx 1 2 y2 z 2 2 is 2 but if x, y, z lies on the plane x 2y z 4, then z 4 x 2y and so we have d s x 1 2 y2 6 x 2y 2. We can minimize d by minimizing the simpler expression d2 f x, y x 1 2 y2 6 x 2y 2 5E-15(pp 0992-1001) 994 ❙❙❙❙ 1/18/06 3:50 PM Page 994 CHAPTER 15 PARTIAL DERIVATIVES By solving the equations fx 2x 1 fy 2y 26 46 2y 4x 4y 2y x x 14 4x 10 y 24 0 0 we find that the only critical point is ( 11, 5 ). Since fxx 4, fx y 4, and fyy 10, we 63 have D x, y fxx fy y fx y 2 24 0 and fxx 0, so by the Second Derivatives Test f has a local minimum at ( 11, 5 ). Intuitively, we can see that this local minimum is actually 63 an absolute minimum because there must be a point on the given plane that is closest to 1, 0, 2 . If x 161 and y 5 , then 3 |||| Example 5 could also be solved using vectors. Compare with the methods of Section 13.5. d sx 1 2 y2 6 The shortest distance from 1, 0, x 2y s(5)2 6 2 2 to the plane x (5)2 3 2y z 5 s6 6 (5)2 6 4 is 5 s6 6. EXAMPLE 6 A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box. SOLUTION Let the length, width, and height of the box (in meters) be x, y, and z, as shown in Figure 10. Then the volume of the box is V z x y x yz We can express V as a function of just two variables x and y by using the fact that the area of the four sides and the bottom of the box is 2xz FIGURE 10 2y z Solving this equation for z, we get z becomes V 12 12 2x xy xy y xy 12 xy 2x 12 xy 2x y , so the expression for V x2y2 y We compute the partial derivatives: V x y 2 12 2 xy 2x y If V is a maximum, then V x must solve the equations 12 2 xy x2 V y 2 Vy x2 0 0, but x 12 x 2 12 2 xy 2x y 0 or y 2 xy y2 2 0 gives V y2 0, so we 0 These imply that x 2 y 2 and so x y. (Note that x and y must both be positive in this problem.) If we put x y in either equation we get 12 3x 2 0, which gives x 2, y 2, and z 12 2 2 2 2 2 1. We could use the Second Derivatives Test to show that this gives a local maximum of V , or we could simply argue from the physical nature of this problem that there must be an absolute maximum volume, which has to occur at a critical point of V, so it must occur when x 2, y 2, z 1. Then V 2 2 1 4, so the maximum volume of the box is 4 m3 . 5E-15(pp 0992-1001) 1/18/06 3:50 PM Page 995 SECTION 15.7 MAXIMUM AND MINIMUM VALUES ❙❙❙❙ 995 A bsolute Maximum and Minimum Values (a) Closed sets For a function f of one variable the Extreme Value Theorem says that if f is continuous on a closed interval a, b , then f has an absolute minimum value and an absolute maximum value. According to the Closed Interval Method in Section 4.1, we found these by evaluating f not only at the critical numbers but also at the endpoints a and b. There is a similar situation for functions of two variables. Just as a closed interval contains its endpoints, a closed set in 2 is one that contains all its boundary points. [A boundary point of D is a point a, b such that every disk with center a, b contains points in D and also points not in D.] For instance, the disk D (b) Sets that are not closed FIGURE 11 x, y x 2 y2 1 which consists of all points on and inside the circle x 2 y 2 1, is a closed set because it contains all of its boundary points (which are the points on the circle x 2 y 2 1). But if even one point on the boundary curve were omitted, the set would not be closed. (See Figure 11.) A bounded set in 2 is one that is contained within some disk. In other words, it is finite in extent. Then, in terms of closed and bounded sets, we can state the following counterpart of the Extreme Value Theorem in two dimensions. 8 Extreme Value Theorem for Functions of Two Variables If f is continuous on a closed, bounded set D in 2, then f attains an absolute maximum value f x 1, y1 and an absolute minimum value f x 2 , y2 at some points x 1, y1 and x 2 , y2 in D. To find the extreme values guaranteed by Theorem 8, we note that, by Theorem 2, if f has an extreme value at x 1, y1 , then x 1, y1 is either a critical point of f or a boundary point of D. Thus, we have the following extension of the Closed Interval Method. 9 To find the absolute maximum and minimum values of a continuous function f on a closed, bounded set D : 1. Find the values of f at the critical points of f in D. 2. Find the extreme values of f on the boundary of D. 3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. EXAMPLE 7 Find the absolute maximum and minimum values of the function f x, y x2 2 xy 2y on the rectangle D x, y 0 x 3, 0 y 2. SOLUTION Since f is a polynomial, it is continuous on the closed, bounded rectangle D, so Theorem 8 tells us there is both an absolute maximum and an absolute minimum. According to step 1 in (9), we first find the critical points. These occur when fx 2x 2y 0 fy 2x 2 0 so the only critical point is 1, 1 , and the value of f there is f 1, 1 1. 5E-15(pp 0992-1001) 996 ❙❙❙❙ 1/18/06 3:50 PM Page 996 CHAPTER 15 PARTIAL DERIVATIVES y (0, 2) L£ (2, 2) L¢ (0, 0) In step 2 we look at the values of f on the boundary of D, which consists of the four line segments L 1 , L 2 , L 3 , L 4 shown in Figure 12. On L 1 we have y 0 and (3, 2) L¡ (3, 0) x2 f x, 0 L™ x 0 x 3 This is an increasing function of x, so its minimum value is f 0, 0 mum value is f 3, 0 9. On L 2 we have x 3 and f 3, y FIGURE 12 9 4y 0 y 2 This is a decreasing function of y, so its maximum value is f 3, 0 value is f 3, 2 1. On L 3 we have y 2 and x2 f x, 2 4x 4 0 0 and its maxi- x 9 and its minimum 3 By the methods of Chapter 4, or simply by observing that f x, 2 x 2 2, we see that the minimum value of this function is f 2, 2 0 and the maximum value is f 0, 2 4. Finally, on L 4 we have x 0 and f 0, y 2y 0 y 2 with maximum value f 0, 2 4 and minimum value f 0, 0 0. Thus, on the boundary, the minimum value of f is 0 and the maximum is 9. In step 3 we compare these values with the value f 1, 1 1 at the critical point and conclude that the absolute maximum value of f on D is f 3, 0 9 and the absolute minimum value is f 0, 0 f 2, 2 0. Figure 13 shows the graph of f . 9 z 0 L¡ D L™ x FIGURE 13 2 30 f(x, y)=≈-2xy+2y y We close this section by giving a proof of the first part of the Second Derivatives Test. Part (b) has a similar proof. Proof of Theorem 3, Part (a) We compute the second-order directional derivative of f in the direction of u h, k . The first-order derivative is given by Theorem 15.6.3: Du f fx h fy k Applying this theorem a second time, we have 2 Du f Du Du f x Du f h fxx h fyx k h fxy h fxx h2 2 fxy hk fyy k 2 y Du f k fyy k k (by Clairaut’s Theorem) 5E-15(pp 0992-1001) 1/18/06 3:50 PM Page 997 ❙❙❙❙ S ECTION 15.7 MAXIMUM AND MINIMUM VALUES 997 If we complete the square in this expression, we obtain 2 Du f 10 fxx h 2 fx y k fxx k2 fxx fyy fxx f2 xy 2 We are given that fxx a, b 0 and D a, b 0. But fxx and D fxx fyy f x y are continuous functions, so there is a disk B with center a, b and radius 0 such that fxx x, y 0 and D x, y 0 whenever x, y is in B. Therefore, by looking at Equation 2 10, we see that Du f x, y 0 whenever x, y is in B. This means that if C is the curve obtained by intersecting the graph of f with the vertical plane through P a, b, f a, b in the direction of u, then C is concave upward on an interval of length 2 . This is true in the direction of every vector u, so if we restrict x, y to lie in B, the graph of f lies above its horizontal tangent plane at P. Thus, f x, y f a, b whenever x, y is in B. This shows that f a, b is a local minimum. |||| 15.7 Exercises 4. f x, y 1. Suppose (1, 1) is a critical point of a function f with contin- x3 3x 2y 2 y4 uous second derivatives. In each case, what can you say about f ? (a) fxx 1, 1 4, fx y 1, 1 1, fyy 1, 1 2 fx y 1, 1 fyy 1, 1 3, 1.5 2 _2.9 _2.7 _2.5 2. Suppose (0, 2) is a critical point of a function t with contin- uous second derivatives. In each case, what can you say about t ? (a) txx 0, 2 1, tx y 0, 2 6, tyy 0, 2 1 (b) txx 0, 2 (c) txx 0, 2 1, 2, tyy 0, 2 tx y 0, 2 4, tx y 0, 2 6, tyy 0, 2 _1 Use the level curves in the figure to predict the location of the critical points of f and whether f has a saddle point or a local maximum or minimum at each of those points. Explain your reasoning. Then use the Second Derivatives Test to confirm your predictions. 4 x3 y3 _1 ■ ■ ■ ■ ■ ■ 5. f x, y 9 2x xy 12 x 4 4 x 8. f x, y e y x2 4y 3 7. f x, y y 2 4 xy 0 1 x 1 xy x 4y 2 y 2x 3 xy2 5x 2 1 12. f x, y 1 4.2 xy 1 2x y x2 x y2 y2 y x 6 x2 y2 13. f x, y e cos y 14. f x, y 15. f x, y 2 5 ■ 2 11. f x, y 4 3.7 3.2 ■ 4y x2 y2 9. f x, y 3.7 ■ 8y 10. f x, y 3.2 ■ |||| Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. 3xy 1 ■ 5–18 6. f x, y _1 x 1 9 |||| 3. f x, y 1.5 1.7 1.9 8 ■ 3–4 1 0 0.5 1 4, _2 _1 . _1 5 (b) fxx 1, 1 y x sin y 16. f x, y 2x x2 18. f x, y x 2 ye 1 x2y2 2y y 2 x2 y2 _1 x2 17. f x, y ■ ■ ■ y2 ey ■ ■ 2 x2 ■ ■ ■ ■ ■ ■ ■ 5E-15(pp 0992-1001) 998 ❙❙❙❙ 1/18/06 3:50 PM Page 998 CHAPTER 15 PARTIAL DERIVATIVES ; 19–22 |||| Use a graph and /or level curves to estimate the local maximum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisely. 2 19. f x, y 3x y y x ye 21. f x, y sin x 2 ,0 x 22. f x, y 0 ■ ■ 3y 2 sin y sin x y2 sin x sin y 4, 0 y x ■ 3x 2 2 ■ y, ■ ■ ■ x3 e 3y 37. Find the shortest distance from the point 2, 1, y, cos x 4 3xe y f x, y has exactly one critical point, and that f has a local maximum there that is not an absolute maximum. Then use a computer to produce a graph with a carefully chosen domain and viewpoint to see how this is possible. x2 y2 20. f x, y 0 3 an absolute maximum. But this is not true for functions of two variables. Show that the function plane x ■ ■ ■ ■ ■ y z 1 to the 1. 38. Find the point on the plane x z y 4 that is closest to the point 1, 2, 3 . ; 23–26 |||| Use a graphing device as in Example 4 (or Newton’s method or a rootfinder) to find the critical points of f correct to three decimal places. Then classify the critical points and find the highest or lowest points on the graph. x4 23. f x, y 5x 2 y2 24. f x, y 5 10 xy 4x 25. f x, y 2x 4x 2 y2 26. f x, y ex y4 3x 2 ■ ■ 27–34 ■ |||| ■ 40. Find the points on the surface x 2 y 2z ■ x4 product is a maximum. y4 42. Find three positive numbers x, y, and z whose sum is 100 such that x a y bz c is a maximum. ■ ■ ■ ■ ■ ■ Find the absolute maximum and minimum values of f on the set D. 1 4 x 5y, D is the closed triangular region with vertices 0, 0 , 2, 0 , and 0, 3 3 x y x 2y, D is the closed triangular region with vertices 1, 0 , 5, 0 , and 1, 4 30. f x, y D 31. f x, y D x2 y2 x2y x, y x 1, y 36 x2 a2 y2 b2 z2 c2 1 octant with three faces in the coordinate planes and one vertex in the plane x 2y 3z 6. 2 y, x 4, 0 y 5 46. Find the dimensions of the rectangular box with largest x4 y4 x, y 0 x 4 xy 3, 0 2 47. Find the dimensions of a rectangular box of maximum volume x y 2, D 33. f x, y 2x 3 y 4, x3 vertices are ■ 2, y x, y x volume if the total surface area is given as 64 cm2 . 0, y x, y x 2 D 0, x 2 y2 y2 such that the sum of the lengths of its 12 edges is a constant c. 3 48. The base of an aquarium with given volume V is made of slate 1 and the sides are made of glass. If slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials. 3x y 3 12y, D is the quadrilateral whose 2, 3 , 2, 3 , 2, 2 , and 2, 2 . 34. f x, y ■ 4z2 45. Find the volume of the largest rectangular box in the first 4, 1 2 36y 2 4x 6y x, y 0 x 32. f x, y ■ parallel to the axes that can be inscribed in the ellipsoid 44. Solve the problem in Exercise 43 for a general ellipsoid 28. f x, y D 43. Find the volume of the largest rectangular box with edges 9x 2 27. f x, y 29. f x, y 1 that are closest to 41. Find three positive numbers whose sum is 100 and whose 4 cos y ■ 1 that are closest to the origin. y4 3y xy the origin. 2 2 xy 2 x3 39. Find the points on the surface z 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ; 35. For functions of one variable it is impossible for a continuous function to have two local maxima and no local minimum. But for functions of two variables such functions exist. Show that the function f x, y x 2 1 2 2 xy x 1 2 has only two critical points, but has local maxima at both of them. Then use a computer to produce a graph with a carefully chosen domain and viewpoint to see how this is possible. ; 36. If a function of one variable is continuous on an interval and has only one critical number, then a local maximum has to be 49. A cardboard box without a lid is to have a volume of 32,000 cm3. Find the dimensions that minimize the amount of cardboard used. 50. A rectangular building is being designed to minimize heat loss. The east and west walls lose heat at a rate of 10 units m2 per day, the north and south walls at a rate of 8 units m2 per day, the floor at a rate of 1 unit m2 per day, and the roof at a rate of 5 units m2 per day. Each wall must be at least 30 m long, the height must be at least 4 m, and the volume must be exactly 4000 m3. (a) Find and sketch the domain of the heat loss as a function of the lengths of the sides. 5E-15(pp 0992-1001) 1/18/06 3:50 PM Page 999 A PPLIED PROJECT DESIGNING A DUMPSTER (b) Find the dimensions that minimize heat loss. (Check both the critical points and the points on the boundary of the domain.) (c) Could you design a building with even less heat loss if the restrictions on the lengths of the walls were removed? ❙❙❙❙ 999 y (x i, yi ) di (⁄, ›) mx i+b 51. If the length of the diagonal of a rectangular box must be L, what is the largest possible volume? 52. Three alleles (alternative versions of a gene) A, B, and O determine the four blood types A (AA or AO), B (BB or BO), O (OO), and AB. The Hardy-Weinberg Law states that the proportion of individuals in a population who carry two different alleles is P 2pq 2pr 2rq where p, q, and r are the proportions of A, B, and O in the population. Use the fact that p q r 1 to show that P is at most 2. 3 53. Suppose that a scientist has reason to believe that two quan- tities x and y are related linearly, that is, y m x b, at least approximately, for some values of m and b. The scientist performs an experiment and collects data in the form of points x 1, y1 , x 2 , y2 , . . . , x n , yn , and then plots these points. The points don’t lie exactly on a straight line, so the scientist wants to find constants m and b so that the line y m x b “fits” the points as well as possible. (See the figure.) 0 x Let di yi m x i b be the vertical deviation of the point x i , yi from the line. The method of least squares determines m and b so as to minimize n 1 d 2, the sum of the squares of i i these deviations. Show that, according to this method, the line of best fit is obtained when n m n xi bn i1 n n x2 i m i1 yi i1 b n xi i1 x i yi i1 Thus, the line is found by solving these two equations in the two unknowns m and b. (See Section 1.2 for a further discussion and applications of the method of least squares.) 54. Find an equation of the plane that passes through the point 1, 2, 3 and cuts off the smallest volume in the first octant. APPLIED PROJECT Designing a Dumpster For this project we locate a trash dumpster in order to study its shape and construction. We then attempt to determine the dimensions of a container of similar design that minimize construction cost. 1. First locate a trash dumpster in your area. Carefully study and describe all details of its construction, and determine its volume. Include a sketch of the container. 2. While maintaining the general shape and method of construction, determine the dimensions such a container of the same volume should have in order to minimize the cost of construction. Use the following assumptions in your analysis: The sides, back, and front are to be made from 12-gauge (0.1046 inch thick) steel sheets, which cost $0.70 per square foot (including any required cuts or bends). The base is to be made from a 10-gauge (0.1345 inch thick) steel sheet, which costs $0.90 per square foot. Lids cost approximately $50.00 each, regardless of dimensions. Welding costs approximately $0.18 per foot for material and labor combined. Give justification of any further assumptions or simplifications made of the details of construction. 3. Describe how any of your assumptions or simplifications may affect the final result. 4. If you were hired as a consultant on this investigation, what would your conclusions be? Would you recommend altering the design of the dumpster? If so, describe the savings that would result. 5E-15(pp 0992-1001) 1000 ❙❙❙❙ 1/18/06 3:51 PM Page 1000 CHAPTER 15 PARTIAL DERIVATIVES DISCOVERY PROJECT Quadratic Approximations and Critical Points The Taylor polynomial approximation to functions of one variable that we discussed in Chapter 12 can be extended to functions of two or more variables. Here we investigate quadratic approximations to functions of two variables and use them to give insight into the Second Derivatives Test for classifying critical points. In Section 15.4 we discussed the linearization of a function f of two variables at a point a, b : L x, y fx a, b x f a, b a fy a, b y b Recall that the graph of L is the tangent plane to the surface z f x, y at a, b, f a, b and the corresponding linear approximation is f x, y L x, y . The linearization L is also called the first-degree Taylor polynomial of f at a, b . 1. If f has continuous second-order partial derivatives at a, b , then the second-degree Taylor polynomial of f at a, b is Q x, y f a, b fx a, b x 1 2 xx f a, b x a a 2 fy a, b y fx y a, b x b ay 1 2 yy b f a, b y b 2 and the approximation f x, y Q x, y is called the quadratic approximation to f at a, b . Verify that Q has the same first- and second-order partial derivatives as f at a, b . 2. (a) Find the first- and second-degree Taylor polynomials L and Q of f x, y ; x2 y2 at (0, 0). (b) Graph f , L , and Q. Comment on how well L and Q approximate f . 3. (a) Find the first- and second-degree Taylor polynomials L and Q for f x, y ; e xe y at (1, 0). (b) Compare the values of L , Q, and f at (0.9, 0.1). (c) Graph f , L , and Q. Comment on how well L and Q approximate f . ax 2 b x y cy 2 (without using the Second Derivatives Test) by identifying the graph as a paraboloid. (a) By completing the square, show that if a 0, then 4. In this problem we analyze the behavior of the polynomial f x, y f x, y ax 2 bx y cy 2 a x b y 2a 2 4ac b 2 2 y 4a 2 (b) Let D 4ac b 2. Show that if D 0 and a 0, then f has a local minimum at (0, 0). (c) Show that if D 0 and a 0, then f has a local maximum at (0, 0). (d) Show that if D 0, then (0, 0) is a saddle point. 5. (a) Suppose f is any function with continuous second-order partial derivatives such that f 0, 0 0 and (0, 0) is a critical point of f . Write an expression for the second-degree Taylor polynomial, Q, of f at (0, 0). (b) What can you conclude about Q from Problem 4? (c) In view of the quadratic approximation f x, y Q x, y , what does part (b) suggest about f ? 5E-15(pp 0992-1001) 1/18/06 3:51 PM Page 1001 SECTION 15.8 LAGRANGE MULTIPLIERS |||| 15.8 ❙❙❙❙ 1001 Lagrange Multipliers In Example 6 in Section 15.7 we maximized a volume function V x y z subject to the constraint 2 x z 2 y z x y 12, which expressed the side condition that the surface area was 12 m2. In this section we present Lagrange’s method for maximizing or minimizing a general function f x, y, z subject to a constraint (or side condition) of the form t x, y, z k. It’s easier to explain the geometric basis of Lagrange’s method for functions of two variables. So we start by trying to find the extreme values of f x, y subject to a constraint of the form t x, y k. In other words, we seek the extreme values of f x, y when the point x, y is restricted to lie on the level curve t x, y k. Figure 1 shows this curve together with several level curves of f . These have the equations f x, y c, where c 7, 8, 9, 10, 11. To maximize f x, y subject to t x, y k is to find the largest value of c such that the level curve f x, y c intersects t x, y k. It appears from Figure 1 that this happens when these curves just touch each other, that is, when they have a common tangent line. (Otherwise, the value of c could be increased further.) This means that the normal lines at the point x 0 , y0 where they touch are identical. So the gradient vectors are parallel; that is, f x 0 , y0 t x 0 , y0 for some scalar . y g(x, y)=k Visual 15.8 animates Figure 1 for both level curves and level surfaces. f(x, y)=11 f(x, y)=10 f(x, y)=9 f(x, y)=8 f(x, y)=7 0 F IGURE 1 x This kind of argument also applies to the problem of finding the extreme values of f x, y, z subject to the constraint t x, y, z k. Thus, the point x, y, z is restricted to lie on the level surface S with equation t x, y, z k. Instead of the level curves in Figure 1, we consider the level surfaces f x, y, z c and argue that if the maximum value of f is f x 0 , y0 , z0 c, then the level surface f x, y, z c is tangent to the level surface t x, y, z k and so the corresponding gradient vectors are parallel. This intuitive argument can be made precise as follows. Suppose that a function f has an extreme value at a point P x 0 , y0 , z0 on the surface S and let C be a curve with vector equation r t x t , y t , z t that lies on S and passes through P. If t0 is the parameter value corresponding to the point P, then r t0 x 0 , y0 , z0 . The composite function ht f x t , y t , z t represents the values that f takes on the curve C. Since f has an extreme value at x 0 , y0 , z0 , it follows that h has an extreme value at t0 , so h t0 0. But if f is differentiable, we can use the Chain Rule to write 0 h t0 fx x 0 , y0 , z0 x t0 f x0 , y0 , z0 fy x 0 , y0 , z0 y t0 fz x 0 , y0 , z0 z t0 r t0 This shows that the gradient vector f x 0 , y0 , z0 is orthogonal to the tangent vector r t0 to every such curve C. But we already know from Section 15.6 that the gradient vector of t, t x 0 , y0 , z0 , is also orthogonal to r t0 . (See Equation 15.6.18.) This means that 5E-15(pp 1002-1011) 1002 ❙❙❙❙ 1/18/06 3:56 PM Page 1002 CHAPTER 15 PARTIAL DERIVATIVES the gradient vectors f x 0 , y0 , z0 and t x 0 , y0 , z0 t x 0 , y0 , z0 0, there is a number such that |||| Lagrange multipliers are named after the French-Italian mathematician Joseph-Louis Lagrange (1736–1813). See page 236 for a biographical sketch of Lagrange. f x 0 , y0 , z0 1 must be parallel. Therefore, if t x 0 , y0 , z0 The number in Equation 1 is called a Lagrange multiplier. The procedure based on Equation 1 is as follows. Method of Lagrange Multipliers To find the maximum and minimum values of f x, y, z subject to the constraint t x, y, z k [assuming that these extreme values exist and t 0 on the surface t x, y, z k]: (a) Find all values of x, y, z, and such that |||| In deriving Lagrange’s method we assumed that t 0. In each of our examples you can check that t 0 at all points where t x, y, z k. f x, y, z t x, y, z and t x, y, z k (b) Evaluate f at all the points x, y, z that result from step (a). The largest of these values is the maximum value of f ; the smallest is the minimum value of f . If we write the vector equation tions in step (a) become fx fy tx t in terms of its components, then the equa- f fz ty t x, y, z tz k This is a system of four equations in the four unknowns x, y, z, and , but it is not necessary to find explicit values for . For functions of two variables the method of Lagrange multipliers is similar to the method just described. To find the extreme values of f x, y subject to the constraint t x, y k, we look for values of x, y, and such that f x, y t x, y and t x, y k This amounts to solving three equations in three unknowns: fx tx fy ty t x, y k Our first illustration of Lagrange’s method is to reconsider the problem given in Example 6 in Section 15.7. EXAMPLE 1 A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box. SOLUTION As in Example 6 in Section 15.7 we let x, y, and z be the length, width, and height, respectively, of the box in meters. Then we wish to maximize V x yz subject to the constraint t x, y, z 2 xz 2yz xy 12 5E-15(pp 1002-1011) 1/18/06 3:56 PM Page 1003 S ECTION 15.8 LAGRANGE MULTIPLIERS Using the method of Lagrange multipliers, we look for values of x, y, z, and V t and t x, y, z 12. This gives the equations Vx Vy tx Vz ty 2 xz tz 2yz ❙❙❙❙ 1003 such that xy 12 which become 2 yz 2z y 3 xz 2z x 4 xy 2x 2y 2 xz 5 2yz xy 12 There are no general rules for solving systems of equations. Sometimes some ingenuity is required. In the present example you might notice that if we multiply (2) by x, (3) by y, and (4) by z, then the left sides of these equations will be identical. Doing this, we have |||| Another method for solving the system of equations (2–5) is to solve each of Equations 2, 3, and 4 for and then to equate the resulting expressions. 6 x yz 2 xz xy 7 x yz 2yz xy 8 x yz 2 xz 2yz We observe that 0 because 0 would imply y z x z x y 0 from (2), (3), and (4) and this would contradict (5). Therefore, from (6) and (7) we have 2 xz which gives x z (8) we have y z. But z 0 (since z 2yz which gives 2 x z we get |||| In geometric terms, Example 2 asks for the highest and lowest points on the curve C in Figure 2 that lies on the paraboloid z x 2 2 y 2 and directly above the constraint circle x 2 y 2 1. z xy 0 would give V 2 xz xy x y and so (since x 4z2 2yz xy 2 z. If we now put x 0) y 4z2 4z2 z=≈+2¥ circle x y 2 z in (5), y 12 1, x EXAMPLE 2 Find the extreme values of the function f x, y 2 y. From (7) and 2yz Since x, y, and z are all positive, we therefore have z 2 0 ), so x 2, and y x2 2 as before. 2y 2 on the 1. SOLUTION We are asked for the extreme values of f subject to the constraint t x, y t x, y x 2 y 2 1. Using Lagrange multipliers, we solve the equations 1, which can be written as fx fy ty 9 2x 2x 10 4y 2y C tx t x, y or as y x FIGURE 2 ≈+¥=1 11 x2 y2 1 1 f t, 5E-15(pp 1002-1011) 1004 ❙❙❙❙ 1/18/06 3:57 PM Page 1004 CHAPTER 15 PARTIAL DERIVATIVES From (9) we have x 0 or 1. If x y 0 from (10), so then (11) gives x at the points 0, 1 , 0, 1 , 1, 0 , and find that |||| The geometry behind the use of Lagrange multipliers in Example 2 is shown in Figure 3. x 2 2y 2 The extreme values of f x, y correspond to the level curves that touch the circle x 2 y 2 1. y f 0, 1 ≈+2¥=2 2 f 0, 0, then (11) gives y 1. If 1, then 1. Therefore, f has possible extreme values 1, 0 . Evaluating f at these four points, we 1 2 f 1, 0 1 f 1, 0 1 Therefore, the maximum value of f on the circle x 2 y 2 1 is f 0, 1 2 and the minimum value is f 1, 0 1. Checking with Figure 2, we see that these values look reasonable. 0 x x2 EXAMPLE 3 Find the extreme values of f x, y ≈+2¥=1 FIGURE 3 2y 2 on the disk x 2 y2 1. SOLUTION According to the procedure in (15.7.9), we compare the values of f at the critical points with values at the points on the boundary. Since fx 2 x and fy 4 y, the only critical point is 0, 0 . We compare the value of f at that point with the extreme values on the boundary from Example 2: f 0, 0 0 f 1, 0 1 f 0, Therefore, the maximum value of f on the disk x 2 minimum value is f 0, 0 0. EXAMPLE 4 Find the points on the sphere x 2 thest from the point 3, 1, y2 y2 1 1 is f 0, z2 1 2 and the 4 that are closest to and far- 1. SOLUTION The distance from a point x, y, z to the point 3, 1, d 2 sx 3 2 y 1 z 2 1 is 1 2 but the algebra is simpler if we instead maximize and minimize the square of the distance: d2 f x, y, z x 3 2 y 1 2 z 1 2 The constraint is that the point x, y, z lies on the sphere, that is, x2 t x, y, z y2 z2 4 According to the method of Lagrange multipliers, we solve 12 2x 3 2y 1 2z 1 4. This gives 2y 14 t, t 2x 13 f 2z x2 15 y2 z2 4 The simplest way to solve these equations is to solve for x, y, and z in terms of from (12), (13), and (14), and then substitute these values into (15). From (12) we have x 3 x or x1 3 or x 3 1 5E-15(pp 1002-1011) 1/18/06 3:57 PM Page 1005 SECTION 15.8 LAGRANGE MULTIPLIERS |||| Figure 4 shows the sphere and the nearest point P in Example 4. Can you see how to find the coordinates of P without using calculus? [Note that 1 give 0 because 1005 1 is impossible from (12).] Similarly, (13) and (14) 1 y z ❙❙❙❙ 1 z 1 1 Therefore, from (15) we have 32 1 11 4 2 which gives 1 12 2 1 ,1 1 y These values of FIGURE 4 1 2 2 4 s11 2, so x (3, 1, _1) 1 2 s11 2 then give the corresponding points x, y, z : 6 2 , , s11 s11 2 s11 6 , s11 and 2 2 , s11 s11 It’s easy to see that f has a smaller value at the first of these points, so the closest point is (6 s11, 2 s11, 2 s11 ) and the farthest is ( 6 s11, 2 s11, 2 s11 ). Two Constraints ±g g h=c C g=k FIGURE 5 P ±f ±h Suppose now that we want to find the maximum and minimum values of a function f x, y, z subject to two constraints (side conditions) of the form t x, y, z k and h x, y, z c. Geometrically, this means that we are looking for the extreme values of f when x, y, z is restricted to lie on the curve of intersection C of the level surfaces t x, y, z k and h x, y, z c. (See Figure 5.) Suppose f has such an extreme value at a point P x0 , y0 , z0 . We know from the beginning of this section that f is orthogonal to C there. But we also know that t is orthogonal to t x, y, z k and h is orthogonal to h x, y, z c, so t and h are both orthogonal to C. This means that the gradient vector f x 0 , y0 , z0 is in the plane determined by t x 0 , y0 , z0 and h x 0 , y0 , z0 . (We assume that these gradient vectors are not zero and not parallel.) So there are numbers and (called Lagrange multipliers) such that 16 f x0 , y0 , z0 t x0 , y0 , z0 h x0 , y0 , z0 In this case Lagrange’s method is to look for extreme values by solving five equations in the five unknowns x, y, z, , and . These equations are obtained by writing Equation 16 in terms of its components and using the constraint equations: fx tx hx fy ty hy fz tz hz t x, y, z k h x, y, z c 5E-15(pp 1002-1011) 1006 ❙❙❙❙ 1/18/06 3:57 PM Page 1006 CHAPTER 15 PARTIAL DERIVATIVES |||| The cylinder x 2 y 2 1 intersects the plane x y z 1 in an ellipse (Figure 6). Example 5 asks for the maximum value of f when x, y, z is restricted to lie on the ellipse. EXAMPLE 5 Find the maximum value of the function f x, y, z x 2y 3z on the curve of intersection of the plane x y z 1 and the cylinder x 2 y 2 1. SOLUTION We maximize the function f x, y, z t x, y, z x f t y x x2 z 1 and h x, y, z h, so we solve the equations 2y y2 3z subject to the constraints 1. The Lagrange condition is 4 17 1 2x 18 2 19 3 20 x 21 x2 3 2 z1 0 _1 Putting gives y _2 _1 0 y 1 2y 1 y2 1 3 [from (19)] in (17), we get 2 x 5 2 . Substitution in (21) then gives 1 FIGURE 6 25 42 2 and so z1 z y 2 x 29 4 , y 1 2, so x 1 2 s29, y 5 s29, and, from (20), s29 2. Then x 7 s29. The corresponding values of f are 2 s29 5 s29 2 7 s29 31 3 Therefore, the maximum value of f on the given curve is 3 |||| 15.8 . Similarly, (18) 1 s29 s29. Exercises 1. Pictured are a contour map of f and a curve with equation t x, y 8. Estimate the maximum and minimum values of f subject to the constraint that t x, y 8. Explain your reasoning. y g(x, y)=8 40 70 60 ; 2. (a) Use a graphing calculator or computer to graph the circle x 2 y 2 1. On the same screen, graph several curves of the form x 2 y c until you find two that just touch the circle. What is the significance of the values of c for these two curves? (b) Use Lagrange multipliers to find the extreme values of f x, y x 2 y subject to the constraint x 2 y 2 1. Compare your answers with those in part (a). 3–17 |||| Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint(s). 50 3. f x, y 0 30 20 10 x x2 y 2; x2 y2 1 4. f x, y 4x 6y; x2 y2 13 5. f x, y 2 x y; x 2 6. f x, y 2 2 x y; 2y x 4 2 6 y 4 1 5E-15(pp 1002-1011) 1/18/06 3:58 PM Page 1007 ❙❙❙❙ S ECTION 15.8 LAGRANGE MULTIPLIERS 7. f x, y, z 2x 8. f x, y, z 8x 9. f x, y, z x y z; 2 4 z; x x2 22 10. f x, y, z x yz; 11. f x, y, z x2 12. f x, y, z 4 x 13. f x, y, z, t x y 2 x1 2 x2 x 2y ; 16. f x, y, z 3x y z 0, 17. f x, y, z ■ ■ yz ■ [Hint: Use Heron’s formula for the area: A ss s x s y s z , where s p 2 and x, y, z are the lengths of the sides.] 35 5 6 2 1 2 z 25–37 |||| Use Lagrange multipliers to give an alternate solution to the indicated exercise in Section 15.7. y4 z4 1 2 2 2 1 x 2 z y x2 z 27. Exercise 39 2 t 2 29. Exercise 41 xn ; 1 x z y y 1, 2 z 2 4 xy; ■ 2x 2 19. f x, y ■ CAS ■ e ■ xy 1 , y 2 ■ 4x 5, x2 4y 2 36. Exercise 48 z 2 1 ■ ■ ■ ■ ■ ■ x2 ■ y2 ■ ■ ■ amount L of labor used and the amount K of capital investment. In Sections 15.1 and 15.3 we discussed how the CobbDouglas model P bL K 1 follows from certain economic 1. assumptions, where b and are positive constants and If the cost of a unit of labor is m and the cost of a unit of capital is n, and the company can spend only p dollars as its total budget, then maximizing the production P is subject to the constraint mL n K p. Show that the maximum production occurs when 1 p n 22. Referring to Exercise 21, we now suppose that the production Q, where Q is a constant. What values of is fixed at bL K 1 L and K minimize the cost function C L, K mL nK ? 23. Use Lagrange multipliers to prove that the rectangle with maximum area that has a given perimeter p is a square. 24. Use Lagrange multipliers to prove that the triangle with maximum area that has a given perimeter p is equilateral. 40. The plane 4 x ■ 21. The total production P of a certain product depends on the K ■ ■ ■ ■ ■ ■ ■ ■ y 2 z 2 intersects the paraboloid z x 2 y 2 in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin. curves, use it to estimate the minimum and maximum values of f x, y x 3 y 3 3xy subject to the constraint 2 y 3 2 9 by graphical methods. x3 (b) Solve the problem in part (a) with the aid of Lagrange multipliers. Use your CAS to solve the equations numerically. Compare your answers with those in part (a). and ■ 39. The plane x 20. (a) If your computer algebra system plots implicitly defined p m ■ whose surface area is 1500 cm2 and whose total edge length is 200 cm. 16 ■ ■ 38. Find the maximum and minimum volumes of a rectangular box ■ 1 ■ L 34. Exercise 46 ■ 1, ■ 3y 2 xy 32. Exercise 44 37. Exercise 51 2z2 18–19 |||| Find the extreme values of f on the region described by the inequality. 18. f x, y 30. Exercise 42 35. Exercise 47 1 28. Exercise 40 31. Exercise 43 2 26. Exercise 38 33. Exercise 45 t; y 25. Exercise 37 3z; x2 ■ z 2 3z 2 x z x 2 z2 x4 z; 2 xn y y 4 15. f x, y, z x 2 x1 y2 10y z 2; y 14. f x 1, x 2 , . . . , x n 2 2y 2 y2 4 x2 10z; 6y 1007 x2 y2 in an ellipse. (a) Graph the cone, the plane, and the ellipse. (b) Use Lagrange multipliers to find the highest and lowest points on the ellipse. ; CAS 5 intersects the cone z 2 8z 3y 41–42 |||| Find the maximum and minimum values of f subject to the given constraints. Use a computer algebra system to solve the system of equations that arises in using Lagrange multipliers. (If your CAS finds only one solution, you may need to use additional commands.) ye x z; 41. f x, y, z 42. f x, y, z ■ ■ x ■ 9x 2 z; y ■ ■ 4y 2 x 2 36 z 2 y ■ 2 z, x ■ ■ yz 36, x y 2 z 2 ■ 1 4 ■ ■ ■ 43. (a) Find the maximum value of f x1, x2 , . . . , xn n sx1 x2 xn given that x1, x2, . . . , xn are positive numbers and x1 x2 xn c, where c is a constant. (b) Deduce from part (a) that if x1, x2, . . . , xn are positive numbers, then n sx1 x2 xn x1 x2 xn n This inequality says that the geometric mean of n numbers is no larger than the arithmetic mean of the numbers. Under what circumstances are these two means equal? 5E-15(pp 1002-1011) 1008 ❙❙❙❙ 1/18/06 3:58 PM Page 1008 CHAPTER 15 PARTIAL DERIVATIVES n i1 44. (a) Maximize and (b) Put n i1 xi y 2 i n i1 x i yi subject to the constraints 1. ai s a j2 and yi x i2 to show that 1 s a j2 s b j2 ai bi bi s b j2 for any numbers a1, . . . , an, b1, . . . , bn. This inequality is known as the Cauchy-Schwarz Inequality. A PPLIED PROJECT Rocket Science Many rockets, such as the Pegasus XL currently used to launch satellites and the Saturn V that first put men on the Moon, are designed to use three stages in their ascent into space. A large first stage initially propels the rocket until its fuel is consumed, at which point the stage is jettisoned to reduce the mass of the rocket. The smaller second and third stages function similarly in order to place the rocket’s payload into orbit about the Earth. (With this design, at least two stages are required in order to reach the necessary velocities, and using three stages has proven to be a good compromise between cost and performance.) Our goal here is to determine the individual masses of the three stages to be designed in such a way as to minimize the total mass of the rocket while enabling it to reach a desired velocity. For a single-stage rocket consuming fuel at a constant rate, the change in velocity resulting from the acceleration of the rocket vehicle has been modeled by V 1 P c ln 1 S Mr Mr where Mr is the mass of the rocket engine including initial fuel, P is the mass of the payload, S is a structural factor determined by the design of the rocket (specifically, it is the ratio of the mass of the rocket vehicle without fuel to the total mass of the rocket with payload), and c is the (constant) speed of exhaust relative to the rocket. Now consider a rocket with three stages and a payload of mass A. Assume that outside forces are negligible and that c and S remain constant for each stage. If Mi is the mass of the i th stage, we can initially consider the rocket engine to have mass M1 and its payload to have mass M2 M3 A; the second and third stages can be handled similarly. 1. Show that the velocity attained after all three stages have been jettisoned is given by vf c ln M1 SM1 M2 M2 M3 M3 A A ln M2 SM2 M3 M3 A A 2. We wish to minimize the total mass M ln M3 SM3 A A M1 M2 M3 of the rocket engine subject to the constraint that the desired velocity vf from Problem 1 is attained. The method of Lagrange multipliers is appropriate here, but difficult to implement using the current expressions. To simplify, we define variables Ni so that the constraint equation may be expressed as vf c ln N1 ln N2 ln N3 . Since M is now difficult to express in terms of the Ni’s, we wish to use a simpler function that will be minimized at the same place as M. Show that M2 M1 M2 M3 M3 M2 A 1 1 S N1 SN1 A 1 1 S N2 SN2 A 1 1 S N3 SN3 A M3 M3 A M3 A and conclude that M A A 1 1 S 3N1 N2 N3 SN1 1 SN2 1 SN3 5E-15(pp 1002-1011) 1/18/06 3:58 PM Page 1009 D ISCOVERY PROJECT HYDRO-TURBINE OPTIMIZATION ❙❙❙❙ 1009 3. Verify that ln M A A is minimized at the same location as M ; use Lagrange multipliers and the results of Problem 2 to find expressions for the values of Ni where the minimum occurs subject to the constraint vf c ln N1 ln N2 ln N3 . [Hint: Use properties of logarithms to help simplify the expressions.] 4. Find an expression for the minimum value of M as a function of vf . 5. If we want to put a three-stage rocket into orbit 100 miles above the Earth’s surface, a final velocity of approximately 17,500 mi h is required. Suppose that each stage is built with a structural factor S 0.2 and an exhaust speed of c 6000 mi h. (a) Find the minimum total mass M of the rocket engines as a function of A. (b) Find the mass of each individual stage as a function of A. (They are not equally sized!) 6. The same rocket would require a final velocity of approximately 24,700 mi h in order to escape Earth’s gravity. Find the mass of each individual stage that would minimize the total mass of the rocket engines and allow the rocket to propel a 500-pound probe into deep space. A PPLIED PROJECT Hydro-Turbine Optimization The Great Northern Paper Company in Millinocket, Maine, operates a hydroelectric generating station on the Penobscot River. Water is piped from a dam to the power station. The rate at which the water flows through the pipe varies, depending on external conditions. The power station has three different hydroelectric turbines, each with a known (and unique) power function that gives the amount of electric power generated as a function of the water flow arriving at the turbine. The incoming water can be apportioned in different volumes to each turbine, so the goal is to determine how to distribute water among the turbines to give the maximum total energy production for any rate of flow. Using experimental evidence and Bernoulli’s equation, the following quadratic models were determined for the power output of each turbine, along with the allowable flows of operation: KW1 18.89 0.1277Q1 2 4.08 10 5Q 1 170 2 1.6 10 6Q T KW2 24.51 0.1358Q2 2 4.69 10 5Q 2 170 2 1.6 10 6Q T KW3 27.02 250 Q1 0.1380Q3 1110, 250 5 3.84 10 Q Q2 2 3 170 1110, 250 2 1.6 10 6Q T Q3 1225 where Qi KWi QT flow through turbine i in cubic feet per second power generated by turbine i in kilowatts total flow through the station in cubic feet per second 1. If all three turbines are being used, we wish to determine the flow Qi to each turbine that will give the maximum total energy production. Our limitations are that the flows must sum to the total incoming flow and the given domain restrictions must be observed. Consequently, use Lagrange multipliers to find the values for the individual flows (as functions of QT ) that maximize the total energy production KW1 K W2 K W3 subject to the constraints Q1 Q2 Q3 QT and the domain restrictions on each Qi. 2. For which values of QT is your result valid? 3. For an incoming flow of 2500 ft3 s, determine the distribution to the turbines and verify (by trying some nearby distributions) that your result is indeed a maximum. 5E-15(pp 1002-1011) 1010 ❙❙❙❙ 1/18/06 3:59 PM Page 1010 CHAPTER 15 PARTIAL DERIVATIVES 4. Until now we assumed that all three turbines are operating; is it possible in some situations that more power could be produced by using only one turbine? Make a graph of the three power functions and use it to help decide if an incoming flow of 1000 ft3 s should be distributed to all three turbines or routed to just one. (If you determine that only one turbine should be used, which one?) What if the flow is only 600 ft3 s? 5. Perhaps for some flow levels it would be advantageous to use two turbines. If the incoming flow is 1500 ft3 s, which two turbines would you recommend using? Use Lagrange multipliers to determine how the flow should be distributed between the two turbines to maximize the energy produced. For this flow, is using two turbines more efficient than using all three? 6. If the incoming flow is 3400 ft3 s, what would you recommend to the company? |||| 15 Review ■ CONCEPT CHECK 1. (a) What is a function of two variables? (b) Describe three methods for visualizing a function of two variables. 2. What is a function of three variables? How can you visualize such a function? 3. What does f x, y lim x, y l a, b L mean? How can you show that such a limit does not exist? 4. (a) What does it mean to say that f is continuous at a, b ? (b) If f is continuous on 2 , what can you say about its graph? 5. (a) Write expressions for the partial derivatives fx a, b and fy a, b as limits. (b) How do you interpret fx a, b and fy a, b geometrically? How do you interpret them as rates of change? (c) If f x, y is given by a formula, how do you calculate fx and fy ? 6. What does Clairaut’s Theorem say? 7. How do you find a tangent plane to each of the following types of surfaces? (a) A graph of a function of two variables, z f x, y (b) A level surface of a function of three variables, F x, y, z k 8. Define the linearization of f at a, b . What is the correspond- ing linear approximation? What is the geometric interpretation of the linear approximation? 9. (a) What does it mean to say that f is differentiable at a, b ? (b) How do you usually verify that f is differentiable? 10. If z f x, y , what are the differentials dx, dy, and dz ? 11. State the Chain Rule for the case where z f x, y and x and y are functions of one variable. What if x and y are functions of two variables? ■ 12. If z is defined implicitly as a function of x and y by an equation of the form F x, y, z 0, how do you find z x and z y ? 13. (a) Write an expression as a limit for the directional derivative a, b . of f at x 0 , y0 in the direction of a unit vector u How do you interpret it as a rate? How do you interpret it geometrically? (b) If f is differentiable, write an expression for Du f x 0 , y0 in terms of fx and fy . 14. (a) Define the gradient vector f for a function f of two or three variables. (b) Express Du f in terms of f . (c) Explain the geometric significance of the gradient. 15. What do the following statements mean? (a) (b) (c) (d) (e) f f f f f has a local maximum at a, b . has an absolute maximum at a, b . has a local minimum at a, b . has an absolute minimum at a, b . has a saddle point at a, b . 16. (a) If f has a local maximum at a, b , what can you say about its partial derivatives at a, b ? (b) What is a critical point of f ? 17. State the Second Derivatives Test. 18. (a) What is a closed set in 2 ? What is a bounded set? (b) State the Extreme Value Theorem for functions of two variables. (c) How do you find the values that the Extreme Value Theorem guarantees? 19. Explain how the method of Lagrange multipliers works in finding the extreme values of f x, y, z subject to the constraint t x, y, z k. What if there is a second constraint h x, y, z c? 5E-15(pp 1002-1011) 1/18/06 3:59 PM Page 1011 CHAPTER 15 REVIEW ■ TRUE-FALSE QUIZ f a, y y lim ylb 3. fxy y 2 and fy x, y x 9. If f x, y y 2. x fz x, y, z 11. If f x, y L. f x, y x, y l a, b EXERCISES 1 |||| 1 ■ ■ x 2 ■ e ■ ■ ■ ■ ■ sz ■ x ■ y ■ y ■ 2 ■ 4. f x, y ■ ■ x2 y2 ■ sx ■ ■ x2 6. f x, y ■ ■ 9. ■ 2 y 2 ■ 1 s2 sin y, then Du f x, y s2. ■ ■ ■ ■ ■ ■ 4y ■ graph is shown. z 2 2 |||| Evaluate the limit or show that it does not exist. lim x2 x, y l 1, 1 ■ 2 xy 2y 2 ■ ■ 10. ■ ■ lim x, y l 0, 0 x2 ■ ■ ■ 2 xy 2y 2 ■ ■ rectangle 0 x 10, 0 y 8, where x and y are measured in meters. The temperature at the point x, y in the plate is T x, y , where T is measured in degrees Celsius. Temperatures at equally spaced points were measured and recorded in the table. (a) Estimate the values of the partial derivatives Tx 6, 4 and Ty 6, 4 . What are the units? i j s2. (b) Estimate the value of Du T 6, 4 , where u Interpret your result. (c) Estimate the value of Txy 6, 4 . y y 0 2 4 6 8 0 30 38 45 51 55 2 52 56 60 62 61 4 78 74 72 68 66 6 98 87 80 75 71 8 96 90 86 80 75 10 92 92 91 87 78 x 8. A contour map of a function f is shown. Use it to make a rough sketch of the graph of f . y 1 ■ 11. A metal plate is situated in the xy-plane and occupies the ■ 7. Make a rough sketch of a contour map for the function whose x sin x 2 Sketch several level curves of the function. 5. f x, y ■ 2. f x, y, z tan y ■ 2 Sketch the graph of the function. ■ 5–6 1 sin x ■ 3. f x, y ■ 2 ■ 9–10 Find and sketch the domain of the function. |||| fx y 2, 1 minimum. ■ 3–4 1 y. 12. If f x, y has two local maxima, then f must have a local at a, b . ■ f x, y f 2, 5 . then f has a saddle point at 2, 1 . 6. If fx a, b and fy a, b both exist, then f is differentiable 1. f x, y f x, y fxx 2, 1 fyy 2, 1 xy through a, b , then lim ■ ln y, then lim x, y l 2, 5 10. If 2, 1 is a critical point of f and 5. If f x, y l L as x, y l a, b along every straight line |||| 0. f 4. Dk f x, y, z 1–2 f a, b 8. If f is a function, then 2. There exists a function f with continuous second-order partial 2 ■ a, b , then f a, b b derivatives such that fx x, y 1011 7. If f has a local minimum at a, b and f is differentiable at Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. fy a, b ❙❙❙❙ 1.5 2 12. Find a linear approximation to the temperature function T x, y 4 x in Exercise 11 near the point (6, 4). Then use it to estimate the temperature at the point (5, 3.8). 5E-15(pp 1012-1015) 1012 ❙❙❙❙ 13–17 1/18/06 Page 1012 CHAPTER 15 PARTIAL DERIVATIVES 1 15. t u, v ■ p ln q ■ e ■ 33. Find the linear approximation of the function f x, y, z x 3sy 2 z 2 at the point (2, 3, 4) and use it 3.97 2. to estimate the number 1.98 3s 3.01 2 x 16. w u tan v 17. T p, q, r e r sin 2 14. u y2 s2 x z y 34. The two legs of a right triangle are measured as 5 m and 12 m r ■ ■ ■ ■ ■ ■ ■ with a possible error in measurement of at most 0.2 cm in each. Use differentials to estimate the maximum error in the calculated value of (a) the area of the triangle and (b) the length of the hypotenuse. ■ 18. The speed of sound traveling through ocean water is a function of temperature, salinity, and pressure. It has been modeled by the function C 1449.2 0.055 T 2 4.6 T 1.34 35. If w 0.01T S 35 cos x y y cos x, where x u 2 v and y use the Chain Rule to find z u and z v. 36. If z 0.016D 4x 3 19. f x, y ■ 20. z x k y lz m 21. f x, y, z ■ xy 2 ■ 22. v ■ ■ 23. If u 24. If sx 2 ■ ■ 1 ln x 2t ■ 2 2 ■ x2 y2 y2 z2 26. z e x cos y, 27. x 2 2y 2 28. x y yz 29. sin x y z ■ ■ 2 x2 42. If yz 4 2, 1 0, 0, 1 3z 2 zx 3, 3, x ■ 1, 2 x, 2, 3z, ■ z y x z x2 2 y2 2 z y2 4uv 2v 1, 1 2, ■ z v z z and . x y e xy z, find x 2z 3 z uv z 2e x sy. 43. Find the gradient of the function f x, y, z 44. (a) When is the directional derivative of f a maximum? (b) When is it a minimum? (c) When is it 0? (d) When is it half of its maximum value? 1, 1, 1 2y ■ x f u, v , where u x y, v y x, and f has continuous second partial derivatives, show that Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point. 3x 2 z x 41. If z 2 |||| 25. z y 2 , where f is differentiable, show that 3 in s, the length y of another side is decreasing at a rate of 2 in s, and the contained angle is increasing at a rate of 0.05 radian s. How fast is the area of the triangle changing when x 40 in, y 50 in, and 6? 2u. 2 f x2 y 40. The length x of a side of a triangle is increasing at a rate of ■ z 2, show that y2 38. Use a tree diagram to write out the Chain Rule for the case where w f t, u, v , t t p, q, r, s , u u p, q, r, s , and v v p, q, r, s are all differentiable functions. y ■ u y f x, y , where x t s, t , y h s, t , t 1, 2 3, ts 1, 2 1, tt 1, 2 4, h 1, 2 6, hs 1, 2 5, h t 1, 2 10, fx 3, 6 7, and fy 3, 6 8. Find z s and z t when s 1 and t 2. 2y r cos s ■ u x x y x y, show that 25–29 xe v 2, u 37. Suppose z 39. If z Find all second partial derivatives of f . |||| s x y 2 z, where x e 2 t, y t 3 4 t, and t 4, use the Chain Rule to find dw dt. 2 z 0.00029T 3 where C is the speed of sound (in meters per second), T is the temperature (in degrees Celsius), S is the salinity (the concentration of salts in parts per thousand, which means the number of grams of dissolved solids per 1000 g of water), and D is the depth below the ocean surface (in meters). Compute C T , C S, and C D when T 10 C, S 35 parts per thousand, and D 100 m. Explain the physical significance of these partial derivatives. 19–22 x 2 tan 1y. 32. Find dz if z Find the first partial derivatives. |||| 13. f x, y ■ 4:03 PM 1, 0 ■ ; 30. Use a computer to graph the surface z ■ ■ ■ ■ ■ 3 2 x y and its x tangent plane and normal line at 1, 2, 5 on the same screen. Choose the domain and viewpoint so that you get a good view of all three objects. y 2 z 2 1 where the tangent plane is parallel to the plane 2 x y 3z 2. 45–46 |||| Find the directional derivative of f at the given point in the indicated direction. 2 s x y 2, 1, 5 , in the direction toward the point 4, 1 45. f x, y x 2 y x s1 z, in the direction of v 2 i j 46. f x, y, z 31. Find the points on the sphere x 2 ■ ■ ■ ■ ■ ■ 1, 2, 3 , 2k ■ ■ ■ ■ ■ ■ 5E-15(pp 1012-1015) 1/18/06 4:03 PM Page 1013 C HAPTER 15 REVIEW x2y at the point 2, 1 . In which direction does it occur? 47. Find the maximum rate of change of f x, y sy 1013 ; 57. Use a graph and or level curves to estimate the local maximum and minimum values and saddle points of f x, y x 3 3x y 4 2y 2. Then use calculus to find these values precisely. xy 48. Find the direction in which f x, y, z ❙❙❙❙ ze increases most rapidly at the point 0, 1, 2 . What is the maximum rate of increase? ; 58. Use a graphing calculator or computer (or Newton’s method or a computer algebra system) to find the critical points of f x, y 12 10 y 2 x 2 8 xy y 4 correct to three decimal places. Then classify the critical points and find the highest point on the graph. 49. The contour map shows wind speed in knots during Hurricane Andrew on August 24, 1992. Use it to estimate the value of the directional derivative of the wind speed at Homestead, Florida, in the direction of the eye of the hurricane. 59–62 |||| Use Lagrange multipliers to find the maximum and minimum values of f subject to the given constraint(s). 59. f x, y 80 75 60. f x, y Homestead 60 ■ 30 52. f x, y x3 6 xy 53. f x, y 3xy x2y 54. f x, y x2 2 z2 3 2 3z ; 2z y ■ ■ 2 ■ ■ ■ ■ ■ ■ 55–56 ■ |||| ■ 65. A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the figure. If the pentagon has fixed perimeter P, find the lengths of the sides of the pentagon that maximize the area of the pentagon. 9x 6y ¨ 10 8y 3 ■ xy 2 ■ ■ ■ ■ ■ ■ ■ Find the absolute maximum and minimum values of f on the set D. 4 xy 2 x 2 y 2 x y 3; D is the closed triangular region in the x y-plane with vertices 0, 0 , 0, 6 , and 6, 0 55. f x, y 56. f x, y ■ ■ e ■ x2 y2 ■ x2 2y 2 ; ■ ■ ■ 2 that are closest to y ey 2 ■ y2 2y x y2 = xy ■ x2 1 U.S. Parcel Post if the sum of its length and girth (the perimeter of a cross-section perpendicular to the length) is at most 108 in. Find the dimensions of the package with largest volume that can be mailed by Parcel Post. 10 20 30 40 (Distance in miles) Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. x2 ■ 1, 1 y2 64. A package in the shape of a rectangular box can be mailed by |||| 51. f x, y ■ 1 x2 the origin. 2, 2, 4 to the curve of intersection of the surface 2 x 2 y 2 and the plane z 4. 51–54 x z y 1 63. Find the points on the surface xy 2z 3 50. Find parametric equations of the tangent line at the point z 2 x 45 40 35 0 x yz; 62. f x, y, z 50 y2 1 ; y 61. f x, y, z 55 Key West 1 x x2 = 60 70 55 65 70 65 x 2 y; D is the disk x 2 ■ ■ ■ y2 ■ 4 ■ ■ 66. A particle of mass m moves on the surface z f x, y . Let x x t , y y t be the x- and y-coordinates of the particle at time t. (a) Find the velocity vector v and the kinetic energy 1 K 2 m v 2 of the particle. (b) Determine the acceleration vector a. (c) Let z x 2 y 2 and x t t cos t, y t t sin t. Find the velocity vector, the kinetic energy, and the acceleration vector. 5E-15(pp 1012-1015) 1/18/06 PROBLEMS PLUS 4:03 PM Page 1014 1. A rectangle with length L and width W is cut into four smaller rectangles by two lines parallel to the sides. Find the maximum and minimum values of the sum of the squares of the areas of the smaller rectangles. 2. Marine biologists have determined that when a shark detects the presence of blood in the water, it will swim in the direction in which the concentration of the blood increases most rapidly. Based on certain tests, the concentration of blood (in parts per million) at a point P x, y on the surface of seawater is approximated by C x, y e x 2 2y 2 10 4 where x and y are measured in meters in a rectangular coordinate system with the blood source at the origin. (a) Identify the level curves of the concentration function and sketch several members of this family together with a path that a shark will follow to the source. (b) Suppose a shark is at the point x 0 , y0 when it first detects the presence of blood in the water. Find an equation of the shark’s path by setting up and solving a differential equation. 3. A long piece of galvanized sheet metal w inches wide is to be bent into a symmetric form with three straight sides to make a rain gutter. A cross-section is shown in the figure. (a) Determine the dimensions that allow the maximum possible flow; that is, find the dimensions that give the maximum possible cross-sectional area. (b) Would it be better to bend the metal into a gutter with a semicircular cross-section than a three-sided cross-section? x ¨ ¨ x w-2x 4. For what values of the number r is the function x f x, y, z continuous on 3 x 0 2 y y2 zr z2 if x, y, z 0 if x, y, z 0 ? 5. Suppose f is a differentiable function of one variable. Show that all tangent planes to the surface z x f y x intersect in a common point. 6. (a) Newton’s method for approximating a root of an equation f x 0 (see Section 4.9) can be adapted to approximating a solution of a system of equations f x, y 0 and t x, y 0. The surfaces z f x, y and z t x, y intersect in a curve that intersects the x y-plane at the point r, s , which is the solution of the system. If an initial approximation x 1, y1 is close to this point, then the tangent planes to the surfaces at x 1, y1 intersect in a straight line that intersects the x y-plane in a point x 2 , y2 , which should be closer to r, s . (Compare with Figure 2 in Section 4.9.) Show that x2 x1 f ty fx ty fy t fy tx and y2 y1 fx t fx ty f tx fy tx where f , t, and their partial derivatives are evaluated at x 1, y1 . If we continue this procedure, we obtain successive approximations x n , yn . 1014 5E-15(pp 1012-1015) 1/18/06 4:04 PM Page 1015 (b) It was Thomas Simpson (1710–1761) who formulated Newton’s method as we know it today and who extended it to functions of two variables as in part (a). (See the biography of Simpson on page 560.) The example that he gave to illustrate the method was to solve the system of equations xx yy xy 1000 yx 100 In other words, he found the points of intersection of the curves in the figure. Use the method of part (a) to find the coordinates of the points of intersection correct to six decimal places. y x x+y y=1000 4 x y+y x=100 2 0 2 x 4 7. (a) Show that when Laplace’s equation 2 2 u x2 2 u y2 u z2 0 is written in cylindrical coordinates, it becomes 2 u r2 1 r u r 2 1 r2 2 u u z2 2 0 (b) Show that when Laplace’s equation is written in spherical coordinates, it becomes 2 u 2 2 u cot u 2 1 2 2 u 2 8. Among all planes that are tangent to the surface xy 2z 2 2 1 sin 2 2 u 2 0 1, find the ones that are farthest from the origin. 9. If the ellipse x 2 a 2 y 2 b 2 1 is to enclose the circle x 2 minimize the area of the ellipse? y2 2y, what values of a and b 1015 ...
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