Section 14.2 (3/23/08) Horizontal cross sections of graphs and level curves Overview: In the last section we analyzed graphs of functions of two variables by studying their vertical cross sections. Here we study horizontal cross sections of graphs and the associated level curves of functions. Topics: Horizontal cross sections of graphs • Level curves • Estimating function values from level curves • Topographical maps and other contour curves • Horizontal cross sections of graphs In Section 14.1 we determined the shape of the surface z = x2 + y2 in Figure 1 by studying its vertical cross sections in planes y = x and x = c perpendicular to the y- and x-axes. In the next example we look at the cross sections of this surface in horizontal planes z = c perpendicular to the z-axis.

FIGURE 1

Example 1 Determine the shape of the graph of z = x2 + y2 in Figure 1 by studying its horizontal cross sections. Solution Horizontal planes have the equations z = c with constant c. Consequently, the horizontal cross sections of the surface z = x2 + y2 are given by the equations,

2 2 z = x + y ( z = c.

Setting z = c in the first of these equation yields the equivalent equations for the cross section,

2 2 c = x + y (1) ( z = c.

288 Section 14.2, Horizontal cross sections of graphs and level curves p. 289 (3/23/08)

If c is positive, then x2 + y2 = c is the circle of radius √c in an xy-plane and the cross section (1) of the surface is the circle of radius √c in the plane z = c with its center at c on the z-axis. If c = 0 then the curve (1) is the origin of xyz-space. The cross section is empty (has no points in it) if c is negative. Since the radius of the circular cross section at z = c increases as c> 0 increases and the plane z = c rises, the surface has the bowl shape in Figure 2. 

y 4

9 6 3

4 x

Level curves of f(x,y)= x2 + y2 FIGURE 2 FIGURE 3

Level curves The horizontal cross sections of the surface in Figure 2 are at z = 0, 1, 2, 3,..., 10. The lowest cross section is the point at the origin. The other cross sections are circles; if we drop them down to the xy-plane, we obtain the ten concentric circles in Figure 3. The point and circles are curves on which f(x,y)= x2 + y2 is constant. They are called level curves or contour curves of the function. The function has the value 0 at the origin and the values 1, 2, 3,..., 10 on the circles in Figure 3. The numbers 3, 6, and 9 on three of the circles indicate the values of the function on those circles. To visualize the surface in Figure 2 from the level curves, imagine that the xy-plane in Figure 3 is horizontal in xyz-space, that the innermost circle in Figure 3 is lifted one unit to z = 1, the next circle is lifted two units to z = 2, the next circle, labeled “3” is lifted to z = 3, and so forth. This gives the horizontal cross sections of the surface which determine its shape in Figure 3. It is because the vertical cross sections of the surface z = x2 + y2 in Figure 2 are parabolas and its horizontal cross sections are circles, that it is called a circular paraboloid. This is a special type of elliptical paraboloid, whose horizontal cross sections are ellipses. p. 290 (3/23/08) Section 14.2, Horizontal cross sections of graphs and level curves

Here is a general definition of level curves.

Definition 1 The level curves (contour curves) of z = f(x,y) are the curves in the xy-plane where the function is constant (Figure 4). They have the equations f(x,y)= c with constants c.

FIGURE 4

Example 2 Determine the shape of the surface z = y2 x2 in Figure 5 by studying its horizontal − cross sections.

FIGURE 5 Section 14.2, Horizontal cross sections of graphs and level curves p. 291 (3/23/08)

Solution The horizontal cross sections of the surface z = y2 x2 are given by the equations, − 2 2 z = y x − ( z = c.

Setting z = c in the first equation gives the equivalent equations,

2 2 c = y x − (2) ( z = c. 2 2 For c = 0, the first of equations (2) reads y x = 0 and gives y = x. The − ± level curve consists of the lines y = x and y = x in Figure 6. −

y2 x2 = c y2 x2 = c y2 x2 = c y − y − y − c = 0 c> 0 c< 0   

x x x

y = x y = √c + x2 x = c + y2 ± ± ± − p FIGURE6 FIGURE7 FIGURE8

2 2 For c > 0, the first of equations (2) gives y = c + x and then y = √c + x2. ± The level curve consists of the two halves of the hyperbola in Figure 7. For c < 0, we obtain x2 = c + y2 and then x = c + y2. The level − ± − curve consists of the two halves of the hyperbola in Figure 8. The level curves for c = 0, 1, 2, 4 and 6 are shown in Figure 9. p ± ± ± ± p. 292 (3/23/08) Section 14.2, Horizontal cross sections of graphs and level curves

The horizontal cross sections on the surface in Figure 10 are obtained from the level curves in a horizontal xy-plane in Figure 9 by raising the level curve y2 x2 = c − for c = 0, 1, 2, 3,... 6 up c units and lowering the level curve for c = 1, 2, 3,..., 6 − − − − down c units. This gives the surface the shape in Figure 10.  | |

y 4

3

0 0

−3 −3 4 x

3

Level curves of z = y2 x2 − FIGURE 9 FIGURE 10

The surface in Figure 10 is called a hyperbolic parabola because its vertical cross sections are parabolas and its horizontal cross sections are hyperbolas. Example 3 Figure 11 shows the “boot-shaped” graph of the function h(x,y)= y 1 y3 1 x2 with − 12 − 4 the vertical cross sections that we found in Section 14.1 and with its horizontal cross sections at the integer values z = 5, 4, 3,..., 4, 5. The corresponding level curves − − − of h are in Figure 11. Explain how the level curves relate to the graph.

y z = 1 6 z = −5

z = −1 2

2 6 x

z = −1 z = 1 z = 5 (To be redrawn)

FIGURE11 FIGURE12 Section 14.2, Horizontal cross sections of graphs and level curves p. 293 (3/23/08)

Solution Imagine that the surface is sitting in a container that is being filled with water. The bottom edge of the surface is at z = 5 and has the shape of the top level curve in − Figure 12. As the water rises to z = 4, z = 3, and z = 2, it reaches the next three − − − cross sections in Figure 12, which have the shape of the next three level curves from the top in Figure 11. At z = 1 the water line is in two parts. One part goes around the “toe” of the − boot; the other is on the “instep” of the boot. The corresponding level curve is in the two parts labeled z = 1 in Figure 12. The next two curves, at z = 0 and z = 1, are − also in two parts. Finally, the highest cross sections, at z = 2, 3, 4, and 5, are only on the “shin” of the boot. They correspond to the bottom four level curves in Figure 11. 

Estimating function values from level curves Level curves of a function, as in Figure 13, show where the function has each of the z-values for the given curves, and we can estimate the function’s values at other points from values on nearby level curves,

y

3

80

63 3 60 6 x Level curves of P − − FIGURE 13 3 40 −

Example 4 The level curves in Figure 13 are of the function z = P (x,y). What are the approximate values of (a) P (0, 1) and (b) P ( 2, 1)? − − Solution (a) The level curves in Figure 14 are where P has the values 40, 50, 60, 70, and 80. The point (0, 1) on the negative y-axis is on the the level curve where P = 50, so − P (0, 1) = 50. − (b) The point ( 2, 1) on the upper left in Figure 15 is about halfway between the − level curves where P = 70 and P = 80. We conclude that P ( 2, 1) 75. (Different − ≈ estimates give different answers.) 

y

3

80

63 3 60 6 x − − FIGURE 14 3 40 − p. 294 (3/23/08) Section 14.2, Horizontal cross sections of graphs and level curves

Topographical maps and other contour curves The next example involves a topographical map showing level curves of the elevation on the mountain. Example 5 Figure 15 shows a topographical map of Mt. Shasta in northern California and Figure 16 is a photograph of the mountain. The numbers on the contour curves are the elevation above sea level. What is the approximate elevation of the highest point on the mountain? Solution The contour curves drawn with heavy lines in Figure 14 are 2,500 feet apart, and the other curves represent elevations 500 feet apart. The summit, therefore, is approxi- mately 14,000 feet above sea level. 

FIGURE15 FIGURE16

Weather maps show isotherms, which are level curves of temperature, and isobars, which are level curves of barometric pressure.

Interactive Examples 14.2 Interactive solutions are on the web page http//www.math.ucsd.edu/˜ashenk/.† 1. Label the level curves of P (x,y)= x2 + 4y2 in Figure 17.

y 4

5 x

FIGURE 17

†In the published text the interactive solutions of these examples will be on an accompanying CD disk which can be run by any computer browser without using an internet connection. Section 14.2, Horizontal cross sections of graphs and level curves p. 295 (3/23/08)

2. Draw and label several level curves of the function B(x,y)= x2y. 3. Figure 18 shows level curves of z = R(x,y). What are the approximate values of (a) R(15, 0) and (b) R(0, 20)?

y

20 1 2

3 4

10 20 x FIGURE 18 −

Exercises 14.2 A O C Answer provided. Outline of solution provided. Graphing calculator or computer required.

CONCEPTS: 1. Draw and label the level curves (a) where f(x,y)= x2 + y2 has the values 1 and 4, (b) where 2 2 g(x,y) = x2 + y2 has the values 1 and 2, and (c) where h(x,y) = ex +y has the values e and e4. (d) How are the three sets of curves related? p 2. Figure 19 shows level curves of a function z = M(x,y). How can you convert it into a sketch of level curves of z = 10M(x,y)?

y

2

4 6 x

FIGURE 19

BASICS: O 1 1 3. Draw the level curves y 2 x = c of N(x,y)= y 2 x for c = 0, 1, and 2. O − − ± ± 4. Draw and label the level curves of S(x,y)= y sin x where it has the values 0, 2, 4. A − ± ± 5. Draw and label the level curves of T (x,y)= √3 x y where it has the values 0, 1, and 2. − ± ± Draw and label level curves of the functions in Exercises 6 through 8. A 6. W (x,y)= y 1 x2. − 2 7. V (x,y) = 4x + y2 8. A(x,y)= x + 3y. p. 296 (3/23/08) Section 14.2, Horizontal cross sections of graphs and level curves

O 9. What are the values of L(x,y)= x + y on its three level curves in Figure 20? | | | | z y

2 2

2 y 2 x

FIGURE20 FIGURE21 1 1 10. What are the values of K(x,y)= 2 + 2 on its eight level curves in Figure 21? x y 11. Figure 22 shows level curves of the function F (x,y)= Ax + By + C. What are the values of the constants A, B, and C? y

2

2 x

4 0 −4 FIGURE 22

EXPLORATION:

A x + 2y 12. Describe the level curve N(x,y)=1 of N(x,y)= . 3x + y A 13. Level curves of G(x,y)= Ay3 cos(Bx) are in Figure 23. What are the constants A and B? − y

0 2 1

1 x −1 −2 FIGURE 23

1 3 14. Match the graphs of (a) z = 9 x sin y in Figure 24 (b) z = sin y in Figure 25, − 1 3 −x/5 (c) z = sin x sin y in Figure 26, (d) z = sin y 9 x in Figure 27, (e) z = 3e sin y in − 1 2 2 − Figure 28, and (f) z = 2 x + sin y in Figure 29 with the level curves in Figures 30 through 35. Section 14.2, Horizontal cross sections of graphs and level curves p. 297 (3/23/08)

FIGURE24 FIGURE25

FIGURE26 FIGURE27

FIGURE28 FIGURE29 p. 298 (3/23/08) Section 14.2, Horizontal cross sections of graphs and level curves

FIGURE30 FIGURE31

FIGURE32 FIGURE33

FIGURE34 FIGURE35 Section 14.2, Horizontal cross sections of graphs and level curves p. 299 (3/23/08)

1 15. (a) Explain why the horizontal cross sections of z = ln(x2 +y2) and of z = are circles. x2 + y2 (b) Match the surfaces in Figures 36 and 37 to their equations in part (a). Explain your choices. p

FIGURE36 FIGURE37 16. Large earthquakes with epicenters near coastlines in an ocean generate tsunami waves which can travel up to 600 miles per hour in very deep water and can be up to 90 feet high near shores. Figure 38 shows level curves of T = T (P ), where for points P in the Pacific Ocean and on its shores, T (P ) is the time it took the tsunami wave created by an earthquake in 1964 to reach P . The dot in the drawing is the epicenter of the earthquake.(1) Approximately how much longer did it take the wave to reach New Zealand than to reach Japan?

FIGURE 38

(1)Adapted from A.Strahler and A.Strahler, Environmental Geoscience, Santa Babara: Hamilton Publishing Co., 1973. p. 300 (3/23/08) Section 14.2, Horizontal cross sections of graphs and level curves

C O 17. Generate the level curves where p(x,y)= xy+ 1 x3 has the values 4, 0, and 4 by solving for y and 4 − generating the graphs of the resulting functions of x. Use the window 5 x 5, 7 y 7. − ≤ ≤ − ≤ ≤ Copy the curves on your paper and label them with the corresponding values of p. C A 18. Generate the level curves where q(x,y) = ey cos(πx) has the values 0, 1, 2, 3, 4 by solving for − y and generating the graphs of the resulting functions of x. Use the window 3 x 3, 2 − ≤ ≤ − ≤ y 2. Copy the curves on your paper and label them with the corresponding values of q. C ≤ 19. Generate the level curves where r(x,y)= ey ex has the values 0, 0.5, 1, 1.5 by solving for − ± ± ± y and generating the graphs of the resulting functions of x. Use a window with equal scales on the axes. Copy the curves on your paper and label those where r = 0, 1. C ± 20 Find the values of a such that the curve with parametric equations x = a cos3 t,y = a sin3 t are the level curves where r(x,y)= x2/3 + y2/3 has the values 1, 2, and 3. Then generate the three curves in a window with equal scales on the axes and copy and label them.

body-mass index w 21. A person’s is the number I(w,h)= 2 , where w is his or her weight, measured h in kilograms, and h is his or her height, measured in meters. (a) What is your body-mass index? (A kilogram is 2.2 pounds and a meter is 39.37 inches.) (b) A study of middle-aged men found that those with a body-mass index of over 29 had twice the risk of death than those whose body-mass index was less than 19. Suppose a man is 1.5 meters tall and has a body-mass index of 29. How much weight would he have to lose to reduce his body-mass index to 19? ln y 22. (a) What is the domain of z = ? (b) Draw some of its level curves. x A 23. (a) What is the domain of z = ln(xy)? (b) How are its level curves related to those of the function xy? y 24. Figure 39 shows the graph of the function f(x,y) = 2 − 2 and Figure 40 shows its level x + y + 1 curves. Show that one of the level curves is a line and the others are circles. (For c = 0, set 6 c = 2/k and complete the square.)

FIGURE39 FIGURE40

(End of Section 14.2)