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# Sol Purcell Ingles

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### Sol Purcell Ingles

1. 1. Derivatives for Functions of CHAPTER 12 Two or More Variables 12.1 Concepts Review 4. a. 6 1. real-valued function of two real variables b. 12 2. level curve; contour map c. 2 3. concentric circles d. (3cos 6)1/ 2 + 1.44 ≈ 3.1372 4. parallel lines 5. F (t cos t , sec2 t ) = t 2 cos 2 t sec2 t = t 2 , cos t ≠ 0 Problem Set 12.1 6. F ( f (t ), g (t )) = F (ln t 2 , et / 2 ) 1. a. 5 = exp(ln t 2 ) + (et / 2 ) 2 = t 2 + et , t ≠ 0 b. 0 7. z = 6 is a plane. c. 6 d. a6 + a 2 e. 2 x2 , x ≠ 0 f. Undefined The natural domain is the set of all (x, y) such that y is nonnegative. 2. a. 4 8. x + z = 6 is a plane. b. 17 17 c. 16 d. 1 + a2 , a ≠ 0 e. x 3 + x, x ≠ 0 f. Undefined The natural domain is the set of all (x, y) such 9. x + 2y + z = 6 is a plane. that x is nonzero. 3. a. sin(2π ) = 0 ⎛π⎞ b. 4sin ⎜ ⎟ = 2 ⎝6⎠ ⎛π⎞ c. 16sin ⎜ ⎟ = 16 ⎝2⎠ d. π2 sin(π2 ) ≈ –4.2469 744 Section 12.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2. 2. 10. z = 6 – x 2 is a parabolic cylinder. 15. z = exp[–( x 2 + y 2 )] 11. x 2 + y 2 + z 2 = 16 , z ≥ 0 is a hemisphere. x2 16. z = ,y>0 y x2 y 2 z 2 12. + + = 1 , z ≥ 0 is a hemi-ellipsoid. 4 16 16 17. x 2 + y 2 = 2 z; x 2 + y 2 = 2k 13. z = 3 – x 2 – y 2 is a paraboloid. 18. x = zy, y ≠ 0 ; x = ky, y ≠ 0 14. z = 2 – x – y 2 19. x 2 = zy, y ≠ 0; x 2 = ky, y ≠ 0 Instructor’s Resource Manual Section 12.1 745 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3. 3. 20. x 2 = –( y – z ); x 2 = –( y – k ) 24. ( x – 2)2 + ( y + 3)2 = 16 V2 25. a. San Francisco and St. Louis had a x2 + 1 temperature between 70 and 80 degrees 21. z = , k = 1, 2, 4 x2 + y 2 Fahrenheit. k = 1: y 2 = 1 or y = ±1 ; b. Drive northwest to get to cooler two parallel lines temperatures, and drive southeast to get k = 2: 2 x 2 + 2 y 2 = x 2 + 1 warmer temperatures. x2 y 2 c. Since the level curve for 70 runs southwest + = 1 ; ellipse 1 1 to northeast, you could drive southwest or 2 northeast and stay at about the same k = 4: 4 x 2 + 4 y 2 = x 2 + 1 temperature. x2 y2 The lowest barometric pressure, 1000 + = 1; ellipse 26. a. 1 1 3 4 millibars and under, occurred in the region of the Great Lakes, specifically near Wisconsin. The highest barometric 22. y = sin x + z; y = sin x + k pressure, 1025 millibars and over, occurred on the east coast, from Massachusetts to South Carolina. b. Driving northwest would take you to lower barometric pressure, and driving southeast would take you to higher barometric pressure. c. Since near St. Louis the level curves run southwest to northeast, you could drive 23. x = 0, if T = 0: southwest or northeast and stay at about the ⎛1 ⎞ y 2 = ⎜ – 1⎟ x 2 , if y ≠ 0 . same barometric pressure. ⎝T ⎠ 27. x 2 + y 2 + z 2 ≥ 16 ; the set of all points on and outside the sphere of radius 4 that is centered at the origin 28. The set of all points inside (the part containing the z-axis) and on the hyperboloid of one sheet; x2 y 2 z 2 + – = 1. 9 9 9 x2 y 2 z 2 29. + + ≤ 1 ; points inside and on the 9 16 1 ellipsoid 746 Section 12.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4. 4. 30. Points inside (the part containing the z-axis) or on x2 y2 37. 4 x 2 – 9 y 2 = k , k in R; – = 1, if k ≠ 0; x2 y 2 z 2 k k the hyperboloid of one sheet, + – = 1, 4 9 9 9 16 2x excluding points on the coordinate planes planes y = ± (for k = 0) and all hyperbolic 3 31. Since the argument to the natural logarithm cylinders parallel to the z-axis such that the ratio function must be positive, we must have ⎛1⎞ ⎛1⎞ a:b is ⎜ ⎟ : ⎜ ⎟ or 3:2 (where a is associated x 2 + y 2 + z 2 > 0 . This is true for all ( x, y , z ) ⎝ 2⎠ ⎝3⎠ except ( x, y, z ) = ( 0, 0, 0 ) . The domain consists with the x-term) 3 all points in except the origin. 2 + y2 + z2 38. e x = k, k > 0 32. Since the argument to the natural logarithm x 2 + y 2 + z 2 = ln k function must be positive, we must have xy > 0 . concentric circles centered at the origin. This occurs when the ordered pair ( x, y ) is in the first quadrant or the third quadrant of the 39. a. All ( w, x, y, z ) except ( 0, 0, 0, 0 ) , which xy-plane. There is no restriction on z. Thus, the would cause division by 0. domain consists of all points ( x, y, z ) such that x and y are both positive or both negative. b. All ( x1 , x2 ,… , xn ) in n-space. 33. x 2 + y 2 + z 2 = k , k > 0; set of all spheres c. All ( x1 , x2 ,… , xn ) that satisfy centered at the origin 2 2 x1 + x2 + 2 + xn ≤ 1 ; other values of 34. 100 x 2 + 16 y 2 + 25 z 2 = k , k > 0; ( x1 , x2 ,… , xn ) would lead to the square root of a negative number. x2 y2 z2 + + = 1; set of all ellipsoids centered k 100 k 16 k 25 40. If z = 0, then x = 0 or x = ± 3 y . at origin such that their axes have ratio ⎛ 1 ⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ : ⎜ ⎟ : ⎜ ⎟ or 2:5:4. ⎝ 10 ⎠ ⎝ 4 ⎠ ⎝ 5 ⎠ x2 y2 z2 35. + – = k ; the elliptic cone 1 1 1 16 4 2 2 x y z2 + = and all hyperboloids (one and two 9 9 16 sheets) with z-axis for axis such that a:b:c is 41. a. AC is the least steep path and BC is the most steep path between A and C since the level ⎛1⎞ ⎛1⎞ ⎛1⎞ curves are farthest apart along AC and ⎜ ⎟ : ⎜ ⎟ : ⎜ ⎟ or 3:3:4. ⎝ 4⎠ ⎝ 4⎠ ⎝ 3⎠ closest together along BC. x2 y2 z2 AC ≈ (5750) 2 + (3000)2 ≈ 6490 ft 36. – – = k ; the elliptical cone b. 1 1 1 9 4 BC ≈ (580) 2 + (3000) 2 ≈ 3060 ft 2 2 y z x2 + = and all hyperboloids (one and two 9 36 4 sheets) with x-axis for axis such that a:b:c is ⎛1⎞ ⎛1⎞ ⎜ ⎟ : ⎜ ⎟ :1 or 2:3:6 ⎝3⎠ ⎝ 2⎠ 42. Completing the squares on x and y yields the equivalent equation f ( x, y ) + 25.25 = ( x – 0.5) 2 + 3( y + 2)2 , an elliptic paraboloid. Instructor’s Resource Manual Section 12.1 747 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5. 5. 43. (2 x – y 2 ) exp(– x 2 – y 2 ) sin 2x 2 + y 2 46. 44. sin x sin y (1 + x 2 + y 2 ) sin( x 2 + y 2 ) x2 + y2 12.2 Concepts Review [( f ( x0 + h, y0 ) – f ( x0 , y0 )] 1. lim ; partial h →0 h derivative of f with respect to x 2. 5; 1 ∂2f 3. 45. ∂ y∂ x 4. 0 Problem Set 12.2 1. f x ( x, y ) = 8(2 x – y )3 ; f y ( x, y ) = –4(2 x – y )3 2. f x ( x, y ) = 6(4 x – y 2 )1/ 2 ; f y ( x, y ) = –3 y (4 x – y 2 )1/ 2 748 Section 12.2 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6. 6. ( xy )(2 x) – ( x 2 – y 2 )( y ) x2 + y2 f s ( s, t ) = –2set 2 – s2 ; f s ( s, t ) = 2tet 2 – s2 3. f x ( x, y ) = = 14. ( xy )2 x2 y ( xy )(−2 y ) − ( x 2 − y 2 )( x) 15. Fx ( x, y ) = 2 cos x cos y; Fy ( x, y ) = –2sin x sin y f y ( x, y ) = ( xy )2 16. f r (r , θ ) = 9r 2 cos 2θ ; fθ (r , θ ) = –6r 3 sin 2θ ( x2 + y2 ) =− xy 2 17. f x ( x, y ) = 4 xy 3 – 3x 2 y5 ; f xy ( x, y ) = 12 xy 2 – 15 x 2 y 4 4. f x ( x, y ) = e x cos y; f y ( x, y ) = – e x sin y f y ( x, y ) = 6 x 2 y 2 – 5 x 3 y 4 ; 5. f x ( x, y ) = e y cos x; f y ( x, y ) = e y sin x f yx ( x, y ) = 12 xy 2 – 15 x 2 y 4 ⎛ 1⎞ 6. f x ( x, y ) = ⎜ – ⎟ (3 x 2 + y 2 ) –4 / 3 (6 x) 18. f x ( x, y ) = 5( x3 + y 2 ) 4 (3 x 2 ); ⎝ 3⎠ f xy ( x, y ) = 60 x 2 ( x3 + y 2 )3 (2 y ) = –2 x(3 x 2 + y 2 ) –4 / 3 ; ⎛ 1⎞ = 120 x 2 y ( x3 + y 2 )3 f y ( x, y ) = ⎜ – ⎟ (3 x 2 + y 2 ) –4 / 3 (2 y ) ⎝ 3⎠ f y ( x, y ) = 5( x3 + y 2 ) 4 (2 y ); ⎛ 2y ⎞ 2 2 –4 / 3 = ⎜– ⎟ (3 x + y ) f yx ( x, y ) = 40 y ( x3 + y 2 )3 (3x 2 ) ⎝ 3 ⎠ = 120 x 2 y ( x3 + y 2 )3 2 2 −1/ 2 7. f x ( x, y ) = x ( x − y ) ; f y ( x, y ) = – y ( x 2 – y 2 ) –1/ 2 19. f x ( x, y ) = 6e2 x cos y; f xy ( x, y ) = –6e2 x sin y f y ( x, y ) = –3e2 x sin y; f yx ( x, y ) = –6e2 x sin y 8. fu (u , v) = ve ; f v (u , v) = ue uv uv 20. f x ( x, y ) = y (1 + x 2 y 2 ) –1; – xy – xy 9. g x ( x, y ) = – ye ; g y ( x, y ) = – xe f xy ( x, y ) = (1 − x 2 y 2 )(1 + x 2 y 2 ) −2 10. f s ( s, t ) = 2 s ( s 2 – t 2 ) –1 ; f x ( x, y ) = x(1 + x 2 y 2 ) –1; ft ( s, t ) = −2t ( s 2 − t 2 )−1 f xy ( x, y ) = (1 − x 2 y 2 )(1 + x 2 y 2 ) −2 11. f x ( x, y ) = 4[1 + (4 x – 7 y ) 2 ]–1; ( xy )(2) – (2 x – y )( y ) y2 1 21. Fx ( x, y ) = = = ; 2 2 2 f y ( x, y ) = –7[1 + (4 x – 7 y ) ] 2 –1 ( xy ) x y x2 1 Fx (3, − 2) = 1 ⎛1⎞ –1 ⎛ w ⎞ 9 12. Fw ( w, z ) = w ⎜ ⎟ + sin ⎜ ⎟ ( xy )(–1) – (2 x – y )( x) –2 x 2 2 1– ( ) w 2 z ⎝ z⎠ ⎝z⎠ Fy ( x, y ) = ( xy ) 2 = 2 2 x y =– x2 ; w ⎛ w⎞ 1 = z + sin –1 ⎜ ⎟ ; Fy (3, − 2) = − 2 ( w) ⎝z⎠ 2 1– z 22. Fx ( x, y ) = (2 x + y )( x 2 + xy + y 2 ) –1; ( w) 2 1 ⎛ w⎞ – Fz = ( w, z ) = w z 2 ⎜– 2 ⎟ = Fx (–1, 4) = ≈ 0.1538 ( ) w 2 ⎝ z ⎠ 1– ( w) 2 1– 13 z z Fy ( x, y ) = ( x + 2 y )( x 2 + xy + y 2 ) –1 ; 13. f x ( x, y ) = –2 xy sin( x 2 + y 2 ); Fy (–1, 4) = 7 ≈ 0.5385 2 2 2 2 2 13 f y ( x, y ) = –2 y sin( x + y ) + cos( x + y ) Instructor’s Resource Manual Section 12.2 749 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7. 7. 23. f x ( x, y ) = – y 2 ( x 2 + y 4 ) –1; 31. P (V , T ) = kT V fx ( ) 5, – 2 = – 4 21 ≈ –0.1905 k P (V , T ) = ; T V f y ( x, y ) = 2 xy ( x 2 + y 4 ) –1; k P (100, 300) = lb/in.2 per degree ( ) 4 5 T fy 5, – 2 = – ≈ –0.4259 100 21 32. V [ P (V , T )] + T [ P (V , T )] V T 24. f x ( x, y ) = e y sinh x; = V (– kTV –2 ) + T (kV –1 ) = 0 f x (–1, 1) = e sinh(–1) ≈ –3.1945 ⎛ kT ⎞ ⎛ k ⎞ ⎛ V ⎞ kT PV f y ( x, y ) = e y cosh x; P VT TP = ⎜ – V ⎟⎜ ⎟⎜ ⎟=– =– = –1 ⎝ V 2 ⎠⎝ P ⎠⎝ k ⎠ PV PV f y (–1, 1) = e cosh(–1) ≈ 4.1945 x2 y 2 33. f x ( x, y ) = 3 x 2 y – y 3 ; f xx ( x, y ) = 6 xy; 25. Let z = f ( x, y ) = + . 9 4 f y ( x, y ) = x3 – 3xy 2 ; f yy ( x, y ) = –6 xy f y ( x, y ) = y Therefore, f xx ( x, y ) + f yy ( x, y ) = 0. 2 The slope is f y (3, 2) = 1. 34. f x ( x, y ) = 2 x( x 2 + y 2 ) –1 ; f xx ( x, y ) = –2( x 2 – y 2 )( x 2 + y 2 ) –1 26. Let z = f ( x, y ) = (1/ 3)(36 – 9 x 2 – 4 y 2 )1/ 2 . ⎛ 4⎞ f y ( x, y ) = 2 y ( x 2 + y 2 ) –1 ; f y ( x, y ) = ⎜ − ⎟ y (36 − 9 x 2 − y 2 ) −1/ 2 ⎝ 3⎠ f yy ( x, y ) = 2( x 2 − y 2 )( x 2 + y 2 ) −1 8 The slope is f y (1, – 2) = ≈ 0.8040. 3 11 35. Fy ( x, y ) = 15 x 4 y 4 – 6 x 2 y 2 ; ⎛1⎞ Fyy ( x, y ) = 60 x 4 y 3 − 12 x 2 y; 27. z = f ( x, y ) = ⎜ ⎟ (9 x 2 + 9 y 2 − 36)1/ 2 ⎝2⎠ Fyyy ( x, y ) = 180 x 4 y 2 – 12 x 2 9x f x ( x, y ) = 2(9 x + 9 y 2 – 36)1/ 2 2 36. f x ( x, y ) = [– sin(2 x 2 – y 2 )](4 x) f x (2, 1) = 3 = –4 x sin(2 x 2 – y 2 ) ⎛5⎞ 28. z = f ( x, y ) = ⎜ ⎟ (16 – x 2 )1/ 2 . f xx ( x, y ) = (–4 x)[cos(2 x 2 – y 2 )](4 x) ⎝4⎠ ⎛ 5⎞ + [sin(2 x 2 – y 2 )](–4) f x ( x, y ) = ⎜ – ⎟ x(16 – x 2 ) –1/ 2 ⎝ 4⎠ f xxy ( x, y ) = –16 x 2 [– sin(2 x 2 – y 2 )](–2 y ) 5 f x (2, 3) = – ≈ –0.7217 – 4[cos(2 x 2 – y 2 )](–2 y ) 4 3 = –32 x 2 y sin(2 x 2 – y 2 ) + 8 y cos(2 x 2 – y 2 ) 29. Vr (r , h) = 2πrh; Vr (6, 10) = 120π ≈ 376.99 in.2 ∂3f 37. a. ∂ y3 2 30. Ty ( x, y ) = 3 y ; Ty (3, 2) = 12 degrees per ft ∂ 3y b. ∂ y∂ x 2 ∂ 4y c. ∂ y 3∂ x 750 Section 12.2 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8. 8. 38. a. f yxx 45. Domain: (Case x < y) The lengths of the sides are then x, y – x, and b. f yyxx 1 – y. The sum of the lengths of any two sides must be greater than the length of the remaining side, leading to three inequalities: c. f yyxxx 1 x + (y – x) > 1 – y ⇒ y > 2 39. a. f x ( x, y, z ) = 6 xy – yz 1 (y – x) + (1 – y) > x ⇒ x < 2 b. f y ( x, y, z ) = 3 x 2 – xz + 2 yz 2 ; 1 x + (1 – y) > y – x ⇒ y < x + f y (0, 1, 2) = 8 2 c. Using the result in a, f xy ( x, y, z ) = 6 x – z. 40. a. 12 x 2 ( x3 + y 2 + z )3 b. f y ( x , y , z ) = 8 y ( x 3 + y 2 + z )3 ; f y (0, 1, 1) = 64 c. f z ( x, y, z ) = 4( x3 + y 2 + z )3 ; The case for y < x yields similar inequalities 2 f zz ( x, y, z ) = 12( x + y + z )2 2 (x and y interchanged). The graph of DA , the domain of A is given above. In set notation it is f x ( x, y, x) = – yze – xyz – y ( xy – z 2 ) –1 ⎧ 1 1 1⎫ 41. D A = ⎨ ( x, y ) : x < , y > , y < x + ⎬ ⎩ 2 2 2⎭ ⎛ 1 ⎞⎛ xy ⎞ –1/ 2 ⎛ y⎞ ⎧ 1 1 1⎫ f x ( x, y, z ) = ⎜ ⎟⎜ ⎟ ∪ ⎨ ( x, y ) : x > , y < , x < y + ⎬ . 42. ⎜ ⎟; ⎩ 2 2 2⎭ ⎝ 2 ⎠⎝ z ⎠ ⎝z⎠ –1/ 2 Range: The area is greater than zero but can be ⎛ 1 ⎞⎛ 1 ⎞ ⎛ 1⎞ 1 arbitrarily close to zero since one side can be f x (–2, – 1, 8) = ⎜ ⎟ ⎜ ⎟ ⎜– ⎟ = – ⎝ 2 ⎠⎝ 4 ⎠ ⎝ 8⎠ 8 arbitrarily small and the other two sides are bounded above. It seems that the area would be 43. If f ( x, y ) = x 4 + xy3 + 12, f y ( x, y ) = 3 xy 2 ; largest when the triangle is equilateral. An 1 f y (1, – 2) = 12. Therefore, along the tangent line equilateral triangle with sides equal to has 3 Δy = 1 ⇒ Δz = 12, so 0, 1, 12 is a tangent 3 ⎛ 3⎤ area . Hence the range of A is ⎜ 0, ⎜ ⎥ . (In vector (since Δx = 0). Then parametric equations 36 ⎝ 36 ⎦ ⎧ x =1 ⎫ Sections 8 and 9 of this chapter methods will be ⎪ ⎪ of the tangent line are ⎨ y = –2 + t ⎬ . Then the presented which will make it easy to prove that ⎪ z = 5 + 12t ⎪ the largest value of A will occur when the triangle ⎩ ⎭ is equilateral.) point of xy-plane at which the bee hits is (1, 0, 29) [since y = 0 ⇒ t = 2 ⇒ x = 1, z = 29]. 46. a. u = cos (x) cos (ct): u x = – sin( x ) cos(ct ) ; ut = – c cos( x) sin(ct ) 44. The largest rectangle that can be contained in the circle is a square of diameter length 20. The edge u xx = – cos( x) cos(ct ) of such a square has length 10 2, so its area is utt = – c 2 cos( x) cos(ct ) 200. Therefore, the domain of A is Therefore, c 2 u xx = utt . 2 2 {( x, y ) : 0 ≤ x + y < 400}, and the range is u = e x cosh(ct ) : (0, 200]. u x = e x cosh(ct ), ut = ce x sinh(ct ) u xx = e x cosh(ct ), utt = c 2 e x cosh(ct ) Therefore, c 2u xx = utt . Instructor’s Resource Manual Section 12.2 751 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9. 9. b. u = e – ct sin( x) : 48. a. sin( x + y 2 ) u x = e – ct cos x u xx = – e – ct sin x ut = – cect sin x Therefore, cu xx = ut . 2 u = t –1/ 2 e – x / 4ct : 2 / 4ct ⎛ x ⎞ u x = t –1/ 2 e – x ⎜– ⎟ b. Dx (sin( x + y 2 )) ⎝ 2ct ⎠ ( x 2 – 2ct ) u xx = 2 / 4ct (4c 2 t 5 / 2 e x ) 2 ( x – 2ct ) ut = 2 (4ct 5 / 2 e x / 4ct ) Therefore, cu xx = ut . 47. a. Moving parallel to the y-axis from the point c. D y (sin( x + y 2 )) (1, 1) to the nearest level curve and Δz approximating , we obtain Δy 4–5 f y (1, 1) = = –4. 1.25 – 1 b. Moving parallel to the x-axis from the point (–4, 2) to the nearest level curve and Δz d. Dx ( D y (sin( x + y ) 2 )) approximating , we obtain Δx 1– 0 2 f x (–4, 2) ≈ = . –2.5 – (–4) 3 c. Moving parallel to the x-axis from the point (–5, –2) to the nearest level curve and Δz approximately , we obtain Δx 1– 0 2 49. a. f y ( x, y , z ) f x (–4, – 5) ≈ = . –2.5 – (–5) 5 f ( x , y + Δy , z ) − f ( x , y , z ) = lim Δy →0 Δy d. Moving parallel to the y-axis from the point (0, –2) to the nearest level curve and b. f z ( x, y , z ) Δz approximating , we obtain f ( x , y , z + Δz ) − f ( x , y , z ) Δy = lim Δz →0 Δz 0 –1 8 f y (0, 2) ≈ = . –19 – (–2) 3 8 c. Gx ( w, x, y, z ) G ( w, x + Δx, y , z ) − G ( w, x, y, z ) = lim Δx →0 Δx 752 Section 12.2 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10. 10. ∂ 4. The limit does not exist because of Theorem A. d. λ ( x, y , z , t ) The function is a rational function, but the limit ∂z of the denominator is 0, while the limit of the λ ( x , y , z + Δz , t ) − λ ( x , y , z , t ) = lim numerator is -1. Δz →0 Δz 5 ∂ 5. – e. S (b0 , b1 , b2 ,… , bn ) = 2 ∂b2 ⎛ S (b0 , b1 , b2 + Δb2 ,… , bn ) ⎞ 6. −1 ⎜ ⎟ − S (b0 , b1 , b2 ,… , bn ) ⎟ 7. 1 = lim ⎜ Δb2 →0 ⎜ Δb2 ⎟ ⎜ ⎟ tan( x 2 + y 2 ) ⎝ ⎠ 8. lim ( x, y )→(0, 0) ( x2 + y2 ) ∂ sin( x 2 + y 2 ) 50. a. ( sin w sin x cos y cos z ) = lim 1 ∂w ( x, y )→(0, 0) ( x + y ) cos( x + y 2 ) 2 2 2 = cos w sin x cos y cos z = (1)(1) = 1 ∂ wyz 9. The limit does not exist since the function is not b. ⎡ x ln ( wxyz ) ⎤ = x ⋅ ⎣ ⎦ + 1 ⋅ ln ( wxyz ) ∂x wxyz defined anywhere along the line y = x. That is, there is no neighborhood of the origin in which = 1 + ln ( wxyz ) the function is defined everywhere except possibly at the origin. c. λt ( x, y, z , t ) (1 + xyzt ) cos x − t ( cos x ) xyz ( x 2 + y 2 )( x 2 – y 2 ) = 10. lim ( x, y )→(0, 0) x2 + y 2 (1 + xyzt )2 cos x = lim ( x2 – y 2 ) = 0 = ( x, y )→(0, 0) (1 + xyzt )2 11. Changing to polar coordinates, xy r cos θ ⋅ r sin θ lim = lim 12.3 Concepts Review ( x, y )→(0,0) x 2 + y 2 r →0 r 1. 3; (x, y) approaches (1, 2). = lim r cos θ ⋅ sin θ = 0 r →0 2. lim f ( x, y ) = f (1, 2) ( x, y )→(1, 2) 12. If ( x, y ) approaches (0, 0) along the line y = x , 3. contained in S x2 1 lim = lim = +∞ 2 2 2 ( x, x )→(0,0) 4 x 2 +x ) ( x, x )→(0,0) ( x 4. an interior point of S; boundary points Thus, the limit does not exist. 13. Use polar coordinates. Problem Set 12.3 r 7 / 3 ( cos θ ) 7/3 x7 / 3 = r1/ 3 ( cos θ ) 7/3 = 1. –18 x +y2 2 r 2 2. 3 r 1/ 3 ( cos θ ) 7/3 → 0 as r → 0 , so the limit is 0. ⎡ 2 ⎛ xy ⎞ ⎤ 14. Changing to polar coordinates, 3. lim ⎢ x cos xy – sin ⎜ 3 ⎟ ⎥ ( x, y )→(2, π) ⎣ ⎝ ⎠⎦ r 2 cos 2 θ − r 2 sin 2 θ lim r 2 cos θ sin θ ⋅ ⎛ 2π ⎞ 3 r →0 r2 = 2 cos 2 2π – sin ⎜ ⎟ = 2 – ≈ 1.1340 ⎝ 3 ⎠ 2 = lim r 2 cos θ sin θ cos 2θ = 0 r →0 Instructor’s Resource Manual Section 12.3 753 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11. 11. r 4 cos 2 θ sin 2 θ 26. Require 4 − x 2 − y 2 − z 2 > 0; 15. f ( x, y ) = r 2 cos 2 θ + r 4 sin 4 θ x 2 + y 2 + z 2 < 4. S is the space in the interior ⎛ cos 2 θ sin 2 θ ⎞ of the sphere centered at the origin with radius 2. = r2 ⎜ ⎟ ⎜ cos 2 θ + r 2 sin 4 θ ⎟ ⎝ ⎠ 27. The boundary consists of the points that form the If cos θ = 0 , then f ( x, y ) = 0 . If cos θ ≠ 0 , outer edge of the rectangle. The set is closed. hen this converges to 0 as r → 0 . Thus the limit is 0. 16. As ( x, y ) approaches (0,0) along x = y2, y4 1 lim = . Along the x-axis, 4 24 y +y ( x, x )→(0,0) 0 however, lim = 0. Thus, the limit does ( x,0)→(0,0) x 2 not exist. 17. f ( x, y ) is continuous for all ( x, y ) since 28. The boundary consists of the points of the circle for all ( x, y ), x 2 + y 2 + 1 ≠ 0. shown. The set is open. 18. f ( x, y ) is continuous for all ( x, y ) since for all ( x, y ), x 2 + y 2 + 1 > 0. 19. Require 1 – x 2 – y 2 > 0; x 2 + y 2 < 1. S is the interior of the unit circle centered at the origin. 20. Require 1 + x + y > 0; y > − x − 1. S is the set of all ( x, y ) above the line y = − x − 1. 29. The boundary consists of the circle and the 2 21. Require y – x ≠ 0. S is the entire plane except origin. The set is neither open (since, for example, (1, 0) is not an interior point), nor the parabola y = x 2 . closed (since (0, 0) is not in the set). 22. The only points at which f might be discontinuous occur when xy = 0. sin( xy ) lim = 1 = f (a, 0) for all nonzero ( x, y )→( a , 0) xy a in , and then sin( xy ) lim = 1 = f (0, b) for all b in . ( x, y )→(0, b ) xy Therefore, f is continuous on the entire plane. 23. Require x – y + 1 ≥ 0; y ≤ x + 1. S is the region below and on the line y = x + 1. 24. Require 4 – x 2 – y 2 > 0; x 2 + y 2 < 4. S is the interior of the circle of radius 2 centered at the origin. 25. f ( x, y, z ) is continuous for all ( x, y, z ) ≠ ( 0, 0, 0 ) since for all ( x, y, z ) ≠ ( 0, 0, 0 ) , x 2 + y 2 + z 2 > 0. 754 Section 12.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12. 12. 30. The boundary consists of the points on the line 0 x = 1 along with the points on the line x = 4. The 35. Along the x-axis (y = 0): lim =0. ( x, y )→(0, 0) x 2 +0 set is neither closed nor open. Along y = x: x2 1 1 lim = = . lim ( x, y )→(0, 0) 2 x 2 ( x, y )→(0, 0) 2 2 Hence, the limit does not exist because for some points near the origin f(x, y) is getting closer to 0, 1 but for others it is getting closer to . 2 0 36. Along y = 0: lim = 0. Along y = x: x→ 0 x 2 +0 31. The boundary consists of the graph of ⎛1⎞ x 2 + x3 1+ x 1 y = sin ⎜ ⎟ along with the part of the y-axis for lim = lim = . x→ 0 x 2 +x 2 x→ 0 2 2 ⎝ x⎠ which y ≤ 1. The set is open. x 2 (mx) mx3 37. a. lim = lim x→ 0 x 4 + (mx)2 x→ 0 x 4 + m2 x 2 mx = lim =0 x→ 0 x 2 + m2 x2 ( x2 ) x4 1 1 b. lim = lim = lim = x→ 0 x 4 + (x )2 2 x→ 0 2 x 4 x→ 0 2 2 32. The boundary is the set itself along with the x2 y origin. The set is neither open (since none of its c. lim does not exist. ( x, y )→( 0, 0) x 4 +y points are interior points) nor closed (since the origin is not in the set). 38. f is discontinuous at each overhang. More interesting, f is discontinuous along the Continental Divide. 39. a. {( x, y, z ) : x 2 + y 2 = 1, z in [1, 2]} [For x 2 + y 2 < 1, the particle hits the hemisphere and then slides to the origin (or bounds toward the origin); for x 2 + y 2 = 1, it 2 2 bounces up; for x 2 + y 2 > 1, it falls straight x – 4y ( x + 2 y )( x – 2 y ) 33. = = x + 2 y (if x ≠ 2y) down.] x – 2y x – 2y If x = 2 y , x + 2 y = 2 x . Take g ( x ) = 2 x . b. {( x, y, z ) : x 2 + y 2 = 1, z = 1} (As one moves at a level of z = 1 from the rim of the bowl 34. Let L and M be the latter two limits. toward any position away from the bowl [ f ( x, y ) + g ( x, y )] – [ L + M ] there is a change from seeing all of the interior of the bowl to seeing none of it.) ε ε ≤ f ( x, y ) – L + f ( x , y ) – M ≤ + 2 2 c. {(x, y, z): z = 1} [f(x, y, z) is undefined for (x, y) in some δ-neighborhood of (a, b). (infinite) at (x, y, 1).] Therefore, lim [ f ( x, y ) + g ( x, y )] = L + M . d. φ (Small changes in points of the domain ( x , y ) →( a , b ) result in small changes in the shortest path from the points to the origin.) Instructor’s Resource Manual Section 12.3 755 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13. 13. 40. f is continuous on an open set D and P0 is in D implies that there is neighborhood of P0 with radius r on which f is continuous. f is continuous at P0 ⇒ lim f ( P) = f ( P0 ). Now let ε = f ( P0 ) which is positive. Then there is a δ P → P0 such that 0 < δ < r and f ( p ) – f ( P0 ) < f ( P0 ) if P is in the δ-neighborhood of P0 . Therefore, – f ( P0 ) < f ( p ) – f ( P0 ) < f ( P0 ), so 0 < f(p) (using the left-hand inequality) in that δ-neighborhood of P0 . ⎧( x 2 + y 2 )1/ 2 + 1 if y ≠ 0 ⎫ ⎪ ⎪ 41. a. f ( x, y ) = ⎨ ⎬ . Check discontinuities where y = 0. ⎪ ⎩ x –1 if y = 0 ⎪ ⎭ As y = 0, ( x 2 + y 2 )1/ 2 + 1 = x + 1, so f is continuous if x + 1 = x – 1 . Squaring each side and simplifying yields x = – x, so f is continuous for x ≤ 0. That is, f is discontinuous along the positive x-axis. b. Let P = (u, v) and Q = (x, y). ⎧ ⎪ OP + OQ if P and Q are not on same ray from the origin and neither is the origin ⎫ ⎪ f (u, v, x, y ) = ⎨ ⎬. ⎪ PQ ⎩ otherwise ⎪ ⎭ This means that in the first case one travels from P to the origin and then to Q; in the second case one travels directly from P to Q without passing through the origin, so f is discontinuous on the set {(u, v, x, y ) : u, v = k x, y for some k > 0, u , v ≠ 0, x, y ≠ 0}. ⎛ hy ( h2 – y 2 ) ⎞ ⎜ 2 2 –0⎟ 2 2 42. a. f x (0, y ) = lim ⎜ h +y ⎟ = lim y (h − y ) = − y h →0 ⎜ h ⎟ h →0 h 2 + y 2 ⎜ ⎟ ⎝ ⎠ ⎛ xh ( x 2 – h 2 ) ⎞ ⎜ –0⎟ x 2 + h2 y( x2 – h2 ) b. f y ( x, 0) = lim ⎜ ⎟ = lim =x h →0 ⎜ h ⎟ h →0 x 2 + y 2 ⎜ ⎟ ⎝ ⎠ f y (0 + h, y ) – f y (0, y ) h–0 c. f yx (0, 0) = lim = lim =1 h →0 h h →0 h f x ( x, 0 + h) – f x ( x, 0) –h – 0 d. f xy (0, 0) = lim = lim = –1 h →0 h h →0 h Therefore, f xy (0, 0) ≠ f yx (0, 0). 43. b. 44. a. 45. 756 Section 12.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14. 14. 46. A function f of three variables is continuous at a point (a, b, c ) if f (a, b, c) is defined and equal to the limit of f ( x, y, z ) as ( x, y, z ) approaches (a, b, c) . In other words, lim f ( x, y, z ) = f (a, b, c) . ( x , y , z ) → ( a ,b , c ) A function of three variables is continuous on an open set S if it is continuous at every point in the interior of the set. The function is continuous at a boundary point P of S if f (Q) approaches f (P) as Q approaches P along any path through points in S in the neighborhood of P. 47. If we approach the point ( 0, 0, 0 ) along a straight path from the point ( x, x, x ) , we have x ( x )( x) x3 1 lim = lim = 3 3 3 3 3 x +x +x ( x, x, x )→(0,0,0) ( x , x, x )→(0,0,0) 3 x Since the limit does not equal to f (0, 0, 0) , the function is not continuous at the point (0, 0, 0) . 48. If we approach the point (0, 0, 0) along the x-axis, we get ( x 2 − 02 ) x2 lim (0 + 1) = lim =1 ( x,0,0)→(0,0,0) ( x 2 + 02 ) ( x,0,0)→(0,0,0) x 2 Since the limit does not equal f (0, 0, 0) , the function is not continuous at the point (0, 0, 0). 12.4 Concepts Review 1. gradient 2. locally linear ∂f ∂f 3. (p)i + (p) j; y 2 i + 2 xyj ∂x ∂y 4. tangent plane Problem Set 12.4 1. 2 xy + 3 y, x 2 + 3 x 2. 3 x 2 y , x3 – 3 y 2 3. ∇f ( x, y ) = ( x)(e xy y ) + (e xy )(1), xe xy x = e xy xy + 1, x 2 4. 2 xy cos y, x 2 (cos y – y sin y ) 5. x( x + y ) –2 y ( x + 2), x 2 6. ∇f ( x, y ) = 3[sin 2 ( x 2 y )][cos( x 2 y )](2 xy ), 3[sin 2 ( x 2 y )][cos( x 2 y )]( x 2 ) = 3x sin 2 ( x 2 y ) cos( x 2 y ) 2 y, x Instructor’s Resource Manual Section 12.4 757 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15. 15. 7. ( x 2 + y 2 + z 2 ) –1/ 2 x, y, z 8. 2 xy + z 2 , x 2 + 2 yz , y 2 + 2 xz 9. ∇f ( x, y ) = ( x 2 y )(e x – z ) + (e x – z )(2 xy ), x 2 e x – z , x 2 ye x – z (–1) = xe x – z y ( x + 2), x, – xy 10. xz ( x + y + z ) –1 + z ln( x + y + z ), xz ( x + y + z ) –1 , xz ( x + y + z ) –1 + x ln( x + y + z ) 11. ∇f ( x, y ) = 2 xy – y 2 , x 2 – 2 xy ; ∇f (–2, 3) = –21, 16 z = f (–2, 3) + –21, 16 ⋅ x + 2, y – 3 = 30 + (–21x – 42 + 16 y – 48) z = –21x + 16y – 60 12. ∇f ( x, y ) = 3x 2 y + 3 y 2 , x3 + 6 xy , so ∇f (2, – 2) = (–12, – 16). Tangent plane: z = f (2, – 2) + ∇ (2, – 2) ⋅ x – 2, y + 2 = 8 + –12, – 16 ⋅ x – 2, y + 2 = 8 + (–12x + 24 – 16y – 32) z = –12x – 16y 13. ∇f ( x, y ) = – π sin(πx) sin(πy ), π cos(πx) cos(πy ) + 2π cos(2πy ) ⎛ 1⎞ ∇f ⎜ –1, ⎟ = 0, – 2π ⎝ 2⎠ ⎛ 1⎞ 1 z = f ⎜ –1, ⎟ + 0, – 2π ⋅ x + 1, y – = –1 + (0 – 2πy + π); ⎝ 2⎠ 2 z = –2 π y + ( π – 1) 2x x2 14. ∇f ( x, y ) = ,− ; ∇f (2, − 1) = −4, − 4 y y2 z = f (2, – 1) + −4, – 4 ⋅ x – 2, y + 1 = –4 + (–4x + 8 –4y – 4) z = –4x – 4y 15. ∇f ( x, y, z ) = 6 x + z 2 , – 4 y, 2 xz , so ∇f (1, 2, – 1) = 7, – 8, – 2 Tangent hyperplane: w = f (1, 2, – 1) + ∇f (1, 2, – 1) ⋅ x – 1, y – 2, z + 1 = –4 + 7, – 8, – 2 ⋅ x – 1, y – 2, z + 1 = –4 + (7x – 7 – 8y + 16 – 2z – 2) w = 7x – 8y – 2z + 3 16. ∇f ( x, y, z ) = yz + 2 x, xz , xy ; ∇f (2, 0, – 3) = 4, – 6, 0 w = f (2, 0, – 3) + 4, – 6, 0 ⋅ x – 2, y, z + 3 = 4 + (4x – 8 – 6y + 0) w = 4x – 6y – 4 ⎛ f ⎞ gf x − fg x , gf y − fg y , gf z − fg z g fx , f y , fz – f gx , g y , gz g ∇f – f ∇g 17. ∇ ⎜ ⎟ = = = ⎝g⎠ g2 g2 g2 18. ∇( f r ) = rf r –1 f x , rf r –1 f y , rf r –1 f z = rf r –1 f x , f y , f z = rf r –1∇f 758 Section 12.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.