Section 14.1 Functions of Several Variables

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Section 14.1 Functions of Several Variables

16. Find and sketch the domain of the function. f (x, y, z) = ln(16 − 4x²− 4y²− z²).

Solution:

SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES ¤ 1343 13.( ) =



 + 1. 

is defined only when  ≥ 0. Further,  is defined only when  + 1 6= 0 ⇔  6= −1. So the domain of  is

{( ) |  ≥ 0,  6= −1}.

14. ( ) = sin⁻¹( + ). sin⁻¹( + )is defined only when

−1 ≤  +  ≤ 1 ⇔ −1 −  ≤  ≤ 1 − . Thus, the domain of  is {( ) | −1 −  ≤  ≤ 1 − }, which consists of those points on or between the parallel lines  = −1 −  and  = 1 − .

15. (  ) =√

4 − ²+

9 − ²+√

1 − ².  is defined only when 4 − ²≥ 0 ⇔ −2 ≤  ≤ 2, and 9 − ²≥ 0 ⇔

−3 ≤  ≤ 3, and 1 − ²≥ 0 ⇔ −1 ≤  ≤ 1. Thus, the domain of  is {(  ) | −2 ≤  ≤ 2 −3 ≤  ≤ 3 −1 ≤  ≤ 1}, which is a solid rectangular box with vertices (±2 ±3 ±1) (all 8 combinations).

16. (  ) = ln(16 − 4²− 4²− ²).  is defined only when

16 − 4²− 4²− ² 0 ⇒ ² 4 +²

4 +²

16 1. Thus,

 =

 (  )



² 4 +²

4 +² 16 1



, that is, the points inside the

ellipsoid ² 4 +²

4 +² 16= 1.

17. (a) (160 70) = 01091(160)^0425(70)^0725≈ 205, which means that the surface area of a person 70 inches (5 feet 10 inches) tall who weighs 160 pounds is approximately 20.5 square feet.

(b) Answers will vary depending on the height and weight of the reader.

18. (120 20) = 147(120)^065(20)^035≈ 942, so when the manufacturer invests $20 million in capital and 120,000 hours of labor are completed yearly, the monetary value of the production is about $94.2 million.

19. (a) From Table 1, (−15 40) = −27, which means that if the temperature is −15^◦Cand the wind speed is 40 kmh, then the air would feel equivalent to approximately −27^◦Cwithout wind.

(b) The question is asking: when the temperature is −20^◦C, what wind speed gives a windchill index of −30^◦C? From Table 1, the speed is 20 kmh.

30. Sketch the graph of the function. f (x, y) =p

4x²+ y². Solution:

1346 ¤ CHAPTER 14 PARTIAL DERIVATIVES

29. = ²+ 4²+ 1, an elliptic paraboloid opening upward with vertex at (0 0 1).

30. =

4²+ ² so 4²+ ²= ² and  ≥ 0, the top half of an elliptic cone.

31. =

4 − 4²− ² so 4²+ ²+ ²= 4 or ²+² 4 +²

4 = 1 and  ≥ 0, the top half of an ellipsoid.

32. (a) ( ) = 1

1 + ²+ ². The trace in  = 0 is  = 1

1 + ², and the trace in  = 0 is  = 1

1 + ². The only possibility is graph III. Notice also that the level curves of  are 1

1 + ²+ ² =  ⇔ ²+ ²= 1

− 1, a family of circles for

  1.

(b) ( ) = 1

1 + ²². The trace in  = 0 is the horizontal line  = 1, and the trace in  = 0 is also  = 1. Both graphs I and II have these traces; however, notice that here   0, so the graph is I.

(c) ( ) = ln(²+ ²). The trace in  = 0 is  = ln ², and the trace in  = 0 is  = ln ². The level curves of  are ln(²+ ²) =  ⇔ ²+ ²= ^, a family of circles. In addition,  is large negative when ²+ ²is small, so this is graph IV.

(d) ( ) = cos

²+ ². The trace in  = 0 is  = cos

² = cos || = cos , and the trace in  = 0 is

 = cos√

²= cos || = cos . Notice also that the level curve ( ) = 0 is cos

²+ ² = 0 ⇔

²+ ²= 2 + 2

, a family of circles, so this is graph V.

(e) ( ) = ||. The trace in  = 0 is  = 0, and the trace in  = 0 is  = 0, so it must be graph VI.

(f ) ( ) = cos(). The trace in  = 0 is  = cos 0 = 1, and the trace in  = 0 is  = 1. As mentioned in part (b), these traces match both graphs I and II. Here  can be negative, so the graph is II. (Also notice that the trace in  = 1 is

 = cos , and the trace in  = 1 is  = cos .)

32. Match the function with its graph (labeled I–VI). Give reasons for your choices.

(a) f (x, y) = _1+x¹2+y² (b) f (x, y) = _1+x¹2y² (c) f (x, y) = ln(x²+y²) (d) f (x, y) = cosp

x²+ y² (e) f (x, y) = |xy|

(f) f (x, y) = cos(xy) 900 Chapter14 Partial Derivatives

z

x y

I z

x y II

III z

x y

x y z IV

y x

z

V VI z

x y

33. A contour map for a function f is shown. Use it to esti mate the values of fs23, 3d and f s3, 22d. What can you say about the shape of the graph?

7et1401x33 04/26/10

MasterID: 01556

y

x 0 1

1 70 60 50 40 30 2010

34. Shown is a contour map of atmospheric pressure in North America on August 12, 2008. On the level curves (called isobars) the pressure is indicated in millibars (mb).

(a) Estimate the pressure at C (Chicago), N (Nashville), S (San Francisco), and V (Vancouver).

(b) At which of these locations were the winds strongest?

C N V

S

7et1401x34 04/26/10

MasterID: 01658

1004 1008 1012 1016

1008 1012 1016

$4.00 for a medium box, and $4.50 for a large box. Fixed costs are $8000.

(a) Express the cost of making x small boxes, y medium boxes, and z large boxes as a function of three variables:

C − fsx, y, zd.

(b) Find fs3000, 5000, 4000d and interpret it.

(c) What is the domain of f ? 9. Let tsx, yd − cossx 1 2yd.

(a) Evaluate ts2, 21d.

(b) Find the domain of t.

(c) Find the range of t.

10. Let Fsx, yd − 1 1s4 2 y². (a) Evaluate Fs3, 1d.

(b) Find and sketch the domain of F.

(c) Find the range of F.

11. Let fsx, y, zd −sx 1sy 1sz 1lns4 2 x²2y²2z²d.

(a) Evaluate fs1, 1, 1d.

(b) Find and describe the domain of f.

12. Let tsx, y, zd − x³y²zs10 2 x 2 y 2 z. (a) Evaluate ts1, 2, 3d.

(b) Find and describe the domain of t.

13–22 Find and sketch the domain of the function.

13. fsx, yd −sx 2 2 1sy 2 1 14. fsx, yd −s⁴x 23y

15. fsx, yd − lns9 2 x²29y²d 16. fsx, yd −sx²1y²24 17. tsx, yd − x 2 y

x 1 y 18. tsx, yd − lns2 2 xd 1 2 x²2y² 19. fsx, yd − sy 2 x²

1 2 x² 20. fsx, yd − sin²¹sx 1 yd

21. fsx, y, zd −s4 2 x²1s9 2 y²1s1 2 z² 22. fsx, y, zd − lns16 2 4x²24y²2z²d

23–31 Sketch the graph of the function.

23. fsx, yd − y 24. fsx, yd − x² 25. fsx, yd − 10 2 4x 2 5y 26. fsx, yd − cos y 27. fsx, yd − sin x 28. fsx, yd − 2 2 x²2y² 29. fsx, yd − x²14y²11 30. fsx, yd −s4x²1y² 31. fsx, yd −s4 2 4x²2y²

32. Match the function with its graph (labeled I–VI). Give reasons for your choices.

(a) fsx, yd − 1

1 1 x²1y² (b) fsx, yd − 1 1 1 x²y² (c) fsx, yd − lnsx²1y²d (d) fsx, yd − cos sx²1y² (e) fsx, yd −

|

^xy

|

^{(f ) f}sx, yd − cossxyd

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Solution:

382 ¤ CHAPTER 14 PARTIAL DERIVATIVES

27.  = sin , a cylinder. 28.  = 2 − ²− ², a circular paraboloid opening downward with vertex at (0 0 2).

29.  = ²+ 4²+ 1, an elliptic paraboloid opening upward with vertex at (0 0 1).

30.  = ^−, a cylinder.

31.  = ²+ 1, a parabolic cylinder

32. (a) ( ) = 1

1 + ²+ ². The trace in  = 0 is  = 1

1 + ², and the trace in  = 0 is  = 1

1 + ². The only possibility is graph III. Notice also that the level curves of  are 1

1 + ²+ ² =  ⇔ ²+ ²= 1

 − 1, a family of circles for

  1.

(b) ( ) = 1

1 + ²². The trace in  = 0 is the horizontal line  = 1, and the trace in  = 0 is also  = 1. Both graphs I and II have these traces; however, notice that here   0, so the graph is I.

(c) ( ) = ln(²+ ²). The trace in  = 0 is  = ln ², and the trace in  = 0 is  = ln ². The level curves of  are ln(²+ ²) =  ⇔ ²+ ²= ^, a family of circles. In addition,  is large negative when ²+ ²is small, so this is graph IV.

SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES ¤ 383 (d) ( ) = cos

²+ ². The trace in  = 0 is  = cos

² = cos || = cos , and the trace in  = 0 is

 = cos√

² = cos || = cos . Notice also that the level curve ( ) = 0 is cos

²+ ² = 0 ⇔

²+ ² =_

2 + 2

, a family of circles, so this is graph V.

(e) ( ) = (²− ²)². The trace in  = 0 is  = ⁴, and in  = 0 is  = ⁴. Notice that the trace in  = 0 is 0 = (²− ²)² ⇒  = ±, so it must be graph VI.

(f ) ( ) = cos(). The trace in  = 0 is  = cos 0 = 1, and the trace in  = 0 is  = 1. As mentioned in part (b), these traces match both graphs I and II. Here  can be negative, so the graph is II. (Also notice that the trace in  = 1 is

 = cos , and the trace in  = 1 is  = cos .)

33. The point (−3 3) lies between the level curves with -values 50 and 60. Since the point is a little closer to the level curve with

 = 60, we estimate that (−3 3) ≈ 56. The point (3 −2) appears to be just about halfway between the level curves with

-values 30 and 40, so we estimate (3 −2) ≈ 35. The graph rises as we approach the origin, gradually from above, steeply from below.

34. (a)  (Chicago) lies between level curves with pressures 1012 and 1016 mb, and since  appears to be located about one-fourth the distance from the 1012 mb isobar to the 1016 mb isobar, we estimate the pressure at Chicago to be about 1013 mb.  lies very close to a level curve with pressure 1012 mb so we estimate the pressure at Nashville to be

approximately 1012 mb.  appears to be just about halfway between level curves with pressures 1008 and 1012 mb, so we estimate the pressure at San Francisco to be about 1010 mb.  lies close to a level curve with pressure 1016 mb but we can’t see a level curve to its left so it is more difﬁcult to make an accurate estimate. There are lower pressures to the right of  and  is a short distance to the left of the level curve with pressure 1016 mb, so we might estimate that the pressure at Vancouver is about 1017 mb.

(b) Winds are stronger where the isobars are closer together (see Figure 13), and the level curves are closer near  than at the other locations, so the winds were strongest at San Francisco.

35. The point (160 10), corresponding to day 160 and a depth of 10 m, lies between the isothermals with temperature values of 8 and 12^◦C. Since the point appears to be located about three-fourths the distance from the 8^◦Cisothermal to the 12^◦C isothermal, we estimate the temperature at that point to be approximately 11^◦C. The point (180 5) lies between the 16 and 20^◦Cisothermals, very close to the 20^◦Clevel curve, so we estimate the temperature there to be about 195^◦C.

36. If we start at the origin and move along the -axis, for example, the -values of a cone centered at the origin increase at a constant rate, so we would expect its level curves to be equally spaced. A paraboloid with vertex the origin, on the other hand, has -values which change slowly near the origin and more quickly as we move farther away. Thus, we would expect its level curves near the origin to be spaced more widely apart than those farther from the origin. Therefore contour map I must correspond to the paraboloid, and contour map II the cone.

37. Near , the level curves are very close together, indicating that the terrain is quite steep. At , the level curves are much farther apart, so we would expect the terrain to be much less steep than near , perhaps almost ﬂat.

54. Sketch both a contour map and a graph of the function and compare them. f (x, y) =p

36 − 9x²− 4y².

Solution: SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES ¤ 387

54. The contour map consists of the level curves  =

36 − 9²− 4² ⇒ 9²+ 4²= 36 − ²,  ≥ 0, a family of ellipses with major axis the

-axis. (Or, if  = 6, the origin.)

The graph of ( ) is the surface  =

36 − 9²− 4², or equivalently the upper half of the ellipsoid 9²+ 4²+ ²= 36. If we visualize lifting each ellipse  =

36 − 9²− 4²of the contour map to the plane  = , we have horizontal traces that indicate the shape of the graph of .

55. The isothermals are given by  = 100(1 + ²+ 2²)or

²+ 2²= (100 − ) [0   ≤ 100], a family of ellipses.

56. The equipotential curves are  = 

²− ²− ² or

²+ ²= ²− 



2

, a family of circles ( ≥ ).

Note: As  → ∞, the radius of the circle approaches .

57. ( ) = ²− ³

The traces parallel to the -plane (such as the left-front trace in the graph above) are parabolas; those parallel to the -plane (such as the right-front trace) are cubic curves. The surface is called a monkey saddle because a monkey sitting on the surface near the origin has places for both legs and tail to rest.

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