14.1 FUNCTIONS OF SEVERAL VARIABLES. In this section we study functions of two or more variables from four points of view:

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1 4 PARTIAL DERIVATIVES Functions of two variables can be visualied b means of level curves, which connect points where the function takes on a given value. Atmospheric pressure at a given time is a function of longitude and latitude and is measured in millibars. Here the level curves are called isobars and those pictured join locations that had the same pressure on arch 7, 7. (The curves labeled 8, for instance, connect points with pressure 8 mb.) Surface winds tend to flow from areas of high pressure across the isobars toward areas of low pressure, and are strongest where the isobars are tightl packed. So far we have dealt with the calculus of functions of a single variable. But, in the real world, phsical quantities often depend on two or more variables, so in this chapter we turn our attention to functions of several variables and etend the basic ideas of differential calculus to such functions. 854

2 4. FUNCTIONS OF SEVERAL VARIABLES In this section we stud functions of two or more variables from four points of view: N verball (b a description in words) N numericall (b a table of values) N algebraicall (b an eplicit formula) N visuall (b a graph or level curves) FUNCTIONS OF TWO VARIABLES The temperature T at a point on the surface of the earth at an given time depends on the longitude and latitude of the point. We can think of T as being a function of the two variables and, or as a function of the pair,. We indicate this functional dependence b writing T f,. The volume V of a circular clinder depends on its radius r and its height h. In fact, we know that V r h. We sa that V is a function of r and h, and we write Vr, h r h. DEFINITION A function f of two variables is a rule that assigns to each ordered pair of real numbers, in a set D a unique real number denoted b f,. The set D is the domain of f and its range is the set of values that f takes on, that is, f,, D. We often write f, to make eplicit the value taken on b f at the general point,. The variables and are independent variables and is the dependent variable. [Compare this with the notation f for functions of a single variable.] A function of two variables is just a function whose domain is a subset of and whose range is a subset of. One wa of visualiing such a function is b means of an arrow diagram (see Figure ), where the domain D is represented as a subset of the -plane. (, ) FIGURE D f f(a, b) f(, ) (a, b) If a function f is given b a formula and no domain is specified, then the domain of f is understood to be the set of all pairs, for which the given epression is a welldefined real number. EXAPLE For each of the following functions, evaluate f 3, and find the domain. (a) SOLUTION f, s (a) f 3, (b) s3 3 f, ln s6 855

3 856 CHAPTER 4 PARTIAL DERIVATIVES ++= The epression for f makes sense if the denominator is not and the quantit under the square root sign is nonnegative. So the domain of f is = D,, The inequalit, or, describes the points that lie on or above the line, while means that the points on the line must be ecluded from the domain. (See Figure.) FIGURE œ ++ Domain of f(, )= - (b) f 3, 3 ln 3 3 ln Since ln is defined onl when, that is,, the domain of f is D,. This is the set of points to the left of the parabola. (See Figure 3.) FIGURE 3 Domain of f(, )= ln( -) = Not all functions are given b eplicit formulas. The function in the net eample is described verball and b numerical estimates of its values. EXAPLE In regions with severe winter weather, the wind-chill inde is often used to describe the apparent severit of the cold. This inde 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 records values of W compiled b the NOAA National Weather Service of the US and the eteorological Service of Canada. TABLE Wind-chill inde as a function of air temperature and wind speed N THE NEW WIND-CHILL INDEX A new wind-chill inde was introduced in November of and is more accurate than the old inde at measuring how cold it feels when it s wind. The new inde is based on a model of how fast a human face loses heat. It was developed through clinical trials in which volunteers were eposed to a variet of temperatures and wind speeds in a refrigerated wind tunnel. Actual temperature ( C) Wind speed (km/h) T v For instance, the table shows that if the temperature is 5C and the wind speed is 5 kmh, then subjectivel it would feel as cold as a temperature of about 5C with no wind. So EXAPLE 3 In 98 Charles Cobb and Paul Douglas published a stud in which the modeled the growth of the American econom during the period The conf 5, 5 5

4 SECTION 4. FUNCTIONS OF SEVERAL VARIABLES 857. Year TABLE P L K sidered a simplified view of the econom in which production output is determined b the amount of labor involved and the amount of capital invested. While there are man other factors affecting economic performance, their model proved to be remarkabl accurate. The function the used to model production was of the form where P is the total production (the monetar value of all goods produced in a ear), L is the amount of labor (the total number of person-hours worked in a ear), and K is the amount of capital invested (the monetar worth of all machiner, equipment, and buildings). In Section 4.3 we will show how the form of Equation follows from certain economic assumptions. Cobb and Douglas used economic data published b the government to obtain Table. The took the ear 899 as a baseline, and P, L, and K for 899 were each assigned the value. The values for other ears were epressed as percentages of the 899 figures. Cobb and Douglas used the method of least squares to fit the data of Table to the function PL, K blk PL, K.L.75 K.5 (See Eercise 75 for the details.) If we use the model given b the function in Equation to compute the production in the ears 9 and 9, we get the values P47, P94, which are quite close to the actual values, 59 and 3. The production function () has subsequentl been used in man 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, K because L and K represent labor and capital and are therefore never negative. + =9 EXAPLE 4 Find the domain and range of t, s9. SOLUTION The domain of t is D, 9, 9 which is the disk with center, and radius 3. (See Figure 4.) The range of t is s9,, D _3 3 Since is a positive square root,. Also 9 9? s9 3 FIGURE 4 Domain of g(, )=œ 9- - So the range is 3, 3

5 858 CHAPTER 4 PARTIAL DERIVATIVES GRAPHS Another wa of visualiing the behavior of a function of two variables is to consider its graph. S {,, f(, )} DEFINITION If f is a function of two variables with domain D, then the graph of f is the set of all points,, in 3 such that f, and, is in D. FIGURE 5 D (,, ) f(, ) Just as the graph of a function f of one variable is a curve C with equation f, so the graph of a function f of two variables is a surface S with equation f,. We can visualie the graph S of f as ling directl above or below its domain D in the -plane. (See Figure 5.) EXAPLE 5 Sketch the graph of the function f, 6 3. SOLUTION The graph of f has the equation 6 3, or 3 6, which represents a plane. To graph the plane we first find the intercepts. Putting in the equation, we get as the -intercept. Similarl, the -intercept is 3 and the -intercept is 6. This helps us sketch the portion of the graph that lies in the first octant. (See Figure 6.) (,, 6) (,, ) (, 3, ) FIGURE 6 The function in Eample 5 is a special case of the function f, a b c which is called a linear function. The graph of such a function has the equation a b c or a b c (,, 3) so it is a plane. In much the same wa that linear functions of one variable are important in single-variable calculus, we will see that linear functions of two variables pla a central role in multivariable calculus. (, 3, ) (3,, ) FIGURE 7 Graph of g(, )= œ 9- - V EXAPLE 6 Sketch the graph of t, s9. SOLUTION The graph has equation s9. We square both sides of this equation to obtain 9, or 9, which we recognie as an equation of the sphere with center the origin and radius 3. But, since, the graph of t is just the top half of this sphere (see Figure 7).

6 SECTION 4. FUNCTIONS OF SEVERAL VARIABLES 859 NOTE An entire sphere can t be represented b a single function of and. As we saw in Eample 6, the upper hemisphere of the sphere 9 is represented b the function t, s9. The lower hemisphere is represented b the function h, s9. EXAPLE 7 Use a computer to draw the graph of the Cobb-Douglas production function PL, K.L.75 K.5. SOLUTION Figure 8 shows the graph of P for values of the labor L and capital K that lie between and 3. The computer has drawn the surface b 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 epected. 3 P FIGURE 8 3 K L 3 V EXAPLE 8 Find the domain and range and sketch the graph of h, 4. SOLUTION Notice that h, is defined for all possible ordered pairs of real numbers,, so the domain is, the entire -plane. The range of h is the set, of all nonnegative real numbers. [Notice that and, so h, for all and.] The graph of h has the equation 4, which is the elliptic paraboloid that we sketched in Eample 4 in Section.6. Horiontal traces are ellipses and vertical traces are parabolas (see Figure 9). FIGURE 9 Graph of h(, )=4 + Computer programs are readil available for graphing functions of two variables. In most such programs, traces in the vertical planes k and k are drawn for equall spaced values of k and parts of the graph are eliminated using hidden line removal.

7 86 CHAPTER 4 PARTIAL DERIVATIVES Figure shows computer-generated graphs of several functions. Notice that we get an especiall 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 ver flat and close to the -plane ecept near the origin; this is because e is ver small when or is large. (a) f(, )=( +3 )e _ _ (b) f(, )=( +3 )e _ _ FIGURE (c) f(, )=sin +sin sin sin (d) f(, )= LEVEL CURVES So far we have two methods for visualiing 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, k, where k is a constant (in the range of f ). A level curve f, 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 the relation between level curves and horiontal traces. The level curves f, k are just the traces of the graph of f in the horiontal plane k projected down to the -plane. So if ou draw the level curves of a function and visualie them being lifted up to the surface at the indicated height, then ou can mentall piece

8 SECTION 4. FUNCTIONS OF SEVERAL VARIABLES LONESOE TN. A 55 B 5 f(, )= k=45 k=4 k=35 k=3 k=5 k= 45 Lonesome Creek FIGURE TEC Visual 4.A animates Figure b showing level curves being lifted up to graphs of functions. FIGURE together a picture of the graph. The surface is steep where the level curves are close together. It is somewhat flatter where the are farther apart. One common eample of level curves occurs in topographic maps of mountainous regions, such as the map in Figure. The level curves are curves of constant elevation above sea level. If ou walk along one of these contour lines, ou neither ascend nor descend. Another common eample 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 3 shows a weather map of the world indicating the average Januar temperatures. The isothermals are the curves that separate the colored bands. The isobars in the atmospheric pressure map on page 854 provide another eample of level curves. FIGURE 3 World mean sea-level temperatures in Januar in degrees Celsius Tarbuck, Atmosphere: Introduction to eteorolog, 4th Edition, 989. Reprinted b permission of Pearson Education, Inc., Upper Saddle River, NJ.

9 86 CHAPTER 4 PARTIAL DERIVATIVES FIGURE EXAPLE 9 A contour map for a function f is shown in Figure 4. Use it to estimate the values of f, 3 and f 4, 5. SOLUTION The point (, 3) lies partwa between the level curves with -values 7 and 8. We estimate that f, 3 73 Similarl, we estimate that f 4, 5 56 EXAPLE Sketch the level curves of the function f, 6 3 for the values k 6,, 6,. SOLUTION The level curves are 6 3 k or 3 k 6 k= k=6 FIGURE 5 Contour map of f(, )=6-3- k= k=_6 This is a famil of lines with slope 3. The four particular level curves with k 6,, 6, and are 3, 3 6, 3, and 3 6. The are sketched in Figure 5. The level curves are equall spaced parallel lines because the graph of f is a plane (see Figure 6). V EXAPLE Sketch the level curves of the function t, s9 for k,,, 3 SOLUTION The level curves are s9 k or 9 k This is a famil of concentric circles with center, and radius s9 k. The cases k,,, 3 are shown in Figure 6. Tr to visualie these level curves lifted up to form a surface and compare with the graph of t (a hemisphere) in Figure 7. (See TEC Visual 4.A.) k=3 k= k= k= (3, ) FIGURE 6 Contour map of g(, )=œ 9- - EXAPLE Sketch some level curves of the function h, 4. SOLUTION The level curves are 4 k or k4 k

10 SECTION 4. FUNCTIONS OF SEVERAL VARIABLES 863 which, for k, describes a famil of ellipses with semiaes sk and sk. Figure 7(a) shows a contour map of h drawn b a computer with level curves corresponding to k.5,.5,.75,..., 4. Figure 7(b) shows these level curves lifted up to the graph of h (an elliptic paraboloid) where the become horiontal traces. We see from Figure 7 how the graph of h is put together from the level curves. TEC Visual 4.B demonstrates the connection between surfaces and their contour maps. FIGURE 7 The graph of h(, )=4 + is formed b lifting the level curves. (a) Contour map (b) Horiontal traces are raised level curves EXAPLE 3 Plot level curves for the Cobb-Douglas production function of Eample 3. SOLUTION In Figure 8 we use a computer to draw a contour plot for the Cobb-Douglas production function PL, K.L.75 K.5 K FIGURE 8 3 L Level curves are labeled with the value of the production P. For instance, the level curve labeled 4 shows all values of the labor L and capital investment K that result in a production of P 4. We see that, for a fied 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 certainl true in Eample 3. (Compare Figure 8 with Figure 8.) It is also true in estimating function values, as in Eample 9.

11 864 CHAPTER 4 PARTIAL DERIVATIVES Figure 9 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 ver steep near the origin. (a) Level curves of f(, )=_e _ _ (b) Two views of f(, )=_e _ _ FIGURE 9 (c) Level curves of f(, )= _3 + + (d) f(, )= _3 + + FUNCTIONS OF THREE OR ORE VARIABLES A function of three variables, f, is a rule that assigns to each ordered triple,, in a domain D 3 a unique real number denoted b f,,. For instance, the temperature T at a point on the surface of the earth depends on the longitude and latitude of the point and on the time t, so we could write T f,, t. EXAPLE 4 Find the domain of f if f,, ln sin SOLUTION The epression for f,, is defined as long as, so the domain of is D,, 3 f This is a half-space consisting of all points that lie above the plane.

12 SECTION 4. FUNCTIONS OF SEVERAL VARIABLES @=9 + +@=4 It s ver difficult to visualie a function f of three variables b its graph, since that would lie in a four-dimensional space. However, we do gain some insight into f b eamining its level surfaces, which are the surfaces with equations f,, k, where k is a constant. If the point,, moves along a level surface, the value of f,, remains fied. EXAPLE 5 Find the level surfaces of the function f,, SOLUTION The level surfaces are k, where k. These form a famil of concentric spheres with radius sk. (See Figure.) Thus, as,, varies over an sphere with center O, the value of f,, remains fied. FIGURE + +@= Functions of an number of variables can be considered. A function of n variables is a rule that assigns a number f,,..., n to an n-tuple,,..., n of real numbers. We denote b n the set of all such n-tuples. For eample, if a compan uses n different ingredients in making a food product, c i is the cost per unit of the ith ingredient, and i units of the ith ingredient are used, then the total cost C of the ingredients is a function of the n variables,,..., n : 3 C f,,..., n c c c n 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 compactl: If,,..., n, we often write f in place of f,,..., n. With this notation we can rewrite the function defined in Equation 3 as f c where c c, c,..., c n and c denotes the dot product of the vectors c and in V n. In view of the one-to-one correspondence between points in n,,..., n and their position vectors,,..., n in V n, we have three was of looking at a function f defined on a subset of : n. As a function of n real variables,,..., n. As a function of a single point variable,,..., n 3. As a function of a single vector variable,,..., n We will see that all three points of view are useful. 4. EXERCISES. In Eample we considered the function W f T, v, where W is the wind-chill inde, T is the actual temperature, and v is the wind speed. A numerical representation is given in Table. (a) What is the value of f 5, 4? What is its meaning? (b) Describe in words the meaning of the question For what value of v is f, v 3? Then answer the question. (c) Describe in words the meaning of the question For what value of T is f T, 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, 5? Describe the behavior of this function.

13 866 CHAPTER 4 PARTIAL DERIVATIVES. The temperature-humidit inde I (or humide, for short) is the perceived air temperature when the actual temperature is T and the relative humidit is h, so we can write I f T, h. The following table of values of I is an ecerpt from a table compiled b the National Oceanic & Atmospheric Administration. Actual temperature ( F) TABLE 3 Apparent temperature as a function of temperature and humidit T h Relative humidit (%) (a) What is the value of f 95, 7? What is its meaning? (b) For what value of h is f 9, h? (c) For what value of T is f T, 5 88? (d) What are the meanings of the functions I f 8, h and I f, h? Compare the behavior of these two functions of h. 3. Verif for the Cobb-Douglas production function discussed in Eample 3 that the production will be doubled if both the amount of labor and the amount of capital are doubled. Determine whether this is also true for the general production function 4. The wind-chill inde W discussed in Eample has been modeled b the following function: WT, v 3..65T.37v Tv.6 Check to see how closel this model agrees with the values in Table 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 Table 4. (a) What is the value of f 4, 5? What is its meaning? (b) What is the meaning of the function h f 3, t? Describe the behavior of this function. (c) What is the meaning of the function h f v, 3? Describe the behavior of this function PL, K blk PL, K.L.75 K Wind speed (knots) TABLE 4 Duration (hours) 6. Let f, ln. (a) Evaluate f,. (b) Evaluate f e,. (c) Find and sketch the domain of f. (d) Find the range of f. 7. Let f, e 3. (a) Evaluate f,. (b) Find the domain of f. (c) Find the range of f. 8. Find and sketch the domain of the function f, s. What is the range of f? 9. Let f,, e s. (a) Evaluate f,, 6. (b) Find the domain of f. (c) Find the range of f.. Let t,, ln5. (a) Evaluate t,, 4. (b) Find the domain of t. (c) Find the range of t. Find and sketch the domain of the function f, s s f, arcsin 9. t f, s f, s f, ln9 9 f, s ln f, s s f, s f,, s. f,, ln

14 SECTION 4. FUNCTIONS OF SEVERAL VARIABLES Sketch the graph of the function.. 3. f, 3 f, f, f, cos 5. f, 6. f, f, 4 f, s6 6 f, s 3. Two contour maps are shown. One is for a function f whose graph is a cone. The other is for a function t whose graph is a paraboloid. Which is which, and wh? I II 3. atch the function with its graph (labeled I VI).Give reasons for our choices. (a) f, (b) f, (c) f, (d) f, (e) f, (f) I II f, sin( ) 33. Locate the points A and B in the map of Lonesome ountain (Figure ). How would ou describe the terrain near A? Near B? 34. ake a rough sketch of a contour map for the function whose graph is shown. III IV V VI A contour map of a function is shown. Use it to make a rough sketch of the graph of f _8 _6 _4 3. A contour map for a function f is shown. Use it to estimate the values of f 3, 3 and f 3,. What can ou sa about the shape of the graph? _3 3

15 868 CHAPTER 4 PARTIAL DERIVATIVES Draw a contour map of the function showing several level curves. 39. f, 4. f, 3 4. f, ln 4. f, e Sketch both a contour map and a graph of the function and compare them. 47. f, f, s A thin metal plate, located in the -plane, has temperature T, at the point,. 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 b 5. If V, is the electric potential at a point, in the -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, csr, where c is a positive constant. ; 5 54 Use a computer to graph the function using various domains and viewpoints. Get a printout of one that, in our opinion, gives a good view. If our software also produces level curves, then plot some contour lines of the same function and compare with the graph. 5. f, e e f, 3 (monke saddle) 54. f, 3 3 (dog saddle) 55 6 atch the function (a) with its graph (labeled A F on page 869) and (b) with its contour map (labeled I VI). Give reasons for our choices. 55. f, e f, 3 e sin sin 58. sin sin f, sec 45. f, s 46. f, T, e cos 6 64 Describe the level surfaces of the function Describe how the graph of t is obtained from the graph of f. 65. (a) t, f, (b) t, f, (c) t, f, (d) t, f, 66. (a) t, f, (b) t, f, (c) t, f 3, 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 valles. Would ou sa the function has a maimum value? Can ou identif an points on the graph that ou might consider to be local maimum points? What about local minimum points? f,, 3 5 f,, 3 5 f,, f,, f, f, e ; 69 7 Use a computer to graph the function using various domains and viewpoints. Comment on the limiting behavior of the function. What happens as both and become large? What happens as, approaches the origin? 69. f, 7. f, ; 7. Use a computer to investigate the famil of functions f, e c. How does the shape of the graph depend on c? ; 7. Use a computer to investigate the famil of surfaces a b e How does the shape of the graph depend on the numbers a and b? ; 73. Use a computer to investigate the famil of surfaces c. In particular, ou should determine the transitional values of c for which the surface changes from one tpe of quadric surface to another.

16 SECTION 4. FUNCTIONS OF SEVERAL VARIABLES 869 I II III IV V VI D E F A B C Graphs and Contour aps for Eercises 55 6

17 87 CHAPTER 4 PARTIAL DERIVATIVES ; 74. Graph the functions f, s f, lns f, e s f, sin(s ) ; 75. (a) Show that, b taking logarithms, the general Cobb- Douglas function P bl K can be epressed as ln P K ln b ln L K and f, s In general, if t is a function of one variable, how is the graph of f, t(s ) obtained from the graph of t? (b) If we let lnlk and lnpk, the equation in part (a) becomes the linear equation ln b. Use Table (in Eample 3) to make a table of values of lnlk and lnpk for the ears Then use a graphing calculator or computer to find the least squares regression line through the points lnlk, lnpk. (c) Deduce that the Cobb-Douglas production function is P.L.75 K LIITS AND CONTINUITY Let s compare the behavior of the functions f, sin and t, as and both approach [and therefore the point, approaches the origin]. TABLE Values of f, TABLE Values of t, Tables and show values of f, and t,, correct to three decimal places, for points, near the origin. (Notice that neither function is defined at the origin.) It appears that as, approaches (, ), the values of f, are approaching whereas the values of t, aren t approaching an number. It turns out that these guesses based on numerical evidence are correct, and we write lim, l, sin and lim, l, does not eist In general, we use the notation lim f, L, l a, b

18 SECTION 4. LIITS AND CONTINUITY 87 to indicate that the values of f, approach the number L as the point, approaches the point a, b along an path that stas within the domain of f. In other words, we can make the values of f, as close to L as we like b taking the point, sufficientl close to the point a, b, but not equal to a, b. A more precise definition follows. DEFINITION Let f be a function of two variables whose domain D includes points arbitraril close to a, b. Then we sa that the limit of f, as, approaches a, b is L and we write lim f, L, l a, b if for ever number there is a corresponding number such that if, D and s a b then f, L Other notations for the limit in Definition are lim f, L l a l b f, L and Notice that is the distance between the numbers f, and L, and s a b is the distance between the point, and the point a, b. Thus Definition sas that the distance between f, and L can be made arbitraril small b making the distance from, to a, b sufficientl small (but not ). Figure illustrates Definition b means of an arrow diagram. If an small interval L, L is given around L, then we can find a disk with center a, b and radius such that f maps all the points in [ecept possibl a, b] into the interval L, L. D D f, l L as, l a, b D (, ) (a, b) f ( ) L- L L+ L+ L L- S (a, b) D FIGURE FIGURE b FIGURE 3 a Another illustration of Definition is given in Figure where the surface S is the graph of f. If is given, we can find such that if, is restricted to lie in the disk and, a, b, then the corresponding part of S lies between the horiontal planes L and L. For functions of a single variable, when we let approach a, there are onl two possible directions of approach, from the left or from the right. We recall from Chapter that if lim l a f lim l a f, then lim l a f does not eist. For functions of two variables the situation is not as simple because we can let, approach a, b from an infinite number of directions in an manner whatsoever (see Figure 3) as long as, stas within the domain of f. D

19 87 CHAPTER 4 PARTIAL DERIVATIVES Definition sas that the distance between f, and L can be made arbitraril small b making the distance from, to a, b sufficientl small (but not ). The definition refers onl to the distance between, and a, b. It does not refer to the direction of approach. Therefore, if the limit eists, then f, must approach the same limit no matter how, approaches a, b. Thus if we can find two different paths of approach along which the function f, has different limits, then it follows that lim, l a, b f, does not eist. If f, l L as, l a, b along a path C and f, l L as, l a, b along a path C, where L L, then lim, l a, b f, does not eist. V EXAPLE Show that lim, l, does not eist. SOLUTION Let f,. First let s approach, along the -ais. Then gives f, for all, so f, l as, l, along the -ais f=_ f= We now approach along the -ais b putting. Then f, for all, so f, l as, l, along the -ais FIGURE 4 (See Figure 4.) Since f has two different limits along two different lines, the given limit does not eist. (This confirms the conjecture we made on the basis of numerical evidence at the beginning of this section.) EXAPLE If f,, does lim f, eist?, l, SOLUTION If, then f,. Therefore f, l as, l, along the -ais f= f= f= = If, then f,, so f, l as, l, along the -ais Although we have obtained identical limits along the aes, that does not show that the given limit is. Let s now approach, along another line, sa. For all, f, Therefore f, l as, l, along FIGURE 5 (See Figure 5.) Since we have obtained different limits along different paths, the given limit does not eist.

20 SECTION 4. LIITS AND CONTINUITY 873 TEC In Visual 4. a rotating line on the surface in Figure 6 shows different limits at the origin from different directions. Figure 6 sheds some light on Eample. The ridge that occurs above the line corresponds to the fact that f, for all points, on that line ecept the origin. FIGURE 6 f(, )= + V EXAPLE 3 If f,, does lim f, eist? 4, l, SOLUTION With the solution of Eample in mind, let s tr to save time b letting, l, along an nonvertical line through the origin. Then m, where m is the slope, and N Figure 7 shows the graph of the function in Eample 3. Notice the ridge above the parabola..5 _.5 FIGURE 7 So Thus f has the same limiting value along ever nonvertical line through the origin. But that does not show that the given limit is, for if we now let, l, along the parabola, we have so f, f, m f, l f, f, 4 4 f, l m m m 3 4 m 4 4 as as, l, along m 4, l, along m m 4 Since different paths lead to different limiting values, the given limit does not eist. Now let s look at limits that do eist. Just as for functions of one variable, the calculation of limits for functions of two variables can be greatl simplified b the use of properties of limits. The Limit Laws listed in Section.3 can be etended 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. lim a, l a, b lim b, l a, b lim c c, l a, b The Squeee Theorem also holds. EXAPLE 4 Find lim, l, 3 if it eists. SOLUTION As in Eample 3, we could show that the limit along an line through the origin is. This doesn t prove that the given limit is, but the limits along the parabolas

21 874 CHAPTER 4 PARTIAL DERIVATIVES and also turn out to be, so we begin to suspect that the limit does eist and is equal to. Let. We want to find such that that is, if if s But since, so and therefore s then then s 3s N Another wa to do Eample 4 is to use the Squeee Theorem instead of Definition. From () it follows that lim 3, l, and so the first inequalit in (3) shows that the given limit is. Thus if we choose and let s, then Hence, b Definition, CONTINUITY 3 3 3s lim, l, 3 Recall that evaluating limits of continuous functions of a single variable is eas. It can be accomplished b direct substitution because the defining propert of a continuous function is lim l a f f a. Continuous functions of two variables are also defined b the direct substitution propert. 4 DEFINITION A function f of two variables is called continuous at a, b if lim f, f a, b, l a, b We sa f is continuous on D if f is continuous at ever point a, b in D. The intuitive meaning of continuit is that if the point, changes b a small amount, then the value of f, changes b 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, ou can see that sums, differences, products, and quotients of continuous functions are continuous on their domains. Let s use this fact to give eamples of continuous functions. A polnomial function of two variables (or polnomial, for short) is a sum of terms of the form c m n, where c is a constant and m and n are nonnegative integers. A rational function is a ratio of polnomials. For instance, f,

22 SECTION 4. LIITS AND CONTINUITY 875 is a polnomial, whereas t, is a rational function. The limits in () show that the functions f,, t,, and h, c are continuous. Since an polnomial can be built up out of the simple functions f, t, and h b multiplication and addition, it follows that all polnomials are continuous on. Likewise, an rational function is continuous on its domain because it is a quotient of continuous functions. V EXAPLE 5 Evaluate lim , l, SOLUTION Since f, is a polnomial, it is continuous everwhere, so we can find the limit b direct substitution: lim , l, EXAPLE 6 Where is the function f, continuous? SOLUTION The function f is discontinuous at, because it is not defined there. Since f is a rational function, it is continuous on its domain, which is the set D,,,. EXAPLE 7 Let t, if if,,,, Here t is defined at, but t is still discontinuous there because lim, l, t, does not eist (see Eample ). N Figure 8 shows the graph of the continuous function in Eample 8. EXAPLE 8 Let 3 f, if if,,,, We know f is continuous for,, since it is equal to a rational function there. Also, from Eample 4, we have lim f,, l, lim, l, 3 f, FIGURE 8 Therefore f is continuous at,, and so it is continuous on. Just as for functions of one variable, composition is another wa 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 b h, t f, is also a continuous function.

23 876 CHAPTER 4 PARTIAL DERIVATIVES FIGURE 9 The function h(, )=arctan(/) is discontinuous where =. _ EXAPLE 9 Where is the function h, arctan continuous? SOLUTION The function f, is a rational function and therefore continuous ecept on the line. The function tt arctan t is continuous everwhere. So the composite function is continuous ecept where. The graph in Figure 9 shows the break in the graph of h above the -ais. FUNCTIONS OF THREE OR ORE VARIABLES Everthing that we have done in this section can be etended to functions of three or more variables. The notation means that the values of f,, approach the number L as the point,, approaches the point a, b, c along an path in the domain of f. Because the distance between two points,, and a, b, c in 3 is given b s a b c, we can write the precise definition as follows: For ever number there is a corresponding number such that if,, is in the domain of f then and The function f is continuous at a, b, c if For instance, the function t f, arctan h, lim f,, L,, l a, b, c lim f,, f a, b, c,, l a, b, c f,, s a b c f,, L is a rational function of three variables and so is continuous at ever point in 3 ecept where. In other words, it is discontinuous on the sphere with center the origin and radius. If we use the vector notation introduced at the end of Section 4., then we can write the definitions of a limit for functions of two or three variables in a single compact form as follows. 5 If f is defined on a subset D of n, then lim l a f L means that for ever number there is a corresponding number such that if D and then a f L Notice that if n, then and a a, and (5) is just the definition of a limit for functions of a single variable. For the case n, we have,, a a, b, and a s a b, so (5) becomes Definition. If n 3, then,,, a a, b, c, and (5) becomes the definition of a limit of a function of three variables. In each case the definition of continuit can be written as lim f f a l a

24 SECTION 4. LIITS AND CONTINUITY EXERCISES. Suppose that lim, l 3, f, 6. What can ou sa about the value of f 3,? What if f is continuous?. Eplain wh each function is continuous or discontinuous. (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 tai ride as a function of distance traveled and time 3 4 Use a table of numerical values of f, for, near the origin to make a conjecture about the value of the limit of f, as, l,. Then eplain wh our guess is correct. 3. f, f, Find h, t f, and the set on which h is continuous. 5. tt t st, f, tt t ln t, ; 7 8 Graph the function and observe where it is discontinuous. Then use the formula to eplain what ou have observed. 7. lim, l, 3 6 f, e f, 8. f, 5 Find the limit, if it eists, or show that the limit does not eist. 5. lim , l, 4 7. lim 8., l, cos. lim., l, e 5. lim 6., l, lim 8., l, s lim, l, lim, l, lim,, l 3,, e sin 3 lim,, l,, lim,, l,, lim,, l,, s ; 3 4 Use a computer graph of the function to eplain wh the limit does not eist lim, l, 3 5 lim, l, e cos lim ln, l, sin lim, l, lim, l, 4 4 lim, l, lim, l, lim, l, sin Determine the set of points at which the function is continuous. 9. F, sin 3. F, e 3. F, arctan( s ) 3. F, e s 33. G, ln G, tan ( ) f, Use polar coordinates to find the limit. [If r, are polar coordinates of the point, with r, note that r l as, l,.] f,, f,, s f, 3 3 lim, l, s lim, l, ln e 4. lim, l, if,, if,, if,, if,,

25 878 CHAPTER 4 PARTIAL DERIVATIVES ; 4. At the beginning of this section we considered the function f, sin and guessed that f, l as, l, on the basis of numerical evidence. Use polar coordinates to confirm the value of the limit. Then graph the function. ; 43. Graph and discuss the continuit of the function sin f, if if 44. Let 4 if or f, if 4 (a) Show that f, l as, l, along an path through, of the form m a with a 4. (b) Despite part (a), show that f is discontinuous at,. (c) Show that f is discontinuous on two entire curves. 45. Show that the function f given b is continuous on n. [Hint: Consider a a.] f a c V n 46. If, show that the function f given b f c is continuous on n. 4.3 PARTIAL DERIVATIVES On a hot da, etreme humidit makes us think the temperature is higher than it reall is, whereas in ver dr air we perceive the temperature to be lower than the thermometer indicates. The National Weather Service has devised the heat inde (also called the temperature-humidit inde, or humide, in some countries) to describe the combined effects of temperature and humidit. The heat inde I is the perceived air temperature when the actual temperature is T and the relative humidit 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 ecerpt from a table compiled b the National Weather Service. TABLE Heat inde I as a function of temperature and humidit T 9 H Relative humidit (%) Actual temperature ( F) If we concentrate on the highlighted column of the table, which corresponds to a relative humidit of H 7%, we are considering the heat inde as a function of the single variable T for a fied value of H. Let s write tt f T, 7. Then tt describes how the heat inde I increases as the actual temperature T increases when the relative humidit is 7%. The derivative of t when T 96F is the rate of change of I with respect to T when T 96F: t96 h t96 t96 lim lim h l h h l f 96 h, 7 f 96, 7 h

26 SECTION 4.3 PARTIAL DERIVATIVES 879 We can approimate t96 using the values in Table b taking h and : t96 t98 t96 f 98, 7 f 96, t96 t94 t96 f 94, 7 f 96, Averaging these values, we can sa that the derivative t96 is approimatel This means that, when the actual temperature is 96F and the relative humidit is 7%, the apparent temperature (heat inde) rises b about 3.75F for ever degree that the actual temperature rises! Now let s look at the highlighted row in Table, which corresponds to a fied temperature of T 96F. The numbers in this row are values of the function GH f 96, H, which describes how the heat inde increases as the relative humidit H increases when the actual temperature is T 96F. The derivative of this function when H 7% is the rate of change of I with respect to H when H 7%: G7 h G7 G7 lim lim h l h h l f 96, 7 h f 96, 7 h B taking h 5 and 5, we approimate G7 using the tabular values: G7 G75 G7 5 f 96, 75 f 96, G7 G65 G7 5 f 96, 65 f 96, B averaging these values we get the estimate G7.9. This sas that, when the temperature is 96F and the relative humidit is 7%, the heat inde rises about.9f for ever percent that the relative humidit rises. In general, if f is a function of two variables and, suppose we let onl var while keeping fied, sa b, where b is a constant. Then we are reall considering a function of a single variable, namel, t f, b. If t has a derivative at a, then we call it the partial derivative of f with respect to at a, b and denote it b f a, b. Thus f a, b ta where t f, b B the definition of a derivative, we have and so Equation becomes ta lim h l ta h ta h f a, b lim h l f a h, b f a, b h

27 88 CHAPTER 4 PARTIAL DERIVATIVES Similarl, the partial derivative of f with respect to at a, b, denoted b f a, b, is obtained b keeping fied a and finding the ordinar derivative at b of the function G f a, : 3 f a, b lim h l f a, b h f a, b h With this notation for partial derivatives, we can write the rates of change of the heat inde I with respect to the actual temperature T and relative humidit H when T 96F and H 7% as follows: f T 96, f H 96, 7.9 If we now let the point a, b var in Equations and 3, f and f become functions of two variables. 4 and If f is a function of two variables, its partial derivatives are the functions defined b f f, lim h l f h, f, h f f, lim h l f, h f, h There are man alternative notations for partial derivatives. For instance, instead of f we can write f or D f (to indicate differentiation with respect to the first variable) or f. But here f can t be interpreted as a ratio of differentials. NOTATIONS FOR PARTIAL DERIVATIVES If f,, we write f, f f f, f D f D f f, f f f, f D f D f To compute partial derivatives, all we have to do is remember from Equation that the partial derivative with respect to is just the ordinar derivative of the function t of a single variable that we get b keeping fied. Thus we have the following rule. RULE FOR FINDING PARTIAL DERIVATIVES OF f,. To find, regard as a constant and differentiate f, with respect to. f. To find, regard as a constant and differentiate f, with respect to. f

28 SECTION 4.3 PARTIAL DERIVATIVES 88 EXAPLE If f, 3 3, find f, and f,. SOLUTION Holding constant and differentiating with respect to, we get f, 3 3 and so f, Holding constant and differentiating with respect to, we get f, 3 4 f, INTERPRETATIONS OF PARTIAL DERIVATIVES To give a geometric interpretation of partial derivatives, we recall that the equation f, represents a surface S (the graph of f ). If f a, b c, then the point Pa, b, c lies on S. B fiing b, we are restricting our attention to the curve C in which the vertical plane b intersects S. (In other words, C is the trace of S in the plane b.) Likewise, the vertical plane a intersects S in a curve C. Both of the curves C and C pass through the point P. (See Figure.) T S C P(a, b, c) T C FIGURE The partial derivatives of f at (a, b) are the slopes of the tangents to C and C. (a, b, ) Notice that the curve C is the graph of the function t f, b, so the slope of its tangent T at P is ta f a, b. The curve C is the graph of the function G f a,, so the slope of its tangent T at P is Gb f a, b. Thus the partial derivatives f a, b and f a, b can be interpreted geometricall as the slopes of the tangent lines at Pa, b, c to the traces C and C of S in the planes b and a. As we have seen in the case of the heat inde function, partial derivatives can also be interpreted as rates of change. If f,, then represents the rate of change of with respect to when is fied. Similarl, represents the rate of change of with respect to when is fied. EXAPLE If f, 4, find f, and f, and interpret these numbers as slopes. SOLUTION We have f, f, f, 4 f, 4

29 88 CHAPTER 4 PARTIAL DERIVATIVES The graph of f is the paraboloid 4 and the vertical plane intersects it in the parabola,. (As in the preceding discussion, we label it C in Figure.) The slope of the tangent line to this parabola at the point,, is f,. Similarl, the curve C in which the plane intersects the paraboloid is the parabola 3,, and the slope of the tangent line at,, is f, 4. (See Figure 3.) =4- - =4- - C = C = (,, ) (,, ) (, ) (, ) FIGURE FIGURE 3 Figure 4 is a computer-drawn counterpart to Figure. Part (a) shows the plane intersecting the surface to form the curve C and part (b) shows C and T. [We have used the vector equations rt t,, t for C and rt t,, t for T.] Similarl, Figure 5 corresponds to Figure FIGURE 4 (a) (b) FIGURE 5

30 f f EXAPLE 3 If f, sin V, calculate and. SOLUTION Using the Chain Rule for functions of one variable, we have f cos SECTION 4.3 PARTIAL DERIVATIVES 883 cos f cos cos N Some computer algebra sstems can plot surfaces defined b implicit equations in three variables. Figure 6 shows such a plot of the surface defined b the equation in Eample 4. V EXAPLE 4 Find and if is defined implicitl as a function of and b the equation SOLUTION To find, we differentiate implicitl with respect to, being careful to treat as a constant: Solving this equation for, we obtain FIGURE 6 Similarl, implicit differentiation with respect to gives FUNCTIONS OF ORE THAN TWO VARIABLES Partial derivatives can also be defined for functions of three or more variables. For eample, if f is a function of three variables,, and, then its partial derivative with respect to is defined as f,, lim h l and it is found b regarding and as constants and differentiating f,, with respect to. If w f,,, then f w can be interpreted as the rate of change of w with respect to when and are held fied. But we can t interpret it geometricall because the graph of f lies in four-dimensional space. In general, if u is a function of n variables, u f,,..., n, its partial derivative with respect to the ith variable is i f h,, f,, h u i lim h l f,..., i, i h, i,..., n f,..., i,..., n h

31 884 CHAPTER 4 PARTIAL DERIVATIVES and we also write u f f i f i D i f i i EXAPLE 5 Find,, and if f,, e ln. f f f SOLUTION Holding and constant and differentiating with respect to, we have f e ln Similarl, f e ln and f e HIGHER DERIVATIVES If f is a function of two variables, then its partial derivatives f and f are also functions of two variables, so we can consider their partial derivatives f, f, f, and f, which are called the second partial derivatives of f. If f,, we use the following notation: Thus the notation f (or f ) means that we first differentiate with respect to and then with respect to, whereas in computing the order is reversed. EXAPLE 6 Find the second partial derivatives of SOLUTION In Eample we found that f f f f f f f f f f f f f f f f f f f f f f, 3 3 f, 3 3 f, 3 4 Therefore f f f f

32 SECTION 4.3 PARTIAL DERIVATIVES 885 _ N Figure 7 shows the graph of the function f in Eample 6 and the graphs of its first- and second-order partial derivatives for,. Notice that these graphs are consistent with our interpretations of f and f as slopes of tangent lines to traces of the graph of f. For instance, the graph of f decreases if we start at, and move in the positive -direction. This is reflected in the negative values of f. You should compare the graphs of f and f with the graph of to see the relationships. f _4 f 4 4 f f _ f f f f FIGURE 7 Notice that f f in Eample 6. This is not just a coincidence. It turns out that the mied partial derivatives f and f are equal for most functions that one meets in practice. The following theorem, which was discovered b the French mathematician Aleis Clairaut (73 765), gives conditions under which we can assert that f f. The proof is given in Appendi F. N Aleis Clairaut was a child prodig in mathematics: he read l Hospital s tetbook on calculus when he was ten and presented a paper on geometr to the French Academ of Sciences when he was 3. At the age of 8, Clairaut published Recherches sur les courbes à double courbure, which was the first sstematic treatise on three-dimensional analtic geometr and included the calculus of space curves. CLAIRAUT S THEORE Suppose f is defined on a disk D that contains the point a, b. If the functions and are both continuous on D, then f f f a, b f a, b Partial derivatives of order 3 or higher can also be defined. For instance, f f f 3 f

33 886 CHAPTER 4 PARTIAL DERIVATIVES and using Clairaut s Theorem it can be shown that f f f if these functions are continuous. V EXAPLE 7 Calculate if f,, sin3. f SOLUTION f 3cos3 f 9sin3 f 9 cos3 f 9cos3 9 sin3 PARTIAL DIFFERENTIAL EQUATIONS Partial derivatives occur in partial differential equations that epress certain phsical laws. For instance, the partial differential equation u u is called Laplace s equation after Pierre Laplace (749 87). Solutions of this equation are called harmonic functions; the pla a role in problems of heat conduction, fluid flow, and electric potential. EXAPLE 8 Show that the function u, e sin is a solution of Laplace s equation. SOLUTION u e sin u e sin u e cos u e sin u u e sin e sin Therefore u satisfies Laplace s equation. The wave equation u t a u FIGURE 8 u(, t) 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, t represents the displacement of a vibrating violin string at time t and at a distance from one end of the string (as in Figure 8), then u, t satisfies the wave equation. Here the constant a depends on the densit of the string and on the tension in the string. EXAPLE 9 Verif that the function u, t sin at satisfies the wave equation. SOLUTION u cos at u t a cos at u sin at u tt a sin at a u So u satisfies the wave equation.

34 SECTION 4.3 PARTIAL DERIVATIVES 887 THE COBB-DOUGLAS PRODUCTION FUNCTION In Eample 3 in Section 4. we described the work of Cobb and Douglas in modeling the total production P of an economic sstem 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 the made about the econom. If the production function is denoted b P PL, K, then the partial derivative PL 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 productivit of labor. Likewise, the partial derivative PK is the rate of change of production with respect to capital and is called the marginal productivit of capital. In these terms, the assumptions made b Cobb and Douglas can be stated as follows. (i) If either labor or capital vanishes, then so will production. (ii) The marginal productivit of labor is proportional to the amount of production per unit of labor. (iii) The marginal productivit of capital is proportional to the amount of production per unit of capital. Because the production per unit of labor is PL, assumption (ii) sas that P L P L for some constant. If we keep K constant K K, then this partial differential equation becomes an ordinar differential equation: 5 dp dl P L If we solve this separable differential equation b the methods of Section 9.3 (see also Eercise 79), we get 6 PL, K C K L Notice that we have written the constant as a function of K because it could depend on the value of K. Similarl, assumption (iii) sas that P K P K and we can solve this differential equation to get C 7 PL, K C L K Comparing Equations 6 and 7, we have 8 PL, K bl K

35 888 CHAPTER 4 PARTIAL DERIVATIVES where b is a constant that is independent of both L and K. Assumption (i) shows that and. Notice from Equation 8 that if labor and capital are both increased b a factor m, then PmL, mk bmlmk mblk mpl, K If, then PmL, mk mpl, K, which means that production is also increased b a factor of m. That is wh Cobb and Douglas assumed that and therefore PL, K blk This is the Cobb-Douglas production function that we discussed in Section EXERCISES. The temperature T at a location in the Northern Hemisphere depends on the longitude, latitude, and time t, so we can write T f,, t. Let s measure time in hours from the beginning of Januar. (a) What are the meanings of the partial derivatives T, T, and Tt? (b) Honolulu has longitude 58 W and latitude N. Suppose that at 9: A on Januar 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 ou epect f 58,, 9, f 58,, 9, and f t58,, 9 to be positive or negative? Eplain.. At the beginning of this section we discussed the function I f T, H, where I is the heat inde, T is the temperature, and H is the relative humidit. Use Table to estimate f T9, 6 and f H9, 6. What are the practical interpretations of these values? 3. The wind-chill inde 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 ecerpt from Table in Section 4.. Actual temperature ( C) T v Wind speed (km/h) (a) Estimate the values of f T5, 3 and f v5, 3. What are the practical interpretations of these values? (b) In general, what can ou sa about the signs of WT and Wv? (c) What appears to be the value of the following limit? 4. 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. Wind speed (knots) t v W lim v l v Duration (hours) (a) What are the meanings of the partial derivatives hv and ht? (b) Estimate the values of f v4, 5 and f t4, 5. What are the practical interpretations of these values? (c) What appears to be the value of the following limit? h lim t l t

36 SECTION 4.3 PARTIAL DERIVATIVES Determine the signs of the partial derivatives for the function f whose graph is shown.. A contour map is given for a function f. Use it to estimate f, and f,. _4 _ (a) f, (b) 6. (a) f, (b) 7. (a) f, (b) 8. (a) f, (b) f, f, f, f,. If f, 6 4, find f, and f, and interpret these numbers as slopes. Illustrate with either hand-drawn sketches or computer plots.. If f, s4 4, find f, and f, and interpret these numbers as slopes. Illustrate with either handdrawn sketches or computer plots. 9. The following surfaces, labeled a, b, and c, are graphs of a function f and its partial derivatives f and f. Identif each surface and give reasons for our choices. ; 3 4 Find f and f and graph f, f, and f with domains and viewpoints that enable ou to see the relationships between them f, e f, Find the first partial derivatives of the function. 5. f, f, f, t e t cos 8. f, t s ln t _4 a _8 _3 3 _ f,.. 3. w sin cos f r, s r lnr s 6. tan f, w e v u v f, t arctan(st ) 4 _4 b _3 3 _ 7. u te wt f,, w ln u sin f,,, t tant u s n f, cost dt f,, sin w e u f,,, t t 8 4 _4 _8 _3 c 3 _ 38. u sin n n 39 4 Find the indicated partial derivatives. 39. f, ln( s ) ; f 3, 4 4. f, arctan; f, 3 4. f,, ; f,,

37 89 CHAPTER 4 PARTIAL DERIVATIVES 4. f,, ssin sin sin ; f,, Use the table of values of f, to estimate the values of f 3,, f 3,., and f 3, Use the definition of partial derivatives as limits (4) to find f, and f,. 43. f, f, Use implicit differentiation to find and ln 47. arctan 48. sin Find and. 49. (a) f t (b) f 5. (a) f t (b) f (c) f Level curves are shown for a function f. Determine whether the following partial derivatives are positive or negative at the point P. (a) f (b) f (c) f (d) f (e) f P 5 56 Find all the second partial derivatives. 5. f, f, sin m n 53. w su v 54. v 55. arctan 56. v e e 57 6 Verif that the conclusion of Clairaut s Theorem holds, that is, u u. 57. u sin 58. u u ln s 6. u e 6 68 Find the indicated partial derivative. 6. f, ;, 6. f, t e ct ;, 63. f,, cos4 3 ;, 64. f r, s, t r lnrs t 3 ;, 65. u e r sin ; 66. usv w ; 3 w 67. w ;, 68. u a b c ; f ttt 3 u r 3 u v w 6 u 3 f t f f rss f f rst 3 w f f 7. Verif that the function u e k t sin k is a solution of the heat conduction equation u t u. 7. Determine whether each of the following functions is a solution of Laplace s equation u u. (a) u (b) u (c) u 3 3 (d) u ln s (e) u sin cosh cos sinh (f) u e cos e cos 73. Verif that the function u s is a solution of the three-dimensional Laplace equation u u u. 74. Show that each of the following functions is a solution of the wave equation u tt a u. (a) u sink sinakt (b) u ta t (c) u at 6 at 6 (d) u sin at ln at 75. If f and t are twice differentiable functions of a single variable, show that the function u, t f at t at is a solution of the wave equation given in Eercise If u e aa an n, where a a an, show that u u u n u 77. Verif that the function lne e is a solution of the differential equations

38 SECTION 4.3 PARTIAL DERIVATIVES 89 and 87. You are told that there is a function f whose partial derivatives are f, 4 and f, 3. Should ou believe it? 78. Show that the Cobb-Douglas production function satisfies the equation 79. Show that the Cobb-Douglas production function satisfies PL, K C K L b solving the differential equation (See Equation 5.) 8. The temperature at a point, on a flat metal plate is given b T, 6, where T is measured in C and, in meters. Find the rate of change of temperature with respect to distance at the point, in (a) the -direction and (b) the -direction. Find RR. L P P K L K P 8. The gas law for a fied mass m of an ideal gas at absolute temperature T, pressure P, and volume V is PV mrt, where R is the gas constant. Show that 83. For the ideal gas of Eercise 8, show that P V dp dl P L R R R R 3 V T T P T 84. The wind-chill inde is modeled b the function W 3..65T.37v Tv.6 where T is the temperature C and v is the wind speed kmh. When T 5C and v 3 kmh, b how much would ou epect the apparent temperature W to drop if the actual temperature decreases b C? What if the wind speed increases b kmh? 85. The kinetic energ of a bod with mass m and velocit v is K mv. Show that K m T P V mr T K v K P bl K 8. The total resistance R produced b three conductors with resistances R, R, R 3 connected in a parallel electrical circuit is given b the formula 86. If a, b, c are the sides of a triangle and A, B, C are the opposite angles, find Aa, Ab, Ac b implicit differentiation of the Law of Cosines. ; 88. The paraboloid 6 intersects the plane in a parabola. Find parametric equations for the tangent line to this parabola at the point,, 4. Use a computer to graph the paraboloid, the parabola, and the tangent line on the same screen. 89. The ellipsoid 4 6 intersects the plane in an ellipse. Find parametric equations for the tangent line to this ellipse at the point,,. 9. In a stud of frost penetration it was found that the temperature T at time t (measured in das) at a depth (measured in feet) can be modeled b the function where and is a positive constant. (a) Find T. What is its phsical significance? (b) Find Tt. What is its phsical significance? (c) Show that T satisfies the heat equation T t kt for a certain constant k. ; (d) If, T, and T, use a computer to graph T, t. (e) What is the phsical significance of the term in the epression sint? 9. Use Clairaut s Theorem to show that if the third-order partial derivatives of f are continuous, then 9. (a) How man nth-order partial derivatives does a function of two variables have? (b) If these partial derivatives are all continuous, how man of them can be distinct? (c) Answer the question in part (a) for a function of three variables. 93. If f, 3 e sin, find f,. [Hint: Instead of finding f, first, note that it s easier to use Equation or Equation.] 94. If f, s 3 3 3, find f,. 95. Let 365. T, t T T e sint 3 3 f, ; (a) Use a computer to graph f. (b) Find f, and f, when,,. (c) Find f, and f, using Equations and 3. (d) Show that f, and f,. CAS (e) Does the result of part (d) contradict Clairaut s Theorem? Use graphs of and to illustrate our answer. f f f f f if,, if,,

39 89 CHAPTER 4 PARTIAL DERIVATIVES 4.4 TANGENT PLANES AND LINEAR APPROXIATIONS One of the most important ideas in single-variable calculus is that as we oom in toward a point on the graph of a differentiable function, the graph becomes indistinguishable from its tangent line and we can approimate the function b a linear function. (See Section 3..) Here we develop similar ideas in three dimensions. As we oom 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 approimate the function b a linear function of two variables. We also etend the idea of a differential to functions of two or more variables. T FIGURE The tangent plane contains the tangent lines T and T. P T C C TANGENT PLANES Suppose a surface S has equation f,, where f has continuous first partial derivatives, and let P,, be a point on S. As in the preceding section, let C and C be the curves obtained b intersecting the vertical planes and with the surface S. Then the point P lies on both C and C. Let T and T be the tangent lines to the curves C and C 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 T and T. (See Figure.) We will see in Section 4.6 that if C is an 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 ou 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 closel approimates the surface S near the point P. We know from Equation.5.7 that an plane passing through the point P,, has an equation of the form A B C B dividing this equation b C and letting a AC and b BC, we can write it in the form a b If Equation represents the tangent plane at P, then its intersection with the plane must be the tangent line T. Setting in Equation gives a and we recognie these as the equations (in point-slope form) of a line with slope a. But from Section 4.3 we know that the slope of the tangent T is f,. Therefore a f,. Similarl, putting in Equation, we get b, which must represent the tangent line, so b f,. T N Note the similarit between the equation of a tangent plane and the equation of a tangent line: f Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface f, at the point P,, is f, f,

40 SECTION 4.4 TANGENT PLANES AND LINEAR APPROXIATIONS 893 V EXAPLE Find the tangent plane to the elliptic paraboloid at the point,, 3. SOLUTION Let f,. Then f, 4 f, f, 4 f, Then () gives the equation of the tangent plane at,, 3 as 3 4 or 4 3 TEC Visual 4.4 shows an animation of Figures and 3. Figure (a) shows the elliptic paraboloid and its tangent plane at,, 3 that we found in Eample. In parts (b) and (c) we oom in toward the point,, 3 b restricting the domain of the function f,. Notice that the more we oom in, the flatter the graph appears and the more it resembles its tangent plane _ (a) (b) (c) FIGURE The elliptic paraboloid = + appears to coincide with its tangent plane as we oom in toward (,, 3). In Figure 3 we corroborate this impression b ooming in toward the point, on a contour map of the function f,. Notice that the more we oom in, the more the level curves look like equall spaced parallel lines, which is characteristic of a plane FIGURE 3 Zooming in toward (, ) on a contour map of f(, )=

41 894 CHAPTER 4 PARTIAL DERIVATIVES LINEAR APPROXIATIONS In Eample we found that an equation of the tangent plane to the graph of the function f, at the point,, 3 is 4 3. Therefore, in view of the visual evidence in Figures and 3, the linear function of two variables L, 4 3 is a good approimation to f, when, is near,. The function L is called the lineariation of f at, and the approimation f, 4 3 is called the linear approimation or tangent plane approimation of f at,. For instance, at the point (.,.95) the linear approimation gives f., which is quite close to the true value of f., But if we take a point farther awa from,, such as, 3, we no longer get a good approimation. In fact, L, 3 whereas f, 3 7. In general, we know from () 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 f a, b f a, b a f a, b b The linear function whose graph is this tangent plane, namel 3 L, f a, b f a, b a f a, b b is called the lineariation of f at a, b and the approimation 4 f, f a, b f a, b a f a, b b is called the linear approimation or the tangent plane approimation of f at a, b. We have defined tangent planes for surfaces f,, where f has continuous first partial derivatives. What happens if f and f are not continuous? Figure 4 pictures such a function; its equation is f, if if,,,, FIGURE 4 f(, )= if (, ) (, ), + f(, )= You can verif (see Eercise 46) that its partial derivatives eist at the origin and, in fact, f, and f,, but f and f are not continuous. The linear approimation would be f,, but f, at all points on the line. So a function of two variables can behave badl even though both of its partial derivatives eist. To rule out such behavior, we formulate the idea of a differentiable function of two variables. Recall that for a function of one variable, f, if changes from a to a, we defined the increment of as f a f a

42 SECTION 4.4 TANGENT PLANES AND LINEAR APPROXIATIONS 895 In Chapter 3 we showed that if f is differentiable at a, then N This is Equation f a where l as l Now consider a function of two variables, f,, and suppose changes from a to a and changes from b to b. Then the corresponding increment of is 6 f a, b f a, b Thus the increment represents the change in the value of f when, changes from a, b to a, b. B analog with (5) we define the differentiabilit of a function of two variables as follows. 7 DEFINITION If f,, then f is differentiable at a, b if can be epressed in the form f a, b f a, b where and l as, l,. Definition 7 sas that a differentiable function is one for which the linear approimation (4) is a good approimation when, is near a, b. In other words, the tangent plane approimates the graph of f well near the point of tangenc. It s sometimes hard to use Definition 7 directl to check the differentiabilit of a function, but the net theorem provides a convenient sufficient condition for differentiabilit. N Theorem 8 is proved in Appendi F. 8 THEORE If the partial derivatives f and f eist near a, b and are continuous at a, b, then f is differentiable at a, b. V EXAPLE Show that f, e is differentiable at (, ) and find its lineariation there. Then use it to approimate f.,.. SOLUTION The partial derivatives are f, e e f, e N Figure 5 shows the graphs of the function f and its lineariation L in Eample. 6 4 _ FIGURE 5 Both f and f are continuous functions, so f is differentiable b Theorem 8. The lineariation is The corresponding linear approimation is so f, L, f, f, f, e f, f.,... Compare this with the actual value of f.,..e

43 896 CHAPTER 4 PARTIAL DERIVATIVES EXAPLE 3 At the beginning of Section 4.3 we discussed the heat inde (perceived temperature) I as a function of the actual temperature T and the relative humidit H and gave the following table of values from the National Weather Service. T 9 H Relative humidit (%) Actual temperature ( F) Find a linear approimation for the heat inde I f T, H when T is near 96F and H is near 7%. Use it to estimate the heat inde when the temperature is 97F and the relative humidit is 7%. SOLUTION We read from the table that f 96, 7 5. In Section 4.3 we used the tabular values to estimate that f T 96, and f H 96, 7.9. (See pages ) So the linear approimation is f T, H f 96, 7 f T 96, 7T 96 f H 96, 7H T 96.9H 7 In particular, f 97, Therefore, when T 97F and H 7%, the heat inde is I 3F DIFFERENTIALS For a differentiable function of one variable, f, we define the differential d to be an independent variable; that is, d can be given the value of an real number. The differential of is then defined as 9 d f d =ƒ d d=î Î (See Section 3..) Figure 6 shows the relationship between the increment and the differential d: represents the change in height of the curve f and d represents the change in height of the tangent line when changes b an amount d. For a differentiable function of two variables, f,, we define the differentials d and d to be independent variables; that is, the can be given an values. Then the differential d, also called the total differential, is defined b a a+î tangent line =f(a)+fª(a)(-a) FIGURE 6 d f, d f, d d d (Compare with Equation 9.) Sometimes the notation df is used in place of d.

44 SECTION 4.4 TANGENT PLANES AND LINEAR APPROXIATIONS 897 If we take d a and d b in Equation, then the differential of is d f a, b a f a, b b So, in the notation of differentials, the linear approimation (4) can be written as f, f a, b d Figure 7 is the three-dimensional counterpart of Figure 6 and shows the geometric interpretation of the differential d and the increment : d represents the change in height of the tangent plane, whereas represents the change in height of the surface f, when, changes from a, b to a, b. surface =f(, ) {a+î, b+î, f(a+î, b+î)} {a, b, f(a, b)} d Î f(a, b) FIGURE 7 f(a, b) (a, b, ) Î=d (a+î, b+î, ) Î=d tangent plane -f(a, b)=f (a, b)(-a)+f (a, b)(-b) V EXAPLE 4 (a) If f, 3, find the differential d. (b) If changes from to.5 and changes from 3 to.96, compare the values of and d. SOLUTION (a) Definition gives d d d 3 d 3 d N In Eample 4, d is close to because the tangent plane is a good approimation to the surface 3 near, 3, 3. (See Figure 8.) 6 4 _ FIGURE 8 (b) Putting, d.5, 3, and d.4, we get d The increment of is f.5,.96 f, Notice that d but d is easier to compute. EXAPLE 5 The base radius and height of a right circular cone are measured as cm and 5 cm, respectivel, with a possible error in measurement of as much as. cm in

45 898 CHAPTER 4 PARTIAL DERIVATIVES each. Use differentials to estimate the maimum error in the calculated volume of the cone. SOLUTION The volume V of a cone with base radius r and height h is V r h3. So the differential of V is dv V r Since each error is at most. cm, we have.,.. 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. and dh. along with r, h 5. This gives dv 5 3 dr V h dh rh 3 r r dr 3 dh h. 3. Thus the maimum error in the calculated volume is about cm cm. FUNCTIONS OF THREE OR ORE VARIABLES Linear approimations, differentiabilit, and differentials can be defined in a similar manner for functions of more than two variables. A differentiable function is defined b an epression similar to the one in Definition 7. For such functions the linear approimation is f,, f a, b, c f a, b, c a f a, b, c b f a, b, c c and the lineariation L,, is the right side of this epression. If w f,,, then the increment of w is The differential dw is defined in terms of the differentials d, d, and d of the independent variables b dw w w w d d d EXAPLE 6 The dimensions of a rectangular bo are measured to be 75 cm, 6 cm, and 4 cm, and each measurement is correct to within. cm. Use differentials to estimate the largest possible error when the volume of the bo is calculated from these measurements. SOLUTION If the dimensions of the bo are,, and, its volume is V and so dv V w f,, f,, d V d V d d d d We are given that.,., and.. To find the largest error in the volume, we therefore use d., d., and d. together with 75, 6, and 4: V dv Thus an error of onl. cm in measuring each dimension could lead to an error of as 3 much as 98 cm in the calculated volume! This ma seem like a large error, but it s onl about % of the volume of the bo.

46 SECTION 4.4 TANGENT PLANES AND LINEAR APPROXIATIONS EXERCISES 6 Find an equation of the tangent plane to the given surface at the specified point.. 4, , 3. s, 4. ln, 5. cos, 6. e, ; 7 8 Graph the surface and the tangent plane at the given point. (Choose the domain and viewpoint so that ou get a good view of both the surface and the tangent plane.) Then oom in until the surface and the tangent plane become indistinguishable. CAS 7. 3, 8. arctan, 9 Draw the graph of f and its tangent plane at the given point. (Use our computer algebra sstem both to compute the partial derivatives and to graph the surface and its tangent plane.) Then oom in until the surface and the tangent plane become indistinguishable. 9.. f,,,, 4,,, 6 Eplain wh the function is differentiable at the given point. Then find the lineariation L, of the function at that point.. f, s,, 4. f, 3 4,, 3. f,,, 4. f, s e 4, 5. f, e cos, 6. f, sin 3,,, 4,,,, 5,, 4 sin,,, 3,, 3,,, f, e (s s s ),,, 3e. 7 8 Verif the linear approimation at, s cos 4 9. Find the linear approimation of the function f, s 7 at, and use it to approimate f.95,.8. ;. Find the linear approimation of the function f, ln 3 at 7, and use it to approimate f 6.9,.6. Illustrate b graphing f and the tangent plane.. Find the linear approimation of the function f,, s at 3,, 6 and use it to approimate the number s 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. Duration (hours) Wind speed (knots) Use the table to find a linear approimation to the wave height function when v is near 4 knots and t is near hours. Then estimate the wave heights when the wind has been blowing for 4 hours at 43 knots. 3. Use the table in Eample 3 to find a linear approimation to the heat inde function when the temperature is near 94F and the relative humidit is near 8%. Then estimate the heat inde when the temperature is 95F and the relative humidit is 78%. 4. The wind-chill inde 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 ecerpt from Table in Section 4.. Actual temperature ( C) t v T v Wind speed (km/h) Use the table to find a linear approimation to the wind-chill

47 9 CHAPTER 4 PARTIAL DERIVATIVES inde function when T is near 5C and v is near 5 kmh. Then estimate the wind-chill inde when the temperature is 7C and the wind speed is 55 kmh. 5 3 Find the differential of the function ln 6. v cos v 7. m p 5 q 3 8. T uvw 9. R cos 3. w e 3. If 5 and, changes from, to.5,., compare the values of and d. 3. If 3 and, changes from 3, to.96,.95, compare the values of and d. 33. The length and width of a rectangle are measured as 3 cm and 4 cm, respectivel, with an error in measurement of at most. cm in each. Use differentials to estimate the maimum error in the calculated area of the rectangle. 34. The dimensions of a closed rectangular bo are measured as 8 cm, 6 cm, and 5 cm, respectivel, with a possible error of. cm in each dimension. Use differentials to estimate the maimum error in calculating the surface area of the bo. 35. Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height cm if the tin is.4 cm thick. 36. Use differentials to estimate the amount of metal in a closed clindrical can that is cm high and 4 cm in diameter if the metal in the top and bottom is. cm thick and the metal in the sides is.5 cm thick. 37. A boundar stripe 3 in. wide is painted around a rectangle whose dimensions are ft b ft. Use differentials to approimate the number of square feet of paint in the stripe. 38. The pressure, volume, and temperature of a mole of an ideal gas are related b the equation PV 8.3T, where P is measured in kilopascals, V in liters, and T in kelvins. Use differentials to find the approimate change in the pressure if the volume increases from L to.3 L and the temperature decreases from 3 K to 35 K. 39. If R is the total resistance of three resistors, connected in parallel, with resistances,,, then If the resistances are measured in ohms as R 5, R 4, and R 3 5, with a possible error of.5% in each case, estimate the maimum error in the calculated value of R. 4. Four positive numbers, each less than 5, are rounded to the first decimal place and then multiplied together. Use differentials to estimate the maimum possible error in the computed product that might result from the rounding. 4. A model for the surface area of a human bod is given b S.9w.45 h.75, where w is the weight (in pounds), h is the height (in inches), and S is measured in square feet. If the errors in measurement of w and h are at most %, use differentials to estimate the maimum percentage error in the calculated surface area. 4. Suppose ou need to know an equation of the tangent plane to a surface S at the point P,, 3. You don t have an equation for S but ou know that the curves both lie on S. Find an equation of the tangent plane at P Show that the function is differentiable b finding values of and that satisf Definition f, 45. Prove that if f is a function of two variables that is differentiable at a, b, then f is continuous at a, b. Hint: Show that 46. (a) The function r t 3t, t, 3 4t t r u u, u 3, u f, 44. lim f a, b f a, b, l, was graphed in Figure 4. Show that f, and f, both eist but f is not differentiable at,. [Hint: Use the result of Eercise 45.] (b) Eplain wh and are not continuous at,. f R R R R R 3 f R R 3 if,, if f, 5,,

48 SECTION 4.5 THE CHAIN RULE THE CHAIN RULE Recall that the Chain Rule for functions of a single variable gives the rule for differentiating a composite function: If f and tt, where f and t are differentiable functions, then is indirectl a differentiable function of t and d dt d d d 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 ) deals with the case where f, and each of the variables and is, in turn, a function of a variable t. This means that is indirectl a function of t, f tt, ht, and the Chain Rule gives a formula for differentiating as a function of t. We assume that f is differentiable (Definition 4.4.7). Recall that this is the case when f and f are continuous (Theorem 4.4.8). THE CHAIN RULE (CASE ) Suppose that f, is a differentiable function of and, where tt and ht are both differentiable functions of t. Then is a differentiable function of t and d dt f d dt f d dt PROOF A change of t in t produces changes of in and in. These, in turn, produce a change of in, and from Definition we have where l and l as, l,. [If the functions and are not defined at,, we can define them to be there.] Dividing both sides of this equation b t, we have If we now let t l, then tt t tt l because t is differentiable and therefore continuous. Similarl, l. This, in turn, means that l and l, so d dt lim t l t f lim f t l t lim t l t lim t l lim t l t lim t l lim t l t f f d dt d dt f f f f t d dt d dt f t d dt f d dt t t t

49 9 CHAPTER 4 PARTIAL DERIVATIVES Since we often write in place of f, we can rewrite the Chain Rule in the form N Notice the similarit to the definition of the differential: d d d d dt d dt d dt EXAPLE If 3 4, where sin t and cos t, find ddt when t. SOLUTION The Chain Rule gives d dt d dt d dt 3 4 cos t 3 sin t It s not necessar to substitute the epressions for and in terms of t. We simpl observe that when t, we have sin and cos. Therefore d dt 3 cos sin 6 t (, ) C The derivative in Eample can be interpreted as the rate of change of with respect to t as the point, moves along the curve C with parametric equations sin t, cos t. (See Figure.) In particular, when t, the point, is, and ddt 6 is the rate of increase as we move along the curve C through,. If, for instance, T, 3 4 represents the temperature at the point,, then the composite function Tsin t, cos t represents the temperature at points on C and the derivative ddt represents the rate at which the temperature changes along C. FIGURE The curve =sin t, =cos t V EXAPLE The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related b the equation PV 8.3T. Find the rate at which the pressure is changing when the temperature is 3 K and increasing at a rate of. Ks and the volume is L and increasing at a rate of. Ls. SOLUTION If t represents the time elapsed in seconds, then at the given instant we have T 3, dtdt., V, dvdt.. Since P 8.3 T V the Chain Rule gives dp dt P T 8.3 dt dt P V dv dt 8.3 V dt dt 8.3T dv V dt The pressure is decreasing at a rate of about.4 kpas. We now consider the situation where f, but each of and is a function of two variables s and t: ts, t, hs, t. Then is indirectl a function of s and t and we

50 SECTION 4.5 THE CHAIN RULE 93 wish to find s and t. Recall that in computing t we hold s fied and compute the ordinar derivative of with respect to t. Therefore we can appl Theorem to obtain t t A similar argument holds for s and so we have proved the following version of the Chain Rule. t 3 THE CHAIN RULE (CASE ) Suppose that f, is a differentiable function of and, where ts, t and hs, t are differentiable functions of s and t. Then s s s t t t EXAPLE 3 If e sin, where st and s t, find s and t. SOLUTION Appling Case of the Chain Rule, we get s t t e st sin s t ste st coss t s t s e sin t e cos st t e sin st e cos s ste st sin s t s e st coss t s t FIGURE s s t s t t Case of the Chain Rule contains three tpes of variables: s and t are independent variables, and are called intermediate variables, and 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. To remember the Chain Rule, it s helpful to draw the tree diagram in Figure. We draw branches from the dependent variable to the intermediate variables and to indicate that is a function of and. Then we draw branches from and to the independent variables s and t. On each branch we write the corresponding partial derivative. To find s, we find the product of the partial derivatives along each path from to s and then add these products: s s s Similarl, we find t b using the paths from to t. Now we consider the general situation in which a dependent variable u is a function of n intermediate variables,..., n, each of which is, in turn, a function of m independent variables t,..., t m. Notice that there are n terms, one for each intermediate variable. The proof is similar to that of Case.

51 94 CHAPTER 4 PARTIAL DERIVATIVES 4 THE CHAIN RULE (GENERAL VERSION) Suppose that u is a differentiable function of the n variables,,..., n and each j is a differentiable function of the m variables,,...,. Then u is a function of,,..., and t t u t i t m for each i,,..., m. u u u t i t i n t t t m n t i w t u v u v u v u v FIGURE 3 V EXAPLE 4 Write out the Chain Rule for the case where w f,,, t and u, v, u, v, u, v, and t tu, v. SOLUTION We appl Theorem 4 with n 4 and m. 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 to u, then the partial derivative for that branch is u. With the aid of the tree diagram, we can now write the required epressions: w u w v w w w u v w w u v w w u t v w t t u t v r s t r s t r s u t V EXAPLE 5 If u 4 3, where rse t, rs e t, and r s sin t, find the value of us when r, s, t. SOLUTION With the help of the tree diagram in Figure 4, we have u s u s u s u s 4 3 re t 4 3 rse t 3 r sin t FIGURE 4 When r, s, and t, we have,, and, so EXAPLE 6 If ts, t f s t, t s and f is differentiable, show that t satisfies the equation t t t s s t SOLUTION Let s t and t s. Then ts, t f, and the Chain Rule gives Therefore t t s t s t t u s f f s t f f s t f f s s f f t t t s t st f f st st f f st

52 SECTION 4.5 THE CHAIN RULE 95 EXAPLE 7 If f, has continuous second-order partial derivatives and r s and rs, find (a) r and (b) r. SOLUTION (a) The Chain Rule gives r r (b) Appling the Product Rule to the epression in part (a), we get r rr s 5 r s r r r r s FIGURE 5 r s r s But, using the Chain Rule again (see Figure 5), we have r r r r r r Putting these epressions into Equation 5 and using the equalit of the mied secondorder derivatives, we obtain r r r s sr 4r 8rs 4s r s r s s IPLICIT DIFFERENTIATION The Chain Rule can be used to give a more complete description of the process of implicit differentiation that was introduced in Sections 3.5 and 4.3. We suppose that an equation of the form F, defines implicitl as a differentiable function of, that is, f, where F, f for all in the domain of f. If F is differentiable, we can appl Case of the Chain Rule to differentiate both sides of the equation F, with respect to. Since both and are functions of, we obtain F d d F d d But dd, so if F we solve for dd and obtain 6 d d F F F F

53 96 CHAPTER 4 PARTIAL DERIVATIVES To derive this equation we assumed that F, defines implicitl as a function of. 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 Fa, b, F a, b, and F and F are continuous on the disk, then the equation F, defines as a function of near the point a, b and the derivative of this function is given b Equation 6. EXAPLE 8 Find if SOLUTION The given equation can be written as F, so Equation 6 gives N The solution to Eample 8 should be compared to the one in Eample in Section 3.5. d d F 3 6 F 3 6 Now we suppose that is given implicitl as a function f, b an equation of the form F,,. This means that F,, f, for all, in the domain of f. If F and f are differentiable, then we can use the Chain Rule to differentiate the equation F,, as follows: F F F But and so this equation becomes F F If F, we solve for and obtain the first formula in Equations 7. The formula for is obtained in a similar manner. 7 F F F F 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 Fa, b, c, F a, b, c, and F, F, and F are continuous inside the sphere, then the equation F,, defines as a function of and near the point a, b, c and this function is differentiable, with partial derivatives given b (7).

54 SECTION 4.5 THE CHAIN RULE 97 EXAPLE 9 Find and if SOLUTION Let F,, Then, from Equations 7, we have N The solution to Eample 9 should be compared to the one in Eample 4 in Section 4.3. F 3 6 F 3 6 F 3 6 F EXERCISES 6 Use the Chain Rule to find ddt or dwdt.., sin t, e t. cos 4, 5t 4, t 3. s, ln t, cos t 4. tan, e t, e t 5. w e, t, t, t 6. w lns, sin t, cos t, 7 Use the Chain Rule to find s and t. 7. 3, s cos t, s sin t 8. arcsin, s t, st st 9. sin cos,, s t tan t 4. Let Ws, t Fus, t, vs, t, where F, u, and v are differentiable, and u, v, 3 u s, v s, 5 u t, 6 v t, 4 F u, 3 F v, 3 Find W s, and W t,. 5. Suppose f is a differentiable function of and, and tu, v f e u sin v, e u cos v. Use the table of values to calculate t u, and t v,. f t f f, , e, st,. e r cos, r st, ts ss t 6. Suppose f is a differentiable function of and, and tr, s f r s, s 4r. Use the table of values in Eercise 5 to calculate t r, and t s,.. tanuv, u s 3t, 3. If f,, where f is differentiable, and tt t3 t3 5 f, 7 6 find ddt when t 3. v 3s t ht h3 7 h3 4 f, Use a tree diagram to write out the Chain Rule for the given case. Assume all functions are differentiable. 7. u f,, where r, s, t, r, s, t 8. R f,,, t, where u, v, w, u, v, w, u, v, w, t tu, v, w 9. w f r, s, t, where r r,, s s,, t t,. t f u, v, w, where u up, q, r, s, v vp, q, r, s, w wp, q, r, s

55 98 CHAPTER 4 PARTIAL DERIVATIVES 6 Use the Chain Rule to find the indicated partial derivatives.. 3, uv w 3, u ve w ;,, when u, v, w u v w. u sr s, r cos t, s sin t; u u u,, when,, t t 3. R lnu v w, u, v, w ; R R, when 4. e, uv, u v, u v;, when u 3, v u v 5. u, pr cos, pr sin, p r; u u u,, when p, r 3, p r 6. Y w tan uv, u r s, v s t, w t r; Y Y Y,, when r, s, t r s t 7 3 Use Equation 6 to find dd. 7. s e 9. cos e 3. sin cos sin cos 3 34 Use Equations 7 to find and arctan cos ln 35. The temperature at a point, is T,, measured in degrees Celsius. A bug crawls so that its position after t seconds is given b s t, 3 t, where and are measured in centimeters. The temperature function satisfies T, 3 4 and T, 3 3. How fast is the temperature rising on the bug s path after 3 seconds? 36. Wheat production W in a given ear depends on the average temperature T and the annual rainfall R. Scientists estimate that the average temperature is rising at a rate of.5 Cear and rainfall is decreasing at a rate of. cmear. The also estimate that, at current production levels, WT and WR 8. (a) What is the significance of the signs of these partial derivatives? (b) Estimate the current rate of change of wheat production, dwdt. 37. The speed of sound traveling through ocean water with salinit 35 parts per thousand has been modeled b the equation 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 leisurel dive into the ocean water; the diver s depth and the 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 eperienced b the diver minutes into the dive. What are the units? D 5 5 C T.55T.9T 3.6D 3 4 t (min) T t (min) 38. The radius of a right circular cone is increasing at a rate of.8 ins while its height is decreasing at a rate of.5 ins. At what rate is the volume of the cone changing when the radius is in. and the height is 4 in.? 39. The length, width w, and height h of a bo change with time. At a certain instant the dimensions are m and w h m, and and w are increasing at a rate of ms while h is decreasing at a rate of 3 ms. 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 4. The voltage V in a simple electrical circuit is slowl decreasing as the batter wears out. The resistance R is slowl 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 4, I.8 A, dvdt. Vs, and drdt.3 s. 4. The pressure of mole of an ideal gas is increasing at a rate of.5 kpas and the temperature is increasing at a rate of.5 Ks. Use the equation in Eample to find the rate of change of the volume when the pressure is kpa and the temperature is 3 K. 4. Car A is traveling north on Highwa 6 and car B is traveling west on Highwa 83. Each car is approaching the intersection of these highwas. At a certain moment, car A is.3 km from the intersection and traveling at 9 kmh while car B is.4 km from the intersection and traveling at 8 kmh. How fast is the distance between the cars changing at that moment? 43. One side of a triangle is increasing at a rate of 3 cms and a second side is decreasing at a rate of cms. If the area of the 8

56 SECTION 4.5 THE CHAIN RULE 99 triangle remains constant, at what rate does the angle between the sides change when the first side is cm long, the second side is 3 cm, and the angle is 6? 44. If a sound with frequenc is produced b a source traveling along a line with speed v s and an observer is traveling with speed v o along the same line from the opposite direction toward the source, then the frequenc of the sound heard b the observer is where c is the speed of sound, about 33 ms. (This is the Doppler effect.) Suppose that, at a particular moment, ou are in a train traveling at 34 ms and accelerating at. ms. A train is approaching ou from the opposite direction on the other track at 4 ms, accelerating at.4 ms, and sounds its whistle, which has a frequenc of 46 H. At that instant, what is the perceived frequenc that ou hear and how fast is it changing? Assume that all the given functions are differentiable. 45. If f,, where r cos and r sin, (a) find r and and (b) show that r r 46. If u f,, where e s cos t and e s sin t, show that u u u e s s u t 47. If f, show that. 48. If f,, where s t and s t, show that s t Assume that all the given functions have continuous second-order partial derivatives. 49. Show that an function of the form f at t at is a solution of the wave equation t a [Hint: Let u at, v at.] f o f s c vo c v s f s 5. If u f,, where e s cos t and e s sin t, show that es u u u s u t 5. If f,, where r s and rs, find r s. (Compare with Eample 7.) 5. If f,, where r cos and r sin, find (a) r, (b), and (c) r. 53. If f,, where r cos and r sin, show that 54. Suppose f,, where ts, t and hs, t. (a) Show that t t t t t (b) Find a similar formula for s t. 55. A function f is called homogeneous of degree n if it satisfies the equation f t, t t n f, for all t, where n is a positive integer and f has continuous second-order partial derivatives. (a) Verif that f, 5 3 is homogeneous of degree 3. (b) Show that if f is homogeneous of degree n, then [Hint: Use the Chain Rule to differentiate f t, t with respect to t.] 56. If f is homogeneous of degree n, show that 57. If f is homogeneous of degree n, show that 58. Suppose that the equation F,, implicitl defines each of the three variables,, and as functions of the other two: f,, t,, h,. If F is differentiable and,, and are all nonero, show that F f f f nn f, F r r F t f f nf, f t, t t n f, t r r

57 9 CHAPTER 4 PARTIAL DERIVATIVES 997 USA Toda 6 FIGURE (, ) 5 San Francisco u Reno 7 cos 4.6 FIGURE A unit vector u=ka, bl=kcos, sin l (Distance in miles) 7 sin Las Vegas 8 Los Angeles DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR The weather map in Figure shows a contour map of the temperature function T, for the states of California and Nevada at 3: P on a da in October. The level curves, or isothermals, join locations with the same temperature. The partial derivative T at a location such as Reno is the rate of change of temperature with respect to distance if we travel east from Reno; T 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 tpe of derivative, called a directional derivative, that enables us to find the rate of change of a function of two or more variables in an direction. DIRECTIONAL DERIVATIVES Recall that if f,, then the partial derivatives and are defined as f, lim h l f, lim h l and represent the rates of change of in the - and -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 at, in the direction of an arbitrar unit vector u a, b. (See Figure.) To do this we consider the surface S with equation f, (the graph of f ) and we let f,. Then the point P,, 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 in the direction of u. f f f h, f, h f, h f, h T P(,, ) TEC Visual 4.6A animates Figure 3 b rotating u and therefore T. FIGURE 3

58 SECTION 4.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 9 If Q,, is another point on C and P, Q are the projections of P, Q on the -plane, then the vector PBQ is parallel to u and so PBQ hu ha, hb for some scalar h. Therefore ha, hb, so ha, hb, and h f ha, hb f, h h If we take the limit as h l, we obtain the rate of change of (with respect to distance) in the direction of u, which is called the directional derivative of f in the direction of u. DEFINITION The directional derivative of f at, in the direction of a unit vector u a, b is D u f, lim h l f ha, hb f, h if this limit eists. B comparing Definition with Equations (), we see that if u i,, then D i f f and if u j,, then D j f f. In other words, the partial derivatives of f with respect to and are just special cases of the directional derivative. EXAPLE Use the weather map in Figure to estimate the value of the directional derivative of the temperature function at Reno in the southeasterl direction. SOLUTION The unit vector directed toward the southeast is u i js, but we won t need to use this epression. We start b drawing a line through Reno toward the southeast. (See Figure 4.) 6 5 Reno San Francisco 7 6 Las Vegas 7 8 FIGURE Los Angeles (Distance in miles) 997 USA Toda We approimate the directional derivative D u T b the average rate of change of the temperature between the points where this line intersects the isothermals T 5 and

59 9 CHAPTER 4 PARTIAL DERIVATIVES T 6. The temperature at the point southeast of Reno is T 6 F and the temperature at the point northwest of Reno is T 5 F. The distance between these points looks to be about 75 miles. So the rate of change of the temperature in the southeasterl direction is D u T Fmi 75 When we compute the directional derivative of a function defined b a formula, we generall use the following theorem. 3 THEORE If f is a differentiable function of and, then f has a directional derivative in the direction of an unit vector u a, b and D u f, f, a f, b PROOF If we define a function t of the single variable h b th f ha, hb then, b the definition of a derivative, we have 4 th t t lim lim h l h h l D u f, f ha, hb f, h On the other hand, we can write th f,, where ha, hb, so the Chain Rule (Theorem 4.5.) gives th f d f dh If we now put h, then,, and d dh f, a f, b 5 t f, a f, b Comparing Equations 4 and 5, we see that D u f, f, a f, b If the unit vector u makes an angle with the positive -ais (as in Figure ), then we can write u cos, sin and the formula in Theorem 3 becomes 6 D u f, f, cos f, sin EXAPLE Find the directional derivative D u f, if f, and u is the unit vector given b angle. What is D u f,?

60 SECTION 4.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 93 N The directional derivative D u f, in Eample represents the rate of change of in the direction of u. This is the slope of the tangent line to the curve of intersection of the surface and the vertical plane through,, in the direction of u shown in Figure 5. SOLUTION Formula 6 gives D u f, f, cos 6 f, sin 3 3 s3 3 8 [3 s3 3 (8 3s3 )] 6 Therefore FIGURE 5 (,, ) π 6 u D u f, [3s3 3 (8 3s3 )] THE GRADIENT VECTOR 3 3s3 Notice from Theorem 3 that the directional derivative can be written as the dot product of two vectors: 7 D u f, f, a f, b f,, f, a, b f,, f, u The first vector in this dot product occurs not onl in computing directional derivatives but in man other contets 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 ). 8 DEFINITION If f is a function of two variables and, then the gradient of is the vector function f defined b f f, f,, f, f f i j EXAPLE 3 If f, sin e, then and f, f, f cos e, e f,, With this notation for the gradient vector, we can rewrite the epression (7) for the directional derivative as 9 D u f, f, u This epresses the directional derivative in the direction of u as the scalar projection of the gradient vector onto u.

61 94 CHAPTER 4 PARTIAL DERIVATIVES N The gradient vector f, in Eample 4 is shown in Figure 6 with initial point,. 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. FIGURE 6 ±f(, _) (, _) v V EXAPLE 4 Find the directional derivative of the function f, 3 4 at the point, in the direction of the vector v i 5j. SOLUTION We first compute the gradient vector at, : Note that v is not a unit vector, but since s9, the unit vector in the direction of v is Therefore, b Equation 9, we have D u f, f, u 4 i 8 j s9 i 5 s9 j f, 3 i 3 4j f, 4 i 8 j u v v v i 5 j s9 s s9 3 s9 FUNCTIONS OF THREE VARIABLES For functions of three variables we can define directional derivatives in a similar manner. Again D u f,, can be interpreted as the rate of change of the function in the direction of a unit vector u. DEFINITION The directional derivative of f at,, in the direction of a unit vector u a, b, c is D u f,, lim h l f ha, hb, hc f,, h if this limit eists. If we use vector notation, then we can write both definitions ( and ) of the directional derivative in the compact form D u f lim h l f hu f h where, if n and,, if n 3. This is reasonable because the vector equation of the line through in the direction of the vector u is given b tu (Equation.5.) and so f hu represents the value of f at a point on this line.

62 SECTION 4.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 95 If f,, is differentiable and u a, b, c, then the same method that was used to prove Theorem 3 can be used to show that D u f,, f,, a f,, b f,, c For a function f of three variables, the gradient vector, denoted b f or grad f, is f,, f,,, f,,, f,, or, for short, 3 f f, f, f f f f i j k Then, just as with functions of two variables, Formula for the directional derivative can be rewritten as 4 D u f,, f,, u V EXAPLE 5 If f,, sin, (a) find the gradient of f and (b) find the directional derivative of f at, 3, in the direction of v i j k. SOLUTION (a) The gradient of f is f,, f,,, f,,, f,, sin, cos, cos (b) At, 3, we have f, 3,,, 3. The unit vector in the direction of v i j k is u s6 i s6 j s6 k Therefore Equation 4 gives D u f, 3, f, 3, u 3k i j k s6 s6 s6 3 s6 3 AXIIZING 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 maimum rate of change? The answers are provided b the following theorem.

63 96 CHAPTER 4 PARTIAL DERIVATIVES TEC Visual 4.6B provides visual confirmation of Theorem 5. 5 THEORE Suppose f is a differentiable function of two or three variables. The maimum value of the directional derivative D is f u f and it occurs when u has the same direction as the gradient vector f. PROOF From Equation 9 or 4 we have where is the angle between f and u. The maimum value of cos is and this occurs when. Therefore the maimum value of D is f u f and it occurs when, that is, when u has the same direction as f. D u f f u f u cos f cos Q ±f(, ) P 3 FIGURE 7 N At, the function in Eample 6 increases fastest in the direction of the gradient vector f,,. Notice from Figure 7 that this vector appears to be perpendicular to the level curve through,. Figure 8 shows the graph of f and the gradient vector. EXAPLE 6 (a) If f, e, find the rate of change of f at the point P, in the direction from P to Q(, ). (b) In what direction does f have the maimum rate of change? What is this maimum rate of change? SOLUTION (a) We first compute the gradient vector: f, f, f e, e f,, The unit vector in the direction of PQ l.5, is u 3 5, 4 5, so the rate of change of f in the direction from P to Q is D u f, f, u, 3 5, 4 5 ( 3 5 ) ( 4 5) 5 5 (b) According to Theorem 5, f increases fastest in the direction of the gradient vector f,,. The maimum rate of change is f,, s5 FIGURE 8 3 EXAPLE 7 Suppose that the temperature at a point,, in space is given b T,, 8 3, where T is measured in degrees Celsius and,, in meters. In which direction does the temperature increase fastest at the point,,? What is the maimum rate of increase? SOLUTION The gradient of T is T T i T j T k 6 3 i 3 3 j 48 3 k 6 3 i j 3 k

64 SECTION 4.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 97 At the point,, the gradient vector is T,, 6 56i j 6 k 5 8i j 6 k B Theorem 5 the temperature increases fastest in the direction of the gradient vector T,, 5 8i j 6 k or, equivalentl, in the direction of i j 6 k or the unit vector i j 6 ks4. The maimum rate of increase is the length of the gradient vector: T,, 5 8 i j 6 k 5 8 s4 Therefore the maimum rate of increase of temperature is 8 s4 4Cm. 5 TANGENT PLANES TO LEVEL SURFACES Suppose S is a surface with equation F,, k, that is, it is a level surface of a function F of three variables, and let P,, be a point on S. Let C be an curve that lies on the surface S and passes through the point P. Recall from Section 3. that the curve C is described b a continuous vector function rt t, t, t. Let t be the parameter value corresponding to P; that is, rt,,. Since C lies on S, an point (t, t, t) must satisf the equation of S, that is, 6 F(t, t, t) k If,, and are differentiable functions of t and F is also differentiable, then we can use the Chain Rule to differentiate both sides of Equation 6 as follows: 7 F d dt F d dt F d dt ±F(,, ) tangent plane But, since F F, F, F and rt t, t, t, Equation 7 can be written in terms of a dot product as F rt In particular, when t t we have rt,,, so 8 F,, rt FIGURE 9 Equation 8 sas that the gradient vector at P, F,,, is perpendicular to the tangent vector rt to an curve C on S that passes through P. (See Figure 9.) If F,,, it is therefore natural to define the tangent plane to the level surface F,, k at P,, as the plane that passes through P and has normal vector F,,. Using the standard equation of a plane (Equation.5.7), we can write the equation of this tangent plane as 9 F,, F,, F,,

65 98 CHAPTER 4 PARTIAL DERIVATIVES 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 b the gradient vector F,, and so, b Equation.5.3, its smmetric equations are F,, F,, F,, In the special case in which the equation of a surface S is of the form f, (that is, S is the graph of a function f of two variables), we can rewrite the equation as and regard S as a level surface (with k ) of F. Then so Equation 9 becomes F,, f, F,, f, F,, f, F,, f, f, which is equivalent to Equation Thus our new, more general, definition of a tangent plane is consistent with the definition that was given for the special case of Section 4.4. V EXAPLE 8 Find the equations of the tangent plane and normal line at the point,, 3 to the ellipsoid SOLUTION The ellipsoid is the level surface (with k 3) of the function N Figure shows the ellipsoid, tangent plane, and normal line in Eample 8. Therefore we have F,, F,, F,, F,, 9 F,, 3 F,, 3 F,, FIGURE Then Equation 9 gives the equation of the tangent plane at,, 3 as 3 3 which simplifies to B Equation, smmetric equations of the normal line are 3 3

66 SECTION 4.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 99 SIGNIFICANCE OF THE GRADIENT VECTOR We now summarie the was in which the gradient vector is significant. We first consider a function f of three variables and a point P,, in its domain. On the one hand, we know from Theorem 5 that the gradient vector f,, gives the direction of fastest increase of f. On the other hand, we know that f,, is orthogonal to the level surface S of f through P. (Refer to Figure 9.) These two properties are quite compatible intuitivel because as we move awa 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 maimum increase. In like manner we consider a function f of two variables and a point P, in its domain. Again the gradient vector f, gives the direction of fastest increase of f. Also, b considerations similar to our discussion of tangent planes, it can be shown that f, is perpendicular to the level curve f, k that passes through P. Again this is intuitivel plausible because the values of f remain constant as we move along the curve. (See Figure.) ±f(, ) P(, ) level curve f(, )=k curve of steepest ascent 3 FIGURE FIGURE If we consider a topographical map of a hill and let f, represent the height above sea level at a point with coordinates,, then a curve of steepest ascent can be drawn as in Figure b making it perpendicular to all of the contour lines. This phenomenon can also be noticed in Figure in Section 4., where Lonesome Creek follows a curve of steepest descent. Computer algebra sstems have commands that plot sample gradient vectors. Each gradient vector f a, b is plotted starting at the point a, b. Figure 3 shows such a plot (called a gradient vector field ) for the function f, superimposed on a contour map of f. As epected, the gradient vectors point uphill and are perpendicular to the level curves. _9 _6 _ FIGURE 3

67 9 CHAPTER 4 PARTIAL DERIVATIVES 4.6 EXERCISES. Level curves for barometric pressure (in millibars) are shown for 6: A on November, 998. A deep low with pressure 97 mb is moving over northeast Iowa. The distance along the red line from K (Kearne, Nebraska) to S (Siou Cit, Iowa) is 3 km. Estimate the value of the directional derivative of the pressure function at Kearne in the direction of Siou Cit. What are the units of the directional derivative? K S 6 From eteorolog Toda, 8E b C. Donald Ahrens (7 Thomson Brooks/Cole). 4. The contour map shows the average maimum temperature for November 4 (in C). Estimate the value of the directional derivative of this temperature function at Dubbo, New South Wales, in the direction of Sdne. What are the units? 7 (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. 7. f, sin 3, P6, 4, 8. f,, P,, u 3 (i s5 j) 9. f,, e, P3,,,. f,, s, P, 3,, u (s3 i j) u 3, 3, 3 u 7, 3 7, Find the directional derivative of the function at the given point in the direction of the vector v.. f, s, 3, 4, v 4, 3. f, ln,,, v, 3. tp, q p 4 p q 3,,, v i 3j 4. tr, s tan rs,,, v 5i j 5. f,, e e e,,,, v 5,, 6. f,, s, 3,, 6, v,, 7. t,, 3 3,,,, v j k 8. Use the figure to estimate D u f,. 3 (Distance in kilometres) 3 Dubbo 4 u (, ) ±f(, ) Copright Commonwealth of Australia. Reproduced b permission. Sdne 9. Find the directional derivative of f, s at P, 8 in the direction of Q5, 4.. Find the directional derivative of f,, at P,, 3 in the direction of Q, 4, A table of values for the wind-chill inde W f T, v is given in Eercise 3 on page 888. Use the table to estimate the value of D u f, 3, where u i js. 4 6 Find the directional derivative of f at the given point in the direction indicated b the angle. 4. f, 3 4,,, 5. f, e,, 4, 6. f, sin,,, Find the maimum rate of change of f at the given point and the direction in which it occurs.. f,,, 4. f p, q qe p pe q,, 3. f, sin,, 4. f,,,,, 5. f,, s, 3, 6, 6. f,, tan 3, 5,,

68 SECTION 4.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 9 7. (a) Show that a differentiable function f decreases most rapidl at in the direction opposite to the gradient vector, that is, in the direction of f. (b) Use the result of part (a) to find the direction in which the function f, 4 3 decreases fastest at the point, Find the directions in which the directional derivative of f, e at the point, has the value. 9. Find all points at which the direction of fastest change of the function f, 4 is i j. 3. Near a buo, the depth of a lake at the point with coordinates, is.. 3, where,, and are measured in meters. A fisherman in a small boat starts at the point 8, 6 and moves toward the buo, which is located at,. Is the water under the boat getting deeper or shallower when he departs? Eplain. 3. The temperature T in a metal ball is inversel proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point,, is. (a) Find the rate of change of T at,, in the direction toward the point,, 3. (b) Show that at an point in the ball the direction of greatest increase in temperature is given b a vector that points toward the origin. 3. The temperature at a point,, is given b T,, e 3 9 where T is measured in C and,, in meters. (a) Find the rate of change of temperature at the point P,, in the direction toward the point 3, 3, 3. (b) In which direction does the temperature increase fastest at P? (c) Find the maimum rate of increase at P. 33. Suppose that over a certain region of space the electrical potential V is given b V,, 5 3. (a) Find the rate of change of the potential at P3, 4, 5 in the direction of the vector v i j k. (b) In which direction does V change most rapidl at P? (c) What is the maimum rate of change at P? 34. Suppose ou are climbing a hill whose shape is given b the equation.5., where,, and are measured in meters, and ou are standing at a point with coordinates 6, 4, 966. The positive -ais points east and the positive -ais points north. (a) If ou walk due south, will ou start to ascend or descend? At what rate? (b) If ou walk northwest, will ou 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 horiontal does the path in that direction begin? 35. Let f be a function of two variables that has continuous partial derivatives and consider the points A, 3, B3, 3, C, 7, and D6, 5. The directional derivative of at in the direction of the vector is 3 and the directional derivative at in the direction of AC l AB l f A A is 6. Find the directional derivative of f at A in the direction of the vector AD l. 36. For the given contour map draw the curves of steepest ascent starting at P and at Q. 37. Show that the operation of taking the gradient of a function has the given propert. Assume that u and v are differentiable functions of and and that a, b are constants. (a) au bv a u b v (b) uv u v v u (c) u v 38. Sketch the gradient vector f 4, 6 for the function f whose level curves are shown. Eplain how ou chose the direction and length of this vector Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point. (d) 39. 3, 4., 4, 7, 3 4., 4. 4 arctan, 43. e cos, v u u v,, 44. ln,,, 6 4 P v _5 _3 _ 6 5 Q (4, 6),,,, 4 u n nu n u , 3, 5

69 9 CHAPTER 4 PARTIAL DERIVATIVES ; Use a computer to graph the surface, the tangent plane, and the normal line on the same screen. Choose the domain carefull so that ou avoid etraneous vertical planes. Choose the viewpoint so that ou get a good view of all three objects , 46. 6,,, If f,, find the gradient vector f 3, and use it to find the tangent line to the level curve f, 6 at the point 3,. Sketch the level curve, the tangent line, and the gradient vector. 48. If t, 4, find the gradient vector t, and use it to find the tangent line to the level curve t, at the point,. Sketch the level curve, the tangent line, and the gradient vector. 5. Find the equation of the tangent plane to the hperboloid a b c at,, and epress it in a form similar to the one in Eercise Show that the equation of the tangent plane to the elliptic paraboloid c a b at the point,, can be written as a,, 49. Show that the equation of the tangent plane to the ellipsoid a b c at the point,, can be written as a b c b 5. At what point on the paraboloid is the tangent plane parallel to the plane 3? 53. Are there an points on the hperboloid where the tangent plane is parallel to the plane? 54. Show that the ellipsoid 3 9 and the sphere are tangent to each other at the point,,. (This means that the have a common tangent plane at the point.) 55. Show that ever plane that is tangent to the cone passes through the origin. c 56. Show that ever normal line to the sphere r passes through the center of the sphere. 57. Show that the sum of the -, -, and -intercepts of an tangent plane to the surface s s s sc is a constant. 58. Show that the pramids cut off from the first octant b an tangent planes to the surface at points in the first octant must all have the same volume. 59. Find parametric equations for the tangent line to the curve of intersection of the paraboloid and the ellipsoid 4 9 at the point,,. 6. (a) The plane 3 intersects the clinder 5 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point,,. ; (b) Graph the clinder, the plane, and the tangent line on the same screen. 6. (a) Two surfaces are called orthogonal at a point of intersection if their normal lines are perpendicular at that point. Show that surfaces with equations F,, and G,, are orthogonal at a point P where F and G if and onl if F G F G F G at P (b) Use part (a) to show that the surfaces and r are orthogonal at ever point of intersection. Can ou see wh this is true without using calculus? 6. (a) Show that the function f, s 3 is continuous and the partial derivatives f and f eist at the origin but the directional derivatives in all other directions do not eist. ; (b) Graph f near the origin and comment on how the graph confirms part (a). 63. Suppose that the directional derivatives of f, are known at a given point in two nonparallel directions given b unit vectors u and v. Is it possible to find f at this point? If so, how would ou do it? 64. Show that if f, is differentiable at,, then lim l f f f [Hint: Use Definition directl.] 4.7 AXIU AND INIU VALUES As we saw in Chapter 4, one of the main uses of ordinar derivatives is in finding maimum and minimum values. In this section we see how to use partial derivatives to locate maima and minima of functions of two variables. In particular, in Eample 6 we will see how to maimie the volume of a bo without a lid if we have a fied amount of cardboard to work with.

70 SECTION 4.7 AXIU AND INIU VALUES 93 absolute minimum FIGURE local maimum local minimum absolute maimum Look at the hills and valles in the graph of f shown in Figure. There are two points a, b where f has a local maimum, that is, where f a, b is larger than nearb values of f,. The larger of these two values is the absolute maimum. Likewise, f has two local minima, where f a, b is smaller than nearb values. The smaller of these two values is the absolute minimum. DEFINITION A function of two variables has a local maimum at a, b if f, f a, b when, is near a, b. [This means that f, f a, b for all points, in some disk with center a, b.] The number f a, b is called a local maimum value. If f, f a, b when, is near a, b, then f has a local minimum at a, b and f a, b is a local minimum value. If the inequalities in Definition hold for all points, in the domain of f, then f has an absolute maimum (or absolute minimum) at a, b. N Notice that the conclusion of Theorem can be stated in the notation of gradient vectors as f a, b. THEORE If f has a local maimum or minimum at a, b and the first-order partial derivatives of f eist there, then f a, b and f a, b. PROOF Let t f, b. If f has a local maimum (or minimum) at a, b, then t has a local maimum (or minimum) at a, so ta b Fermat s Theorem (see Theorem 4..4). But ta f a, b (see Equation 4.3.) and so f a, b. Similarl, b appling Fermat s Theorem to the function G f a,, we obtain f a, b. If we put f a, b and f a, b in the equation of a tangent plane (Equation 4.4.), we get. Thus the geometric interpretation of Theorem is that if the graph of f has a tangent plane at a local maimum or minimum, then the tangent plane must be horiontal. A point a, b is called a critical point (or stationar point) of f if f a, b and f a, b, or if one of these partial derivatives does not eist. Theorem sas that if f has a local maimum 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 maima or minima. At a critical point, a function could have a local maimum or a local minimum or neither. EXAPLE Let f, 6 4. Then f, f, 6 (, 3, 4) FIGURE = These partial derivatives are equal to when and 3, so the onl critical point is, 3. B completing the square, we find that f, 4 3 Since and 3, we have f, 4 for all values of and. Therefore f, 3 4 is a local minimum, and in fact it is the absolute minimum of f. This can be confirmed geometricall from the graph of f, which is the elliptic paraboloid with verte, 3, 4 shown in Figure.

71 94 CHAPTER 4 PARTIAL DERIVATIVES EXAPLE Find the etreme values of f,. SOLUTION Since f and f, the onl critical point is,. Notice that for points on the -ais we have, so f, (if ). However, for points on the -ais we have, so f, (if ). Thus ever disk with center, contains points where f takes positive values as well as points where f takes negative values. Therefore f, can t be an etreme value for f, so f has no etreme value. FIGURE 3 = - Eample illustrates the fact that a function need not have a maimum or minimum value at a critical point. Figure 3 shows how this is possible. The graph of f is the hperbolic paraboloid, which has a horiontal tangent plane ( ) at the origin. You can see that f, is a maimum in the direction of the -ais but a minimum in the direction of the -ais. Near the origin the graph has the shape of a saddle and so, is called a saddle point of f. We need to be able to determine whether or not a function has an etreme 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. 3 SECOND DERIVATIVES TEST Suppose the second partial derivatives of f are continuous on a disk with center a, b, and suppose that f a, b and f a, b [that is, a, b is a critical point of f ]. Let D Da, b f a, b f a, b f a, b (a) If D and f a, b, then f a, b is a local minimum. (b) If D and f a, b, then f a, b is a local maimum. (c) If D, then f a, b is not a local maimum or minimum. NOTE 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. NOTE If D, the test gives no information: f could have a local maimum 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: D f f f f f f f V EXAPLE 3 Find the local maimum and minimum values and saddle points of f, SOLUTION We first locate the critical points: f Setting these partial derivatives equal to, we obtain the equations 3 To solve these equations we substitute one. This gives f and 3 3 from the first equation into the second

72 SECTION 4.7 AXIU AND INIU VALUES 95 so there are three real roots:,,. The three critical points are,,,, and,. Net we calculate the second partial derivatives and D, : f f f 4 FIGURE 4 =$+$-4+ D, f f f 44 6 Since D, 6, it follows from case (c) of the Second Derivatives Test that the origin is a saddle point; that is, f has no local maimum or minimum at,. Since D, 8 and f,, we see from case (a) of the test that f, is a local minimum. Similarl, we have D, 8 and f,, so f, is also a local minimum. The graph of f is shown in Figure 4. N A contour map of the function f in Eample 3 is shown in Figure 5. The level curves near, and, are oval in shape and indicate that as we move awa from, or, in an direction the values of f are increasing. The level curves near,, on the other hand, resemble hperbolas. The reveal that as we move awa from the origin (where the value of f is ), 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 Eample 3. _ FIGURE 5 TEC In odule 4.7 ou can use contour maps to estimate the locations of critical points. EXAPLE 4 Find and classif the critical points of the function f, Also find the highest point on the graph of f. SOLUTION The first-order partial derivatives are f 4 3 f So to find the critical points we need to solve the equations From Equation 4 we see that either or 5 In the first case ( ), Equation 5 becomes 4, so and we have the critical point,.

73 96 CHAPTER 4 PARTIAL DERIVATIVES In the second case 5, we get and, putting this in Equation 5, we have So we have to solve the cubic equation Using a graphing calculator or computer to graph the function _3.7 FIGURE 6 as in Figure 6, we see that Equation 7 has three real roots. B ooming in, we can find the roots to four decimal places:.545 t (Alternativel, we could have used Newton s method or a rootfinder to locate these roots.) From Equation 6, the corresponding -values are given b s5.5 If.545, then has no corresponding real values. If.6468, then If.8984, then.644. So we have a total of five critical points, which are analed in the following chart. All quantities are rounded to two decimal places. Critical point Value of f D Conclusion f,.64,.9.86, local maimum local maimum 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 epression for f, : The dominant terms are 4 4 when and are large.] Comparing the values of f at its local maimum points, we see that the absolute maimum value of f is f.64, In other words, the highest points on the graph of f are.64,.9, 8.5. TEC Visual 4.7 shows several families of surfaces.the surface in Figures 7 and 8 is a member of one of these families. FIGURE 7 FIGURE 8

74 SECTION 4.7 AXIU AND INIU VALUES 97 N The five critical points of the function f in Eample 4 are shown in red in the contour map of f in Figure 9. _3 _.48 _3 7 3 _ _ FIGURE 9 V EXAPLE 5 Find the shortest distance from the point,, to the plane 4. SOLUTION The distance from an point,, to the point,, is d s but if,, lies on the plane 4, then 4 and so we have d s 6. We can minimie d b minimiing the simpler epression d f, 6 B solving the equations f f ( 6, 5 3) we find that the onl critical point is. Since f 4, f 4, and f, we have D, f f f 4 and f, so b the Second Derivatives Test f has a local minimum at ( 6, 5 3). Intuitivel, we can see that this local minimum is actuall an absolute minimum because there must be a point on the given plane that is closest to,,. If and 5 6 3, then N Eample 5 could also be solved using vectors. Compare with the methods of Section.5. d s 6 s( 5 6) ( 5 3) ( 5 6) 5 6 s6 The shortest distance from,, to the plane 4 is 6s6. 5 V EXAPLE 6 A rectangular bo without a lid is to be made from m of cardboard. Find the maimum volume of such a bo. SOLUTION Let the length, width, and height of the bo (in meters) be,, and, as shown in Figure. Then the volume of the bo is V We can epress V as a function of just two variables and b using the fact that the area of the four sides and the bottom of the bo is FIGURE

75 98 CHAPTER 4 PARTIAL DERIVATIVES Solving this equation for, we get, so the epression for V becomes V We compute the partial derivatives: V V If V is a maimum, then V V, but or gives V, so we must solve the equations These impl that and so. (Note that and must both be positive in this problem.) If we put in either equation we get 3, which gives,, and. We could use the Second Derivatives Test to show that this gives a local maimum of V, or we could simpl argue from the phsical nature of this problem that there must be an absolute maimum volume, which has to occur at a critical point of V, so it must occur when,,. Then V 4, so the maimum volume of 3 the bo is 4 m. ABSOLUTE AXIU AND INIU VALUES (a) Closed sets For a function f of one variable the Etreme Value Theorem sas that if f is continuous on a closed interval a, b, then f has an absolute minimum value and an absolute maimum value. According to the Closed Interval ethod in Section 4., we found these b evaluating f not onl 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 is one that contains all its boundar points. [A boundar point of D is a point a, b such that ever disk with center a, b contains points in D and also points not in D.] For instance, the disk D, FIGURE (b) Sets that are not closed which consists of all points on and inside the circle, is a closed set because it contains all of its boundar points (which are the points on the circle ). But if even one point on the boundar curve were omitted, the set would not be closed. (See Figure.) A bounded set in is one that is contained within some disk. In other words, it is finite in etent. Then, in terms of closed and bounded sets, we can state the following counterpart of the Etreme Value Theorem in two dimensions. 8 EXTREE VALUE THEORE FOR FUNCTIONS OF TWO VARIABLES If f is continuous on a closed, bounded set D in, then f attains an absolute maimum value f, and an absolute minimum value f, at some points, and, in D.

76 SECTION 4.7 AXIU AND INIU VALUES 99 To find the etreme values guaranteed b Theorem 8, we note that, b Theorem, if f has an etreme value at,, then, is either a critical point of f or a boundar point of D. Thus we have the following etension of the Closed Interval ethod. 9 To find the absolute maimum and minimum values of a continuous function f on a closed, bounded set D:. Find the values of f at the critical points of f in D.. Find the etreme values of f on the boundar of D. 3. The largest of the values from steps and is the absolute maimum value; the smallest of these values is the absolute minimum value. EXAPLE 7 Find the absolute maimum and minimum values of the function f, on the rectangle D, 3,. SOLUTION Since f is a polnomial, it is continuous on the closed, bounded rectangle D, so Theorem 8 tells us there is both an absolute maimum and an absolute minimum. According to step in (9), we first find the critical points. These occur when f f (, ) L (, ) (3, ) L L (, ) L (3, ) FIGURE 9 so the onl critical point is,, and the value of f there is f,. In step we look at the values of f on the boundar of D, which consists of the four line segments,,, shown in Figure. On we have and L L L 3 This is an increasing function of, so its minimum value is f, and its maimum value is f 3, 9. On we have 3 and This is a decreasing function of, so its maimum value is f 3, 9 and its minimum value is f 3,. On we have and L 3 f, B the methods of Chapter 4, or simpl b observing that f,, we see that the minimum value of this function is f, and the maimum value is f, 4. Finall, on we have and L 4 L 4 f, 3 L L f 3, 9 4 L 3 D FIGURE 3 f(, )= -+ L f, with maimum value f, 4 and minimum value f,. Thus, on the boundar, the minimum value of f is and the maimum is 9. In step 3 we compare these values with the value f, at the critical point and conclude that the absolute maimum value of f on D is f 3, 9 and the absolute minimum value is f, f,. Figure 3 shows the graph of f.

77 93 CHAPTER 4 PARTIAL DERIVATIVES We close this section b giving a proof of the first part of the Second Derivatives Test. Part (b) has a similar proof. PROOF OF THEORE 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 b Theorem 4.6.3: D u f f h f k Appling this theorem a second time, we have D u f D u D u f D u f h D u f k f h f kh f h f kk f h f hk f k (b Clairaut s Theorem) If we complete the square in this epression, we obtain D u f f h f f k k f f f f We are given that f a, b and Da, b. But f and D f f f are continuous functions, so there is a disk B with center a, b and radius such that f, and D, whenever, is in B. Therefore, b looking at Equation, we see that D u f, whenever, is in B. This means that if C is the curve obtained b intersecting the graph of f with the vertical plane through Pa, b, f a, b in the direction of u, then C is concave upward on an interval of length. This is true in the direction of ever vector u, so if we restrict, to lie in B, the graph of f lies above its horiontal tangent plane at P. Thus f, f a, b whenever, is in B. This shows that f a, b is a local minimum. 4.7 EXERCISES. Suppose, is a critical point of a function f with continuous second derivatives. In each case, what can ou sa about f? (a) f, 4, f,, f, (b) f, 4, f, 3, f,. Suppose (, ) is a critical point of a function t with continuous second derivatives. In each case, what can ou sa about t? (a) t,, (b) t,, (c) t, 4, t, 6, t,, t, 6, t, t, 8 t, 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 maimum or minimum at each critical point. Eplain our reasoning. Then use the Second Derivatives Test to confirm our predictions. 3. f, _ _

78 SECTION 4.7 AXIU AND INIU VALUES f, _.9 _.7 _.5.5 _ f, e f, sin sin sin,, 4. f, sin sin cos, 4, 4 ; 5 8 Use a graphing device as in Eample 4 (or Newton s method or a rootfinder) to find the critical points of f correct to three decimal places. Then classif the critical points and find the highest or lowest points on the graph. 5. f, Find the local maimum and minimum values and saddle point(s) of the function. If ou have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function f, f, f, e cos 5. f, f, f, f, e 4 9. f,. f, 3 5. f, f, 3. f, e cos 4. f, cos 5. f, e 6. f, e 7. f, cos, 7 8. f, sin sin,, 9. Show that f, 4 4 has an infinite number of critical points and that D at each one. Then show that f has a local (and absolute) minimum at each critical point.. Show that f has maimum values at (, s ), e and minimum values at (, s ). Show also that f has infinitel man other critical points and D at each of them. Which of them give rise to maimum values? inimum values? Saddle points? ; 4 Use a graph and/or level curves to estimate the local maimum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisel.. f, 9 36 Find the absolute maimum and minimum values of f on the set D. 9. f, 4 5, D is the closed triangular region with vertices,,,, and, 3 3. f, 3, D is the closed triangular region with vertices,, 5,, and, 4 3. f, 4, D,, 3. f, 4 6, D, 4, f, 4, D, 4 4 3, 34. f,, 35. f, 3 4, D,,, 3 D, 36. f, 3 3 3, D is the quadrilateral whose vertices are, 3,, 3,,, and,. ; 37. For functions of one variable it is impossible for a continuous function to have two local maima and no local minimum. But for functions of two variables such functions eist. Show that the function f, has onl two critical points, but has local maima at both of them. Then use a computer to produce a graph with a carefull chosen domain and viewpoint to see how this is possible. ; 38. If a function of one variable is continuous on an interval and has onl one critical number, then a local maimum has to be

79 93 CHAPTER 4 PARTIAL DERIVATIVES an absolute maimum. But this is not true for functions of two variables. Show that the function has eactl one critical point, and that f has a local maimum there that is not an absolute maimum. Then use a computer to produce a graph with a carefull chosen domain and viewpoint to see how this is possible. 39. Find the shortest distance from the point,, to the plane. 4. Find the point on the plane 4 that is closest to the point,, Find the points on the surface 9 that are closest to the origin. 43. f, 3e 3 e 3 4. Find the points on the cone that are closest to the point 4,,. Find three positive numbers whose sum is and whose product is a maimum. 44. Find three positive numbers whose sum is and the sum of whose squares is as small as possible. 45. Find the maimum volume of a rectangular bo that is inscribed in a sphere of radius r. 46. Find the dimensions of the bo with volume cm 3 that has minimal surface area. 47. Find the volume of the largest rectangular bo in the first octant with three faces in the coordinate planes and one verte in the plane Find the dimensions of the rectangular bo with largest volume if the total surface area is given as 64 cm. (b) Find the dimensions that minimie heat loss. (Check both the critical points and the points on the boundar of the domain.) (c) Could ou design a building with even less heat loss if the restrictions on the lengths of the walls were removed? 53. If the length of the diagonal of a rectangular bo must be L, what is the largest possible volume? 54. Three alleles (alternative versions of a gene) A, B, and O determine the four blood tpes A (AA or AO), B (BB or BO), O (OO), and AB. The Hard-Weinberg Law states that the proportion of individuals in a population who carr two different alleles is where p, q, and r are the proportions of A, B, and O in the population. Use the fact that p q r to show that P is at most Suppose that a scientist has reason to believe that two quantities and are related linearl, that is, m b, at least approimatel, for some values of m and b. The scientist performs an eperiment and collects data in the form of points,,,,..., n, n, and then plots these points. The points don t lie eactl on a straight line, so the scientist wants to find constants m and b so that the line m b fits the points as well as possible. (See the figure.) (, ) P pq pr rq d i m i +b ( i, i ) 49. Find the dimensions of a rectangular bo of maimum volume such that the sum of the lengths of its edges is a constant c. 5. The base of an aquarium with given volume V is made of slate 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 minimie the cost of the materials. 5. A cardboard bo without a lid is to have a volume of 3, cm 3. Find the dimensions that minimie the amount of cardboard used. 5. A rectangular building is being designed to minimie heat loss. The east and west walls lose heat at a rate of unitsm per da, the north and south walls at a rate of 8 unitsm per da, the floor at a rate of unitm per da, and the roof at a rate of 5 unitsm per da. Each wall must be at least 3 m long, the height must be at least 4 m, and the volume must be eactl 4 m 3. (a) Find and sketch the domain of the heat loss as a function of the lengths of the sides. Let d i i m i b be the vertical deviation of the point i, i from the line. The method of least squares determines and so as to minimie n m b i d i, the sum of the squares of these deviations. Show that, according to this method, the line of best fit is obtained when m n i m n i i b n i bn n Thus the line is found b solving these two equations in the two unknowns m and b. (See Section. for a further discussion and applications of the method of least squares.) 56. Find an equation of the plane that passes through the point,, 3 and cuts off the smallest volume in the first octant. i i n i i i i i

80 DISCOVERY PROJECT QUADRATIC APPROXIATIONS AND CRITICAL POINTS 933 APPLIED PROJECT DESIGNING A DUPSTER For this project we locate a trash dumpster in order to stud its shape and construction. We then attempt to determine the dimensions of a container of similar design that minimie construction cost.. First locate a trash dumpster in our area. Carefull stud and describe all details of its construction, and determine its volume. Include a sketch of the container.. While maintaining the general shape and method of construction, determine the dimensions such a container of the same volume should have in order to minimie the cost of construction. Use the following assumptions in our analsis: N The sides, back, and front are to be made from -gauge (.46 inch thick) steel sheets, which cost $.7 per square foot (including an required cuts or bends). N The base is to be made from a -gauge (.345 inch thick) steel sheet, which costs $.9 per square foot. N Lids cost approimatel $5. each, regardless of dimensions. N Welding costs approimatel $.8 per foot for material and labor combined. Give justification of an further assumptions or simplifications made of the details of construction. 3. Describe how an of our assumptions or simplifications ma affect the final result. 4. If ou were hired as a consultant on this investigation, what would our conclusions be? Would ou recommend altering the design of the dumpster? If so, describe the savings that would result. DISCOVERY PROJECT QUADRATIC APPROXIATIONS AND CRITICAL POINTS The Talor polnomial approimation to functions of one variable that we discussed in Chapter can be etended to functions of two or more variables. Here we investigate quadratic approimations to functions of two variables and use them to give insight into the Second Derivatives Test for classifing critical points. In Section 4.4 we discussed the lineariation of a function f of two variables at a point a, b: L, f a, b f a, b a f a, b b Recall that the graph of L is the tangent plane to the surface f, at a, b, f a, b and the corresponding linear approimation is f, L,. The lineariation L is also called the first-degree Talor polnomial of f at a, b.. If f has continuous second-order partial derivatives at a, b, then the second-degree Talor polnomial of f at a, b is Q, f a, b f a, b a f a, b b f a, b a f a, b a b f a, b b and the approimation f, Q, is called the quadratic approimation to f at a, b. Verif that Q has the same first- and second-order partial derivatives as f at a, b.

81 934 CHAPTER 4 PARTIAL DERIVATIVES. (a) Find the first- and second-degree Talor polnomials and of f, e L Q at (, ). ; (b) Graph f, L, and Q. Comment on how well L and Q approimate f. 3. (a) Find the first- and second-degree Talor polnomials L and Q for f, e at (, ). (b) Compare the values of L, Q, and f at (.9,.). ; (c) Graph f, L, and Q. Comment on how well L and Q approimate f. 4. In this problem we anale the behavior of the polnomial f, a b c (without using the Second Derivatives Test) b identifing the graph as a paraboloid. (a) B completing the square, show that if a, then f, a b c a b a 4ac b 4a (b) Let D 4ac b. Show that if D and a, then f has a local minimum at (, ). (c) Show that if D and a, then f has a local maimum at (, ). (d) Show that if D, then (, ) is a saddle point. 5. (a) Suppose f is an function with continuous second-order partial derivatives such that f, and (, ) is a critical point of f. Write an epression for the second-degree Talor polnomial, Q, of f at (, ). (b) What can ou conclude about Q from Problem 4? (c) In view of the quadratic approimation f, Q,, what does part (b) suggest about f? g(, )=k FIGURE 4.8 f(, )= f(, )= f(, )=9 f(, )=8 f(, )=7 TEC Visual 4.8 animates Figure for both level curves and level surfaces. LAGRANGE ULTIPLIERS In Eample 6 in Section 4.7 we maimied a volume function V subject to the constraint, which epressed the side condition that the surface area was m. In this section we present Lagrange s method for maimiing or minimiing a general function f,, subject to a constraint (or side condition) of the form t,, k. It s easier to eplain the geometric basis of Lagrange s method for functions of two variables. So we start b tring to find the etreme values of f, subject to a constraint of the form t, k. In other words, we seek the etreme values of f, when the point, is restricted to lie on the level curve t, k. Figure shows this curve together with several level curves of f. These have the equations f, c, where c 7, 8, 9,,. To maimie f, subject to t, k is to find the largest value of c such that the level curve f, c intersects t, k. It appears from Figure that this happens when these curves just touch each other, that is, when the have a common tangent line. (Otherwise, the value of c could be increased further.) This means that the normal lines at the point, where the touch are identical. So the gradient vectors are parallel; that is, f, for some scalar. This kind of argument also applies to the problem of finding the etreme values of f,, subject to the constraint t,, k. Thus the point,, is restricted to lie on the level surface S with equation t,, k. Instead of the level curves in Figure, we consider the level surfaces f,, c and argue that if the maimum value of f is f,, c, then the level surface f,, c is tangent to the level surface t,, k and so the corresponding gradient vectors are parallel. t,

82 SECTION 4.8 LAGRANGE ULTIPLIERS 935 This intuitive argument can be made precise as follows. Suppose that a function f has an etreme value at a point P,, on the surface S and let C be a curve with vector equation rt t, t, t that lies on S and passes through P. If t is the parameter value corresponding to the point P, then rt,,. The composite function ht f t, t, t represents the values that f takes on the curve C. Since f has an etreme value at,,, it follows that h has an etreme value at t, so ht. But if f is differentiable, we can use the Chain Rule to write ht f,, t f,, t f,, t f,, rt This shows that the gradient vector f,, is orthogonal to the tangent vector rt to ever such curve C. But we alread know from Section 4.6 that the gradient vector of t, t,,, is also orthogonal to rt for ever such curve. (See Equation ) This means that the gradient vectors f,, and t,, must be parallel. Therefore, if t,,, there is a number such that N Lagrange multipliers are named after the French-Italian mathematician Joseph-Louis Lagrange (736 83). See page 83 for a biographical sketch of Lagrange. f,, t,, The number in Equation is called a Lagrange multiplier. The procedure based on Equation is as follows. N In deriving Lagrange s method we assumed that t. In each of our eamples ou can check that t at all points where t,, k. See Eercise for what can go wrong if t. ETHOD OF LAGRANGE ULTIPLIERS To find the maimum and minimum values of f,, subject to the constraint t,, k [assuming that these etreme values eist and t on the surface t,, k]: (a) Find all values of,,, and such that f,, and t,, k t,, (b) Evaluate f at all the points,, that result from step (a). The largest of these values is the maimum value of f ; the smallest is the minimum value of f. in terms of its components, then the equa- If we write the vector equation f tions in step (a) become t f t f t f t t,, k This is a sstem of four equations in the four unknowns,,, and, but it is not necessar to find eplicit values for. For functions of two variables the method of Lagrange multipliers is similar to the method just described. To find the etreme values of f, subject to the constraint t, k, we look for values of,, and such that f, t, and t, k

83 936 CHAPTER 4 PARTIAL DERIVATIVES This amounts to solving three equations in three unknowns: f t f t t, k Our first illustration of Lagrange s method is to reconsider the problem given in Eample 6 in Section 4.7. V EXAPLE A rectangular bo without a lid is to be made from m of cardboard. Find the maimum volume of such a bo. SOLUTION As in Eample 6 in Section 4.7, we let,, and be the length, width, and height, respectivel, of the bo in meters. Then we wish to maimie subject to the constraint V t,, Using the method of Lagrange multipliers, we look for values of,,, and V and t,,. This gives the equations t such that V t V t V t which become There are no general rules for solving sstems of equations. Sometimes some ingenuit is required. In the present eample ou might notice that if we multipl () b, (3) b, and (4) b, then the left sides of these equations will be identical. Doing this, we have N Another method for solving the sstem of equations ( 5) is to solve each of Equations, 3, and 4 for and then to equate the resulting epressions We observe that because would impl from (), (3), and (4) and this would contradict (5). Therefore, from (6) and (7), we have which gives. But (since would give V ), so. From (7) and (8) we have which gives and so (since ). If we now put in (5), we get Since,, and are all positive, we therefore have and so and. This agrees with our answer in Section 4.7.

84 SECTION 4.8 LAGRANGE ULTIPLIERS 937 N In geometric terms, Eample asks for the highest and lowest points on the curve C in Figure that lies on the paraboloid and directl above the constraint circle. = + V EXAPLE Find the etreme values of the function f, on the circle. SOLUTION We are asked for the etreme values of f subject to the constraint t,. Using Lagrange multipliers, we solve the equations f and t,, which can be written as f t f t t, or as t C 9 4 FIGURE + = From (9) we have or. If, then () gives. If, then from (), so then () gives. Therefore f has possible etreme values at the points,,,,,, and,. Evaluating f at these four points, we find that N The geometr behind the use of Lagrange multipliers in Eample is shown in Figure 3. The etreme values of f, correspond to the level curves that touch the circle. + = f, f, f, f, Therefore the maimum value of f on the circle is f, and the minimum value is f,. Checking with Figure, we see that these values look reasonable. EXAPLE 3 Find the etreme values of f, on the disk. SOLUTION According to the procedure in (4.7.9), we compare the values of f at the critical points with values at the points on the boundar. Since f and f 4, the onl critical point is,. We compare the value of f at that point with the etreme values on the boundar from Eample : FIGURE 3 + = f, f, f, Therefore the maimum value of f on the disk is f, and the minimum value is f,. EXAPLE 4 Find the points on the sphere 4 that are closest to and farthest from the point 3,,. SOLUTION The distance from a point,, to the point 3,, is d s 3 but the algebra is simpler if we instead maimie and minimie the square of the distance: d f,, 3 The constraint is that the point,, lies on the sphere, that is, t,, 4

85 938 CHAPTER 4 PARTIAL DERIVATIVES t According to the method of Lagrange multipliers, we solve f, t 4. This gives The simplest wa to solve these equations is to solve for,, and in terms of from (), (3), and (4), and then substitute these values into (5). From () we have 3 or 3 or 3 [Note that because is impossible from ().] Similarl, (3) and (4) give N Figure 4 shows the sphere and the nearest point P in Eample 4. Can ou see how to find the coordinates of P without using calculus? Therefore, from (5), we have 3 s which gives 4,, so 4 s P (3,, _) FIGURE 4 These values of then give the corresponding points,, : and 6 s, s, s 6 s, s s, It s eas to see that f has a smaller value at the first of these points, so the closest point is (6s, s, s) and the farthest is (6s, s, s). TWO CONSTRAINTS h=c ±g C P g=k FIGURE 5 ±f ±h Suppose now that we want to find the maimum and minimum values of a function f,, subject to two constraints (side conditions) of the form t,, k and h,, c. Geometricall, this means that we are looking for the etreme values of f when,, is restricted to lie on the curve of intersection C of the level surfaces t,, k and h,, c. (See Figure 5.) Suppose f has such an etreme value at a point P,,. We know from the beginning of this section that f is orthogonal to C at P. But we also know that t is orthogonal to t,, k and h is orthogonal to h,, c, so t and h are both orthogonal to C. This means that the gradient vector f,, is in the plane determined b t,, and h,,. (We assume that these gradient vectors are not ero and not parallel.) So there are numbers and

86 SECTION 4.8 LAGRANGE ULTIPLIERS 939 (called Lagrange multipliers) such that 6 f,, t,, h,, In this case Lagrange s method is to look for etreme values b solving five equations in the five unknowns,,,, and. These equations are obtained b writing Equation 6 in terms of its components and using the constraint equations: f t h f t h f t h t,, k h,, c N The clinder intersects the plane in an ellipse (Figure 6). Eample 5 asks for the maimum value of f when,, is restricted to lie on the ellipse. 4 3 _ V EXAPLE 5 Find the maimum value of the function f,, 3 on the curve of intersection of the plane and the clinder. SOLUTION We maimie the function f,, 3 subject to the constraints t,, and h,,. The Lagrange condition is f, so we solve the equations 7 8 t h 9 3 FIGURE 6 3 Putting [from (9)] in (7), we get, so. Similarl, (8) gives 5. Substitution in () then gives 5 4 s9 and so 9 4,. Then s9, 5s9, and, from (), 7s9. The corresponding values of f are s9 5 s9 3 7 s9 3 s9 Therefore the maimum value of f on the given curve is 3 s9.

87 94 CHAPTER 4 PARTIAL DERIVATIVES 4.8 EXERCISES. Pictured are a contour map of f and a curve with equation t, 8. Estimate the maimum and minimum values of f subject to the constraint that t, 8. Eplain our reasoning. 5. f,, ;, 6. f,, 3 3;, 4 7. f,, ;, ;. (a) Use a graphing calculator or computer to graph the circle. On the same screen, graph several curves of the form c until ou 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 etreme values of f, subject to the constraint. Compare our answers with those in part (a). 3 7 Use Lagrange multipliers to find the maimum and minimum values of the function subject to the given constraint(s). 3. f, ; 4. f, 4 6; 5. f, ; 6. f, e ; 7. f,, 6 ; 8. f,, 8 4; 9. f,, ;. f,, ;. f,, ;. f,, ; f,,, t t; f,,..., n n ; n 7 g(, )= t 8 9 Find the etreme values of f on the region described b the inequalit. 8. f, 3 4 5,. Consider the problem of maimiing the function f, 3 subject to the constraint s s 5. (a) Tr using Lagrange multipliers to solve the problem. (b) Does f 5, give a larger value than the one in part (a)? ; (c) Solve the problem b graphing the constraint equation and several level curves of f. (d) Eplain wh the method of Lagrange multipliers fails to solve the problem. (e) What is the significance of f 9, 4? CAS 9. f, e,. Consider the problem of minimiing the function f, on the curve 4 3 (a piriform). (a) Tr using Lagrange multipliers to solve the problem. (b) Show that the minimum value is f, but the Lagrange condition f, t, is not satisfied for an value of. (c) Eplain wh Lagrange multipliers fail to find the minimum value in this case.. (a) If our computer algebra sstem plots implicitl defined curves, use it to estimate the minimum and maimum values of f, subject to the constraint b graphical methods. (b) Solve the problem in part (a) with the aid of Lagrange multipliers. Use our CAS to solve the equations numericall. Compare our answers with those in part (a). 3. The total production P of a certain product depends on the amount L of labor used and the amount K of capital investment. In Sections 4. and 4.3 we discussed how the Cobb- Douglas model P bl K follows from certain economic 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 compan can spend onl p dollars as its total budget, then maimiing the production P is subject to the constraint ml nk p. Show that the maimum production occurs when L p m 4 and 6 K p n

88 APPLIED PROJECT ROCKET SCIENCE Referring to Eercise 3, we now suppose that the production is fied at bl K Q, where Q is a constant. What values of L and K minimie the cost function CL, K ml nk? 5. Use Lagrange multipliers to prove that the rectangle with maimum area that has a given perimeter p is a square. 6. Use Lagrange multipliers to prove that the triangle with maimum area that has a given perimeter p is equilateral. Hint: Use Heron s formula for the area: A sss s s where s p and,, are the lengths of the sides Use Lagrange multipliers to give an alternate solution to the indicated eercise in Section Eercise Eercise 4 9. Eercise 4 3. Eercise 4 3. Eercise Eercise Eercise Eercise Eercise Eercise Eercise Eercise Eercise Find the maimum and minimum volumes of a rectangular bo whose surface area is 5 cm and whose total edge length is cm. 4. The plane intersects the paraboloid in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin. 4. The plane intersects the cone 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 Find the maimum and minimum values of f subject to the given constraints. Use a computer algebra sstem to solve the sstem of equations that arises in using Lagrange multipliers. (If our CAS finds onl one solution, ou ma need to use additional commands.) 43. f,, e ; , 44. f,, ;, (a) Find the maimum value of given that,,..., n are positive numbers and n c, where c is a constant. (b) Deduce from part (a) that if,,..., n are positive numbers, then This inequalit sas 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? 46. (a) aimie n i i subject to the constraints n i i i and n i i. (b) Put i to show that f,,..., n s n n s n n ai s a j n n and i a ib i s aj s b j bi s b j for an numbers a,..., a n, b,..., b n. This inequalit is known as the Cauch-Schwar Inequalit. APPLIED PROJECT ROCKET SCIENCE an rockets, such as the Pegasus XL currentl 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 initiall 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 similarl in order to place the rocket s paload into orbit about the earth. (With this design, at least two stages are required in order to reach the necessar 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, which are to be designed in such a wa as to minimie the total mass of the rocket while enabling it to reach a desired velocit.

89 94 CHAPTER 4 PARTIAL DERIVATIVES For a single-stage rocket consuming fuel at a constant rate, the change in velocit resulting from the acceleration of the rocket vehicle has been modeled b V c ln Sr P r where r is the mass of the rocket engine including initial fuel, P is the mass of the paload, S is a structural factor determined b the design of the rocket (specificall, it is the ratio of the mass of the rocket vehicle without fuel to the total mass of the rocket with paload), and c is the (constant) speed of ehaust relative to the rocket. Now consider a rocket with three stages and a paload of mass A. Assume that outside forces are negligible and that c and S remain constant for each stage. If i is the mass of the ith stage, we can initiall consider the rocket engine to have mass and its paload to have mass 3 A; the second and third stages can be handled similarl.. Show that the velocit attained after all three stages have been jettisoned is given b 3 A 3 A 3 A v f cln S 3 A ln S 3 A ln S 3 A Courtes of Orbital Sciences Corporation. We wish to minimie the total mass 3 of the rocket engine subject to the constraint that the desired velocit v f from Problem is attained. The method of Lagrange multipliers is appropriate here, but difficult to implement using the current epressions. To simplif, we define variables N i so that the constraint equation ma be epressed as v f cln N ln N ln N 3. Since is now difficult to epress in terms of the N i s, we wish to use a simpler function that will be minimied at the same place as. Show that and conclude that 3 A 3 A A A 3 A 3 A 3 A A SN SN SN SN SN3 SN 3 S 3 N N N 3 SN SN SN 3 3. Verif that ln AA is minimied at the same location as ; use Lagrange multipliers and the results of Problem to find epressions for the values of N i where the minimum occurs subject to the constraint v f cln N ln N ln N 3. [Hint: Use properties of logarithms to help simplif the epressions.] 4. Find an epression for the minimum value of as a function of. 5. If we want to put a three-stage rocket into orbit miles above the earth s surface, a final velocit of approimatel 7,5 mih is required. Suppose that each stage is built with a structural factor S. and an ehaust speed of c 6 mih. (a) Find the minimum total mass of the rocket engines as a function of A. (b) Find the mass of each individual stage as a function of A. (The are not equall sied!) 6. The same rocket would require a final velocit of approimatel 4,7 mih in order to escape earth s gravit. Find the mass of each individual stage that would minimie the total mass of the rocket engines and allow the rocket to propel a 5-pound probe into deep space. v f