CHAPTER 13 Functions of Several Variables

CHAPTER Functions o Several Variables Section. Section. Introduction to Functions o Several Variables...67 Limits and Continuit...7 Section. Partial Derivatives...8 Section. Dierentials...97 Section.5 Chain Rules or Functions o Several Variables...0 Section.6 Directional Derivatives and Gradients... Section.7 Tangent Planes and Normal Lines... Section.8 Etrema o Functions o Two Variables...6 Section.9 Applications o Etrema o Functions o Two Variables...6 Section.0 Lagrange Multipliers...56 Review Eercises...69 Problem Solving...8 00 Brooks/Cole, Cengage Learning

CHAPTER Functions o Several Variables Section. Introduction to Functions o Several Variables. No, it is not the graph o a unction. For some values o and ( ) or eample,, 0, 0, there are -values.. Yes, it is the graph o a unction.. + 0 0 + 0 + Yes, is a unction o and.. + No, is not a unction o and. For eample,,,0 corresponds to both ±. 5. + + 9 No, is not a unction o and. For eample,, 0,0 corresponds to both ±. 6. + ln 8 0 ( 8) ln ln 8 Yes, is a unction o and. 7. (, ) (a) (, ) 6 (b) (,) ( ) (c) ( 0, 5) 0( 5) 50 (d) ( 5, ) 5 (e) (,) ( ) ( 5, t) 5t 8. (, ) (a) ( 0, 0) (b) ( 0, ) 0 0 (c) (, ) 6 6 (d) (, ) (e),0 0 ( ) t, t t 9. (, ) e 0 (a) e 5, 0 5 5 (b), e (c), e e 5, 5e (d) (e) (,) e t ( ) ( t, t) te 0. g(, ) ln + (a) g (, 0) ln + 0 0 (b) g( 0, ) ln 0 ln 0 (c) g( 0, e) ln 0 + e (d) g (,) ln+ ln (e) e e e g e, ln e + ln ln + ln e ln + ln ln ( ) g (, 5) ln + 5 ln 7. h (,, ) (a) (,, 9) h 9 (b) 0 h, 0, 0 (c) h(,, ) ( ) (d) ( 5,, 6) 5 0 h 6. (,, ) + + (a) ( 0, 5, ) 0 + 5 + (b) ( 6, 8, ) 6 + 8 (c) (, 6, ) + 6 + (d) ( 0,, ) 0 00 Brooks/Cole, Cengage Learning 67

68 Chapter Functions o Several Variables, sin. π π (a), sin (b) (, ) sin( ) (c) π π, sin π π (d), sin. V( r, h) π r h V, 0 π 0 90π (a) ( ) (b) ( ) (c) ( ) (d) ( ) V 5, π 5 50π V, 8 π 8 8π V 6, π 6 π 5. (, ) ( ) g t dt t t + (a) g (, 0) 0 6 + (b) g (, ) ( ) 6 + 6 9 9 5 (c) (d) (,0) 0 6. g, 6 + 9 9 9 g + g, dt ln t ln ln ln t g, ln ln (a) g 6, ln ln 6 (b) 5 g, 5 ln (c) (d) g 7,7 ln ln 7. (, ) + (a) (b) ( + ) + ( + ) +,,, 0 (, + ) (, ) + ( + ) + ( ) +, 0 8., (a) (b) ( +, ) (, ) ( + ) ( ) 6 + ( ) 6 +, 0 ( + ) ( ), +,, 0 9. (, ) + Domain:, : is an real number, is an real number { } Range: 0 0. (, ) e Domain: Entire -plane Range: > 0. g(, ) Domain: {(, ) : 0} Range: all real numbers. (, ). Domain: {(, ) : > 0} Range: all real numbers + Domain: {(, ) : 0and 0} Range: all real numbers 00 Brooks/Cole, Cengage Learning

Section. Introduction to Functions o Several Variables 69., : Domain: { } Range: all real numbers, 5. Domain: 0 +, : + { } Range: 0, 6. Domain: 0 + +, : + ( ) Range: 0 7. (, ) arccos( + ) Domain: {(, ) : + } Range: 0 π 8. ( ), arcsin Domain: ( ), : π π Range: 9. (, ) ln( ) Domain: > 0 + <, : < + { } Range: all real numbers 0. (, ) ln( 6) Domain: 6 > 0 > 6 {(, ) : > 6} Range: all real numbers + +. (, ) (a) View rom the positive -ais: ( 0, 0, 0 ) (b) View where is negative, and are positive: 5,0, 0 (c) View rom the irst octant: ( 0,5, 5 ) (d) View rom the line ( 0, 0, 0 ). (a) Domain: in the -plane: {(, ) : is an real number, is an real number} Range: (b) 0 when 0 which represents points on the -ais. (c) No. When is positive, is negative. When is negative, is positive. The surace does not pass through the irst octant, the octant where is negative and and are positive, the octant where is positive and and are negative, and the octant where, and are all negative.. (, ) Plane:., 6 Plane Domain: entire -plane Range: < < 5. (, ) 5 Because the variable is missing, the surace is a clinder with rulings parallel to the -ais. The generating curve is. The domain is the entire -plane and the range is 0. 5 5 6 00 Brooks/Cole, Cengage Learning

70 Chapter Functions o Several Variables Plane: 6. g(, ) 7. Paraboloid Domain: entire -plane Range: 0, 6 9 Semi-ellipsoid Domain: set o all points ling on or inside the ellipse + 9 6 Range: 0.. ( ) (, ) e 8. + Cone Domain o : entire -plane Range: 0, sin. 9. (, ) e Because the variable is missing, the surace is a clinder with rulings parallel to the -ais. The generating curve is e. The domain is the entire -plane and the range is > 0. 8 6 5. e 0. (, ), 0, 0 0, else where Domain o : entire -plane Range: 0 5 0 5 0 5 Level curves: c e ln c ln + c Circles centered at ( 0, 0 ) 5 Matches (c). + Hperbolic paraboloid Domain: entire -plane Range: < < 6. e + Level curves: + c e ln c + c ln Hperbolas centered at ( 0, 0 ) Matches (d) 00 Brooks/Cole, Cengage Learning

Section. Introduction to Functions o Several Variables 7 7. ln Level curves: c ln c e c e ± ± Parabolas Matches (b) + 8. cos Level curves: c + cos cos c + + cos c Ellipses Matches (a) 9. + Level curves are parallel lines o the orm + c., 6 50. The level curves are o the orm 6 cor + 6 c. So, the level curves are straight lines with a slope o. 5. + The level curves are ellipses o the orm + c ( ecept + 0 is the point ( 0, 0 )). c 0 c c c c c 0 c c 0 c c c 0 c c c 6 c 8, 9 5. The level curves are o the orm c 9 + 9 c,circles. ( + 0 is the point ( 0, 0 ). ) 5. (, ) The level curves are hperbolas o the orm c. 5., e The level curves are o the orm e c, or ln c. So, the level curves are hperbolas. + The level curves are o the orm c + + 0 c +. c c 55. (, ) So, the level curves are circles passing through the origin ± c,0. and centered at 56. (, ) ln( ) The level curves are o the orm c ln( ) c e c e. So, the level curves are parallel lines o slope passing through the ourth quadrant. 57. ( ), + 6 c 6 c c c c c c c 6 c 5 c c c c c c c c c 5 c 6 c c c c c 0 c c c c c c c 6 c c c c ± c c 0 c 9 9 6 00 Brooks/Cole, Cengage Learning

7 Chapter Functions o Several Variables 58. (, ) 6 6 65. The surace is sloped like a saddle. The graph is not unique. An vertical translation would have the same level curves. One possible unction is,. 8 + + 59. g (, ) 6 6 66. The surace could be an ellipsoid centered at ( 0,, 0 ). One possible unction is ( ), +. 60. h (, ) sin( + ) 67. V( I R) ( R) 0 + 0.06, 000 + I Inlation Rate 6. The graph o a unction o two variables is the set o all points (,, ) or which (, ) and (, ) is in the domain o. The graph can be interpreted as a surace in space. Level curves are the scalar ields, c, where c is a constant. 6. No, the ollowing graphs are not hemispheres. ( + ) e + 6. (, ) The level curves are the lines c or. c These lines all pass through the origin. 6.,, 0, 0 (a) 5 0 5 68. A( rt) Ta Rate 0 0.0 0.05 69. 0 790.85.56 099. 0.8 56. 5.80 97.09 0.5 66.07 090.90 900.0, 5000e rt,, +, c +,Plane Number o Year Rate 5 0 5 0 0.0 555.85 607.0 679.9 759. 0.0 5809.7 679.9 78.56 90.59 0.0 607.0 759. 90.59,7.70 0.05 60. 8.6 0,585.00,59. 0 5 5 (b) g is a vertical translation o three units downward. (c) g is a relection o in the -plane. (d) The graph o g is lower than the graph o. I (, ) is on the graph o, then is on the graph o g. (e) 5,, + + 70. c + + Plane 0 5 0 5 5 00 Brooks/Cole, Cengage Learning

Section. Introduction to Functions o Several Variables 7 7. (,, ) + + c 9 9 + + Sphere 78. V(, ) 5 5 + + 5 c c c 5 5 5 7. (,, ) + c + Elliptic paraboloid Verte: ( 0, 0, ) 7. (,, ) + c 0 0 + Elliptic cone 7.,, sin c 0 0 sin or sin 8 5 5 79., 00 0.6 0. (, ) 00( ) ( ) 00 00 0.6 0. 0.6 0.6 0. 0. 0.6 0. 0.6 0. 0.6 0. 00, a a 80. C ln ln C + aln + ( a) ln ln ln ln C + aln aln ln ln C + aln 8..0 + ( 0.75) + ( 0.75) C base ront and back ends.0 +.50 + 75. N( d, L) d L (a) N, board-eet (b) N 0 0, 507 board-eet 76. w, < (a) w ( 5, 9) h 5 9 6 0 min (b) w ( 5,) h 5 0 min (c) w (, 7) h 7 5 min (d) w ( 5, ) h 5 0 min 77. T 600 0.75 0.75 The level curves are o the orm c 600 0.75 0.75 600 c +. 0.75 The level curves are circles centered at the origin. c 600 c 500 c 00 0 0 0 c 00 c 00 c 00 c 0 0 πr 8. V πr l + πr ( l + r) 8. PV kt (a) 6( 000) k( 00) k 50 kt 50 T (b) P V V The level curves are o the orm 50 T c, V or 50 V T. c These are lines through the origin with slope 50. c l r 00 Brooks/Cole, Cengage Learning

7 Chapter Functions o Several Variables, 0.06 + 0.6 + 5.0 8. (a) Year 00 00 00 005 006 007 5. 9.5.6 9. 5. 6.6 Model 5. 9..0 9. 5. 6.8 (b) has the greater inluence because its coeicient (0.6) is greater than s coeicient (0.06)., 95 0.06 + 0.6 95 + 5.0 0.06 + 5.06 (c) This gives the shareholder s equit in terms o net sales, assuming total assets o 95 ( billion ). 85. (a) Highest pressure at C (b) Lowest pressure at A (c) Highest wind velocit at B 86. Southwest 87. (a) No; the level curves are uneven and sporadicall spaced. (b) Use more colors. 88. (a) The dierent colors represent various amplitudes. (b) No, the level curves are uneven and sporadicall spaced. Section. Limits and Continuit. lim (, ) (,0),, L We need to show that or all ε > 0, there eist a -neighborhood, 0 such that δ about (, ) L < ε Whenever (, ) (,0) lies in the neighborhood. From 0 < + 0 < δ, it ollows that + 0 < δ. So, choose δ ε and the limit is veriied.. Let ε > 0 be given. We need to ind δ > 0 such that, L < ε whenever 0 < a + b + + < δ. Take δ ε. Then i ( ) 0 < + + < δ ε, we have < ε < ε. 89. False. Let (, ),,, but. 90. False. Let (, ) 5. Then, (,) 5 (, ). 9. True 9. False. I there were a point (, ) on the level curves (, ) C and ( ) C,, then C C.. lim.,, L (, ) (, ) We need to show that or all ε > 0, there eists a -neighborhood, such that δ about (, ) L + < ε whenever (, ) (, ) lies in the neighborhood. From 0 < + + < δ it ollows that + + + + < δ. So, choose δ ε and the limit is veriied.. Let ε > 0 be given. We need to ind δ > 0 such that (, ) L b < ε whenever δ ε. 0 < a + b < δ. Take Then i 0 < a + b < δ ε, we have ( b) < ε b <. ε 00 Brooks/Cole, Cengage Learning

Section. Limits and Continuit 75 lim, g, lim, lim g,, a, b (, ) ( a, b) (, ) ( a, b) 5. ( ) 6. lim (, ) ( a, b ) g 5 lim (, ) 5, (, ) ( a, b) 5( ) 0, lim g (, ) (, ) ( a, b) 7. lim, g, lim (, ) lim g(, ) ( ), a, b (, ) ( a, b) (, ) ( a, b) ( ) g( ) 8. lim (, ) ( a, b ) ( ) lim, + lim g,, +, (, ) ( a, b) (, ) ( a, b) + 7, lim (, ) (, ) ( a, b) 9. ( ) lim + 8 + 9,, Continuous everwhere 0. ( ) lim + + 0 + 0 +, 0,0 Continuous everwhere. lim e e e,, Continuous everwhere + + 6. lim (, ) (, ) + + 5 Continuous everwhere 0. lim 0 (, ) ( 0,) Continuous or all 0 + +. lim (, ) (,) Continuous or all. 5. lim (, ) (, ) + Continuous ecept at ( 0, 0 ) 6. lim (, ) (,) + + Continuous or + > 0 π lim cos cos 0, π, Continuous everwhere 7. π 8. lim sin sin (, ) ( π,) Continuous or all 0 arcsin arcsin 0 9. lim 0 (, ) ( 0,) Continuous or, arccos arccos 0 0. lim π (, ) ( 0,) + Continuous or, 0, 0 π. lim + + (,, ) (,,) + + Continuous or + + 0 0. lim e e,,,,0 Continuous everwhere. lim 0 (, ) (,) + +. lim (, ) (, ) + + 5. lim does not eist (, ) ( 0,0) + Because the denominator + approaches 0 as (, ) ( 0,0 ). 6. lim does not eist because the denominator (, ) ( 0,0 ) approaches 0 as (, ) ( 0,0 ). ( )( + ) 7. lim lim lim +,,,, ( ) (, ) (,) ( )( + ) 8. lim lim + + lim 0, 0,0, 0,0 ( ) (, ) ( 0,0) 00 Brooks/Cole, Cengage Learning

76 Chapter Functions o Several Variables 9. lim (, ) ( 0,0) does not eist because ou can t approach ( 0, 0) rom negative values o and. 0. lim (, ) (, ) lim (, ) (,) + + ( )( + ) ( ) ( ) (, ) (,) lim +. The limit does not eist because along the line 0 ou have + lim lim lim (, ) ( 0,0 ) (,0) ( 0,0 ) + (,0) ( 0,0) which does not eist.. The limit does not eist because along the line ou have ( lim ) ( lim, 0,0, ) ( 0,0 ) ( lim, ) ( 0,0) 0. Because the denominator is 0, the limit does not eist. 0 lim 0. (, ) ( 0,0 ) ( )( + + ) ()(). lim ln (, 0,0 ) because ln( ) + does not eist + as (, ) ( 0,0 ). 5. The limit does not eist because along the path 0, 0, ou have + + 0 lim lim 0 (,, ) ( 0, 0, 0) ( 0, 0, ) ( 0, 0, 0) + + whereas along the path, ou have + + + + lim lim (,, ) ( 0, 0, 0 ) (,, ) ( 0, 0, 0) + + + + 6. The limit does not eist because along the path 0, ou have + + 0 lim lim 0 (,, ) ( 0, 0, 0 ) (, 0, 0) ( 0, 0, 0) + + However, along the path 0,, ou have + + lim lim (,, ) ( 0, 0, 0 ) (,, 0) ( 0, 0, 0) + + + 7. lim e (, ) ( 0,0) Continuous everwhere cos( + ) 8. lim, 0,0 + The limit does not eist. Continuous ecept at ( 0, 0 ) + 9. (, ) Continuous ecept at ( 0, 0 ) Path: 0 (, ) (, 0 ) ( 0.5, 0 ) ( 0., 0 ) ( 0.0, 0 ) ( 0.00, 0 ) (, ) 0 0 0 0 0 Path: (, ) (, ) ( 0.5, 0.5 ) ( 0., 0. ) ( 0.0, 0.0 ) ( 0.00, 0.00 ) (, ) The limit does not eist because along the path 0 the unction equals 0, whereas along the path the unction equals. 00 Brooks/Cole, Cengage Learning

Section. Limits and Continuit 77 + 0. (, ) Continuous ecept at ( 0, 0 ) Path: Path: 0 The limit does not eist because along the path 0 the unction equals 0, whereas along the path the unction tends to ininit.. (, ) + Continuous ecept at ( 0, 0 ) Path: Path: The limit does not eist because along the path the unction equals whereas along the path the unction equals.. (, ) (, ) (, ) ( 0.5, 0.5 ) ( 0., 0. ) ( 0.0, 0.0 ) ( 0.00, 0.00 ) (, ) + Continuous ecept at ( 0, 0 ) Path: 0 5 50 500 (, ) (, 0 ) ( 0.5, 0 ) ( 0., 0 ) ( 0.0, 0 ) ( 0.00, 0 ) (, ) 0 0 0 0 0 (, ) (, ) ( 0.5, 0.5 ) ( 0.0, 0. ) ( 0.000, 0.0 ) ( 0.00000, 0.00 ) (, ) (, ) (, ) ( 0.5, 0.5) ( 0.0, 0.) ( 0.000, 0.0) ( 0.00000, 0.00) (, ) (, ) (, 0 ) ( 0.5, 0 ) ( 0.0, 0 ) ( 0.00, 0 ) ( 0.00000, 0 ) (, ) 00 000,000,000, Path: (, ) (, ) ( 0.5, 0.5 ) ( 0.0, 0.0 ) ( 0.00, 0.00 ) ( 0.000, 0.000 ) (, ).7.95.995.0 The limit does not eist because along the 0 the unction tends to ininit, whereas along the line the unction tends to. 00 Brooks/Cole, Cengage Learning

78 Chapter Functions o Several Variables ( + )( ). lim lim lim + + 0, 0,0, 0,0, 0,0 ( ) So, is continuous everwhere, whereas g is continuous everwhere ecept at ( 0, 0 ). g has a removable discontinuit at ( 0, 0 ). ( + )( ). lim lim lim 0 + +, 0,0, 0,0, 0,0 ( ) So, g is continuous everwhere, whereas is continuous everwhere ecept ( 0, 0 ). has a removable discontinuit at ( 0, 0 ). 5. lim 0 (, ) ( 0,0 ) + So, lim (, ) lim g(, ) 0., 0,0 (, ) ( 0,0) is continuous at ( 0, 0 ), whereas g is not continuous at ( 0, 0 ). + + lim, lim, 0,0 (, ) ( 0,0) + lim + (, ) ( 0,0) + 6. ( ) (same limit or g) So, is not continuous at ( 0, 0 ), whereas g is continuous at ( 0, 0 ). 5 5. lim (, ) ( 0,0 ) + Does not eist. Use the paths 0 and. 6 5. lim 0 (, ) ( 0,0 ) + + 7. lim sin + sin 0 (, ) ( 0,0) 8. lim sin + cos (, ) ( 0,0) Does not eist 9. lim (, ) ( 0,0 ) + Does not eist. Use the paths 0 and. 6 6 ( r cos θ )( r sin θ ) 5. lim lim + r, 0,0 r 0 r 0 ( r θ θ) lim cos sin 0 + r ( cos θ + sin θ) 5. lim lim + r, 0,0 r 0 r 0 ( θ θ) lim r cos + sin 0 + 50. lim (, ) ( 0,0 ) 8 r cos θ sin θ 55. lim lim (, ) ( 0,0 ) r 0 + r lim r cos θ sin θ 0 r 0 Does not eist 00 Brooks/Cole, Cengage Learning

Section. Limits and Continuit 79 56. r cos θ, r sin θ, + r, r ( cos θ sin θ ) r ( cos θ sin θ) r( θ θ) lim lim lim cos sin 0 (, ) ( 0,0) r 0 r r 0 + 57. ( ) ( r ) lim cos + lim cos cos 0, 0,0 r 0 58. ( r) lim sin + lim sin sin 0 0, 0,0 r 0 59. + r sin + sin r lim lim (, ) ( 0,0) r 0+ + r ( + ) 60. sin sin r r cos r + r r lim lim lim lim cos r, 0,0 r 0 r 0 r 0 6. + r cos( + ) cos( r ) lim lim 0 (, ) ( 0,0) 0 + r 6. + r lim ( + ) ln ( + ) limr ln ( r ) lim r ln ( r ) +, 0,0 r 0 r 0 ( r) ln r B L Hôpital s Rule, lim r ln( r) lim lim lim ( r ) 0 r 0 r 0 r + + r 0 + r r 0 +,, 68. For, the unction is clearl continuous. + + For, let. Then Continuous ecept at ( 0, 0, 0 ) sin ( ) lim 0,, + implies that is continuous or all,. Continuous or +. 69. ( t) t, g, 6. 6. sin,, e + e Continuous everwhere 65. ( ),, sin 66. Continuous everwhere 67. For 0, the unction is clearl continuous. For 0, let. Then sin lim 0 implies that is continuous or all,. ( (, )) ( ) ( ) g Continuous everwhere 70. () t t g, + ( (, )) g + Continuous ecept at (0, 0) t, g, t 7. () ( (, )) ( ) g Continuous or all + 00 Brooks/Cole, Cengage Learning

80 Chapter Functions o Several Variables t, g, + t 7. () ( (, )) g + Continuous or + 7., (a) (b) (, ) (, ) ( ) + + ( ) + ( ) lim lim lim lim ( + ) 0 0 0 0 ( + ) ( ), +, lim lim lim lim ( ) 0 0 0 0 7. (, ) + (a) (b) (, ) (, ) ( ) + + + ( + ) + ( ) lim lim lim lim ( + ) 0 0 0 0 (, ) (, ) ( ) + + + ( + ) + ( ) lim lim lim lim ( + ) 0 0 0 0 75. (, ) (a) (b) 76. (, ) (a) + ( +, ) (, ) lim lim lim lim 0 0 0 0 lim lim lim lim lim (, + ) (, ) + ( + ) ( + ) ( + ) ( + ) 0 0 0 0 0 + lim lim lim ( +, ) (, ) + + + ( + ) ( + + ) ( + + )( + ) 0 0 0 (b) B smmetr, 77. 0, + (a) (b) lim lim lim ( + + )( + ) ( + + )( + ) ( + ) 0 0, +, lim. + ( +, ) (, ) ( + ) + ( + ) ( + ) lim + lim lim ( + ) + 0 0 0 0 lim (, + ) (, ) + ( + ) ( + ) ( + ) lim 0 0 lim lim ( ) 0 0 00 Brooks/Cole, Cengage Learning

Section. Limits and Continuit 8 78. (, ) ( + ) (a) (b) ( +, ) (, ) ( + ) ( + ) lim lim 0 0 0 lim (, + ) (, ) ( + ) + ( + ) ( + ) lim 0 0 + + lim + lim 0 0 + ( L Hôpital's Rule) + 79. True. Assuming (,0) eists or 0., +. See Eercise 9. 80. False. Let ( ) 8. False. Let ( ) 8. True ( ) ( ) ln +,, 0, 0,. 0, 0, 0 + 8. lim (, ) ( 0,0 ) (a) Along a: + a + a lim lim ( a, ) ( 0,0) a 0 a + a, a 0 a I a 0, then 0 and the limit does not eist. (b) Along (, ) ( 0,0) + + : lim lim 0 Limit does not eist. (c) No, the limit does not eist. Dierent paths result in dierent limits. 8. (, ) + (a) a: (, a) ( a) a a + I a 0, (b) : a lim 0. ( a, ) ( 0,0) + a (, ) lim (, ) ( ) + + a (c) No, the limit does not eist. approaches dierent numbers along dierent paths. ( ρ sin φ cos θ)( ρ sin φ sin θ)( ρ cos φ) 85. lim lim + + ρ ρ lim ρ sin φ cos θ sin θ cos φ 0,, 0,0,0 0+ ρ 0+ 86. lim tan lim tan (,, ) ( 0,0,0 ) + + ρ 0 + ρ π 00 Brooks/Cole, Cengage Learning

8 Chapter Functions o Several Variables + and + ( ) 0. 87. As (, ) ( 0,, ) + π lim tan. + So, (, ) ( 0, ) ( ) r cos θ r sin θ lim, lim cos sin lim cos sin cos sin 0 r 88. ( ) ( r θ)( r θ) r θ θ( θ θ), 0,0 r 0 r 0 So, deine ( 0, 0) 0. 89. Because lim,, then or ε > 0, there corresponds 0 L, a, b 0 < a + b < δ. Because g L, a, b lim,, then or ε > 0, there corresponds 0 0 < a + b < δ. δ > such that δ > such that δ B the triangle inequalit, whenever, L < ε whenever g, L < ε whenever Let δ be the smaller o δ and. a + b < δ, we have ε ε (, ) + g(, ) ( L + L) ( (, ) L) + ( g(, ) L) (, ) L + g(, ) L < + ε. So, ( ) + g( ) L + L lim,,., a, b 90. Given that (, ) is continuous, then ( ) ( a b), a, b a δ > 0 such that (, ) ( a, b) < ε whenever 0 < ( a) + ( b) < δ. Let ε ( a, b), then lim,, < 0, which means that or each ε > 0, there corresponds, < 0or ever point in the corresponding δ neighborhood because ab, ab, ab, (, ) ab (, ) < < (, ) ( ab, ) < ( a, b) < (, ) < ( a, b) < 0. 9. See the deinition on page 899. Show that the value o lim (, ) is not the same or two dierent paths to ( 0, 0). 9. See the deinition on page 90., 0, 0 9. (a) True (b) False. The convergence along one path does not impl convergence along all paths. (c) False. Let ( ) (d) True ( ) ( ) ( ) ( ),. + 9. (a) No. The eistence o (, ) has no bearing on the eistence o the limit as (, ) (, ). (b) No, (, ) can equal an number, or not even be deined. 00 Brooks/Cole, Cengage Learning

Section. Partial Derivatives 8 Section. Partial Derivatives, < 0.., < 0, > 0., 0. 5. No, onl occurs in the numerator. 6. Yes, occurs in both the numerator and denominator. 7. Yes, occurs in both the numerator and denominator. 8. No, onl occurs in the numerator., 5 + 9.,, 5 0. ( ), +,,. (, ) (, ) (, )., (, ) (, ) 8.. 5. + + 6 6. 7. e 8. e e e e e e 9. e 0. e e e e e e e + e e +. ln ln ln. ln ln 00 Brooks/Cole, Cengage Learning

8 Chapter Functions o Several Variables. ln( + ) + + +. ln ln( + ) ln( ) 5. + + + + + + 6 + 6. (, ) (, ) (, ) 7. h (, ) + ( ) + ( + ) ( + ) ( ) + ( ) h, e h, e e + ( + ) ( + ) ( ) + ( ) + 8. g(, ) ln + ln( + ) g g + + (, ) + + (, ), + 9. (, ) ( + ) (, ) ( + ) + +, + 0. ( + ) + ( ) ( + ) +. cos sin sin. sin( + ) cos( + ) cos + ( ). tan( ) sec sec ( ) ( ). sin 5cos 5 5 cos 5 cos 5 5sin 5 sin 5 5. e sin e cos e sin + e cos ( cos sin ) e + 6. cos( + ) sin + sin + 7. sinh( + ) cosh + cosh + ( ) ( ) 00 Brooks/Cole, Cengage Learning

Section. Partial Derivatives 85 8. cosh sinh sinh 9. (, ) ( ) t dt t t, +, [You could also use the Second Fundamental Theorem o Calculus. ] 0. (, ) ( + ) + ( ) ( + ) ( ) dt [ t] t dt t dt ( ) ( ),, t dt t dt, +. ( +, ) (, ) lim 0 lim 0 lim 0 lim 0 ( + ) + ( + ) lim 0 (, + ) (, ) + + + lim 0. (, ) + ( ) ( +, ) (, ) lim 0 ( + ) ( + ) + + lim lim ( + ) ( ) 0 0 (, + ) (, ) lim 0 ( + ) + ( + ) + lim lim ( + + ) ( ) 0 0. (, ) + ( +, ) (, ) lim 0 + + + lim 0 lim lim 0 0 + + + + + lim ( + + + )( + + + + ) ( + + + + ) (, ) (, ) + + + + lim 0 0 lim 0 ( + + + )( + + + + ) ( + + + + ) lim 0 + + + + + 00 Brooks/Cole, Cengage Learning

86 Chapter Functions o Several Variables. (, ) + ( +, ) (, ) + + + lim lim lim + + + + 0 0 0 (, + ) (, ) + + + lim lim lim + + + + 0 0 0 5. (, ) e sin (, ) cos At (, ) sin At ( π,0, ) e π,0, π,0. e π,0 0. 6. At At, e cos, e cos 0, 0, 0, 0., e sin 0, 0, 0, 0 0. 7. (, ) cos( ) (, ) sin ( ) π π π π π π At,,, sin., sin π π At,, 8. ( ), sin, cos π π π π, sin. π π π π At,,, cos 0., cos π At,, π π, cos 0. 9. ( ), arctan (, ) + At (, ): (, ) (, ) ( ) + + At (, ): (, ) ( ) + 50. (, ) arccos (, ) At (, ), is undeined. (, ) At (, ), is undeined. 5. (, ) (, ) ( ) At (, ): (, ) (, ) ( ) + At (, ): (, ) 00 Brooks/Cole, Cengage Learning

Section. Partial Derivatives 87 5. (, ) (, ) + 5 0 ( + 5 ) At 0 0,,,. 9 7 (, ) 8 ( + 5 ) At (, ), 5. 8 8,. 9 7 g, g, At g (, ) At, : g,, : g, 5. h (, ) h (, ) At (, ): h(, ) At h h,, :, 55. 9,, (,, 6) Intersecting curve: 5 5 56. +,, (,,8) Intersecting curve: +,, 8 : At 57. 9,, (,, 0 ) Intersecting curve: 9 9 8,, 0 : 8 8 At () 58. 9,, (,,0) Intersecting curve: 9,, 0 : 6 At () 59. H(,, ) sin( + + ) H(,, ) cos( + + ) H(,, ) cos( + + ) H (,, ) cos( + + ),, 5 + 0 60.,, 6 5,, 5 + 0,, 5 + 0 0 60 0 At (,, 6 ): 0 8 5 9 8 w + + 6. w + + w + + w + + 7 + 6. w 7( + ) ( + )( 7) 7 7 w 7 + w + + w 7 + 00 Brooks/Cole, Cengage Learning

88 Chapter Functions o Several Variables 6. F(,, ) ln + + ln( + + ) (,, ) F F (,, ) (,, ) F 6. G (,, ) G G G (,, ) (,, ) (,, ) 65. (,, ) (,, ) + + + + + + ( ) ( ) ( ),,,,,,,,,, 66.,,,,,, + 0,, +,, 8 6,,,, 7 67. (,, ) (,, ),, (,, ),, (,, ) + +,, 68. (,, ) (,, ),, 0 (,, ) (,, ) (,, ) (,, ) + + ( + + ) + ( + + ) ( + + ) ( + + ) + ( + + ) ( + + ) 9 ( + + )( 0) ( + + ) ( + + ) 9 69. (,, ) sin( + ) (,, ) cos( + ) π π 0,, cos 0 (,, ) cos( + ) π π 0,, cos 0 (,, ) sin( + ) π π 0,, sin 70. + (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) 7. 6 + + 6 5 + 5 5 + + 5 5 5 + + 5 5 5, 0, 6 6, 6, 6 00 Brooks/Cole, Cengage Learning

Section. Partial Derivatives 89 7. +,, 6, 6, 7. + + 6 6 7. 75. + 6 6 6 + 6 + + + + + + 0 0 76. ln( ) ( ) ( ) ( ) ( ) So,. 77. e tan 78. e e e e e tan tan sec sec sec tan sec e e e e + e e e + e e e e e + e + + 00 Brooks/Cole, Cengage Learning

90 Chapter Functions o Several Variables 79. cos sin, 80. cos cos sin sin, cos cos sin arctan + ( ) + + ( + ) + ( ) ( ) + + + ( ) + + ( + ) + +, + 0 8., + + 0 + 0 + ( ) + 0 + 6 0, Point: (, ), 5 0 8., + + 0 5 0 5 + ( 5) + 0 9 0, Point: (, ) 8., 0, + 0 0 + 0 0, 0 Point: ( 0, 0 ) + + 85. (, ), (, ) 0: 0 + and + 0 and Points: (, ) 86. (, ) 9, (, ) + 0: 9 0 0 Solving or in the second equation,, ou obtain. or 6 0 Points: 6 0 or 0, 0,, 87. + +, + e 0 + +, + e 0 + 0 + 0 0 0 Point: ( 0, 0 ), 0 0 + + (, ) 0 0 + + 88. Points: ( 0, 0 ) 8. (, ) +, 0: + + 6 Solving or and, 6 and., + + 6 00 Brooks/Cole, Cengage Learning

Section. Partial Derivatives 9 89. sec sec 0 sec tan sec tan sec sec + tan So, sec tan There are no points or which 0, because sec 0. 90. 5 5 5 ( 5 ) ( 5 ) 5 5 ( 5 ) ( 5 ) 0 i 0 + 9. ln ln ln( + ) + + + ( ) + + + + There are no points or which 0. 9. ( ) ( ) ( ) ( ) + ( )( ) ( ) + ( ) ( ) ( ) There are no points or which 0. 9. (,, ) (,, ) (,, ) ( ) (,, ) (,, ) ( ) ( ) ( ),, 0,, 0,, 0,, 0 So, 0. 00 Brooks/Cole, Cengage Learning

9 Chapter Functions o Several Variables 9. ( ) ( ) ( ) ( ) ( ) ( ),, + +,,,, +,, 0,,,,,, 0,, 0,, 0 So, 0. 95.,, e sin,, e sin,, e cos,, sin e,, e cos,, e cos,, sin e,, sin e,, sin e So,. 96. (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) + ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) 97. 5 5 0 5 0 So, + 0 + 0 0. 98. e e sin e e cos e e sin e + e sin e e sin So, e e e e + sin sin 0. + 99. e sin e sin e sin e cos e sin So, + e sin e sin 0. 00. arctan From Eercise 80, we have + + 0. + + 00 Brooks/Cole, Cengage Learning

Section. Partial Derivatives 9 0. sin( ct) ccos ct t sin t cos( ct) c ct sin So, t ( ct) c. 0. cos( + ct) csin + ct t t c + ct 6 cos ( ct) sin + 6 cos + t ( ct) + c ( 6 cos ( ct) ) c 0. ln( + ct) c t + ct c + t ct + ct + ct c c t + ( ct) 0. sin( wct) sin( w) t sin wc cos wct w sin t wsin wct w sin So, t sin wc wct w cos sin w wct w c. 05. 06. t e cos c t e cos t c t e sin c c t e cos c c So, t c. t e sin c t e sin t c t e cos c c t e sin c c So, t c. 07. Yes. The unction (, ) cos( ) both equations. satisies 08. A unction (, ) with the given partial derivatives does not eist.,, 0 There are no s in the epression. 09. ( ),, sinh 0. ( ). I ( ). ( ),, then to ind ou consider constant and dierentiate with respect to. Similarl, to ind, ou consider constant and dierentiate with respect to. ( 0, 0, 0 ) Plane: 0 denotes the slope o surace in the -direction. denotes the slope o the surace in the -direction. ( 0, 0, 0 ) Plane: 0 00 Brooks/Cole, Cengage Learning

9 Chapter Functions o Several Variables. The plane (, ) < 0 and > 0. + satisies. The plane (, ) > 0 and > 0. + satisies 5. In this case, the mied partials are equal,. See Theorem.. 6. (, ) sin( ) (, ) cos( ) (, ) sin( ) (, ) sin( ) (, ) cos( ) (, ) sin( ) (, ) sin( ) So, (, ) (, ). 7. R 00 + 00 8 r (a) 00 8 8 At ( ),,, R (b) 00 8 8 R,,, 7. At 6 8 6 R 00 96 7. 8. (a) C + 75 + 05 + 050 C 6 + 75 C ( 80, 0) 6 + 75 8 C 6 + 05 C 6 + 05 7 ( 80, 0) (b) The ireplace-insert stove results in the cost C C increasing at a aster rate because >. 9. IQ( M C) IQ IQ M c, 00 M C 00, IQM(,0 ) 0 C 00M, IQ c(,0) C When the chronological age is constant, IQ increases at a rate o 0 points per mental age ear. When the mental age is constant, IQ decreases at a rate o points per chronological age ear. 0.7 0. 0. (, ) 00 (a) (b) 0. 0. 0 0 At (, ) ( 000, 500 ), 0. 0. 0. 500 0 0.7. 000 0.7 0.7 60 60 At (, ) ( 000, 500 ), 0.7 000 0.7 60 60 97.7. 500 0.7. An increase in either price will cause a decrease in demand.. V( I R) I V I ( I R) R ( I R) ( R) 0 + 0.06, 000 + I 9 0 + 0.06 ( R) + 0.06 ( R) ( + 0.06 ( R) ), 0,000 0,000 + I ( + I) ( + I) V 0.0, 0.8,07.0 V 9 ( R) ( + 0.06 ( R) ) 0 I I ( + I) + 0.06 0.06, 0,000 600 + + V 0.0, 0.8 65.6 R The rate o inlation has the greater negative inluence. 9 00 Brooks/Cole, Cengage Learning

Section. Partial Derivatives 95. T 500 0.6.5 T T., (, ). m T T (,) 9 m. A 0.885t.h +.0th 0.5 A (a) 0.885 +.0h t A ( 0, 0.80) 0.885 +.0( 0.80).85 t A. +.0t h A ( 0, 0.80). +.0( 0 ).6 h (b) The humidit has a greater eect on A because its coeicient. is larger than that o t. 5. PV n RT B PV T V T n P n R R B B P n n RT RT B P B V V V V n n RT R B V B P T P n n RT R T P V V B B n P V T R V P B n n RT RT B B VP n RT B 6. U 5 + (a) U 0 + (b) U 6 (c) U (, ) 7 and U, 6. The person should consume one more unit o because the rate o decrease o satisaction is less or. (d) 7. 0.9 +.0 + 0.0 (a) 0.9.0 (b) As the consumption o lavored milk () increases, the consumption o plain light and skim milk () decreases. As the consumption o plain reduced-at milk () decreases, the consumption o plain light and skim milk decreases. 8..5 + 0.0096 + 7.8. 5.65 (a).5 + 7.8.5 0.09. 0.09 (b) Concave downward < 0 The rate o increase o Medicare epenses ( ) is declining with respect to worker s compensation epenses ( ). (c) Concave upward > 0 The rate o increase o Medicare epenses ( ) is increasing with respect to public assistance epenses. 00 Brooks/Cole, Cengage Learning

96 Chapter Functions o Several Variables 9. False Let + +. 0. True. True. True. (, ) ( ),, 0,0 + 0,, 0, 0 (a) (, ) (, ) (b) ( ) ( ) + + + + + + + 0,0 0,0 0 0, 0 lim lim 0 0 0 0 0, 0, 0 0 0, 0 lim lim 0 0 0 (c) (d) ( ) ( ) ( ) 0, 0, 0 0, 0 lim lim lim ( ) 0 0 0 ( 0, 0),0 0,0 ( 0, 0) lim lim lim 0 0 0 ( 0, 0) or or both are not continuous at ( 0, 0 ).. (, ) + t dt B the Second Fundamental Theorem o Calculus, d d + t dt + t dt + d d d + t dt + d. 5. (, ) ( + ) (a) 0, 0 lim 0 lim 0 0, 0 lim 0 lim 0 (b) (, ) and (, ), 0. ( 0 +,0) ( 0,0) ( 0, 0 + ) ( 0, 0) ail to eist or 6. (, ) ( + ) For (, ) ( 0,0 ),, +. ( ) ( ) For (, ) ( 0,0 ), use the deinition o partial derivative. ( 0 + ) ( 0,0) ( ) ( + ) 0, 0 lim lim lim ( ) 0 0 0 0 00 Brooks/Cole, Cengage Learning

Section. Dierentials 97 Section. Dierentials 6. d d + d. d d d +.. d d + d w ( + ) ( + ) + ( d d ) + + + + dw d + d d 5. cos cos ( cos sin ) ( sin cos ) ( cos sin ) ( sin cos ) d + d + d + d + d 6. ( + e e ) 7. + e + e d d + e + e + d + ( e e )( d d) + + e sin d e d + e d ( sin ) ( cos ) 8. w e cos + dw e sin d + e cos d + d 9. 0. w sin cos sin 6 sin dw d + d + d w + sin ( cos ) d dw d + + cos d + +,. (a) (, ) (.,.05).05.,.05, 0.05 0. 0.05 0.05 (b) d d d. (, ) + (a) (,) 5 (.,.05) 5.55.,.05, 0.55 (b) d d + d 0. + 0.05 0.5. (, ) 6 (a) (,) (.,.05) 0.875.,.05. 0.55 (b) d d d. (, ) 0. 0.05 0.5 (a) (, ) 0.5 (.,.05) 0.5.,.05, 0 + 0. + 0.05 0 (b) d d d 5. (, ) e (a) (,) e 7.89. e (b) 6..,.05.05 8.575.,.05,.85 d e d + e d e e +, cos 0. 0.05.08 (a) (,) cos.0806.,.05. cos.05.09.,.05, 0.057 (b) d cos d sin d cos 0. sin 0.05 0.00 00 Brooks/Cole, Cengage Learning

98 Chapter Functions o Several Variables 7. Let,, 9, d 0.0, d 0.0. Then: d d + d.0 9.0 9 9 0.0 + 0.0 0. 8. Let +, 5,, d 0.05, d 0.. Then: d d + + + d 5 0.55 5.05 +. 5 + 0.05 + 0. 0.09 5 + 5 +,, 9. Let 6, d 0.05, d 0.05. Then: ( ) d d + d ( 5.95) () ( ).05 0.05 0.05 0.0 6 6 6 0. Let ( sin + ),, d 0.05, d 0.05. Then: cos( + ) + cos( + ) sin (.05) + ( 0.95) sin ( ) cos( + )( 0.05) + ( ) cos( + )( 0.05) 0 d d d. See the deinition on page 98.. In general, the accurac worsens as and increases.. The tangent plane to the surace (, ) P is a linear approimation o.. I ( ) at the point,, then dis the propagated error, d and is the relative error. 5. A lh da l dh + h dl A ( + dl)( h + dh) lh h dl + l dh + dl dh A da dl dh 6. V πr h dv πrh dr + πr dh h πrhdr h h da da l l πrdh A r da V dv 7. πrh V, r, h 8 πrh πr πr π dv dr + dh h dr + r dh dr + dh ( ) ( 6 ) π π V r + r h + h r h ( + r) ( 8 + h) 8 r r h dv V V dv 0. 0. 8.776 8.56 0.686 0. 0. 5.065 5.055 0.000 0.00 0.00 0.005 0.006 0.000 0.000 0.000 0.00 0.00 0.0000 00 Brooks/Cole, Cengage Learning

Section. Dierentials 99 8. S πr r + h, r 6, h 6 ds r h π( r + h ) + πr ( r + h ) π dr r + h ds rh π dh r + h + π π ds r + h dr + rh dh dr + dh r + h 9 S ( 6, 6).05 ( ) [ 8 96 ] π S π r + r r + r + h + h 6 + r 6 + r + 6 + h.05 r h ds S S ds 0. 0. 7.795 7.875 0.0 0. 0..65.56 0.009 0.00 0.00 0.0956 0.0956 0.0000 0.000 0.000 0.005 0.005 0.0000 9. 0.9 +.0 + 0.0 (a) d 0.9 d +.0 d 0.9 ± 0.5 +.0 ± 0.5 0. ± 0.575 (b) d [ ] [ ] Maimum error: ± 0.875 Relative error: 0. d ± 0.875 0.08 8.% 0.9.9 +.0 7.5 + 0.0, 7.,.5, d 0.05, d 0.05 r + dr dr.77( 0.05) 0.06 7..5 d + d d + d + + 7. +.5 7. +.5 0.97 d + 0.80 d arctan θ dθ d + d 0.00 d + 0.9 d + + Using the worst case scenario, d 0.05 and d 0.05, ou see that dθ 0.008.. π ( π ) + ( π ) V r h dv rh dr r dh dv dr dh + ( 0.0) + ( 0.0) 0.0 0% V r h. A absin C ( sin ) ( sin ) ( cos ) da b C da + a C db + ab C dc sin 5 ± + sin 5 ± + cos 5 ± 0.0 ± 0. in. 6 6 00 Brooks/Cole, Cengage Learning

00 Chapter Functions o Several Variables C 5.7 + 0.65T 5.75v + 0.75Tv C 0.6 0.65 + 0.75v T C 0.8 0.8 5.7v + 0.068Tv v C C dc dt + dv + ± + + ± T v ±.75 ±. ±.8 Maimum propagated error. 0.6 0.6. dc C ±.8 ± 0.9 or 9%.6807 0.6 0.8 0.8 ( 0.65 0.75( ) )( ) ( 5.7( ) 0.068( 8)( ) )( ) a v r da v v dv dr r r da dv dr a v r ( 0.0) ( 0.0) 0.08 8% Note: The maimum error will occur when dv and dr dier in signs. θ θ V bhl 8 sin 8 cos 6,0 sin θ in. 8 sin θ t 5. (a) V is maimum when sin θ or θ π. b s (b) V ( sin θ ) l h 8 θ 8 s s dv s( sin θ) l ds + l( cos θ) dθ + ( sin θ) dl π 8 π π 8 π 8 sin ( 6)( ) + ( 6)( ) cos + sin 809 in..07 t 90 6. (a) Using the Law o Cosines: a b + c bc cos A 0 + 0 0 0 cos 9 a 07. t. (b) + a b 0 b 0 cos θ 0 t 0 t 7. P + θ ( θ) + θ θ da b 0 80b cos b 80 cos db 80b sin d 0 0 80 0 cos 0 80 cos 6 80 0 sin π π π π + 0 + 0 0 80 [ 5.79 ] [ ± 77.79 ] ± 8.7 t E de, R E % 0.0, dr % 0.0 R E E dp de dr R R P R R R R E R dp E E E E de dr P de dr E R de dr de dr Using the worst case scenario, 0.0 and 0.0: E dp 0.0 0.0 0.0 0%. P R 9 00 Brooks/Cole, Cengage Learning

Section. Dierentials 0 8. + R R R R RR R + R dr R 0.5 dr R R R R R R dr dr + dr R + R R R R + R R + R ( ) ( ) When R 0 and R 5, we have 5 0 R 0.5 + 0. ohm. 0 + 5 0 + 5 9. L dl L h 0.000 ln 0.75 r ( 00) ( 6) dh dr ± ± 0.000 0.000 ( 6.6) 0 h r ± 00 6 0.000 ln 00 0.75 dl 8.096 0 6.6 0 micro henrs ± ± 6 L 0. T π g dg..09 0. dl.8.50 0.0 T T π L π T dt dg + dl dg + dl g L g g Lg When g.09 and.50,, +. π.5 π 0. + 0.0 0.008 seconds..09.09.5.09 L T ( ) +, +, + + + + + (, ) (, ) ε ε where ε and ε 0. + + + + + 0 + + + As ( ) ε ε, 0,0, 0and 0.. (, ) + (, ) (, ) ( ) + + + + + + + + + + +, +, + ε + ε whereε and ε. As ( ) ε ε, 0,0, 0and 0.. (, ) ( +, + ) (, ) ( + ( ) + ( ) )( + ) + + + + + + + +, +, + + where and +. ε ε ε ε As ( ) ε, 0,0, 0 and ε 0. 00 Brooks/Cole, Cengage Learning

0 Chapter Functions o Several Variables. (, ) 5 0 + ( +, + ) (, ) 5 + 5 0 0 + + + + 5 0 + 5 + 0 + 0 + + ε ε ε ε, +, + + where 0and +. As ( ) ε 5. (, ), 0,0, 0 and ε 0., (, ) ( 0,0) + 0, (, ) ( 0, 0) 0 0 (,0) ( 0,0) ( ) 0, 0 lim lim 0 0 0 0 0 ( 0, ) ( 0, 0) ( ) 0, 0 lim lim 0 0 0 So, the partial derivatives eist at ( 0, 0 ). Along the line : lim (, ) lim lim 0 (, ) ( 0,0) 0 0 + + Along the curve : lim, (, ) ( 0,0) is not continuous at ( 0, 0 ). So, is not dierentiable at ( 0, 0 ). ( See Theorem.5 ) 6. (, ) 5,, 0,0 + 0,, 0, 0 ( ) ( ),0 0,0 0 0 0, 0 lim lim 0 0 0 0, 0, 0 0 0 0, 0 lim lim 0 0 0 So, the partial derivatives eist at ( 0, 0 ). 5 5 Along the line : lim (, ) lim. (, ) ( 0,0) 0 Along the line 0, lim (, ) 0. (, ) ( 0,0) So, is not continuous at ( 0, 0 ). Thereore is not dierentiable at ( 0, 0 ). 7. I is dierentiable at ( 0, 0), (, ) eists, which implies that (, ) 0 0 then the partial derivative dierentiable at (, ). I ( ) 0 0 0 is, +, then,0. Because is not dierentiable at 0,, + is not dierentiable., + 8. (a), + 5.6.05,..05 +..79 (b).05,., 0.8 (c) d d + d + + ( 0.5) + ( 0.) 0.8 5 5 The results are the same. 00 Brooks/Cole, Cengage Learning

Section.5 Chain Rules or Functions o Several Variables 0 Section.5 Chain Rules or Functions o Several Variables.. w +, t t dw w d w d + + dt dt dt + 6 8t + 8t 6t ( ) w + cos t, t e dw w d w d + dt dt dt t ( sin t) + e + + t sin t + e cos t sin t + e t + cos t + e. w sin. t e, π t dw w d w d t + sin e + cos ( ) dt t dt t t t t sin π te e cos π t e sin t+ e cos t w ln cos t sin t dw + dt tan t + cot t sin t cos t ( sin t) ( cos t) t 5., t t w e, e dw w d w d (a) + dt dt dt e + e e e ee e (b) w e e e dw t e dt t t t t t t t t t t 6. w cos, t, dw dt t sin t sin t (a) sin( )( t) + sin( )( 0) dw dt (b) w cos( t ), t sin( t ) 7. w + +, cos t, sin t, (a) dw w d w d w d + + dt dt dt dt sint + cost + e t e t cos t sin t + sin t cos t + e e t t (b) w cos t + sin t + e + e dw e dt t t t 8. w cos t t arccos t dw ( cos )( ) + ( cos )( t) + ( sin ) t () t + t()( t t) t( t ) t (a) dt t t t + t + t t dw (b) w t, t dt 9. w + +, t, t, t dw w d w d w d dt dt dt dt (a) + + ( + ) + ( + )( t) + ( + ) ( t + t) + ( t + t)( t) + ( t + t ) ( t ) (b) w ( t )( t ) + ( t ) t + ( t ) t dw tt + t + t + t t dt 00 Brooks/Cole, Cengage Learning

0 Chapter Functions o Several Variables 0. w + +, t, t, dw w d w d w d (a) + + dt dt dt dt + t + + + + 0 t + t t + t + t + 8 (b) + + + dw t + 8 dt w t t t t 6t 8t. Distance () t ( ) + ( ) ( t t) + ( t t) 0 cos 7 cos 6 sin sin t 0 cos t 7 cos t + 6 sin t sin t () ( 0 cos t 7 cos t)( 0 sin t 7 sin t) ( 6 sin t sin t)( cos t cos t) + + π 9 ( 0) + ( 0)( 7) ( ( ) ( ) ( 6) ( ).0 + 9 9. Distance (). () 8 8 6 () t + 8 + + 8t 8t 8 6 t ln, t t w + e, e dw w d dw d + dt dt dt t t t t e e ( e ) + ( e ) t t + + e + e d w t ( e t t + e )( e t t + e ) ( e t t e )( e t e ) t t ( + ) dt e e At t 0, t t ( e + e ) d w dt. ( + ). w t t + t dw w d w d + ( t) + () dt dt dt t ( t) t ( t + )( t ) t t + t t + t + t + t + d w dt At t : ( t + ) ( t + t ) ( t + t ) ( t + ) ( t + ) d w 7 68.5 dt 6 6 5. w + s + t, s t w s () + () ( s + t) + ( s t) s w t () + ( ) ( s + t) ( s t) t w w When s and t 0, and s t 0. 6. w s t e, e w 6( e ) + ( s )( 0) 6e e e 6e w t ( 6)( 0) + ( t ) e ( t e s t e ) e t e s t e w w 6 When s and t, 6and e. s t 7. w sin( + ) s + t s t s s t s s+ t w s cos + + cos + 5cos + 5cos 5 ( ) ( ) ( ) ( s t) w t cos + cos + cos + cos 5 ( ) ( ) ( ) ( s t) π w w When s 0 and t, 0 and 0. s t 00 Brooks/Cole, Cengage Learning

Section.5 Chain Rules or Functions o Several Variables 05 8. w scos t ssin t w s cost sint s t s t s t cos sin cos w ( ssin t) ( scos t) s sin t t π w w When s and t, 0 and 8. s t 9. w, θ, r + θ, r θ w + r + + r θ (a) ( 0) () () w ( θ ) + () + ( ) θ + + θ + θ θ ( r θ )( r θ) ( r θ) ( r θ) ( θ r ) r θ θ θ r + θ r θ r (b) w θ θ w r r θ w θ r θ 0. w +, r + θ, r θ w + + 0 r (a) ( ) ( ) w ( )( ) + ( + )( ) ( ) ( r + θ ) ( r θ) 8θ θ + θ + θ θ + θ + θ + θ θ + θ + θ θ (b) w ( r ) ( r )( r ) ( r ) ( r r ) ( r ) ( r r ) w r 0 w 8θ θ. w arctan, r cos θ, r sin θ (a) w r sin θ cos θ rcos θ sin θ cos θ + sin θ + 0 r + + r r ( r sin θ)( r sin θ) ( r cos θ)( r cos θ) w ( r sin θ) + ( r cos θ) + θ + + r r rsin θ θ r cos θ (b) w arctan arctan( tan θ ) w r 0 w θ. w 5 5 5, r cos θ, r sin θ (a) w 5 5 5r cos θ 5r sin θ 5r cos θ + sin θ r 5 5 5 5 5 5 5 5 5 5 5r w 5 5 5r sin θ cosθ 5r sinθ cosθ ( r sin θ) + ( r cos θ) 0 θ 5 5 5 5 5 5 5 5 5 00 Brooks/Cole, Cengage Learning

06 Chapter Functions o Several Variables (b) w 5 5r w 5r w ; 0 r 5 5r θ. w, s + t, s t, st w s () () + + t ( s tst ) ( s tst ) ( s t)( s tt ) st st t st t t( s t ) + + + + + w ( ) + ( ) + ( st) ( s t) st ( s + t) st + ( s + t)( s t)( st) st + s t st s t st t ( t ) st s. w + +, t sin s, t cos s, st w + cos s + ( t sin s) + ( t ) s t sin scos s t sin s cos s + st st w sin s + cos s + ( st) t t sin s + t cos s + s t t + s t 5. w e, s t, s + t, st w e () + e () + e () t s ( s t)( s+ t e ) ( s + t) st + ( s t) st + t ( s t)( s+ t) s t e s t + t te ( s + ) w e ( ) + e ( ) + e ( s) t ( s t)( s+ t e ) ( s + t)( st) + ( s t) st + s e st + s se t ( s t)( s+ t) s t ( ) 6. w cos, s, t, s t w cos sin 0 sin s cos sin ( )( s) ( ) ( ) ( st t ) s s t ( st t ) w cos( )( 0) sin( )( t) sin( )( ) t + s ts ( t) sin( st t) st sin( st t) ( 6st st) sin( st t) 7. + + 0 d F (, ) + d F, + + + 8. sec + tan + 5 0 9. F, sec tan + sec F, sec tan + sec d d ln + + + ln ( + ) + + 0 F (, ) (, ) ( sec tan sec ) ( sec tan + sec ) + + d + d F + + + 0. 6 0 + (, ) (, ) d F d F + ( ) ( + ) ( ) ( + ) + + 5 + + +. F( ),, + + F, F, F F F F F + + 00 Brooks/Cole, Cengage Learning

Section.5 Chain Rules or Functions o Several Variables 07. F(,, ) + + F F + + F + F + F + F + F +. F( ) F (,, ) F (,, ),, + + 0 F,, + + F,, + +. ( ) + sin + 0 + cos + 0implies sec ( + ). cos + (i) ( ) + cos + 0implies. (ii) ( ) 5. F( ) ( ) ( ),, tan + + tan + sec sec ( + ) sec sec ( + ) + sec ( + ) sec sec ( + ) + sec ( + ) F sec + F sec + + + F sec + F F + F F + 6. F(,, ) e sin( + ) F e sin( + ) F e cos( + ) F e cos( + ) F e sin( + ) F e cos( + ) F e cos( + ) F e cos( + ) 7. F (,, ) (,, ) F ( ) (,, ) F,, e + 0 e + F e,, e F e e 8. ln + + 8 0 F (,, ) ln (i) F (,, ) + + F (,, ) + (ii) F,, + + 9. F(,,, w) + w + w s F + w F + F w Fw + w F + w + w Fw + w F + + Fw + w F w w F + w 0. + 5w + 0w F(,,, w) F, F 5 w, F, Fw 5 + 0w w F Fw 5 + 0w 5 0w w F 5w Fw 0w 5 w F F 5 0w w F,,, w cos + sin + w 0 w F sin Fw w F sin cos Fw w F cos + w F. w 00 Brooks/Cole, Cengage Learning

08 Chapter Functions o Several Variables. F,,, w w 0 ( ) w F F w w F ( ) + ( ) + F w w F F w. (a) (, ) + ( t)( t) ( t) ( t) ( t, t) t t, + + (b) Degree:,, + +, + + +. (a) (, ) + (, ) + ( + ) (, ) t t t t t t t t Degree: (b) (, ) + (, ) ( ) + ( 6 + ) 9 + (, ) 5. (a) (, ) e t t ( t, t) e e (, ) Degree: 0, +, e + e 0 (b) 6. (a) (, ) + ( t) ( t) ( t) ( t, t) t t, + + (b) Degree: + + +, +, +, ( + ) ( + ) ( + ) ( + ) + dw w d w d dg dh 7. + + dt dt dt dt dt At t,,,, 5 and dw So, ( 5)( ) + ( 7)( 6) 7 dt, 7. 8. w w w + s s s g h + ( 5)( ) + ( 7)( 5) 50 s s w w w + t t t g h + ( 5)( ) + ( 7)( 8) 66 t t 00 Brooks/Cole, Cengage Learning

Section.5 Chain Rules or Functions o Several Variables 09 dw w d w d dt dt dt 9. + ( Page 95) 50. 5. w w w + s s s w w w + t t t (, ) (, ) (,, ) (,, ) (,, ) (,, ) d d ( Page 95) 5. ( ),,, t,, t d d d d (a) + + dt dt dt dt t e t ( ) t + + e t t t t te + te te te t (b) ( t ) t t t e t e d dt te te te t t t t + 6 The results are the same. 5. V πr h dv dr dh dr dh π + π + π + ( ) π dt dt dt dt dt rh r r h r 6 6 608 in. min S π r r + h ds dr dh π ( r h) r π π dt + + + + dt dt 6 6 6 in. min 5. V πr h dv dr dh π rh r π + + π dt dt dt 6 6 56 in. min S πr r + h + πr ( Surace area includes base. ) ds r dr rh dh π r + h + + r + dt dt dt r + h r + h 6 π + + + + + 6 + 6 6 6 6 68 6 π + + + + + 0 0 0 5 π π π 0 6 in. min 0 9 0 in. min 55. pv T mrt mr ( pv) dt dp dv V p dt mr + dt dt 56. θ θ A bh sin cos sin θ θ θ + sin θ + cos θ dt dt θ dt dt dt da A d A d d d π π π π + + 90 0 6 6 sin cos m hr.566 m h h θ b 00 Brooks/Cole, Cengage Learning

0 Chapter Functions o Several Variables di dr dr m r r m ( 6) ( 8) 8m cm sec dt + + dt dt 57. I m( r + r ) π V r rr R h 58. ( + + ) dv π dr dr dh ( r R) h ( r R) h ( r rr R ) dt + + + + + + dt dt dt π ( 5) 5 ( 0)( ) 5 ( 5) ( 0)( ) ( 5) ( 5)( 5) ( 5) ( ) + + + + + + π ( 9,500) 6,500 cm min π ( R r) S π R + r R r + h ds ( R r) dr ( R r) dr π ( R r) + h ( R + r) + ( R r) + h + ( R + r) dt ( R r) h dt ( R r) + + h dt + + dt h dh R r + h 5 5 π ( 5 5) + 0 ( 5 + 5) ( 5 5) + 0 5 5 + ( 5 5) + 0 + ( 5 + 5) ( 5 5) + 0 0 cm min π 59. w (, ) u v v u w w d w d w w + u du du w wd wd w w + + v dv dv w w + 0 u v 60. w ( ) sin( ) w + w w w + 0 ( ) cos( ) sin( ) ( ) cos( ) sin( ) 0 + 5 + 5 ( 5 5) + 0 00 Brooks/Cole, Cengage Learning

Section.5 Chain Rules or Functions o Several Variables 6. w ( ),, r cos θ, r sin θ w w cos θ + w sin θ r w w w + θ ( r sin θ ) ( r cos θ ) w w w (a) r cos θ r cos θ + r sin θ cos θ r w w w sin θ ( r sin θ) r sin θ cos θ θ w w w r cos θ sin θ ( r cos θ + r sin θ) r θ w w w r ( r cos θ) sin θ r θ w w wsin θ cos θ r θ r w w w r sin r sin cos + r sin r θ θ θ θ ( First Formula) cos θ w w w ( r sin cos ) ( r cos ) θ θ θ + θ w w w r sin θ + cos θ r sin θ + r cos θ r θ w w w r r sin θ + cos θ r θ w w wcos θ sin θ + r θ r ( Second Formula) (b) + cos + sin cos + sin + sin w w w w w w w θ θ θ θ θ r r θ w w w w w sinθ cosθ + cos θ + 6. w arctan, r cos θ, r sin θ r sin θ π π arctan arctan( tan θ) θor < θ < r cos θ w +, w +, w w 0, r θ w w + + + + + r ( ) ( ) w w + 0 + () r r θ r r w w w w So, + +. r r θ 00 Brooks/Cole, Cengage Learning

Chapter Functions o Several Variables 6. Given u v and u v, r cos θ and r sin θ. u u u v v cos θ + sin θ cos θ sin θ r v v v v v ( r sin θ ) + ( r cos θ) r cos θ sin θ θ u v So,. r r θ v v v u u cos θ + sin θ cos θ + sin θ r u u u u u ( r sin θ ) + ( r cos θ) r cos θ + sin θ θ v u So,. r r θ 6. Note irst that u v + u v. + u cos sin cos sin r θ r θ θ + θ + r + + r r v sin cos ( sin ) ( cos ) r θ r r r θ θ + θ + θ + + r So, u v. r r θ v r sin θ cos θ r sin θ cos θ cos θ + sin θ + 0 r + + r u sin cos sin cos ( sin ) ( cos ) r θ θ r r r θ θ θ + θ + 0 θ + + r So, v u. r r θ Section.6 Directional Derivatives and Gradients., + 9 v i + j 5 5 (, ) ( ) i ( 9 ) (, ) 5i 5j + j + v u i + j v 5 5 D,, u + u. (, ), v ( i + j), i j, 8i 7j D u v u i + j v (, ) (, ) u 7 00 Brooks/Cole, Cengage Learning

Section.6 Directional Derivatives and Gradients. (, ) v ( i + j) (, ) i + j 0, i v u i + j v D 0, 0, u u. (, ) v j (, ) i j (, ) i j v u j v D,, u u 5. h, e sin v i h e sin i + e cos j π h, ei u v v i π π Dh u, h, u e 6. g( ), arccos v j g (, ) i + g(, 0) j v u j v Dg, 0 g, 0 u u 7. (, ) g + v i j g i + j + + g(, ) i + j 5 5 u v i j v 5 5 7 Dg u (, ) g(, ) u 5 j 8. h (, ) ( ) e v i + j ( 0, 0) 0 h ( ) ( ) h e i e h Dh0, 0 0, 0 u 0 u 9. (,, ) + + + v i j k ( + ) + +,, i + j + k,, i + j + k u v i j + k v Du (,, ) (,, ) u 0. (,, ) + + v i + j k (,, ) (,, ) i k + i + + j + + k + v u ( i + j k) v 6 D u. h (,, ) (,, ) (,, ) u 6 6 6 6 v,, h i + j + k h(,,) i + j + k u v i + j + k v 8 Dh u (,,) h(,,) u. h,, arctan v,, h (,, ) arctan i + j + k + ( ) + ( ) h(,,) π i + j + k u v,, v 6 6 6 π + 8 ( π + 8) 6 Dh u (,,) h(,,) u 6 j 00 Brooks/Cole, Cengage Learning

Chapter Functions o Several Variables. (, ) +. (, ) u i + j i + j Du u + + + u i j i + ( + ) ( + ) Du u ( + ) ( + ) ( ) ( + ) + 5. (, ) sin( + ) j u i + j cos + + cos + j i Du u cos + + cos + 6. g (, ) 7. e u i + j g e e i + + cos + j ( ) ( ) e Dg u e + e, + v i + j i + 6 j,, i + 6j v u i + j v 5 5 6 Du (, ) (, ) u + 6 5 5 8. (, ) cos( + ) π v i π j sin + + j i sin v u i j v 5 5 Du sin + + + 5 5 At 9. g (,, ) sin 5 sin + + 5 5 0, π, D 0. At u e ( ) sin( ) v i j g e i + e j + e k,, 0, g i + j + 8 k. v u i j v 5 5 8 Dg u g u 5 5 5 0. h (,, ) ln( + + ) At v i + j + k h + + + +, 0, 0, h i + j + k. ( i j k) v u i + j + k v 9 ( ) 7 7 9 Dh u h u 9 9. ( ), + 5 +, i + 0j, i + 0j. g, e g (, ) e + e i + e g, 0 i + j. ln( ) (, ) i, i j j j 00 Brooks/Cole, Cengage Learning

Section.6 Directional Derivatives and Gradients 5. cos( + ) (, ) sin( + ) i sin( + ) 5., 6 sin 5i + 8 sin 5j 0.79i.0588j w 5 + w,, 6i 0j + k w,, 6i 0j 8k 6. w tan( + ) w (,, ) tan( + ) i + sec ( + ) + sec ( + ) k w(,, ) tan + sec + sec ( π ) i j k 7. PQ i + j, u i + j g (, ) i + j, g(,) i + j Dg u g u + 8. PQ i + j, u i + j 5 5 6i j, (,) 6i 8j 8 Du u 5 5 5 9. PQ i + j, u i + j 5 5 e cos i + e sin j ( 0, 0) i 5 Du u 5 5 π 0. PQ i + π j, u i + j 5 5 cos cos i sin sin j,0 i 5 Du u 5 5. (, ) + (, 0) i + j, + i + j, 0 j j. (, ) + + (, ) i + + + ( ) ( 0, ) i + j ( 0,) + 5 6. h (, ) tan (, ) tan i + sec h j π h, i + j π h, 7. h (, ) cos( ) h (, ) sin( ) i cos( ) sin( ) π π π h 0, i + j 6 6 π π 9 6 π + π h 0, + 6 6 5. g (, ) (, ) g e 0, 5 5i + j g 0, 5 6 ( π π + ) 6 + + j i + g e e 6. g(, ) ln + ln( + ) (, ) i j ( i j) (, ) j j g (, ) i j + + + g + + 5 5 5 g 7. (,, ) 5 5 + + (,, ) ( i j k) (,, ) ( i j k) + + + + + +,, 00 Brooks/Cole, Cengage Learning

6 Chapter Functions o Several Variables 8. 9. w w + + ( i j k ) ( 0, 0, 0) w 0 w 0, 0, 0 0 w w i + j + k w,, i + j + k w,, e e i + e j + e k, 0, i 8j 0. (,, ) (,, ), 0, 65 For eercises 6, (, ) Dθ (, ) cosθ sin θ.,. 9. (a) D (, ) π (b) Dπ (, ) (c) D (, ) π (d) D (, ) π 6 (,, ) 6 and 5 + 5 Du u. (a) u ( i + j). 5. 6. (b) v i j v 9 + 6 5 u i j 5 5 Du u + 5 5 5 (c) v i + j v 9 + 6 5 u i + j 5 5 Du u 5 5 5 (d) v i + j v u 0 i + j 0 0 Du u 0 6 0 60 i j + 9 6 i j ( i j) So, u ( )( i j) and Du, u 0. is the direction o greatest rate o change o. So, in a direction orthogonal to, the rate o change o is 0. 7. (a) In the direction o the vector i + j (b) ( ) i + 0 0( + ) (, ) ( ) () ( Same direction as in part ( a) ) 0 0 5 0 i + j i + j (c) 5 i 0 j, the direction opposite that o the gradient j 00 Brooks/Cole, Cengage Learning

Section.6 Directional Derivatives and Gradients 7 8. (a) In the direction o the vector i + j (b) i + j i + j (, ) i + j Same direction as in part a (c) ( ) i j, the direction opposite that o the gradient 9. (, ),(,,7) (a) 8, + + 50. (a) ( ) ( ) + + + + + + Circle: center: ( 0, ), radius: 6 8 + 8 8 (b) + j i ( + + ) ( + + ) (, ) i (b) u Du θ D,, u cosθ sinθ D u 8, 8 cos + 6 sin θ (c) The directional derivative o is 0 in the direction ± j. (d) 6 π π θ 6 8 (c) Zeros: θ., 5.6 These are the angles θ or which D (, ) ero. g θ Du, 8 cos θ + 6 sin θ g θ 8sinθ + 6cosθ (d) Critical numbers: θ 0.6,.79 These are the angels or which D (, ) maimum ( 0.6) and minimum (.79 ). (e) i j the maimum value o D (, ), θ u u is a equals, 6 + 6 0, u at 0.6., 7 ( ) curve at (, )., 8i + 6jis perpendicular to the level 6 Generated b Mathematica 6 6 6 Generated b Mathematica 5., 6 c 6, P 0, 0, i j 6 6 0 + 0, 0 i j 5. (, ) + c 5, P,, i + j + 5, 6i + 8j 5. (, ) (, ) 6 6 c, P, i + j, i j 00 Brooks/Cole, Cengage Learning

8 Chapter Functions o Several Variables + c, P (, ) 5. (, ) (, ) + + 0 (, ) j 55., (a) ( ) i ( + ) ( + ), 8i j,0 6i j (b) 6i j 57 ( 6i j) is a unit vector normal to the level 57 curve 6, 0. at (c) The vector i + 6jis tangent to the level curve. 6 Slope 6 0 6( ) 6 Tangent line (d) 5 0 5 5 0 56. (, ) (a) ( ) i i, j, + j (b) 5 0, 5 ( ) 5 i + j is a unit vector normal to the level curve at (, ). (c) The vector i jis tangent to the level curve. Slope. + ( ) + Tangent line 5 j (d) 57., (a) 6i j, 6i j (b), 6 + 6 ( ) i j is a unit vector normal to the level curve at (, ). (c) The vector i + jis tangent to the level curve. (d) Slope. ( ) 58. (, ) 9 + (a) 8i + 8j, 6i 8j (b) tangent line, 60 85 ( 9 ) 85 i j is a unit vector normal to the level curve 9 + 0at (, ). (c) The vector i + 9jis tangent to the level curve. (d) 6 (, ) Slope 9. 9 + ( ) 9 0 Tangent line (, ) 00 Brooks/Cole, Cengage Learning

Section.6 Directional Derivatives and Gradients 9 59. See the deinition, page 9. 60. Let (, ) be a unction o two variables and u cos θi + sin θ j a unit vector. (a) I θ 0, then Du. (b) I θ 90, then Du. 6. See the deinition, pages 96 and 97. 6. 65. T + T i ( + ) ( + ) 7 T 65 65 65 66. 67. (, ) i j ( 7i j) h, 5000 0.00 0.00 h 0.00i 0.008j i h ( i j) h 500, 00.jor 5 5 + j P 5 67 B 800 99 6. The gradient vector is normal to the level curves. See Theorem.. 6. (, ) 9 and Dθ (, ) cosθ sinθ ( cosθ + sinθ ) (a) (, ) 9 (b) D π, (c) Dπ (, ) + ( + ), i j (d), + 6 0 5 (e), i j,, 5 ( i j) Thereore, u ( 5)( i j) + and D,, u 0. u 9 (,, ) 68. The wind speed is greatest at B. 69. T(, ) 00, P ( 0,0) d d dt dt t Ce t t () Ce t C 0 ( 0) C 0 0 t 0e t t () 0e t 0 () 0 t 00e t 70. T(, ) 00, P (, ) d d dt dt t Ce t t () Ce t C ( 0) C 0 t e t t () e t A e u 6 6 800 t 00 Brooks/Cole, Cengage Learning

0 Chapter Functions o Several Variables 7. (a) 500 7. (a) 00 D 00 6 6 (b) ( ) + T, 00e i 7 T, 5 00e i j j There will be no change in directions perpendicular ± i 6j to the gradient: (c) The greatest increase is in the direction o the gradient: i j (b) The graph o D 50 0 50 sin( π ) would model the ocean loor. π, 0.5 50 + 0 + 50 sin 5. t (c) D D (d) 60 D, 0.5 60 and D π (e) 5π cos and D π (, 0.5) 5π cos 55.5 ( ) π D 60i + 5π cos j D, 0.5 60i + 55.5j 7. True 7. False π π Du (, ) > when u cos i + sin j. 75. True 76. True 77. Let (,, ) e cos + + C. Then (,, ) e cos i e sin j + k. 78. We cannot use Theorem.9 because is not a dierentiable unction o and. So, we use the deinition o directional derivatives. ( + cos θ, + sin θ) (, ) t t Du (, ) lim t 0 t D u t t t t 0,0 ( 0,0 + + ) t ( 0, 0) lim lim lim lim which does not eist. t 0 t 0 t 0 t t t t t t t 0 t + t t 0 +,0 + t 0, 0 lim lim 0 t 0 t 0 t t t which implies that the directional derivative eists. I ( 0, 0), then Du 79. (a) (, ) is the composition o two continuous unctions, h (, ) thereore continuous b Theorem.. and g( ), and 00 Brooks/Cole, Cengage Learning

Section.7 Tangent Planes and Normal Lines (b) (c) 0 +,0 0,0 0 0 0, 0 lim lim 0 0 0 0, 0 + 0, 0 0 0 0, 0 lim lim 0 0 0 π π Let u cos θi + sin θ j, θ 0,, π,. Then ( + + ) 0 t cos θ,0 t sinθ 0,0 t cos θ sin θ cos θ sin θ Du ( 0, 0) lim lim lim, does not eist. t 0 t 0 t 0 t t t Section.7 Tangent Planes and Normal Lines F,, 5 + 5 0 5 + 5 Plane.. F( ),, + + 5 0 + + 5 Sphere, radius 5, centered at origin.. F( ),, + 9 0. + 9 Elliptic cone F,, 6 9 + 6 0 6 9 6 0 5. + Hperbolic paraboloid F,, + + 0 F i + j + k, F 9 + 6 + F n i + j + k F F,, + + F i + j + k 6. F n i + j + k i + j + k F 7. F( ),, + + 6 F i + j + k F,, i + j + k F,, + + 6 6 F n i + j + k F 6 6 6 8. (,, ) + F F(,, ) i + j k + + F(,, 5) i + j k 5 5 F 5 n i + j k F 5 5 5 ( i + j 5k) 5 9. (,, ) F F i + 5 0 F,, 8 i k F,, 8 5 ( i j k) F n i k F 5 k ( ) 00 Brooks/Cole, Cengage Learning

Chapter Functions o Several Variables 0. (,, ) F F F,, + i j k,,6 i + j k n F F ( i + j k). F 09 09 + 09 F ( i j k),, + + 9 F,, + + i j k,, i + j + k F n i + j + k F. F,, + ( ) ( ) ( 6 ) F + i + j + k F,, i + 5j 7k F,, 0 F n i + j k F 0 ( 5 7 ) F(,, ) i j + k. F(,, ) ln ln ln( ) (,, ) F i j + k F n i j + k i j + k F. F( ),, e F,, e i e j + e k F,, i j + k F n i j + k F 7 5. F,, sin + F,, sin i cos j + k π F 6,, 7 i j + k 6 n F F i j + k ( i 6 j + k ) ( i 6 j k) + 6. F(,, ) sin( ) F(,, ) cos( ) cos( ) i j k π π F,, i j k 6 n F F i j k 0 ( i j k ) 0 ( i j k) 0 0 7. (, ) + +, (,,8) F,, + + (,, ) F (,, ) F ( ) F F (,, 8) F (,, 8) F ( ) ( ) ( ) + 8 0 + 8. (, ),(,,) (,, ) F F,,,, 8 (,, ) F (,, ) F ( ) F (,, ) F (,, ) F ( ) ( ) ( ) + 0 + + 0 +,,,, 00 Brooks/Cole, Cengage Learning

Section.7 Tangent Planes and Normal Lines 9. (, ) +,(,,5) F(,, ) + F (,, ) + 5 ( ) + ( ) ( 5) 0 5 5 ( ) + ( ) 5( 5) 0 + 5 0 F (,, ) + F,, F (,, 5) F (,, 5) F G (,, ) arctan 0. g (, ) arctan,(,0,0) G (,, ) ( ) ( ) + + 5 G (,, ),, 5 + ( ) + G,, G (, 0, 0) 0 G (, 0, 0) 0. g (, ) +, (,,) G (,, ) + G(,, ) G(,, ) G(,, ) G(,, ) ( ) ( + ) ( ) 0 G,, G,,. +,(,,) F(,, ) + F (,, ) F (,, ) + F ( ) F (,, ) F (,, ) F ( ) + ( ) ( ) 0 + 0 +. (, ), (,,) F,, F (,, ), F (,, ), F ( ) ( ) ( + ) ( ) 0 + 0 + + 6,,,,,, G, 0, 0 00 Brooks/Cole, Cengage Learning

Chapter Functions o Several Variables π sin +, 0,,. e ( ) (,, ) ( sin ) F e + (,, ) ( sin + ) F (,, ) e cos F ( ) F e,, F π 0,, F π 0,, 0 F π 0,, 5. h (, ) ln +,(,,ln5) H(,, ) ln + ln( + ) H H (,, ) (,, ln 5) 5 + H H (,, ) (,, ln 5) ( ) + ( ) ( ln5) 0 5 5 + 5 ln5 0 ( ) ( ) ( ) 5 + + 5 5 ln 5 H H ( ),,,, ln 5 6. h π, cos, 5,, H,, cos H ( ),, 0 H,, sin π π H 5,, 0 H 5,, π 0 π + + 0 8 ( π ) + 8 + H ( ),, π H 5,, 7. + + 6, (,, ) F,, + + 6 F,, F,, ( ) F,, 8 F,, 6 ( ) ( ) ( ) ( ) ( ) ( ) 6 + + 8 0 + + 0 + 8 F,, F,, 8 00 Brooks/Cole, Cengage Learning

Section.7 Tangent Planes and Normal Lines 5 8. +,(,, ) F,, + F,, F,, ( ) F,, F,, 6 ( ) ( ) ( ) ( ) ( ) ( ) 6 8 + 0 + 0 0 F,, F,, 8 9. + 8, (,,) F,, + 8 F,, + F,, ( ) ( ) ( ) ( ) F,, F,, 6 6 + 0 0. (, ) (,,) F,, + F ( ),, F,, 6 6 F,, + F,, ( ) ( ) ( ) 8 0 8 6 + + 8 6 F,, F,, F,, F,, 8. + + 9, (,, ) F(,, ) + + 9 F (,, ) F ( ) F (,, ) F,,,, ( ) ( ) ( ) F,, F,, + + 0 + + 9 same plane! Direction numbers:,, Line:. + + + 9, (,, ) F( ) + + F (,, ) F (,, ) F,, 9 F,,,, F,, F,, Direction numbers:,, Plane: ( ) + ( ) + ( ) 0, + + 9 Line: 00 Brooks/Cole, Cengage Learning

6 Chapter Functions o Several Variables. + + 9, (,, ) F(,, ) + + 9 F (,, ) F (,, ) F F,,,, F,, F,, Direction numbers:,, Plane: ( ) + ( ) + ( ) 0, + + Line:. 6, (,, 8) F(,, ) 6 F (,, ) F (,, 8) F ( ) ( ) ( ) F,,,, 8 8 0 + + Direction numbers:,, Line: 8 5.,(,,5) F(,, ) F (,, ) F (,, 5) 6 F ( ) ( ) ( ) F,,,, 5 6 5 0 6 5 Direction numbers: 6,, 5 Line: 6 6. 0, (,, 6) F(,, ) F (,, ) F (,, 6) (,, ) F F,, 6 F,, F,, 8 F,, F,, 5 F ( ),, F,, 6 Direction numbers:,, Plane: ( + ) + ( + ) + ( 6) 0, + + 6 6 Line: + + 7. 0, (,, 5) F( ) F (,, ) F (,, 5) 0,, 0 F (,, ) F,, 5 5 (,, ) F F,, 5 Direction numbers: 0, 5, Plane: 0( ) + 5( ) + ( 5) 0, 0 + 5 + 0 5 Line: 0 5 00 Brooks/Cole, Cengage Learning

Section.7 Tangent Planes and Normal Lines 7 8. e,( 0,,) (,, ) (,, ) (,, ) ( + ) ( 0,, ) 8 F ( 0,,) ( ) + ( ) ( ) F e F e F 8 0 0 8 + 0 Direction number: 8,, Line: 8 F e F,, F 0,, π 9. arctan,,, F(,, ) arctan F (,, ) F (,, ) F (,, ) + + π π π F,, F,, F,, Direction numbers:,, Plane: ( ) ( ) π π + 0, + Line: ( ) π 0. ln( ), ( e,,) [ ] (,, ) F,, ln + ln F e (,,) F e e F,, ln + ln F e,, ( e) ( ) ( ) + + 0 Direction numbers:,, e e e + + 8 e. F(,, ) + 5 (,, ) F(,, ) i + j (,, ) F(,, ) i + j (,, ) (a) F G i j k G G i k G i k F (,, ) F e,, 0 i + j k ( i j + k) 0 Direction numbers:,, Line: 00 Brooks/Cole, Cengage Learning

8 Chapter Functions o Several Variables (b) F G cos θ F G ( ) Not orthogonal G,,. F(,, ) + F(,, ) i + j k (,, ) F(,, 5) i j k (,, 5) i j k G j k G j k (a) F G i + j k 0 (b). Direction numbers:,,. + 5 F G cos θ ; not orthogonal F G F F i + k F,, 6i + 8k,, + 5 i j k G G j + k G,, 6j + 8k,, + 5 (a) F G 6 0 8 8i 8j + 6k ( i + j k) (b) 0 6 8 Direction numbers:,,. F G 6 6 cos θ ; not orthogonal F G 0 0 5. F(,, ) + F(,, ) i + j k + + 5 5 G,, 5 + F(,, 5) i + j k (a) (b) i j k 6 F G 5 5 i j k 5 5 5 5 Direction numbers:, 7, 5 Tangent line 7 F G ( 85) 8 cos θ F G 8 5 76 5. F(,, ) + + (,, ) F(,, ) i + j + k (,, ) F(,, ) 6i + j + k (,, ) G,, 5i j + k G,, 5 5i j + k Not orthogonal G G i j k G i j k 00 Brooks/Cole, Cengage Learning

Section.7 Tangent Planes and Normal Lines 9 (a) F G i j k 6 i + 0j 8k [ i + 5j k] Direction numbers:, 5, Line: 5 F G (b) cos θ 0 orthogonal F G 6. F(,, ) + F(,, ) i + j k F(,, 5) i + j k G,, + + 6 G,, i + j + 6k G,, 5 i + j + 6k i j k (a) F G 5i j k 6 Direction numbers: 5,,. 5 5 (b) cos θ F G F G 0; orthogonal 7. F(,, ) + 5, (,,5) F,, 6i + j k F,, 5 i + 8j k F F,, 5 k cos θ,, 5 09 θ arccos 86.0 09 8. F(,, ), (,, ) F + i j k F,, i + j k F F,, k cos θ,, 76 θ arccos 5. 9. F(,, ) +, (,,) F,, i j + k F,, i j + k F F,, k cos θ,, θ arccos 77.0 50. F(,, ) + 5, (,,) F(,, ) i + j F(,, ) i + j F(,, ) k cos θ 0 F(,, ) θ arccos 0 90 5. (,, ) 6 F(,, ) ( 6) F + i + + j k 0, 0 + 6 0, 0 + 6( ) ( 0,,) ( verte o paraboloid ) 5. F,, + + 5 (,, ) ( 6 ) ( ) F i + + j k 6 0, + 0, (,, ) + + 5 00 Brooks/Cole, Cengage Learning

0 Chapter Functions o Several Variables 5. F(,, ) + F(,, ) ( ) ( ) i + + j k 0 + 0 + ( ) 6 0, Point: (,, ) 5. (,, ) 8 5 F(,, ) ( 8 8) ( 5) F + + + + i + j k 8 + + 8 0 5 0 Adding, + 0, and 5 Point: (,, 5) F,, 5 55. F,, 5i + 5j k 5 0 5 0 Point: ( 0, 0, 0 ) 56. (,, ) 0 F + + F(,, ) i j k +,, Point: (,, ) 57. F(,, ) + +, (,, 0) F (,, ) F (,, ) F ( ) F (,, 0) F (,, 0) F ( + ) + ( ) + 0( 0) 0 + 6 +,, 6,, 0 0 (,, ) + + + 6 0 +, (,,0) (,, ) + 6 G(,, ) 0 (,, 0) G(,, 0) 8 G ( + ) ( ) + ( ) G G G 8 0 0 0 8 + 0 + The tangent planes are the same. G,,,, 0 0 58. F(,, ) + + 8 + +, (,, ) F (,, ) 8 F (,, ) F ( ) F (,, ) F (,, ) 6 F ( ) ( ) ( ) ( ) 6 + 0 6 + 0 0 + + 0 (,, ) + + 7, (,, ) G G(,, ) ( (,, ) G ) ( ) ( ) ( ) G G,,,, 6 + 6 + + 0 + 6 + 0 0 + + 0 The tangent planes are the same. G G ( ) ( ),,,,,, +,, 00 Brooks/Cole, Cengage Learning

Section.7 Tangent Planes and Normal Lines 59. (a) F(,, ), F G (,, ) 8 5 8+, So, (,, ) lies on both suraces. (b) F i + j k, G 6i 0j 8 k, F G 6 + ( 0) + ( )( 8) 0 The tangent planes are perpendicular at (,, ).,, 0 G,, 8 5 6 + 0 F,, i + j k G,, 6i 0j 8k 60. (a) F(,, ) 0 G (,, ) + + 6 G(,, ) 0 So, (,, ) lies on both suraces. (b) F ( + ) + ( ) + ( ) F(,, ) i 8j k F,, + + + i j k G 8i + j + k G(,, ) 8i j + k F G + 6 0 The planes are perpendicular at (,, ). 6. F( ),, + + 9 F i + 8j + k This normal vector is parallel to the line with direction number, 8,. So, t t 8 8t t t t + + 9 t + t + t 9 0 t ± There are two points on the ellipse where the tangent plane is perpendicular to the line: (,, ) ( t ),, t 6. F( ),, + F i + 8j k The normal to the plane, n i + j k must be parallel to F. So, t t 8 t t t t + t t + t t t ±. Two points:,, ( t ) and,, t 00 Brooks/Cole, Cengage Learning

Chapter Functions o Several Variables,, +,, +,, 0 6. F( 0 0 0)( 0) F( 0 0 0)( 0) F( 0 0 0)( 0) ( Theorem. ) 6. For a sphere, the common object is the center o the sphere. For a right circular clinder, the common object is the ais o the clinder. 65. Answers will var. 66. (a) + 0, ( 5,, ) F(,, ) + F (,, ) F ( 5,, ) 0 F ( ) Direction numbers: 5,, F,, 5,, 6 5 5 + 0 5 0 Plane: ( ) ( ) ( ) (b) Line: 67. 5 + 5 ( + )( + ),,,0 (a) Let F(,, ) (b) (c) ( + )( + ) F,, F,, + + F(,, ) i + j k i + j k + ( ) + ( ) + + + + + + (,, ) F k Direction numbers: 0, 0, Line:,, t 0 + 0 0 Tangent plane: ( ) 6,, 0 + 5 ( 5) 5 F i j k j k 6 Line:, + t, t 5 5 6 0 + + + 0 5 5 6 5 0 0 6 5 0 Plane: 00 Brooks/Cole, Cengage Learning

Section.7 Tangent Planes and Normal Lines sin,,,0 π sin Let F(,, ) sin cos F(,, ) i + j k 68. (a) π F,, i k (b) (c) Direction numbers:,0, or, 0, Line: + t, π, + t Tangent plane: π + 0 + 0 + 0 π 9 F,, i k Direction numbers: 9,0, or 9, 0, Line: + 9, t π, t π Tangent plane: 9 + + 0 0 9 + 0 π, 6, g, + 69. F(,, ) i j k (a) F(,, ) + + 6 (b) + + F(,, ) i + j + k G,, G,, i j + k G,, i j + k The cross product o these gradients is parallel to the curve o intersection. i j k G F,,,, i j Using direction numbers,, 0, ou get + t, t,. F G + cos θ θ 8. F G 6 6 6 8 (,, ) 8 6 00 Brooks/Cole, Cengage Learning

Chapter Functions o Several Variables 70. (a), 6 + g(, ) + + 6 +, g, 6 + + + 6 + + 8 + + 6 + + + + + + + (b) ( ) + ( + ) To ind points o intersection, let. Then + ( ) + ± (, + ) j, g(, ) ( ) j. ( ) j k, which are orthogonal. Similarl, (, ) j and g(, ) ( ) ( ) j k, which are also orthogonal. + The normals to and g at this point are j k and j and the normals are j k and (c) No, showing that the suraces are orthogonal at points does not impl that the are orthogonal at ever point o intersection. 5 5 g 5,, + + a b c F (,, ) a F (,, ) b F (,, ) c 7. F( ) 0 + 0 + 0 a b c 0 0 0 Plane: ( ) ( ) ( ) 7. F( ) F F 0 0 0 0 0 0 + + + + a b c a b c,, + a b c a (,, ) b (,, ) c (,, ) F + a b c 0 0 0 Plane: ( ) ( ) ( ) 0 0 0 0 0 0 0 0 0 + + a b c a b c 0 0 7. F(,, ) a + b (,, ) F a (,, ) (,, ) F b F Plane: a + b 0 0 0 0 0 0 0 a+ b a + b 0 0 0 0 0 0 0 So, the plane passes through the origin. 00 Brooks/Cole, Cengage Learning

Section.7 Tangent Planes and Normal Lines 5 7. F(,, ) F (,, ) + F (,, ) F,, Tangent plane at ( 0 0 0),, : 0 0 0 0 ( 0) + ( 0) ( 0) 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 + 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 So, the plane passes through the origin (,, ) ( 0, 0, 0 ). 0 0 0 75. (, ) e (, ) e, (, ) e (, ) e, (, ) e, (, ) e (a) P (, ) ( 0,0) + ( 0,0) + ( 0,0) + (b) P, 0,0 + 0,0 + 0,0 + 0,0 + 0,0 + 0,0 + + + (c) I 0, P ( 0, ) +. This is the second-degree Talor polnomial or e. I 0, P (,0) + +. This is the second-degree Talor polnomial or e. (d) (e) P (, ) P( ), 0 0, 0 0. 0.908 0.9000 0.9050 0. 0..05.000.050 0. 0.5 0.708 0.7000 0.750 0.5.687.5000.650 P P 76. (, ) cos( + ) (, ) sin ( + ), (, ) sin( + ) (, ) cos ( + ), (, ) cos ( + ), (, ) cos( + ) (a) P, 0,0 + 0,0 + 0,0 (b) (, ) ( 0, 0) + ( 0, 0) + ( 0, 0) + ( 0, 0) + ( 0, 0) + ( 0, 0) P (c) I 0, P ( 0, ). This is the second-degree Talor polnomial or cos. I 0, P (,0). This is the second-degree Talor polnomial or cos. 00 Brooks/Cole, Cengage Learning

6 Chapter Functions o Several Variables (d) (, ) P( ) P ( ),, (e) 5 0 0 0 0. 0.9950 0.9950 0. 0. 0.955 0.9950 5 5 0. 0.5 0.768 0.7550 0.5 0.0707 0.50 77. Given ( ),, then: F(,, ) (, ) 0 ( 0, 0, 0) ( 0, 0) ( 0, 0) F( 0, 0, 0) k cos θ F (,, ) k F i + j k 0 0 0 (, ) (, ) ( ) 0 0 + 0 0 + 0 0 0 0, +, + 78. Given w F(,, ) where F is dierentiable at ( 0, 0, 0) and F( 0, 0, 0) 0, the level surace o F at ( 0, 0, 0) is o the orm F(,, ) C or some constant C. Let G(,, ) F(,, ) C 0. Then G( 0, 0, 0) F( 0, 0, 0) where G ( 0, 0, 0) is normal to F(,, ) C 0 at ( 0, 0, 0). F( 0, 00) is normal to the level surace through ( 0, 0, 0). Section.8 Etrema o Functions o Two Variables So,. g ( ) ( ), + 0 Relative minimum: (,, 0 ) Check: g 0 g 0 g, g, g 0, d 0 > 0 At critical point (, ), 0 minimum at (,, 0 ). d > and g > 0 relative. g s ( ) ( ), + 5 Relative maimum: (,, 5) Check: g 0 g + 0 g, g, g 0 d 0 > 0 At critical point maimum at (,, 5 ).,, d > 0 and g < 0 relative 00 Brooks/Cole, Cengage Learning

Section.8 Etrema o Functions o Two Variables 7. ( ), + + Relative minimum: ( 0, 0, ) Check: 0 0 + + 0 0 + + + ( + + ) + ( + + ) ( + + ) At the critical point 0, 0, > 0 and > 0. So, ( 0, 0,) is a relative minimum.. Relative maimum: (, 0, 5 ), 5 5 Check: 0 5 ( ) 0 0 5 ( ) 5 5 ( ) 5 5 ( ) ( ) ( ) ( ) 5 At the critical point, 0, < 0 and > 0. So, (, 0, 5) is a relative maimum. 5. ( ) ( ) ( ), + + 6 + 6 + + Relative minimum: (,, ) Check: + 0 6 0,, 0 At the critical point (, ), > 0 and > 0. So, (,, ) 6., + 0 + 6 is a relative minimum. ( ) ( ) ( ) ( ) 0 + 5 + 6 + 5 + 6 6 5 6 Relative maimum: ( 5, 6, ) Check: + 0 0 5 + 0 6,, 0, d 0 > 0 At critical point ( 5, 6 ), 0 d > and 0 < relative maimum at ( 5, 6, ). 7., + 6 + 6 6 6 0, 0 6,, 0, d 6 0 > 0. At the critical point (, ), d > 0 and 0 (,,) > is a relative minimum. 8., + + 5 6 + 0when 0 when. 6,, 0 At the critical point and > 0. So, (,, ).,, 0 < is relative maimum. 00 Brooks/Cole, Cengage Learning

8 Chapter Functions o Several Variables 9., 5 + 0 0 8 + 0 0 0 0 0 5,, 0, 0, d 0 > 0. At the critical number and 0 ( 5,,) 5,, d > 0 < is a relative maimum. 0., + + + + + 0 Solving simultaneousl + 0 ields and.,, At the critical point,, > 0 and > 0. So, (,, ) is a relative minimum.., + + + + 0 Solving simultaneousl + + 0 ields,,,, d > 0. At the critical point and 0 (,, 5),, d > 0 > is a relative minimum.., 5 + + 6 + 0 0 + + 6 0 Solving simultaneousl 0 ields 8 and 6.. h ( ) h h, + + 0 ( ) 0 ( ) + 0, 0 + Because h (, ) or all (, ), minimum. 5. g, 0, 0, is a relative ( 0, 0) is the onl critical point. Because all (, ), ( 0, 0, ) is a relative maimum. 6. (, ) + Since (, ) or all ( ) o consist o all points (, ) 7. + + Relative minimum: (, 0, ) Relative maimum: (, 0, ) 8. g, or,, the relative minima satising + 0., + Relative maimum: ( 0, 0, ) Saddle points: ( 0,, ), ( ±,, ) 5 0 0 0,, At the critical point 8, 6, < 0 and > 0. So, ( 8,6, 7) is a relative maimum.. (, ) + 0 + 0 + 0 9. ( + ) e Relative minimum: ( 0, 0, 0 ) Relative maima: ( 0, ±, ) Saddle points: ( ±, 0, ) 0. e Saddle point: ( 0, 0, ) 00 6 5 Because (, ) 0or all (, ) and ( 0,0) 0, ( 0,0,0) is a relative minimum. 00 Brooks/Cole, Cengage Learning

Section.8 Etrema o Functions o Two Variables 9. h (, ) 80+ 80. g(, ) h 80 0 0 h 80 0 h, h, h 0, d 0 > 0 At the critical point ( 0, 0 ), d > 0 and 0 ( 0, 0, 00) h < is a relative maimum.. g, g 0 g 0 g, g, g 0, d 0 < 0 At the critical point (,, ) d < 0 (,,0) is a saddle point. g 0 and 0 g g 0, g 0, g At the critical point So, ( 0, 0, 0) is a saddle point.. h (, ) 0, 0, g g g < 0. h 0 Solving simultaneousl h 0 ields 0 and 0. h, h, h At the critical point h h ( h ) So, ( 0, 0, 0) is a saddle point. 0, 0, < 0. 5., 0 0 Solving simultaneousl ields,.,, d 5 < 0 At the critical point (, ), d 0 (,, ), + + 6. ( ) ( ) < is a saddle point. Solving b substitution ields critical points: ( 0, 0 ), (, ), (, ) 6, 6, At ( 0, 0 ), ( ) 0 ( 0, 0,) < saddle point. At (, ), ( ) > 0 and 0 (,,) < relative maimum. At (, ), > 0 and 0 (,,) 7., e sin < relative maimum. sin 0 e Because e > 0 or all and sin and cos are never e cos 0 both ero or a given value o, there are no critical points. 00 Brooks/Cole, Cengage Learning

0 Chapter Functions o Several Variables + e 8. (, ) e 0 6 Solving ields the critical points ( 0, 0 ), 0,,, 0. ± 0 ± + e ( ) ( 8 ) ( ) + + e + + e + + e At the critical point ( 0, 0 ), ( ) < 0. So, ( 0, 0, e ) 0,, 0 points points ± < and ( ) 0. ± 6,0, > 0 and ( ) 0. is a saddle point. At the critical > So, ( 0,, e) > So, ( 6,0, ee) ± are relative maima. At the critical ± are relative minima. 9. 0. ( ) 0. 0 i 0. + Relative minimum at all points (, ), 0. ( ) 0. 0 i 0. + Relative minima at all points (, ) and 5. ( ) Insuicient inormation.. 0 60 0 9 6 0 < and ( ),, 0. > has a relative maimum at (, ). ( ) 0 0 < 9 6 0 0 has a saddle point at (, ). 5 0 0 8 0 5. d 8 6 > 0 < 6 < < 6. d < 0 i and have opposite ab,, ab, is a saddle point. For eample, signs. So, ( ) consider (, ) and ( ab, ) ( 0,0 ). 7., + (a) 0 0 0 Critical point: ( 0, 0 ) (b) 6, 6, 0 ( 0, 0, 0) is a saddle point. (c) Test ails at ( 0, 0 ). (d) At 0, 0, 0. Saddle point (0, 0, 0) > and ( ) has a relative minimum at ( 0, 0). 0 > 5 8 0 0 00 Brooks/Cole, Cengage Learning

Section.8 Etrema o Functions o Two Variables 8., + 6 + 9 + + 7 + 9 (a) + 0 Solving ields 8 7 0 and. + + (b) 6, 6 + 8, 0 At (,, 0) is a saddle point. (c) Test ails at (, ). (d),, 0. 9. (, ) ( ) ( + ) 0 (a) ( )( + ) 0 critical points: ( ) ( ) 0 (, a) and ( b, ) + (b) ( + ) ( ) ( )( + ) At both (, a) and ( b, ), 0. Because (, ) 0, there are absolute minima at (, a, 0) and ( b,,0 ). (c) Test ails at (, a) and ( b, ). (d) Absolute minimum (b,, 0) (,, 0) 6 6 Absolute minimum (, a, 0) 0. ( ) ( ) ( ) (a) (b), + + 0 0 ( ) + ( + ) Solving ields + and. 0 ( ) ( ) + + ( + ) ( ) + ( + ) ( ) ( ) + ( + ) ( )( + ) ( ) + ( + ) At (, ), (,, 0) is an absolute minimum. (c) Test ails at (, ). (d). ( ) (a) is undeined., + 0 are undeined and at 0 and 0. Critical point: ( 0, 0),, 0 9 9 (b) At ( 0, 0 ), ( 0, 0, 0) is an absolute minimum. (c) Test ails at ( 0, 0 ). (d) 6 (,, 0) is undeined. 6 6 6 Absolute minimum (0, 0, 0) 00 Brooks/Cole, Cengage Learning

Chapter Functions o Several Variables. ( ) ( ) (a) (b), + 0 ( + ) and are undeined at 0, 0. Critical Point: ( 0, 0) ( + ) ( + ) ( + ) ( + ) ( + ) 9 9 9 8 ( + ) At ( 0, 0 ), ( 0, 0, 0) is an absolute minimum. (c) Test ails at ( 0, 0 ). (d) 5 is undeined. (0, 0, 0). ( ) ( ) ( ),, + + + 0 0 ( ) 0 Solving ields the critical point ( 0,, ). ( + ) 0 Absolute minimum: 0 at ( 0,, ). ( ) ( )( ),, 9 + 9 The absolute maimum value o is 9, and realied at all points where ( ) So, the critical points are o the orm ( 0, ab, ), ( c,, d), ( e,, ) where abcde,,,,, are real numbers. 5. (, ) + 5, R {(, ) :,0 } 0 0 (not in region R) 0 0, : + 5,, 0 6,, 0. Along Along, : 8 + 5, 8 0,,,. Along Along So, the maimum is (, 0, ) and the minimum is (,, )., 0 : + 6,, 0 6,,., 0 : 6,, 0,,. + 0. 00 Brooks/Cole, Cengage Learning

Section.8 Etrema o Functions o Two Variables 6. (, ) +, R {(, ) :, } + 0 0 ( 0, 0) 0 0 Along,, +, + 0. Thus, (, ), (,) and, 6. Along,,, 0. Thus,, 6,,,,. Along,, + 0. Along,, 0. So, the maima are (,) 6 and (, ) 6 and the minima are (,) and,. 7., has no critical points. On the line +, 0,, + 5 + 0 and the maimum is0, the minimum is 5. On the line +,,, + + and the maimum is 6, the minimum is 5. On the line +, 0,, + + 0 and the maimum is0, the minimum is 6. Absolute maimum: 0 at ( 0, ) Absolute minimum: 5 at (, ) + (, ) + (0, ) (, 0) + 8. (, ) ( ) 0 0 On the line +, 0, (, ) ( ( + ) ) ( ) and the maimum is, the minimum is 0. On the line +, 0, 5 (, ) ( ( + )) ( ) and the maimum is6, the minimum is 0. On the line +,, (, ) ( ( + ) ) ( ) and the maimum is6, the minimum is 0. Absolute maimum: 6 at (, 0 ) Absolute Minimum: 0 at, and along the line. (, ) (0, ) (, 0) 00 Brooks/Cole, Cengage Learning

Chapter Functions o Several Variables 9., + 6 0 0 ( 0,) 0 On the line,,, + 6 + 6 and the maimum is 8, the minimum is6. On the curve,, (, ) + ( ) ( ) and the maimum is 8, the minimum is. 8 Absolute maimum: 8 at ( ±, ) Absolute minimum: at ( 0, ) (, ) (, ) 50. (, ) + 0 (, ) 0 On the line,,, +. (, ) (, ) On the curve,, + + and the maimum is, the minimum is. 6 Absolute maimum: at (,) and on Absolute minimum: 6 0.6875 at (, ) 5. (, ) + +, R {(, ) :, } + 0 + 0, + 0 Along,, + +, + 0,,,, 0,, 9. Along,, +, 0,, 9,, 0,,. Along,, + +, + 0. Along,, +, 0. So, the maima are (, ) 9 and (,) 9, and the minima are, 0,. 00 Brooks/Cole, Cengage Learning

5. { } R, + +,, : + 8 + 0 + 0, + 0 Section.8 Etrema o Functions o Two Variables 5 On the boundar + 8, we have 8 and ± 8. Thus, ± 8 + 8 8 ± 8 6 ± 8 + 8 ±. 8 Then, 0 implies6 or ±. (, ) (, ) 6 and (, ) (, ) 0 So, the maima are (, ) 6 and (, ) 6, and the minima are, 0,.,, R, :0,0 5. ( + )( + ) ( ) ( + )( + ) 0 ( ) ( + )( + ) 0 { } or or 0 For 0, 0, also, and ( 0, 0) 0. For,, (, ). The absolute maimum is (,. ) The absolute minimum is 0 ( 0,0 ).( In act, ( 0, ) (,0) 0. ) R,, R, : 0, 0, + 5. ( + )( + ) ( ) ( + )( + ) 0 ( ) ( + )( + ) 0 For 0, 0, also, and { } 0, 0 0. or 0 or 0 For and, the point (, ) is outside R. +,,,, and the maimum occurs at + For ( ) ( ) Absolute maimum is 8,. 9 0 0,0. In act, 0,,0 0 The absolute minimum is ( ) R,. 00 Brooks/Cole, Cengage Learning

6 Chapter Functions o Several Variables 55. In this case, the point A will be a saddle point. The,. unction could be 56. A and B are relative etrema. C and D are saddle points. 57. 75 60 5 0 No etrema 58. 60. (, ), g(, ) + (a) 0, 0 ( 0,0) is a critical point. g 0, g 0 ( 0,0) is a critical point. (b),, 0 d 0 < 0 0, 0 is a saddle point. g, g, g 0 d 0 > 0 0, 0 is a relative minimum. 6. False.,. Let ( 0, 0,) is a relative maimum, but ( 0, 0) ( 0, 0) do not eist. and Etrema at all (, ) 6. False. Consider ( ),. Then ( 0, 0) ( 0, 0) 0, but point. 0, 0, 0 is a saddle 59. 7 6 6. False. Let 6. False., (See Eample on page 958). Let ( ) +,. 6 Relative minima: ( ±, 0, ) Saddle point: ( 0, 0, 0 ) Saddle point Section.9 Applications o Etrema o Functions o Two Variables. A point on the plane is given b,,,, +. The square o the distance rom ( 0, 0, 0) to this point is S + + ( + ). S ( + ) S + ( + ) From the equations S 0 and S 0 we obtain 6 + 6. Solving simultaneousl, we have,,. So, the distance is + +.. A point on the plane is given b,,,, +. The square o the distance rom (,, ) to this point is S ( ) + ( ) + ( + ) ( ) + ( ) + ( ). S ( ) ( ) S ( ) + ( ) From the equation S 0 and S 0 we obtain +. Solving simultaneousl, we have, 5, 0. So, the distance is 5 5 + +. 00 Brooks/Cole, Cengage Learning

Section.9 Applications o Etrema o Functions o Two Variables 7. A point on the surace is given b (,, ) (,, ). rom (,, 0) to a point on the surace is given b The square o the distance S + + + + 0 + + + +. S ( + ) S + From the equations S 0 and S 0, we obtain + 0, 5. + 0 So, the distance is + + + + 5 7.. A point on the surace is given b (,, ) (,, ). The square o the distance rom ( 0, 0, ) to a point on the surace is given b ( ) S + +. S + + ( ) S + + From the equation S 0 and S 0, we obtain. So, + 0. Using a graphing utilit, 0.. The minimum distance is S ( ) + + 0.667. 5. Let,, and be the numbers. Because 7, Z 7. S + + + 7 +. S 7 7 0, S 0. 7 7 So,. 6. Because + +,. So, P P 0 P 6 6 0. Ignoring the solution 0 and substituting into P 0, we have 6 0 8 0. So, 8, 6, and 8. 7. Let,,and be the numbers and let S + +. Because + + 0, we have S + + 0 S + 0 0 + 0 S + 0 0 + 0. Solving simultaneousl ields 0, 0, and 0. 8. Let,, and be the numbers. Because,. + + + + S S 0, S 0 ( ) ( ) So,. 00 Brooks/Cole, Cengage Learning

8 Chapter Functions o Several Variables 668.5 9. The volume is 668.5. 668.5 668.5 C 0.06 + + 0. 0. + + 0. 80.9 80.9 C + + 0. 80.9 C + 0. 0 8.9 C + 0. 0 Solving simultaneousl, 9 and 8.5. 80.9 80.9 Minimum cost: + + 0. $6.7 9 9 0. Let,, and be the length, width, and height, respectivel. Then C0.5 + + and The volume is given b C 0.5 V ( + ) ( C0 6) V ( + ) ( C0 6) V. + In solving the sstemv 0 andv 0, we note b the smmetr o the equations that. Substituting into V 0 ields ( 0 9 ) C 0 C0 C0 C0 C 0, 9,,, and. 6 C0.5. ( + ) π abc. Let a + b + c k. Then V πab( k a b) π( kab a b ab ) π Va ( kb ab b ) 0 kb ab b 0 π Vb ( ka a ab) 0 ka a ab 0 Solving this sstem simultaneousl ields a b and substitution ields b k. So, the solution is a b c k.. Consider the sphere given b r Then the volume is given b V r 8 r + + and let a verte o the rectangular bo be (,, r ). V V 8 8 + r ( r ) 0 r r 8 8 + r ( r ) 0. r r Solving the sstem + r + r ields the solution r. 00 Brooks/Cole, Cengage Learning

Section.9 Applications o Etrema o Functions o Two Variables 9. Let,, and be the length, width, and height, respectivel and letv0 be the given volume. ThenV 0 and V0. The surace area is V0 V0 S + + + + V0 S 0 V 0 0 V0 S 0 V0 0. Solving simultaneousl ields V, V,and V.. 0 0 0 0 0 cos sin 0 sin sin sin cos A + + θ θ θ θ + θ θ A 0 sin θ sin θ + sin θ cos θ 0 A 0 cos θ cos θ + ( cos θ ) 0 θ A 5 From 0 we have5 + cos θ 0 cos θ. A 5 5 5 From 0 we obtain 0 + 0 θ 0( 5) ( 5) + ( 5) 0 0 0 0. Thencos θ θ 60. 5. R, 5 8 + + 0 R 0 + 0, 5 + R 6 + 0 0, + 8 5 Solving this sstem ields and 6. R R 0 R 6 R < and 0 R R R 0 > So, revenue is maimied when and 6. 6. (, ) 5 P + C C 5 + 5 0.0 + + 500 0.05 + + 75 0.0 0.05 + + 775 P 0.0 + 0, 75 P 0.0 + 0, 0 P P 0.0 0 P 0.0 P < and P P ( P ) 0 > 0 So, proit is maimied when 75 and 0. 00 Brooks/Cole, Cengage Learning

50 Chapter Functions o Several Variables P p, q, r pq + pr + qr. p + q + r implies that r p q. P( p, q) pq + p( p q) + q( p q) pq + p p pq + q pq q pq + p + q p q P P q + p; p + q p q P P Solving 0 gives q + p p q p + q 7. and so p 6 q and,. P + + 9 9 9 9 8. (a) H ln ln ln, + + ln ln ( ) ln( ) H H ln + + ln( ) 0 ln + + ln( ) 0 ln ln ln. So, ( ) ln ln. (b) ( ) ( ) ( ) H ln ln ln ln ln 9. The distance rom P toq is. + The distance rom Q to R is ( ). ( ) + + + + C k k k 0 C k + k 0 + + C k k 0 ( ) + ( ) + k + k 0 + + + 9 + ( ) ( ) + ( ) ( ) ( ) + + + 6 + So, 0.707 km and.8 km. 6 + The distance rom R tos is0. 00 Brooks/Cole, Cengage Learning

Section.9 Applications o Etrema o Functions o Two Variables 5 0. S d + d + d ( 0 0) + ( 0) + ( 0 ) + ( ) + ( 0 + ) + ( ) ds d ( ) + + ( ) + ( ) + 0when The sum o the distance is minimied when. (a) (, ) S d + d + d 6. ( ) 0.85. ( 0) ( 0) ( ) ( ) ( ) ( ) + + + + + + ( ) + + + + + + From the graph we see that the surace has a minimum. + S, + + + + + + S(, ) + + + + + + (b) (c) S(, ) S (, ) S (, ) i j i j 0 0 tan θ θ 86.07 5, S, t, S t t, + t 0 (d) ( ) ( ( ) (, ) ) 0 5 S t, + t + 0 5 t + + t 5 5 0 5 + 0 + t t 5 + + 5 5 0 5 + 0 t + + t 5 5 5 Using a computer algebra sstem, we ind that the minimum occurs when t.. So (, ) ( 0.05, 0.90 ). (e) (, ) ( S(, ) t, S(, ) t) ( 0.05 + 0.0 t, 0.90 0.6t) S( 0.05 + 0.0 t, 0.90 0.6t) ( 0.05 + 0.0t) + ( 0.90 0.6t) + (.05 + 0.0t) + (.0 0.6t) + (.95 + 0.0t) + (.0 0.6t) 6 8 Using a computer algebra sstem, we ind that the minimum occurs when t.78. So (, ) ( 0.0, 0. ). (, ) ( S(, ) t, S(, ) t) ( 0.0 0.09 t, 0. 0.0t) S( 0.0 0.09 t, 0.5 0.0t) ( 0.0 0.09t) + ( 0.5 0.0t) + (.0 0.09t) + (.55 0.0t) + (.90 0.09t) + (.55 0.0t) Using a computer algebra sstem, we ind that the minimum occurs when t 0.. So, (, ) ( 0.06, 0. ). Note: The minimum occurs at (, ) ( 0.0555, 0.99) ( ) S (, ) points in the direction that S decreases most rapidl. You would use S (, ) 0 S 6 or maimiation problems. 8 00 Brooks/Cole, Cengage Learning

5 Chapter Functions o Several Variables. (a) S ( + ) + + ( ) + ( 6) + ( ) + ( ) The surace appears to have a minimum near (, ) (,5 ). 0 (b) S S + + + ( + ) + ( ) + ( 6) ( ) + ( ) 6 + + ( + ) + ( ) + ( 6) ( ) + ( ) (c) Let (, ) (,5 ). Then S, 5 0.58i + 0.0j. Direction 6.6 6 8 (d) t 0.9,., 5.0 (e) t.56,., 5.06, t.0,., 5.06 Note: Minimum occurs at (, ) (.5, 5.069) ( ) S (, ) points in the direction that S decreases most rapidl.. Write the equation to be maimied or minimied as a unction o two variables. Set the partial derivatives equal to ero (or undeined) to obtain the critical points. Use the Second Partials Test to test or relative etrema using the critical points. Check the boundar points, too.. See pages 96 and 965. 5. (a) 0 0 0 0 0 (b) 6. (a) 6 i 0 i i i 6 () i 6 0 a, b ( 0 ), + 8 0 8 S + 0 + + + 6 0 0 9 6 9 i 0 i i i 6 i 0 (b) a 6 0, ( 0), 0 0 0 b 0 0 + 7 9 S 0 + + + 0 0 0 0 5 00 Brooks/Cole, Cengage Learning

Section.9 Applications o Etrema o Functions o Two Variables 5 7. (a) 0 0 0 0 0 i i 8 i i i 6 8 a, b 8 + ( ), + 6 (b) S + + + 0 0 8. (a) 0 0 9 0 0 0 0 9 6 8 6 5 0 5 (b) 6 6 i 8 i 8 i i 7 i 6 87 8 8 7 a, b 8 ( 8 ), 8 6 8 8 5 5 7 9 S 0 + 0 + 0 + + + + + 9. ( 0,0, ) (,, ) (,, ) (,, ) ( 5,5 ) i, i, i i 6, i 5 a 55 ( ) 56 7 7 86 7 7 b ( ) 5 7 7 + 7 7 + 7 0. (, 0 ), (, ), ( 5, 6 ) i a 5 ( 9) b 9 ( 9) 9, i 9, i i 9, i 5 9 99 6 9 6 7 (5, 5) (, ) (, ) (, ) (0, 0) 0 6 00 Brooks/Cole, Cengage Learning

5 Chapter Functions o Several Variables. ( 0,6, ) (,, ) ( 5,0, ) ( 8,, ) ( 0, 5) i i i i i 7, 0, a 5( 05) ( 7) b 0 ( 7) 70, 05 5 70 7 0 50 75 96 8 75 95 5 8 8 75 95 + 8 8 8 (0, 6) (, ) (5, 0) 8 (8, ) (0, 5) 6. ( 6,, ) (,, ) (,, ) ( 8,6, ) (,8, ) (,8; ) n 6 i i i i i a 6( 00) ( ) 75 00 6 75 9 0.57 5 9 5 b 6 5 8.65 9 5 + 5 8. (a).6 + 8 (b) 50 0 50 75 95 + 8 8 (c) For each one-ear increase in age, the pressure changes b approimatel.6, the slope o the line.. (a) Using a graphing utilit, 00 + 8. (b) When.59, 00.59 + 8 55. 5. (.0, ), (.5, ), (.0, 8 ), (.5, 5 ) i 7, i 7, i i, i.5 a, b 9, + 9 When.6,. bushels per acre. 80 9 6. (a).09 60. (b) For each one-point increase in the percent, increases b about.09 million. n 7. Sabc (,, ) ( ) i ai bi c i n i ( i ai bi c) 0 i n i( i ai bi c) 0 i n S a S b S c ( i ai bi c) 0 i n n n n i + i + i i i i i i i n n n n i + i + i i i i i i i n n n i + i + i i i i a b c a b c a b cn 8. Let,, and be the length, width, and height, respectivel. Then the sum o the length and girth is + + 08 given b or 08. The volume is given b V 08 V 08 ( 08 ) 0 V 08 08 0. Solving the sstem + 08 and + 08, we obtain the solution 6 inches, 8 inches, and 8 inches. 9. (, 0 ), (, 0 ), ( 0, ), (, ),(, 5) i 0 i 8 i 0 i 0 i i i i i a + 0c,0b,0a + 5c 8 a, b, c, + + 9 (, 0) (, 0) 6 6 6 6 7 5 5 7 5 5 8 (0, ) (, 5) (, ) 6 00 Brooks/Cole, Cengage Learning

Section.9 Applications o Etrema o Functions o Two Variables 55 0. (, 5 ), (, 6 ), (, 6) (, ) i 0 i 9 i 0 i 0 i 5 i i i i 60 5a + 0c 60, 0b, 0a + c 9 a, b, c, + 5 5 0 6 0 6. ( 0,0, ) (,, ) (,6, ) (, ) i 9 i 0 i 9 i 99 i 5 i i 70 i i 5 5a + 99b + 9c 5 99a + 9b + 9c 70 9a + 9b + c 0 a, b, c 0,. ( 0,0, ) (,9, ) (,6, ) (,0 ) i 6 i 5 i i 6 i 98 i i i i 98a + 6b + c 6a + b + 6c a + 6b + c 5 a, b, c, + + 5 9 99 5 9 99 0 0 0 0 9 5 9 (, 6) (, 5) (0, 0) (0, 0) 8 (, 6) (, ) (, 6) (, ) (, 9) (, 6) (, ) (, 0) 9 7 9. (a) ( 0, 0 ), (,5 ), (, 0 ), ( 6, 50 ), ( 8, 65 ), ( 0, 70) i 0 i 0 i 0 i 800 i 5,66 i i 670 i i,500 (b) 5,66a + 800b + 0c,500 800a + 0b + 0c 670 0a + 0b + 6c 0 5 5 5 + 0. + 9.66.79 56 0 0. (a) 0.075 + 5. (b) 0.00 + 0. + 5. (c) 7 5 8 5.5 (d) For 0, and 0.075( ) + 5. 7. (billion) 0.00 + 0. + 5. 6.8 (billion). The quadratic model is less accurate because o the negative coeicient. 00 Brooks/Cole, Cengage Learning

56 Chapter Functions o Several Variables 5. (a) ln P 0.99h + 9.08 (b) ln P 0.99h + 9.08 P e 0,957.7e (c),000 0.99h + 9.08 0.99h 6. (a) a + b 0.007 + 0.5 0.007 + 0.5 (b) 0,000 0 0 60 (d) Same answers (c) No. For 70,, which is nonsense. 000 which seems inaccurate. n 7. Sab (, ) ( a ) i + b i i n n n a(, ) i + i i i i i i S a b a b S a, b a + nb b i i i i n i i S a, b aa S a, b n bb S a, b ab n n i i n S a, b > 0as long as 0 or all i. (Note: I 0 or all i, then 0 is the least squares regression line.) aa i i n n n n d SaaSbb Sab n i i n i i 0 i i i i As long as d 0, the given values or a and b ield a minimum. since n n n i i. i i Section.0 Lagrange Multipliers. Maimie ( ),. Constraint: + 0 λ g i + j λ i + j λ λ + 0 5 ( 5, 5) 5. Maimie ( ),. Constraint: + λ g i + j λi + λ j λ λ + λ λ,, (, ) 0 8 Level curves 6 Constraint 6 8 0 00 Brooks/Cole, Cengage Learning

Section.0 Lagrange Multipliers 57. Minimie ( ) +,. Constraint: + λ g i + j λi + λ j λ λ + (, ) 8. Minimie ( ) +,. Constraint: + 5 λ g i + j λi + λ j λ λ λ λ + 5 0λ 5 λ,, 5 (,) 5. Minimie, +. Constraint: + 5 0 λ g i + j λ i + j λ λ λ λ + 5 0 λ + λ 5 λ,, (, ) 5 Constraint Level curves 6. Maimie ( ),. Constraint: 0 λ g i j λi + λ j λ 0 or λ I 0, then 0 I λ, and 0, 0 0. λ. (,), Maimum 7. Maimie, + +. Constraint: + 00 λ g + i + + j λi + λ j + λ λ λ + λ + 00 00 5, 50 ( 5, 50) 600 8. Minimie, + + 0. Constraint: 6 λ g i + j λi + λ j λ λ λ λ ( 0) 6 6, 9 0, + 00 Brooks/Cole, Cengage Learning

58 Chapter Functions o Several Variables, 6 is maimum when 9. Note: g(, ) is maimum. Maimie g( ) Constraint: + λ λ, 6. + g,, 0. Note: (, ) + is minimum when g(, ) is minimum. g, +. Minimie Constraint: + 5 λ λ + 5 0 5, 5, g,. Minimie ( ) + +,,. Constraint: + + 9 λ λ λ + + 9 (,, ) 7. Maimie ( ),,. Constraint: + + λ λ λ + + (,, ). Minimie ( ) + +,,. Constraint: + + λ λ λ + + (,, ). Minimie, 0 + + 8. Constraint: + 0 ( 0 λ) ( λ) 0 λ + λ + + 0 ( λ) ( λ) 0 + + + 0 λ, 6 (, 6) 6 0 + 6 8 + 8 5. Maimie or minimie ( ) + + Constraint: + Case : On the circle + + λ + λ,. + ±, ± Maima: 5, ± ± Minima:, ± Case : Inside the circle + 0 0 + 0,,, 0 Saddle point: ( 0, 0) 0 B combining these two cases, we have a maimum o 5 at ±, ± and a minimum o at ±,. 00 Brooks/Cole, Cengage Learning

Section.0 Lagrange Multipliers 59, e. 6. Maimie or minimie Constraint: + Case : On the circle + ( ) e λ ( ) e λ + ± 8 Maima: ±, e. 8 Minima: ±, ± e 0.885 Case : Inside the circle ( ) e 0 ( ) e 0 0 e, e, e 6 6 6 0, 0, < 0. At Saddle point: ( 0, 0) Combining the two cases, we have a maimum o e 8 at ±, and a minimum 8 o e at ±, ±. 7. Maimie ( ),,. Constraints: + + + 0 λ g + µ h i + j + k λ i + j + k + µ i j + k λ + µ λ µ λ + µ + + + 8 + 0 6 ( 8,6, 8) 0 8. Minimie ( ) + +,,. Constraints: + 6 + λ g + µ h i + j + k λ i + k + µ i + j λ + µ µ + λ 6 + 6 + 9 ( ) + 7 6 6, 0 ( 6, 6, 0) 7 9. (, ) ( 0) ( 0) + + Constraint: + λ λ λ λ + Minimum distance: + 0. Minimie the square o the distance subject to the constraint +. λ λ +,, + The point on the line is, and the desired distance is d +. 00 Brooks/Cole, Cengage Learning

60 Chapter Functions o Several Variables. (, ) + ( ) Constraint: λ λ λ ( ) λ λ λ λ 6, Minimum distance:, +. Constraint: + λ + ( ) λ λ λ + λ + + ( λ) λ + + 6λ 6 7λ λ 7 9 8, 7 7 Minimum distance: + 9 8 5 7 + 7 7 7. (, ) + ( ) Constraint: 0 λ λ I 0, 0, and 0, 0 9 distance. I 0, λ, 5, ± 5 ± 5,5 5+ < Minimum distance:., + + Constraint: 0 ( ) + λ λ λ ( + ) ( ) + + 6 0 Minimum distance: + + 0,, 5. (, ) ( ) + ( ) + Constraint: 0 ( ) ( ) λ λ + λ + + 0 0.855.86 Minimum distance: ( 0.855,.86) 0.88 00 Brooks/Cole, Cengage Learning

Section.0 Lagrange Multipliers 6 6. Minimie the square o the distance (, ) + ( 0) subject to the constraint λ 0 5 + 0 ( 0) λ 5 ( ) + ( 8 + 6) + 50 + 00 9 58 + 0 Using a graphing utilit, we obtain.57 and.78 or b the Quadratic Formula, 58 58 9 58 9 9 ± ± ±. 9 9 9 Using the smaller value, we have 9 9 and 0 9.8570. 9 +. 9 0 9 The point on the circle is, 9 and the desired distance is d 9 The larger -value does not ield a minimum. 9 0 9 6 + 0 8.77. 9 9 7. Minimie the square o the distance (,, ) ( ) + ( ) + ( ) subject to the constraint + +. ( ) ( ) λ λ and λ + + +, 0 The point on the plane is (, 0, 0) and the desired distance is d ( ) + ( 0 ) + ( 0 ). 8. Minimie the square o the distance,, + + subject to the constraint + 0. ( ) λ λ + ( ) λ λ + λ + 0,, 0, The point on the plane is (, 0, ) and the desired distance is d + 0 +. 9. Maimie (,, ) subject to the constraints + 0 and +. 0 λ + µ 0 λ 0 λ + µ + 0 + ( ) + 0 0 6 + 6 0 ( )( ) 0 or The maimum value o occurs when at the point o (, 0, ). 0. Maimie (,, ) subject to the constraints + + 6 and +. 0 λ + µ 0 λ + µ λ µ + + 6 + + 5 + + ( 5 ) 6 0 0 0 5 0 6 0 5 ± 65 5 Choosing the positive value or we have the point 0 + 65 5 + 65 + 65,,. 5 5 00 Brooks/Cole, Cengage Learning

6 Chapter Functions o Several Variables. Optimiation problems that have restrictions or constraints on the values that can be used to produce the optimal solution are called contrained optimiation problems.. See eplanation at the bottom o page 97.. Minimie ( ) + + Constraint: g ( ),,. λ λ λ λ λ λ + λ λ λ + λ λ,, +,, + + Minimum distance. Minimie ( ) ( ) + ( ) + ( ) Constraint: g (,, ) +,,. + λ ( ) λ λ ( ) λ 6 + λ ( ) λ + + λ λ 6 + λ + λ + 6 λ 5 0,, Minimum distance + + 5. Minimie Constraint: g ( ),, + +.,, 7 λ λ λ λ λ λ 7 7 6. Maimie P ( ) Constraint: λ λ λ + +,,. g,, + + 0 + + 8 6 8 7. Minimie ( ) ( + ) + Constraint: g (,, ) 668.5 0. + 0. λ 0. + 0. λ 0.( + ) λ 668.5,, 0.06 0.. 0. + 0. λ 0. + 0. 0. λ λ 0. 0. 0. 0. + 0. 0. 0. 668.5 9, 9, 9, $6.7 00 Brooks/Cole, Cengage Learning

Section.0 Lagrange Multipliers 6 8. Maimie (,, ) Constraint:.5λ + λ.5λ + λ λ + λ (volume). g,,.5 + + C.5 + + C [.5 λ λ] [.5 λ λ] + + λ λ (also b smmetr). λ + λ λ.5 + C 9.5 + + C C, C C, 9. Maimie ( ) πabc,,. Constraint: a + b + c K, constant π bc λ πbc πac K a b a b c π ac λ π ab λ So, ellipse is a sphere: + + a 0. Consider the sphere given b + + r and let a verte o the rectangular bo be (,, ). Maimie V (,, ) ( ) 8. Constraint: g,, + + r 8 λ λ 8 λ λ 8 λ λ. Maimie Constraint: P p, q, r pq + pr + qr. g p, q, r p + q + r q + r λ p + r λ p q r p + q λ p + q + r p p and 9 P,,.. Maimie Constraint: H,, ln ln ln. g,, + + (a) ln λ ln λ ln λ + + (b) H,, ln ln + + r So, the rectangular bo is a cube. 00 Brooks/Cole, Cengage Learning

6 Chapter Functions o Several Variables V,, 8 subject to the constraint. Maimie 8 8 8 λ a λ b a b c λ c + +,, a b c a b c a b c,, So, the dimensions o the bo are a b c. a b c + +.. (a) Yes. Lagrange multipliers can be used. (b) Maimie V(,, ) subject to the constraint + + 08. λ λ and λ + + 08 6 08, 8 6, 8 Volume is maimum when the dimensions are 6 8 8 inches. 5. Minimie C(,, ) 5 ( ) 8 + 6 λ 8 + 6 λ, 6 + 6 λ + + + subject to the constraint 80. 80 80 60, 60 Dimensions: 60 60 60 eet. 6. (a) Maimie P (,, ) subject to the constraint + + S. λ λ λ + + S S So, S S S,,, > 0 S 7 S + +. 00 Brooks/Cole, Cengage Learning

Section.0 Lagrange Multipliers 65 (b) Maimie P subject to the constraint n n λ n λ n λ n n λ n i i S. n i i S So, n S S S S, i n n n n 0 n S n n n n S n n n S n + + + + n n n., + subject to the constraint π rh V 0. 7. Minimie A( π r) πrh πr πh + πr πrhλ h πr πr λ r πrh V πr V 0 0 Dimensions: r V 0 and h π V0 π,, 00 + + subject to the constraints + + 50 and 0. 8. Maimie T λ + µ λ 0 λ µ I 0, then λ and µ 0, 0. So, 0 and 50. T ( 0, 50, 0) 00 + 50 50 I 0 then + 50and 50. T 50 50 50, 0, 00.5 + So, the maimum temperature is50. 00 Brooks/Cole, Cengage Learning

66 Chapter Functions o Several Variables Distance 9. Using the ormula Time, Rate λ minimie T (, ) v d + λ v d + v d + v d + a + Because sin θ d + and sin θ, d + d + we have v sin θ sin θ. v v 50. Case : Minimie d + or v πl the constraint lh + 8 πl Pl, h h+ l + subject to A. π πl + h + λ π h π lλ λ, + + l l l h Case : Minimie Al (, h) πl lh + 8 subject to πl the constraint h + l + P. πl π h + + λ l λ λ l πl l πl, h + + h l or l h d + d + + subject to the constraint + a. v v P d a Medium Medium θ θ, 00 subject to the constraint 7 + 60 50,000. 0.5 0.75 5. Maimie P 0.75 0.75 5 7 0.5 0.5 75 60 0.75 0.5 7λ 75 5 60λ d Q 0.75 λ 0.5 λ 8 5 8 5 7λ 5 60λ 75 8 5,65 7 + 60 88 50,000 5 8 5 565 P, 5 6,869 8 h l 00 Brooks/Cole, Cengage Learning

Section.0 Lagrange Multipliers 67, 00 subject to the 0. 0.6 5. Maimie P constraint 7 + 60 50,000. 0.6 0.6 0 7 0. 0. 60 60 0.6 0. 0.6 λ 0. 7λ 0 60λ λ λ 60 7λ 0 λ 9 9 5 5 9 5,000 7 + 60 80 50,000 5 9 500 5,000 P, 500 96,99 9 5. Minimie C, 7+ 60subject to the 0.5 0.75 constraint00 50,000. 0.75 0.75 0.75 7 5 0.5 0.5 0.5 60 75 0.75 0.5 λ λ 7 5λ 60 75λ 7 75λ 5λ 60 8 8.6 5 5 0.5 00 (.6 ) 0.75 50,000 500 9. 0.75.6.6 688.77 C( 9., 688.77) 55,097.97 5. Minimie 0.6 0. C, 7+ 60subject to the constraint00 50,000. 0. 0. 0. 7 60 7 60λ 0.6 0.6 0.6 60 0 0. 0.6 λ λ 7 λ 60λ 5 5 0. 0.6 00 50,000 C 5 500 ( 5) 0. 00 ( 5) 0. 500 00 0. 0. ( 5) ( 5), $65,60.7 55. (a) Maimie 60 0λ λ g α, β, λ cosα cos β cosγ subject to the constraint α + β + γ π. sin α cos β cos γ λ cos α sin β cos γ λ tan α tan β tan γ α β γ cos α cos β sin γ λ α + β + γ π α β γ g π, π π 8 π 00 Brooks/Cole, Cengage Learning

68 Chapter Functions o Several Variables (b) α + β + γ π γ π ( α + β) g( α + β ) cos α cos β cos( π ( α + β )) cos α cos β cos π cos( α β ) sin π sin( α β ) cos α cos β cos( α + β ) + + + γ α β 56. Let r radius o clinder, and h height o clinder height o cone. S rh r h r π + π + constant surace area πrh 5πrh V πr h + volume We maimie ( r, h) r h subject to g( r, h) rh + r h + r C. ( C rh) r ( h + r ) C Crh r h C r Cr 5 C r Cr r ( r, h) F( r) r Cr C C 5r F ( r) 0 C C 5r C r 5 F ( r) 0r C C r C C 5 h Cr CC 5 So, ( 5) C ( C 5) C 5 r 5r ( 5r ) 5 5 h r r 5. 5 B the Second Derivative Test, this is a maimum. 00 Brooks/Cole, Cengage Learning

Review Eercises or Chapter 69 Review Eercises or Chapter. ( ),, + + Elliptic paraboloid. ( ),, + 0 Elliptic cone.. (, ) (a) (b) g(, ) (, ) + is the graph o shited units back on the -ais. The hemisphere has equation (c) g(, ) (, ) (d) + + +, 0. is the graph o relected in the -plane, and shited units upward. ( 0, ) (,0). (, ) + (a) 5 5., e + The level curves are o the orm + c e c + ln. The level curves are circles centered at the origin. (b) g ( ),, + is a vertical translation o two units upward. (c) g(, ) (, ) (d) is a horiontal translation o two units to the right. The verte moves rom ( 0, 0, 0) to 0,, 0. 5 (, ) (, ) (, ) (,) 5 c 0 c Generated b Mathematica 6. ( ), ln The level curves are o the orm c ln e c. The level curves are hperbolas c c c c 0 00 Brooks/Cole, Cengage Learning

70 Chapter Functions o Several Variables 7. (, ) The level curves are o the orm c. c c The level curves are hperbolas. c c c. lim (, ) (, ) + Continuous ecept at ( 0, 0 ).. lim (, ) (, ) Does not eist. Continuous ecept when ±. c + e 0 + 0. lim (, ) ( 0,0 ) + + 0 0 Continuous everwhere. + The level curves are o the orm c + c. c 8. (, ) The level curves are passing through the origin with slope c. c 9. ( ), e + g + 0. (, ) Generated b Mathematica c c c c c c c c. lim (, ) ( 0,0 ) + For,. + + For 0, 0 + or 0. The limit does not eist. Continuous to all (, ) ( 0,0) 5., e cos e cos e sin 6. (, ) 7. + ( + ) ( + ) ( + ) ( + ) e + e e, e 8. ln( + + ) + + + + 60 5 5 00 Brooks/Cole, Cengage Learning

Review Eercises or Chapter 7 9. g (, ) g g + ( ) ( ) ( + ) + + +. 5. ut, csin akcoskt u u t ( cos ) akc ak cos kt ( sin ) kc ak sin kt 0. w w ( ) w w. ( ). (,, ),, arctan + + ( ) ( ) + + arctan + + + ( ) + + + + + + ( ) ( + + + ) + + + ( ) + + + ( ) n t. ut (, ) ce sin( n) u u t cne n t cn e cos n t ( n) sin ( n) 6. ln( + ) ln( + ). At (, 0, 0 ), 0. Slope in -direction. +. Slope in -direction., 0, 0,. At, + 7. 6 + 6 8. h (, ) h h h h h h 6 + ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) + ( + ) ( + ) ( + ) 00 Brooks/Cole, Cengage Learning

7 Chapter Functions o Several Variables 9. h, sin + cos h sin sin h cos + cos h cos h sin h cos sin h cos sin 0. g(, ) cos( ) g sin g sin g cos g cos g cos g cos. So, + 0.. 6 6 6 So, + 0.. + + + + + + ( + ) ( + ) ( ) ( )( + ) + + + + So, + 0.. e sin e cos e sin e sin e sin So, + e sin + e sin 0. 00 Brooks/Cole, Cengage Learning

Review Eercises or Chapter 7 5. 6. sin d d + d ( cos + sin ) d + ( cos ) d + d d + d + + + + d + d d + d + + ( + ) ( + ) + 7. d d + d 5 7 d d + d + 6 d 7 6 Percentage error: 0.050 5% 0.65 cm 8. From the accompaning igure we observe h tan θ or h tan θ h h dh d + dθ tan θ d + sec θ dθ. θ π π Letting 00, d ±, θ, and dθ ±. 60 80 (Note that we epress the measurement o the angle in radians.) The maimum error is approimatel π π π dh ± + ± ± ± ± 60 60 80 tan 00 sec 0.7.8.8 eet. h θ 9. V πr h dv πrh dr + πr dh π 5 ± + π ± ± π ± π ± π in. 5 8 8 6 6 0. A πr r + h πr πrh r + h r + h π( r + h ) πrh π( 8 + 5) 0π π dr + dh ± + ± ± r + h r + h 9 8 9 8 8 9 da π r + h + dr dh + w ln +, t, t dw w d w d (a) Chain Rule: + dt dt dt + + + 8t t + t. (b) Substitution: w ln( + ) ln( t + t) dw dt t + t ( 8t ). u, cos t, sint du u u (a) Chain Rule: + dt t t sint + cost sin t + ( sin t) cos t sin t + cos t (b) Substitution: u sin t cos t du sint cost + sint dt sin t + cos ( t) 00 Brooks/Cole, Cengage Learning

7 Chapter Functions o Several Variables. w, r + t, rt, r t w w w w (a) Chain Rule: + + r r r r + () t rt r + t t r + t rt + r t r t ( r t) rt rt t ( r t) w w w w + + t t t t () + ( r) ( ) rt rt + r ( r t) (b) Substitution: w ( r + t)( rt) r t + rt r t r t w r t rt t r ( r t) w r t rt + r t r t. u + +, r cos t, r sin t, t u u u u (a) Chain Rule: + + r r r r cost + sint + ( 0) ( r cos t + r sin t) r u u u u + + t t t t r sin t + r cos t + r sin t cos t + r sin t cos t + t t ur, t r cos t+ r sin t+ t r + t u r r u t t (b) Substitution: 5. + + + + 0 + + + 0 + + + + + 0 + 6. sin 0 + cos 0 cos cos sin 0 sin cos 00 Brooks/Cole, Cengage Learning

Review Eercises or Chapter 75 7. (, ) i + j 5, 5 50i + 5 j u i j unit vector 5 5 ( 5, 5) ( 5, 5) Du u 8. (, ) 0 0 50 i+ j, i + j D u 5 5 u v i + j 5 5 5 (, ) (, ) u 5 5 5 + 5 5 5 9. w + w i + j + k w(,, ) i + j + k u v i j + k Dw u (,,) w(,,) u + 50. w 5 + w ( 0 + ) i + ( 6) j k w(, 0, ) 0i + j u ( i + j k) Dw u (, 0, ) w(, 0, ) u 0 + 5., i + j i + j, 5. e cos e cos i e sin j π 0, i j, π 0, 5. 5. + + i ( + ) ( + ) (, ) i, 0 (, ) +, j, i ( ) ( ) 55. (, ) 9, c 65, P(,) (a) (, ) 8i 8j (, ) 5i 6j 5i 6 j 5i 6 j 79 (b) Unit normal: ( 7i 8j) (c) Slope (d) 56. 7. 8 7 8 ( ) 7 65 Tangent line 8 8 6 6 6 Unit normal vector 6 Tangent line, sin, c, P π,, cos i + sin j π, j (a) (b) Unit normal vector: j (c) Tangent line horiontal: (d) π ( π, π ( π j j 00 Brooks/Cole, Cengage Learning

76 Chapter Functions o Several Variables 57. F,, 0 F i + j k F,, i + j k So, the equation o the tangent plane is + 0or + 8, and the equation o the normal line is t +, t +, t +. F + F j + k F + + 58.,, 5 0 (,, ) 6 j 8k ( j k) So, the equation o the tangent plane is ( ) + ( ) 0 or + 5, and the equation o the normal line is,. F + + + + F ( ) i + ( + 6) j + k F k 59.,, 6 9 0 (,, ) So, the equation o the tangent plane is 0 or, and the equation o the normal line is,, + t. F,, + + 9 0 F i + j + k F + + + + 60. (,, ) i j k ( i j k) So, the equation o the tangent plane is + + 0 or + + 9, and the equation o the normal line is., cos + sin, 0,0 sin, ( 0, 0) 0 cos, ( 0, 0) P(, ) + (b) cos, ( 0, 0) sin, ( 0, 0) 0 0, 0, 0 0 6. (a) 6. F (,, ) + 9 0, (,,5) G (,, ) F j + k G i j F,, 5 j + k i j k F G 0 i + j k 0 5 Tangent line: 6. G ( ),, 0 F,, 0 F i j k G k F,, i j k i j k ( i j) F G + 0 0 So, the equation o the tangent line is,. 6. + +,,,, i + j + k,, i + j + 6k Normal vector to plane. n k 6 cos θ n 56 θ 6.7 (, ) + P 00 Brooks/Cole, Cengage Learning

Review Eercises or Chapter 77 (c) I 0, ou obtain the nd degree Talor polnomial or cos. (d) (, ) P( ) P ( ),, 0 0.0.0.0 0 0..0998.. 0. 0..0799..095 0.5 0..7..75 0.5.097.5.0 (e) The accurac lessens as the distance rom ( 0, 0) increases. 65., + 6 + 9 + 8 + + 6 + 8 0 6 + 8 0, + 6 8, 8 6 ( ) So, (,, ) 8 6 6 > 0. is a relative minimum. 66., + + 5 + 5 0 + 0 + 5 9 0,,,, d 9 < 0 (, ) is a saddle point. 67. (, ) + + 0, 0, So, or and substitution ields the critical point (, ). At the critical point 0. >,, > 0 and So, (,, ) is a relative minimum. 0 0 (,, ) 00 Brooks/Cole, Cengage Learning

78 Chapter Functions o Several Variables 68. 50( + ) ( 0. + 0 + 50) ( 0.05 + 0.6 + 5) 50 0. 0 0, ± 0 50 0.5 0.6 0, ± Critical Points: ( 0, ), ( 0, ), ( 0, ), ( 0, ) 0.6, 0., 0 At 0,, 6. 0 > 0, < 0. ( 0,,99.) is a relative maimum. At ( ) ( 0,, 9.) is a saddle point. 0,, 6. 0 < 0. At ( ) < 0,, 6. 0 0. ( 0,, 00.6) is a saddle point. At > < 0,, 6. 0 0, 0. ( 0,, 79.) is a relative minimum. 69. The level curves are hperbolas. There is a critical point at ( 0, 0 ), but there are no relative etrema. The gradient is normal to the level curve at an given point ( 0, 0). 70. The level curves indicate that there is a relative etremum at A, the center o the ellipse in the second quadrant, and that there is a saddle point at B, the origin. 7. (, ) P R C C 5 0.( + ) ( + ) ( 0.05 + 5 + 500) ( 0.0 + 5 + 600) 0.5 0. 0.8 + 0 + 0,500 P 0.9 0.8 + 0 0 0.9 + 0.8 0 P 0.86 0.8 + 0 0 0.8 + 0.86 0 Solving this sstem ields 9 and 57. P P 0.9 0.8 P 0.86 P < 0 P P P > 0 So, proit is maimum when 9 and 57. 00 Brooks/Cole, Cengage Learning

Review Eercises or Chapter 79 7. Minimie C, 0.5 + 0 + 0.5 + subject to the constraint + 000. 0.50 + 0 λ 5 0.0 + λ 0 + 000 + 5 000 0 8 00 77.5 6.5 C ( 77.5, 6.5) 0,997.50 7. Maimie, + + subject to the constraint 0 + 000. + 0λ 5 + λ 6 0 + 000 5 + 500 5 6 0 9 9. 5 ( 9., 5),0.8 7. Minimie the square o the distance: ( ) + ( ) + ( + ),, 0. + + + + 0 0 + + + + 0 0 Clearl and hence: + 0. Using a computer algebra sstem, 0.689. So, distance 0.689 + 0.689 + 0.689.89. Distance.08 75. (a) 0.00 + 0.07 + 9. (b) When 76. (a).9t +.0 0 80, 0.00 80 + 0.07 80 + 9. 50.6 Kg. 0 0 9 (b) Yes, the data appear linear. (c). + 8.7 ln t (d) 0 5 00 Brooks/Cole, Cengage Learning

80 Chapter Functions o Several Variables 77. Optimie (,, ) + + subject to the constraint + +. + λ + λ + λ + + Maimum: (,, ) 78. Optimie (, ) subject to the constraint +. λ λ + 0 or I 0,. I,then,. 6 Maimum: (, ) 7 Minimum: ( 0,) 0 80., a + b,, > 0 79. PQ +, QR +, RS ; + + 0 C + + + + Constraint: + + 0 C λ g i + j + k λ i + j + k + + λ + λ + λ 9 + + So, 0.707 km, 0.577 km, 0 8.76 km. [ ] Constraint: + 6 6 (a) Level curves o Using, + are lines o orm + C. +., ou obtain 7,, and ( 7, ) 8 + 9 7. 8 0 0 Constraint is an ellipse. (b) Level curves o Using 8, + 9are lines o orm + C. 9 + 7, ou obtain, 5., and (, 5.) 6.8. 9 00 Brooks/Cole, Cengage Learning

Problem Solving or Chapter 8 Problem Solving or Chapter. (a) The three sides have lengths 5, 6, and 5. 6 Thus, s 8 and A 8 ()()(). (b) Let ( abc) ss ( a)( s b)( s c),, area, subject to the constraint a + b + c constant (perimeter). Using Lagrange multipliers, ss ( b)( s c) λ ss ( a)( s c) λ ss a s b λ. From the irst equations s b s a a b. Similarl, b cand hence a b c which is an equilateral triangle. abc,, a+ b+ c, subject (c) Let to ( Area) s( s a)( s b)( s c) constant. Using Lagrange multipliers, λs( s b)( s c) λs( s a)( s c) λss a s b So, s a s b a band a b c.. V πr + πr h Material M π r + π rh 000 πr V 000 h πr πr 000 So, M r r πr 000 8 πr + πr r dm 000 6 8πr πr 0 dr r 6 000 8πr πr r 8 r π 000 750 6 r r 5. π π 000 ( ) π( 750 π) Then, h 0. πr The tank is a sphere o radius r π + π 6 5. π. (a) F( ),, 0 F, F, F Tangent plane: 0 0 0 + 0 0 0 + 0 0 0 0 0 0 + 0 0 + 0 0 0 0 0 V base height 9 0 0 0 0 0 0 (b). (a) As, ( ) ± and hence g g lim lim 0. (b) Let 0, 0 be a point on the graph o. The line through this point perpendicular to g is + 0 + 0. This line intersects g at the point 0 0, 0 0 + +. The square o the distance between these two points is h 0 0 0. h is a maimum or 0. So, the point on arthest rom g is,. 6 0 0 0 0 Base Tangent plane 0 0 6 ( 00 Brooks/Cole, Cengage Learning