Section 14.1 Functions of two variables

Transcription

1 CHAPTER 14 Derivatives with Two or More Variables (3/22/08) Man mathematical models involve functions of two or more variables. The elevation of a point on a mountain, for eample, is a function of two horiontal coordinates; the densit of the earth at points in its interior is a function of three coordinates; the pressure in a gas-filled balloon is a function of its temperature and volume; and if a store sells a thousand items, its profit might be studied as a function of the amounts of each of the thousand items that it sells. This chapter deals with the differential calculus of such functions. We stud functions of two variables in Sections 14.1 through We discuss vertical cross sections of graphs in Section 14.1, horiontal cross sections and level curves in Section 14.2, partial derivatives in Section 14.3, Chain Rules in Section 14.4, directional derivatives and gradient vectors in Section 14.5, and tangent planes in Section Functions with three variables are covered in Section 14.7 and functions with more than three variables in Section Section 14.1 Functions of two variables Overview: In this section we discuss domains, ranges and graphs of functions with two variables. Topics: The domain, range, and graph of = f(,) Fiing or : vertical cross sections of graphs Drawing graphs of functions The domain, range, and graph of = f(,) The definitions and notation used for functions with two variables are similar to those for one variable. Definition 1 A function f of the two variables and is a rule that assigns a number f(,) to each point (, ) in a portion or all of the -plane. f(, ) is the value of the function at (,), and the set of points where the function is defined is called its domain. The range of the function is the set of its values f(,) for all (,) in its domain. If a function = f(, ) is given b a formula, we assume that its domain consists of all points (,) for which the formula makes sense, unless a different domain is specified. Eample 1 Solution (a) What is the domain of f(,) = 2 + 2? (b) What are the values f(2, 3) and f( 2, 3) of this function at (2, 3) and ( 2, 3)? (c) What is its range? (a) Because the epression is defined for all and, the domain of f is the entire -plane. (b) f(2,3) = = 13 and f( 2, 3) = ( 2) 2 + ( 3) 2 = 13. (c) The values of the function are all nonnegative and for ever 0 it has the value at all points (, ) on the circle =. Consequentl, the range of f is the closed infinite interval [0, ). Recall that the graph of a function f of one variable is the curve = f() in an -plane consisting of the points (, ) with in the domain of the function and = f(). The graph of a function of two variables is a surface in three-dimensional space. 270

2 Section 14.1, Functions of two variables p. 271 (3/22/08) Definition 2 The graph of a function f with the two variables and is the surface = f(, ) formed b the points (,,) in -space with (,) in the domain of the function and = f(,). For a point (, ) in the domain of the function, its value f(,) at (,) is determined b moving verticall (parallel to the -ais) from (, ) in the -plane to the graph and then horiontall (parallel to the -plane) to f(,) on the -ais, as is shown in Figure 1. FIGURE 1 Fiing or : vertical cross sections of graphs One wa to stud the graph = f(,) of a function of two variables is to stud the graphs of the functions of one variable that are obtained b holding or constant. To understand this process, we need to look at the geometric significance of setting or equal to a constant. The meaning of the equation = c, with c a constant, depends on the contet in which it is used. If we are dealing with points on an -ais, then the equation = c denotes the set : = c} consisting of the one point with -coordinate c (Figure 2). If we are talking about points in an -plane, then the equation = c denotes the vertical line (, ) : = c} in Figure 3 consisting of all points (,) with = c. Notice that this line is perpendicular to the -ais and intersects it at = c. FIGURE 2 FIGURE 3 FIGURE 4 We are using here set-builder notation P : Q} for the set of points P such that condition Q is satisfied.

3 p. 272 (3/22/08) Section 14.1, Functions of two variables When we are dealing with -space, the equation = c denotes the vertical plane (,,) : = c} consisting of all points (,,) with = c. This is the plane perpendicular to the -ais (parallel to the -plane) that intersects the -ais at = c. With the aes oriented as in Figure 4, this plane is in front of the -plane if c is positive and behind the -plane if c is negative. (c is negative in Figure 4.) Similarl, = c denotes the point : = c} on a -ais (Figure 5), denotes the horiontal line (, ) : = c} perpendicular to the -ais at = c in an -plane (Figure 6), and denotes the vertical plane (,,) : = c} perpendicular to the -ais (parallel to the -plane) at = c in -space. With aes oriented as in Figure 7, this plane is to the right of the -plane for positive c and to the left of the -plane for negative c. (c is positive in Figure 7.) FIGURE 5 FIGURE 6 FIGURE 7 In the net eamples the shapes of surfaces are determined b studing their cross sections in vertical planes = c and = c. Eample 2 Solution Determine the shape of the surface = (a) b studing its cross sections in the planes = c perpendicular to the -ais and (b) b studing the cross sections in the planes = c perpendicular to the -ais. (a) The intersection of the surface = with the plane = c is determined b the simultaneous equations, = = c. Replacing b c in the first equation ields the equivalent pair of equations, = c = c. These equations show that the intersection is a parabola in the plane = c that opens upward and whose verte is at the origin if c = 0 and is c 2 units above the -plane if c 0. It has the shape of the curve = c in the -plane of Figure 8.

4 Section 14.1, Functions of two variables p. 273 (3/22/08) c 2 FIGURE 8 The curve drawn with a heav line in the picture of -space in Figure 9 is the cross section in the plane = c for one positive c. The other curves are the cross sections for other values of c. (In the drawings the cross sections are cut off at a positive value of ; the in fact, etend infinitel high.) Notice how the cross sections change as c changes in Figure 9. For positive c, the plane = c is in front of the -plane and the verte (lowest point) of the cross section is above the -plane. As c decreases to 0, the plane moves back toward the -plane and the verte moves to the origin. Then as c decreases through negative values, the plane = c moves further toward the back and the verte rises from the -plane. This gives the surface the bowl shape shown in Figure 10. c 2 = c FIGURE 9 FIGURE 10 (b) The intersection of the surface = with the plane = c is determined b the simultaneous equations, = = c. We replace b c in the first equation to obtain the equivalent equations, = 2 + c 2 = c. These show that the intersection is a parabola in the plane = c that opens upward and whose verte is at the origin if c = 0 and is c 2 units above the -plane if c 0. The curve for one positive c is the heav curve in Figure 11. As c decreases toward 0, the plane = c moves to the left toward the -plane and the verte moves down to the origin. Then as c decreases farther through negative values, the plane moves farther to the left and rises above the -plane. This gives the surface the bowl shape in Figure 10. (Figure 12 shows the cross sections from Figures 10 and 11 together.)

5 p. 274 (3/22/08) Section 14.1, Functions of two variables c 2 = c FIGURE 11 FIGURE 12 The surface in Figure 12 is called a circular paraboloid because its vertical cross sections are parabolas and, as we will see in the net section, its horiontal cross sections are circles. Eample 3 Solution Determine the shape of the surface = 2 2 (a) b studing its cross sections in the planes = c perpendicular to the -ais and (b) b studng the cross sections in the planes = c perpendicular to the -ais. (a) We follow the procedure from Eample 2. The intersection of the surface = 2 2 with the plane = c is determined b the simultaneous equations, = 2 2 = c which are equivalent to = 2 c 2 = c. The intersection is a parabola in the plane = c that opens upward and whose verte is at = c 2, as shown in Figure 13 for a positive value of c. As c decreases to 0, the plane moves toward the -plane and the verte rises to the origin. Then as c decreases through negative values, the plane = c moves toward the back and the verte drops below the -plane. This gives the surface the saddle shape shown in Figure 14. FIGURE 13 FIGURE 14

6 Section 14.1, Functions of two variables p. 275 (3/22/08) (b) The intersection of the surface = 2 2 with the plane = c is determined b the equations, = 2 2 = c which are equivalent to = c 2 2 = c. The intersection is a parabola in the plane = c that opens downward and whose verte is at the origin if c = 0, and is c 2 units above the -plane if c 0 (Figure 15). The curve for one positive c is the heav curve in Figure 16. As c decreases toward 0, the plane = c moves to the left toward the -plane and the verte moves down to the origin. Then as c decreases further through negative values, the plane moves farther to the left and the verte rises above the -plane. This again gives the surface the saddle shape in Figure 16. Figure 17 shows the two tpes of cross sections together. c 2 FIGURE 15 = c FIGURE 16 FIGURE 17 The surface in Figure 17 is called a hperblic paraboloid because its vertical cross sections are parabolas and, as we will see in the net section, its horiontal cross sections are hperbolas.

7 p. 276 (3/22/08) Section 14.1, Functions of two variables Eample 4 Solution Determine the shape of the surface = b analing its cross sections in the planes = c, perpendicular to the -ais. The cross section of the surface in the vertical plane = c has the equations = = c which are equivalent to = c2 = c. (1) We first consider the case of c = 0, where the cross section (1) is the curve = in the -plane. The derivative = d d ( ) = = 1 4 (4 2 ) is ero at = ±2, negative for < 2 and > 2, and positive for 2 < < 2. Consequentl, this curve has the shape shown in Figure 18. The cross section (1) with = c with nonero c has the shape of the curve in Figure 18 lowered c 2 units. As c increases from 0 through positive values the cross section moves down and in the positive -direction and, and as c decreases from 0 through negative values it moves down and in the negative -direction. This gives the surface the boot-like shape in Figure = FIGURE 18 FIGURE 19 We could also determine the shape of the surface in Figure 19 b studing the cross sections of the surface in the plane = c perpendicular to the -ais. The have the equations, = = c which are equivalent to = c 1 12 c = c. (2) For c = 0, this cross section is is the parabola = in the -plane of Figure 20. It passes through the origin and opens downward.

8 Section 14.1, Functions of two variables p. 277 (3/22/08) 2 1 = FIGURE 20 3 The cross section (2) in the plane = c has the same parabolic shape for nonero c. It is to the right of the origin if c is positive and to the left of the origin if c is negative. The height of its verte is the value of c 1 12 c3. Hence, the cross section moves up and to the right as c increases from 0 to 2 and moves down and to the right as c increases beond 2. It moves down and to the left as c decreases from 0 to 2 and moves up and to the left as c decreases beond 2. This gives the surface the shape in Figure 19. Another wa to visualie the surface in Figure 19 is to imagine that wire hoops with the shape of the parabola in Figure 19 are hung in perpendicular planes on the curve in Figure 17. Eample 5 Determine the shape of the surface h(,) = Solution The graph = is especiall eas to anale because the equation does not involve the variable. This implies that if one point ( 0, 0, 0 ) with -coordinate 0 and -coordinate 0 is on it, then the line parallel to the -ais formed b the points (, 0, 0 ) as ranges over all numbers is on the surface. Consequentl, the surface consists of lines parallel to the -ais. In this case, the intersection of the surface with the -plane, where = 0, is the curve = 1 8 3, and the surface in Figure 21 consists of this curve and all lines through it parallel to the -ais. FIGURE 21

9 p. 278 (3/22/08) Section 14.1, Functions of two variables We will find in Chapter 15 that the surfaces = k with nonero constants k are important in the stud of maima and minima of functions with two variables. We cannot determine the shapes of the graphs of these functions from their cross sections in planes = c and = c since those intersections are lines while the surfaces are curved. These surfaces are best understood b rotating coordinates as in Figure 22, which shows shows the result of establishing and -aes b rotating the - and -aes 45 counterclockwise. According to Theorem 3 of Section 12.2, the original coordinates (,) can be calculated from the new coordinates (, ) b the formulas, = 1 2 ( ), = 1 2 ( + ). (3) 45 FIGURE 22 Eample 6 Solution Use -coordinates as in Figure 22 to anale the surface = 2. Equations (3) transform the equation = 2 into ] 1 2 = 2[ ( )][ 1 2 ( + ). This simplifies to = ( )( + ) and then to = ( ) 2 ( ) 2. (4) This is the equation of Eample 3 with and replaced b and. Consequentl, the graph is the surface of Figure 17 rotated 45 as in Figure 23. Notice that the vertical cross sections of the graph in the planes = c perpendicular to the -ais are parabolas that open upward, and the vertical cross sections in the planes = c perpendicular to the -ais are parabolas that open downward, as can be seen from equation (4). The primes on the variables and here are just to distinguish them from and. The do not denote derivatives.

10 Section 14.1, Functions of two variables p. 279 (3/22/08) (To be redrawn) FIGURE 23 Drawing graphs of functions In earlier chapters when we were dealing with functions of one variable, we could rel etensivel on hand-drawn sketches or graphs generated b calculators or computers to guide our reasoning and check our work. In dealing with functions of two variables, we generall have to reason more abstractl because it is difficult to draw good pictures of most surfaces, and drawings of surfaces generated b graphing calculators or computers are frequentl difficult to interpret. Moreover, even when we have a good sketch of the graph of a function with two variables, we often cannot determine the function s values from it because we cannot tell from the two-dimensional picture where vertical lines intersect the graph. You should, however, be able to sketch paraboloids and other surfaces given b simple equations that are eas to anale. Draw a portion of the surface in an imaginar bo with sides parallel to the coordinate planes. Draw an profiles of the surface that are inside the bo and an curves that are obtained b chopping off the surface at the sides of the bo. Then add coordinate aes to match our drawing. For eample, to draw a circular or elliptic paraboloid = a + b 2 + c 2 or = a b 2 c 2 with b > 0, c > 0, as in Figure 24, first draw a horiontal circle to represent the circle or ellipse where the surface is chopped off. Add a parabola for the profile of the surface. Then draw coordinate aes as appropriate. FIGURE 24 To draw a hperbolic paraboloid = a + b 2 c 2 or = a b 2 + c 2 with b > 0, c > 0 or = k with k 0, as in Figure 25, first draw the saddle seat of the surface with part of a parabola that opens upwards. Add portions of horiontal hperbolas where the surface is chopped off at the top and bottom, and draw parabolas or vertical lines to represent where the surface is chopped off at the sides. Then draw coordinate aes as appropriate. Man graphs of functions f(, ) = G() or f(,) = G() that depend on onl one of the variables or are eas to sketch because their cross sections in one direction are horiontal lines and in the other

11 p. 280 (3/22/08) Section 14.1, Functions of two variables FIGURE 25 direction have the shape of the graph of the function G of one variable. For eample, to draw the surface = a + b 2, = a + b 2, = a b 2, or = a b 2 with b > 0, as in Figure 26, draw parts of vertical parabolas to represent parabolic ends of the surface where it is cut off b vertical planes and draw three parallel lines to represent its profile and where is it chopped off b a horiontal plane. FIGURE 26 Interactive Eamples 14.1 Interactive solutions are on the web page http// ashenk/. 1. Add aes to the surface in Figure 27 so that it represents the surface = FIGURE 27 In the published tet the interactive solutions of these eamples will be on an accompaning CD disk which can be run b an computer browser without using an internet connection.

12 Section 14.1, Functions of two variables p. 281 (3/22/08) 2. Label the positive ends of the - and -aes in Figure 28 so that the surface is = 2 2. FIGURE What is f(1, 1) for f(,) = 2 e 3 e? 4. Draw the surface =. 5. Match the equations (a) = sin and (b) = sin to their graphs in Figures 29 and 30 b studing their cross sections in the vertical planes = c perpendicular to the -ais. FIGURE 29 FIGURE Match the equations (a) = sin and (b) = sin to their graphs in Figures 29 and30 b studing their cross sections in the vertical planes = c perpendicular to the -ais. Eercises 14.1 A Answer provided. CONCEPTS: O Outline of solution provided. C Graphing calculator or computer required. 1. (a) Does increasing c move the plane = c toward the front or to the back if the aes are oriented as in Figure 4? (b) Does increasing c move the plane = c toward the left or to the right if the aes are oriented as in Figure 7? 2. How is the graph of = f(,) related (a) to the graph of = f(,), (b) to the graph of = 2f(, ), and (c) to the graph of = f(2,2)?

13 p. 282 (3/22/08) Section 14.1, Functions of two variables 3. How is the graph of = f(, ) related (a) to the graph of = f(, ) + 1, (b) to the graph of = f( + 1, ), and (c) to the graph of = f(, + 1)? 4. How is the graph of = f(,) related to the graph of = f(,)? BASICS: Find the indicated values of the functions in Eercises 5 through O Calculate f(2, 3) and f( 1,10) for f(,) = A g(1, 3) for g(, ) = + 7. A = ln( + 3) at = 4, = 5 8. h(3, 2) for h(,) = +? 9. O (a) Describe the intersections of the graph = 2 with the planes = c and = c for constants c. (b) Cop the surface in Figure 31, add aes, and label them so the drawing represents the graph from part (a). FIGURE O (a) Figure 32 shows the surface = Eplain wh it is the lower half of the sphere = 1 of radius 1 with its center at the origin. (b) Draw the graph of the function g(,) = FIGURE 32

14 Section 14.1, Functions of two variables p. 283 (3/22/08) 11. A Label the positive ends of the - and -aes in Figure 33 so that the surface is the graph of Q(,) = 1 2. FIGURE Label the positive ends of the - and -aes in Figure 34 so that the surface has the equation =. FIGURE O Draw the graph of the function L(, ) = Draw the graph of V (,) = A (a) What is the domain of (,) =? (b) For what values of and is the function from part (a) positive? 16. O (a) Is g(,) = 3 an increasing or decreasing function of for = 10? (b) Is g(,) = 3 an increasing or decreasing function of for = 2? 17. O What is the global minimum of h(,) = ( ) and at what values of and does it occur? What is the global maimum of k(,) = and at what point (,) does it occur? A Use a calculator to determine which is the greatest and which is the least of the numbers M(1,2), M(2,1), and M(2,2) for M(,) = 1 + cos 2 + sin.

15 p. 284 (3/22/08) Section 14.1, Functions of two variables 20. A The total resistance R = R(r 1, r 2 ) of an electrical circuit consisting of resistances of r 1 and r 2 ohms (Figure 35) is determined b the equation 1 R = 1 r r 2. (1) Does R(r 1,r 2 ) increase or decrease as r 1 or r 2 increases? r 1 FIGURE 35 r When a ver small spherical pebble falls under the force of gravit in a deep bod of still water, it quickl approaches a constant speed, called its terminal speed. B Stokes law the terminal speed is v(r,ρ) = 21800(ρ 1)r 2 centimeters per second if the radius of the pebble is r centimeters and its densit is ρ 1 grams per cubic centimeter. (2) (a) What is the terminal speed of a quart pebble of densit 2.6 grams per cubic centimeter if its radius is 0.01 centimeters? (b) Which of two pebbles has the greater terminal speed if the have the same densit and one is larger than the other? (c) What happens to the pebble if its densit is 1 gram per centimeter, the densit of water? EXPLORATION: 22. The table below gives the equivalent human age A = A(t,w) of a dog that is t ears old and weighs w pounds. (3) (a A ) What does A(11, 50) represent and, based on the table, what is its approimate value? (b) What does A(14, 70) represent and what is its approimate value? A(t, w) = equivalent human age t = 6 t = 8 t = 10 t = 12 t = 14 t = 16 w = w = w = (1) CRC Handbook of Chemistr and Phsics, Boca Raton FL: CRC Press, Inc., 1981, p. F-112. (2) CRC Handbook of Chemistr and Phsics, Boca Raton FL: CRC Press, Inc., 1981, p. F-115. (3) Data from Senior-Care Health Report b Pfier, Inc., based on a chart developed b F. Menger, State College, PA.

16 Section 14.1, Functions of two variables p. 285 (3/22/08) In Eercises 23 through 26 match the functions to their graphs in Figures 36 through 39 and eplain how the shapes of the surfaces are determined b their equations. 23. O = sin 24. A = sin 25. = sin sin 26. = 3e /5 sin FIGURE 36 FIGURE 37 FIGURE 38 FIGURE A person s bod-mass inde is the number I(w,h) = w, where w is his or her weight, measured 2 h in kilograms, and h is his or her height, measured in meters. (a) What is our bod-mass inde? (A kilogram is 2.2 pounds and a meter is inches.) (b) A stud of middle-aged men found that those with a bod-mass inde of over 29 had twice the risk of death than those whose bod-mass inde was less than 19. Suppose a man is 1.5 meters tall and has a bod-mass inde of 29. How much weight would he have to lose to reduce his bod-mass inde to 19? 28. Sketch the graph of H(,) = 29. (a) What is the domain of = points does it occur? for < 1 2 ( ) for ? (b) What is its global maimum value and at what

17 p. 286 (3/22/08) Section 14.1, Functions of two variables 30. (a) What is the domain of sin 1 ( )? (b) What are its global maimum and minimum values and at what points do the occur? 31. A Figure 40 shows the graph of g(, ) = 10cos(). Find, without using derivatives, the global maimum of = g(,) and the points (,) where it occurs. FIGURE 40 FIGURE 41 3cos( + ) 32. Find, without using derivatives, the global maimum of h(,) =. and the values ( ) of (,) where the occur. The graph of = h(,) is in Figure 41.

18 Section 14.1, Functions of two variables p. 287 (3/22/08) 33. Match the functions (a A ) A(,) = 2 cos( + ), (b) B(,) = 2 sin( 2 8 ) , (c) C(,) = 4 cos(+)+ 3 1 (2 + 2 ), and (d) D(,) = 2sin(2π)+ sin(π) to their graphs in Figures 42 through 45 and eplain in each case how the shape of the surface is determined b the formula for the function. FIGURE 42 FIGURE 43 FIGURE 43 FIGURE Describe and draw the graph of G(,) = 1 ( + 1) 2 ( 1) (a) What is the domain of = points does it occur? 1 1? (b) What is its global maimum value and at what 36. (a) What is the domain of sin 1 ( )? (b) What are its global maimum and minmimum values and at what points do the occur? 37. Sketch the graph of H(,) = for < 1 2 ( ) for (End of Section 14.1)