 Calculus Volume 3

# Chapter 4

### Checkpoint

4.1

The domain is the shaded circle defined by the inequality $9x2+9y2≤36,9x2+9y2≤36,$ which has a circle of radius $22$ as its boundary. The range is $[0,6].[0,6].$ 4.2

The equation of the level curve can be written as $(x−3)2+(y+1)2=25,(x−3)2+(y+1)2=25,$ which is a circle with radius $55$ centered at $(3,−1).(3,−1).$ 4.3

$z=3−(x−1)2.z=3−(x−1)2.$ This function describes a parabola opening downward in the plane $y=3.y=3.$

4.4

$domain ( h ) = { ( x , y , t ) ∈ ℝ 3 | y ≥ 4 x 2 − 4 } domain ( h ) = { ( x , y , t ) ∈ ℝ 3 | y ≥ 4 x 2 − 4 }$

4.5

$(x−1)2+(y+2)2+(z−3)2=16(x−1)2+(y+2)2+(z−3)2=16$ describes a sphere of radius $44$ centered at the point $(1,−2,3).(1,−2,3).$

4.6

$lim ( x , y ) → ( 5 , −2 ) x 2 − y y 2 + x − 1 3 = 3 2 lim ( x , y ) → ( 5 , −2 ) x 2 − y y 2 + x − 1 3 = 3 2$

4.7

If $y=k(x−2)+1,y=k(x−2)+1,$ then $lim(x,y)→(2,1)(x−2)(y−1)(x−2)2+(y−1)2=k1+k2.lim(x,y)→(2,1)(x−2)(y−1)(x−2)2+(y−1)2=k1+k2.$ Since the answer depends on $k,k,$ the limit fails to exist.

4.8

$lim ( x , y ) → ( 5 , −2 ) 29 − x 2 − y 2 lim ( x , y ) → ( 5 , −2 ) 29 − x 2 − y 2$

4.9
1. The domain of $ff$ contains the ordered pair $(2,−3)(2,−3)$ because $f(a,b)=f(2,−3)=16−2(2)2−(−3)2=3f(a,b)=f(2,−3)=16−2(2)2−(−3)2=3$
2. $lim(x,y)→(a,b)f(x,y)=3lim(x,y)→(a,b)f(x,y)=3$
3. $lim(x,y)→(a,b)f(x,y)=f(a,b)=3lim(x,y)→(a,b)f(x,y)=f(a,b)=3$
4.10

The polynomials $g(x)=2x2g(x)=2x2$ and $h(y)=y3h(y)=y3$ are continuous at every real number; therefore, by the product of continuous functions theorem, $f(x,y)=2x2y3f(x,y)=2x2y3$ is continuous at every point $(x,y)(x,y)$ in the $xy-plane.xy-plane.$ Furthermore, any constant function is continuous everywhere, so $g(x,y)=3g(x,y)=3$ is continuous at every point $(x,y)(x,y)$ in the $xy-plane.xy-plane.$ Therefore, $f(x,y)=2x2y3+3f(x,y)=2x2y3+3$ is continuous at every point $(x,y)(x,y)$ in the $xy-plane.xy-plane.$ Last, $h(x)=x4h(x)=x4$ is continuous at every real number $x,x,$ so by the continuity of composite functions theorem $g(x,y)=(2x2y3+3)4g(x,y)=(2x2y3+3)4$ is continuous at every point $(x,y)(x,y)$ in the $xy-plane.xy-plane.$

4.11

$lim ( x , y , z ) → ( 4 , −1 , 3 ) 13 − x 2 − 2 y 2 + z 2 = 2 lim ( x , y , z ) → ( 4 , −1 , 3 ) 13 − x 2 − 2 y 2 + z 2 = 2$

4.12

$∂ f ∂ x = 8 x + 2 y + 3 , ∂ f ∂ y = 2 x − 2 y − 2 ∂ f ∂ x = 8 x + 2 y + 3 , ∂ f ∂ y = 2 x − 2 y − 2$

4.13

$∂ f ∂ x = ( 3 x 2 − 6 x y 2 ) sec 2 ( x 3 − 3 x 2 y 2 + 2 y 4 ) ∂ f ∂ y = ( −6 x 2 y + 8 y 3 ) sec 2 ( x 3 − 3 x 2 y 2 + 2 y 4 ) ∂ f ∂ x = ( 3 x 2 − 6 x y 2 ) sec 2 ( x 3 − 3 x 2 y 2 + 2 y 4 ) ∂ f ∂ y = ( −6 x 2 y + 8 y 3 ) sec 2 ( x 3 − 3 x 2 y 2 + 2 y 4 )$

4.14

Using the curves corresponding to $c=−2andc=−3,c=−2andc=−3,$ we obtain

$∂ f ∂ y | ( x , y ) = ( 0 , 2 ) ≈ f ( 0 , 3 ) − f ( 0 , 2 ) 3 − 2 = −3 − ( −2 ) 3 − 2 · 3 + 2 3 + 2 = − 3 − 2 ≈ −3.146 . ∂ f ∂ y | ( x , y ) = ( 0 , 2 ) ≈ f ( 0 , 3 ) − f ( 0 , 2 ) 3 − 2 = −3 − ( −2 ) 3 − 2 · 3 + 2 3 + 2 = − 3 − 2 ≈ −3.146 .$

$∂ f ∂ y | ( x , y ) = ( 0 , 2 ) = ( −2 y | ( x , y ) = ( 0 , 2 ) = −2 2 ≈ −2.828 . ∂ f ∂ y | ( x , y ) = ( 0 , 2 ) = ( −2 y | ( x , y ) = ( 0 , 2 ) = −2 2 ≈ −2.828 .$

4.15

$∂ f ∂ x = 4 x − 8 x y + 5 z 2 − 6 , ∂ f ∂ y = −4 x 2 + 4 y , ∂ f ∂ z = 10 x z + 3 ∂ f ∂ x = 4 x − 8 x y + 5 z 2 − 6 , ∂ f ∂ y = −4 x 2 + 4 y , ∂ f ∂ z = 10 x z + 3$

4.16

$∂ f ∂ x = 2 x y sec ( x 2 y ) tan ( x 2 y ) − 3 x 2 y z 2 sec 2 ( x 3 y z 2 ) ∂ f ∂ y = x 2 sec ( x 2 y ) tan ( x 2 y ) − x 3 z 2 sec 2 ( x 3 y z 2 ) ∂ f ∂ z = −2 x 3 y z sec 2 ( x 3 y z 2 ) ∂ f ∂ x = 2 x y sec ( x 2 y ) tan ( x 2 y ) − 3 x 2 y z 2 sec 2 ( x 3 y z 2 ) ∂ f ∂ y = x 2 sec ( x 2 y ) tan ( x 2 y ) − x 3 z 2 sec 2 ( x 3 y z 2 ) ∂ f ∂ z = −2 x 3 y z sec 2 ( x 3 y z 2 )$

4.17

$∂ 2 f ∂ x 2 = −9 sin ( 3 x − 2 y ) − cos ( x + 4 y ) ∂ 2 f ∂ x ∂ y = 6 sin ( 3 x − 2 y ) − 4 cos ( x + 4 y ) ∂ 2 f ∂ y ∂ x = 6 sin ( 3 x − 2 y ) − 4 cos ( x + 4 y ) ∂ 2 f ∂ y 2 = −4 sin ( 3 x − 2 y ) − 16 cos ( x + 4 y ) ∂ 2 f ∂ x 2 = −9 sin ( 3 x − 2 y ) − cos ( x + 4 y ) ∂ 2 f ∂ x ∂ y = 6 sin ( 3 x − 2 y ) − 4 cos ( x + 4 y ) ∂ 2 f ∂ y ∂ x = 6 sin ( 3 x − 2 y ) − 4 cos ( x + 4 y ) ∂ 2 f ∂ y 2 = −4 sin ( 3 x − 2 y ) − 16 cos ( x + 4 y )$

4.19

$z = 7 x + 8 y − 3 z = 7 x + 8 y − 3$

4.20

$L(x,y)=6−2x+3y,L(x,y)=6−2x+3y,$ so $L(4.1,0.9)=6−2(4.1)+3(0.9)=0.5L(4.1,0.9)=6−2(4.1)+3(0.9)=0.5$ $f(4.1,0.9)=e5−2(4.1)+3(0.9)=e−0.5≈0.6065.f(4.1,0.9)=e5−2(4.1)+3(0.9)=e−0.5≈0.6065.$

4.21

$f ( −1 , 2 ) = −19 , f x ( −1 , 2 ) = 3 , f y ( −1 , 2 ) = −16 , E ( x , y ) = −4 ( y − 2 ) 2 . f ( −1 , 2 ) = −19 , f x ( −1 , 2 ) = 3 , f y ( −1 , 2 ) = −16 , E ( x , y ) = −4 ( y − 2 ) 2 .$

$lim ( x , y ) → ( x 0 , y 0 ) E ( x , y ) ( x − x 0 ) 2 + ( y − y 0 ) 2 = lim ( x , y ) → ( −1 , 2 ) −4 ( y − 2 ) 2 ( x + 1 ) 2 + ( y − 2 ) 2 ≤ lim ( x , y ) → ( −1 , 2 ) −4 ( ( x + 1 ) 2 + ( y − 2 ) 2 ) ( x + 1 ) 2 + ( y − 2 ) 2 = lim ( x , y ) → ( 2 , −3 ) − 4 ( x + 1 ) 2 + ( y − 2 ) 2 = 0. lim ( x , y ) → ( x 0 , y 0 ) E ( x , y ) ( x − x 0 ) 2 + ( y − y 0 ) 2 = lim ( x , y ) → ( −1 , 2 ) −4 ( y − 2 ) 2 ( x + 1 ) 2 + ( y − 2 ) 2 ≤ lim ( x , y ) → ( −1 , 2 ) −4 ( ( x + 1 ) 2 + ( y − 2 ) 2 ) ( x + 1 ) 2 + ( y − 2 ) 2 = lim ( x , y ) → ( 2 , −3 ) − 4 ( x + 1 ) 2 + ( y − 2 ) 2 = 0.$

4.22

$d z = 0.18 Δ z = f ( 1.03 , −1.02 ) − f ( 1 , −1 ) = 0.180682 d z = 0.18 Δ z = f ( 1.03 , −1.02 ) − f ( 1 , −1 ) = 0.180682$

4.23

$d z d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t = ( 2 x − 3 y ) ( 6 cos 2 t ) + ( −3 x + 4 y ) ( −8 sin 2 t ) = −92 sin 2 t cos 2 t − 72 ( cos 2 2 t − sin 2 2 t ) = −46 sin 4 t − 72 cos 4 t . d z d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t = ( 2 x − 3 y ) ( 6 cos 2 t ) + ( −3 x + 4 y ) ( −8 sin 2 t ) = −92 sin 2 t cos 2 t − 72 ( cos 2 2 t − sin 2 2 t ) = −46 sin 4 t − 72 cos 4 t .$

4.24

$∂ z ∂ u = 0 , ∂ z ∂ v = −21 ( 3 sin 3 v + cos 3 v ) 2 ∂ z ∂ u = 0 , ∂ z ∂ v = −21 ( 3 sin 3 v + cos 3 v ) 2$

4.25

$∂ w ∂ u = 0 ∂ w ∂ v = 15 − 33 sin 3 v + 6 cos 3 v ( 3 + 2 cos 3 v − sin 3 v ) 2 ∂ w ∂ u = 0 ∂ w ∂ v = 15 − 33 sin 3 v + 6 cos 3 v ( 3 + 2 cos 3 v − sin 3 v ) 2$

4.26

$∂w∂t=∂w∂x∂x∂t+∂w∂y∂y∂t∂w∂u=∂w∂x∂x∂u+∂w∂y∂y∂u∂w∂v=∂w∂x∂x∂v+∂w∂y∂y∂v∂w∂t=∂w∂x∂x∂t+∂w∂y∂y∂t∂w∂u=∂w∂x∂x∂u+∂w∂y∂y∂u∂w∂v=∂w∂x∂x∂v+∂w∂y∂y∂v$ 4.27

$dydx=2x+y+72y−x+3|(3,−2)=2(3)+(−2)+72(−2)−(3)+3=−114dydx=2x+y+72y−x+3|(3,−2)=2(3)+(−2)+72(−2)−(3)+3=−114$
Equation of the tangent line: $y=−114x+254y=−114x+254$

4.28

$D u f ( x , y ) = ( 6 x y − 4 y 3 − 4 ) ( 1 ) 2 + ( 3 x 2 − 12 x y 2 + 6 y ) 3 2 D u f ( 3 , 4 ) = 72 − 256 − 4 2 + ( 27 − 576 + 24 ) 3 2 = −94 − 525 3 2 D u f ( x , y ) = ( 6 x y − 4 y 3 − 4 ) ( 1 ) 2 + ( 3 x 2 − 12 x y 2 + 6 y ) 3 2 D u f ( 3 , 4 ) = 72 − 256 − 4 2 + ( 27 − 576 + 24 ) 3 2 = −94 − 525 3 2$

4.29

$∇ f ( x , y ) = 2 x 2 + 2 x y + 6 y 2 ( 2 x + y ) 2 i − x 2 + 12 x y + 3 y 2 ( 2 x + y ) 2 j ∇ f ( x , y ) = 2 x 2 + 2 x y + 6 y 2 ( 2 x + y ) 2 i − x 2 + 12 x y + 3 y 2 ( 2 x + y ) 2 j$

4.30

The gradient of $gg$ at $(−2,3)(−2,3)$ is $∇g(−2,3)=i+14j.∇g(−2,3)=i+14j.$ The unit vector that points in the same direction as $∇g(−2,3)∇g(−2,3)$ is $∇g(−2,3)‖∇g(−2,3)‖=1197i+14197j=197197i+14197197j,∇g(−2,3)‖∇g(−2,3)‖=1197i+14197j=197197i+14197197j,$ which gives an angle of $θ=arcsin((14197)/197)≈1.499rad.θ=arcsin((14197)/197)≈1.499rad.$ The maximum value of the directional derivative is $‖∇g(−2,3)‖=197.‖∇g(−2,3)‖=197.$

4.31

$∇f(x,y)=(2x−2y+3)i+(−2x+10y−2)j∇f(x,y)=(2x−2y+3)i+(−2x+10y−2)j$
$∇f(1,1)=3i+6j∇f(1,1)=3i+6j$
Tangent vector: $6i−3j6i−3j$ or $−6i+3j−6i+3j$ 4.32

$∇ f ( x , y , z ) = 2 x 2 + 2 x y + 6 y 2 − 8 x z − 2 z 2 ( 2 x + y − 4 z ) 2 i − x 2 + 12 x y + 3 y 2 − 24 y z + z 2 ( 2 x + y − 4 z ) 2 j + 4 x 2 − 12 y 2 − 4 z 2 + 4 x z + 2 y z ( 2 x + y − 4 z ) 2 k . ∇ f ( x , y , z ) = 2 x 2 + 2 x y + 6 y 2 − 8 x z − 2 z 2 ( 2 x + y − 4 z ) 2 i − x 2 + 12 x y + 3 y 2 − 24 y z + z 2 ( 2 x + y − 4 z ) 2 j + 4 x 2 − 12 y 2 − 4 z 2 + 4 x z + 2 y z ( 2 x + y − 4 z ) 2 k .$

4.33

$D u f ( x , y , z ) = − 3 13 ( 6 x + y + 2 z ) + 12 13 ( x − 4 y + 4 z ) − 4 13 ( 2 x + 4 y − 2 z ) D u f ( 0 , −2 , 5 ) = 384 13 D u f ( x , y , z ) = − 3 13 ( 6 x + y + 2 z ) + 12 13 ( x − 4 y + 4 z ) − 4 13 ( 2 x + 4 y − 2 z ) D u f ( 0 , −2 , 5 ) = 384 13$

4.34

$( 2 , −5 ) ( 2 , −5 )$

4.35

$(43,13)(43,13)$ is a saddle point, $(−32,−38)(−32,−38)$ is a local maximum.

4.36

The absolute minimum occurs at $(1,0):(1,0):$ $f(1,0)=−1.f(1,0)=−1.$

The absolute maximum occurs at $(0,3):(0,3):$ $f(0,3)=63.f(0,3)=63.$

4.37

$ff$ has a maximum value of $976976$ at the point $(8,2).(8,2).$

4.38

A maximum production level of $1389013890$ occurs with $56255625$ labor hours and $55005500$ of total capital input.

4.39

$f ( 3 3 , 3 3 , 3 3 ) = 3 3 + 3 3 + 3 3 = 3 f ( − 3 3 , − 3 3 , − 3 3 ) = − 3 3 − 3 3 − 3 3 = − 3 . f ( 3 3 , 3 3 , 3 3 ) = 3 3 + 3 3 + 3 3 = 3 f ( − 3 3 , − 3 3 , − 3 3 ) = − 3 3 − 3 3 − 3 3 = − 3 .$

4.40

$f(2,1,2)=9f(2,1,2)=9$ is a minimum.

### Section 4.1 Exercises

1 .

$17 , 72 17 , 72$

3 .

$20π.20π.$ This is the volume when the radius is $22$ and the height is $5.5.$

5 .

All points in the $xy-planexy-plane$

7 .

$x < y 2 x < y 2$

9 .

All real ordered pairs in the $xy-planexy-plane$ of the form $(a,b)(a,b)$

11 .

${ z | 0 ≤ z ≤ 4 } { z | 0 ≤ z ≤ 4 }$

13 .

The set $ℝℝ$

15 .

$y2−x2=4,y2−x2=4,$ a hyperbola

17 .

$4=x+y,4=x+y,$ a line; $x+y=0,x+y=0,$ line through the origin

19 .

$2x−y=0,2x−y=−2,2x−y=2;2x−y=0,2x−y=−2,2x−y=2;$ three lines

21 .

$x x + y = −1 , x x + y = 0 , x x + y = 2 x x + y = −1 , x x + y = 0 , x x + y = 2$

23 .

$e x y = 1 2 , e x y = 3 e x y = 1 2 , e x y = 3$

25 .

$x y − x = −2 , x y − x = 0 , x y − x = 2 x y − x = −2 , x y − x = 0 , x y − x = 2$

27 .

$e −2 x 2 = y , y = x 2 , y = e 2 x 2 e −2 x 2 = y , y = x 2 , y = e 2 x 2$

29 .

The level curves are parabolas of the form $y=cx2−2.y=cx2−2.$

31 .

$z=3+y3,z=3+y3,$ a curve in the $zy-planezy-plane$ with rulings parallel to the $x-axisx-axis$ 33 .

$x 2 25 + y 2 4 ≤ 1 x 2 25 + y 2 4 ≤ 1$

35 .

$x 2 9 + y 2 4 + z 2 36 < 1 x 2 9 + y 2 4 + z 2 36 < 1$

37 .

All points in $xyz-spacexyz-space$

39 . 41 . 43 . 45 . 47 .

The contour lines are circles.

49 .

$x2+y2+z2=9,x2+y2+z2=9,$ a sphere of radius $33$

51 .

$x2+y2−z2=4,x2+y2−z2=4,$ a hyperboloid of one sheet

53 .

$4 x 2 + y 2 = 1 , 4 x 2 + y 2 = 1 ,$

55 .

$1 = e x y ( x 2 + y 2 ) 1 = e x y ( x 2 + y 2 )$

57 .

$T ( x , y ) = k x 2 + y 2 T ( x , y ) = k x 2 + y 2$

59 .

$x2+y2=k40,x2+y2=k40,$ $x2+y2=k100.x2+y2=k100.$ The level curves represent circles of radii $10k/2010k/20$ and $k/10k/10$

### Section 4.2 Exercises

61 .

2.0

63 .

$2 3 2 3$

65 .

$1 1$

67 .

$1 2 1 2$

69 .

$− 1 2 − 1 2$

71 .

$e −32 e −32$

73 .

$11.0 11.0$

75 .

$1.0 1.0$

77 .

The limit does not exist because when $xx$ and $yy$ both approach zero, the function approaches $ln0,ln0,$ which is undefined (approaches negative infinity).

79 .

every open disk centered at $(x0,y0)(x0,y0)$ contains points inside $RR$ and outside $RR$

81 .

$0.0 0.0$

83 .

$0.00 0.00$

85 .

The limit does not exist.

87 .

The limit does not exist. The function approaches two different values along different paths.

89 .

The limit does not exist because the function approaches two different values along the paths.

91 .

The function $ff$ is continuous in the region $y>−x.y>−x.$

93 .

The function $ff$ is continuous at all points in the $xy-planexy-plane$ except at $(0,0).(0,0).$

95 .

The function is continuous at $(0,0)(0,0)$ since the limit of the function at $(0,0)(0,0)$ is $0,0,$ the same value of $f(0,0).f(0,0).$

97 .

The function is discontinuous at $(0,0).(0,0).$ The limit at $(0,0)(0,0)$ fails to exist and $g(0,0)g(0,0)$ does not exist.

99 .

Since the function $arctanxarctanx$ is continuous over $(−∞,∞),(−∞,∞),$ $g(x,y)=arctan(xy2x+y)g(x,y)=arctan(xy2x+y)$ is continuous where $z=xy2x+yz=xy2x+y$ is continuous. The inner function $zz$ is continuous on all points of the $xy-planexy-plane$ except where $y=−x.y=−x.$ Thus, $g(x,y)=arctan(xy2x+y)g(x,y)=arctan(xy2x+y)$ is continuous on all points of the coordinate plane except at points at which $y=−x.y=−x.$

101 .

All points $P(x,y,z)P(x,y,z)$ in space

103 .

The graph increases without bound as $xandyxandy$ both approach zero. 105 .

a. b. The level curves are circles centered at $(0,0)(0,0)$ with radius $9−c.9−c.$ c. $x2+y2=9−cx2+y2=9−c$ d. $z=3z=3$ e. ${(x,y)∈ℝ2|x2+y2≤9}{(x,y)∈ℝ2|x2+y2≤9}$ f. ${z|0≤z≤3}{z|0≤z≤3}$

107 .

$1.0 1.0$

109 .

$f(g(x,y))f(g(x,y))$ is continuous at all points $(x,y)(x,y)$ that are not on the line $2x−5y=0.2x−5y=0.$

111 .

$2.0 2.0$

### Section 4.3 Exercises

113 .

$∂ z ∂ y = −3 x + 2 y ∂ z ∂ y = −3 x + 2 y$

115 .

The sign is negative.

117 .

The partial derivative is zero at the origin.

119 .

$∂ z ∂ y = −3 sin ( 3 x ) sin ( 3 y ) ∂ z ∂ y = −3 sin ( 3 x ) sin ( 3 y )$

121 .

$∂ z ∂ x = 6 x 5 x 6 + y 4 ; ∂ z ∂ y = 4 y 3 x 6 + y 4 ∂ z ∂ x = 6 x 5 x 6 + y 4 ; ∂ z ∂ y = 4 y 3 x 6 + y 4$

123 .

$∂ z ∂ x = y e x y ; ∂ z ∂ y = x e x y ∂ z ∂ x = y e x y ; ∂ z ∂ y = x e x y$

125 .

$∂ z ∂ x = 2 sec 2 ( 2 x − y ) , ∂ z ∂ y = − sec 2 ( 2 x − y ) ∂ z ∂ x = 2 sec 2 ( 2 x − y ) , ∂ z ∂ y = − sec 2 ( 2 x − y )$

127 .

$f x ( 2 , −2 ) = 1 4 = f y ( 2 , −2 ) f x ( 2 , −2 ) = 1 4 = f y ( 2 , −2 )$

129 .

$∂ z ∂ x = − cos ( 1 ) ∂ z ∂ x = − cos ( 1 )$

131 .

$f x = 0 , f y = 0 , f z = 0 f x = 0 , f y = 0 , f z = 0$

133 .

a. $V(r,h)=πr2hV(r,h)=πr2h$ b. $∂V∂r=2πrh∂V∂r=2πrh$ c. $∂V∂h=πr2∂V∂h=πr2$

135 .

$f x y = 1 ( x − y ) 2 f x y = 1 ( x − y ) 2$

137 .

$∂ 2 z ∂ x 2 = 2 , ∂ 2 z ∂ y 2 = 4 ∂ 2 z ∂ x 2 = 2 , ∂ 2 z ∂ y 2 = 4$

139 .

$f x y y = f y x y = f y y x = 0 f x y y = f y x y = f y y x = 0$

141 .

$d 2 z d x 2 = − 1 2 ( e y − e − y ) sin x d 2 z d y 2 = 1 2 ( e y − e − y ) sin x d 2 z d x 2 + d 2 z d y 2 = 0 d 2 z d x 2 = − 1 2 ( e y − e − y ) sin x d 2 z d y 2 = 1 2 ( e y − e − y ) sin x d 2 z d x 2 + d 2 z d y 2 = 0$

143 .

$f x y z = 6 y 2 x − 18 y z 2 f x y z = 6 y 2 x − 18 y z 2$

145 .

$( 1 4 , 1 2 ) , ( 1 , 1 ) ( 1 4 , 1 2 ) , ( 1 , 1 )$

147 .

$( 0 , 0 ) , ( 0 , 2 ) , ( 3 , −1 ) , ( − 3 , −1 ) ( 0 , 0 ) , ( 0 , 2 ) , ( 3 , −1 ) , ( − 3 , −1 )$

149 .

$∂ 2 z ∂ x 2 + ∂ 2 z ∂ y 2 = e x sin ( y ) − e x sin y = 0 ∂ 2 z ∂ x 2 + ∂ 2 z ∂ y 2 = e x sin ( y ) − e x sin y = 0$

151 .

$c 2 ∂ 2 z ∂ x 2 = e − t cos ( x c ) c 2 ∂ 2 z ∂ x 2 = e − t cos ( x c )$

153 .

$∂ f ∂ y = −2 x + 7 ∂ f ∂ y = −2 x + 7$

155 .

$∂ f ∂ x = y cos x y ∂ f ∂ x = y cos x y$

159 .

$∂ F ∂ θ = 6 , ∂ F ∂ x = 4 − 3 3 ∂ F ∂ θ = 6 , ∂ F ∂ x = 4 − 3 3$

161 .

$δfδxδfδx$ at $(500,1000)=172.36,(500,1000)=172.36,$ $δfδyδfδy$ at $(500,1000)=36.93(500,1000)=36.93$

### Section 4.4 Exercises

163 .

$( 145 145 ) ( 12 i − k ) ( 145 145 ) ( 12 i − k )$

165 .

Normal vector: $i+j,i+j,$ tangent vector: $i−ji−j$

167 .

Normal vector: $7i−17j,7i−17j,$ tangent vector: $17i+7j17i+7j$

169 .

171 .

$−36 x − 6 y − z = −39 −36 x − 6 y − z = −39$

173 .

$z = 0 z = 0$

175 .

$5 x + 4 y + 3 z − 22 = 0 5 x + 4 y + 3 z − 22 = 0$

177 .

$4 x − 5 y + 4 z = 0 4 x − 5 y + 4 z = 0$

179 .

$2 x + 2 y − z = 0 2 x + 2 y − z = 0$

181 .

$−2 ( x − 1 ) + 2 ( y − 2 ) − ( z − 1 ) = 0 −2 ( x − 1 ) + 2 ( y − 2 ) − ( z − 1 ) = 0$

183 .

$x = 20 t + 2 , y = −4 t + 1 , z = − t + 18 x = 20 t + 2 , y = −4 t + 1 , z = − t + 18$

185 .

$x = 0 , y = 0 , z = t x = 0 , y = 0 , z = t$

187 .

$x − 1 = 2 t ; y − 2 = −2 t ; z − 1 = t x − 1 = 2 t ; y − 2 = −2 t ; z − 1 = t$

189 .

The differential of the function $z(x,y)=dz=fxdx+fydyz(x,y)=dz=fxdx+fydy$

191 .

Using the definition of differentiability, we have $exyx≈x+y.exyx≈x+y.$

193 .

$Δz=2xΔx+3Δy+(Δx)2.Δz=2xΔx+3Δy+(Δx)2.$ $(Δx)2→0(Δx)2→0$ for small $ΔxΔx$ and $zz$ satisfies the definition of differentiability.

195 .

$Δz≈1.185422Δz≈1.185422$ and $dz≈1.108.dz≈1.108.$ They are relatively close.

197 .

$1616$ cm3

199 .

$Δz=Δz=$ exact change $=0.6449,=0.6449,$ approximate change is $dz=0.65.dz=0.65.$ The two values are close.

201 .

$13 % or 0.13 13 % or 0.13$

203 .

$0.025 0.025$

205 .

$0.3 % 0.3 %$

207 .

$2 x + 1 4 y − 1 2 x + 1 4 y − 1$

209 .

$1 2 x + y + 1 4 π − 1 2 1 2 x + y + 1 4 π − 1 2$

211 .

$3 7 x + 2 7 y + 6 7 z 3 7 x + 2 7 y + 6 7 z$

213 .

$z=0z=0$ ### Section 4.5 Exercises

215 .

$d w d t = y cos z + x cos z ( 2 t ) − x y sin z 1 − t 2 d w d t = y cos z + x cos z ( 2 t ) − x y sin z 1 − t 2$

217 .

$∂w∂s=−30x+4y,∂w∂s=−30x+4y,$ $∂w∂t=10x−16y∂w∂t=10x−16y$

219 .

$∂ f ∂ r = r sin ( 2 θ ) ∂ f ∂ r = r sin ( 2 θ )$

221 .

$d f d t = 2 t + 4 t 3 d f d t = 2 t + 4 t 3$

223 .

$d f d t = −1 d f d t = −1$

225 .

$d f d t = 1 d f d t = 1$

227 .

$dwdt=2e2tdwdt=2e2t$ in both cases

229 .

$d u d t = 2 ( π -4 d u d t = 2 ( π -4$

231 .

$d y d x = − 3 x 2 + y 2 2 x y d y d x = − 3 x 2 + y 2 2 x y$

233 .

$d y d x = y − x − x + 2 y 3 d y d x = y − x − x + 2 y 3$

235 .

$d y d x = − y x 3 d y d x = − y x 3$

237 .

$d y d x = − y e x y x e x y + e y ( 1 + y ) d y d x = − y e x y x e x y + e y ( 1 + y )$

239 .

$d z d t = 42 t 13 d z d t = 42 t 13$

241 .

$d z d t = − 10 3 t 7 / 3 × e 1 − t 10 / 3 d z d t = − 10 3 t 7 / 3 × e 1 − t 10 / 3$

243 .

$∂z∂u=−2sinu3sinv∂z∂u=−2sinu3sinv$ and $∂z∂v=−2cosucosv3sin2v∂z∂v=−2cosucosv3sin2v$

245 .

$∂z∂r=3e3,∂z∂r=3e3,$ $∂z∂θ=(2−43)e3∂z∂θ=(2−43)e3$

247 .

$∂ w ∂ t = cos ( x y z ) × y z × ( −3 ) − cos ( x y z ) x z e 1 − t + cos ( x y z ) x y × 4 ∂ w ∂ t = cos ( x y z ) × y z × ( −3 ) − cos ( x y z ) x z e 1 − t + cos ( x y z ) x y × 4$

249 .

$f(tx,ty)=t2x2+t2y2=t1f(x,y),f(tx,ty)=t2x2+t2y2=t1f(x,y),$ $∂f∂y=x12(x2+y2)−1/2×2x+y12(x2+y2)−1/2×2y=1f(x,y)∂f∂y=x12(x2+y2)−1/2×2x+y12(x2+y2)−1/2×2y=1f(x,y)$

251 .

$34 π 3 34 π 3$

253 .

$d V d t = 1066 π 3 cm 3 / min d V d t = 1066 π 3 cm 3 / min$

255 .

$d A d t = 12 in . 2 / min d A d t = 12 in . 2 / min$

257 .

$2 ° C/sec 2 ° C/sec$

259 .

$∂ u ∂ r = ∂ u ∂ x ( ∂ x ∂ w ∂ w ∂ r + ∂ x ∂ t ∂ t ∂ r ) + ∂ u ∂ y ( ∂ y ∂ w ∂ w ∂ r + ∂ y ∂ t ∂ t ∂ r ) + ∂ u ∂ z ( ∂ z ∂ w ∂ w ∂ r + ∂ z ∂ t ∂ t ∂ r ) ∂ u ∂ r = ∂ u ∂ x ( ∂ x ∂ w ∂ w ∂ r + ∂ x ∂ t ∂ t ∂ r ) + ∂ u ∂ y ( ∂ y ∂ w ∂ w ∂ r + ∂ y ∂ t ∂ t ∂ r ) + ∂ u ∂ z ( ∂ z ∂ w ∂ w ∂ r + ∂ z ∂ t ∂ t ∂ r )$

### Section 4.6 Exercises

261 .

$- 4 3 + 3 2 - 4 3 + 3 2$

263 .

$−1 −1$

265 .

$2 6 2 6$

267 .

$3 3$

269 .

$−1.0 −1.0$

271 .

$22 25 22 25$

273 .

$2 3 2 3$

275 .

$− 2 ( x + y ) 2 ( x + 2 y ) 2 − 2 ( x + y ) 2 ( x + 2 y ) 2$

277 .

$e x ( y + 3 ) 2 e x ( y + 3 ) 2$

279 .

$1 + 2 3 2 ( x + 2 y ) 1 + 2 3 2 ( x + 2 y )$

281 .

$〈 5 , 4 , 3 〉 〈 5 , 4 , 3 〉$

283 .

$−320 −320$

285 .

$3 11 3 11$

287 .

$31 255 31 255$

289 . 291 .

$4 3 i − 3 j 4 3 i − 3 j$

293 .

$2 i + 2 j + 2 k 2 i + 2 j + 2 k$

295 .

$1.6 ( 10 19 ) 1.6 ( 10 19 )$

297 .

$5 2 99 5 2 99$

299 .

$2 , 〈 1 , -1 〉 2 , 〈 1 , -1 〉$

301 .

$13 2 , 〈 −3 , −2 〉 13 2 , 〈 −3 , −2 〉$

303 .

a. $x+y+z=3,x+y+z=3,$ b. $x−1=y−1=z−1x−1=y−1=z−1$

305 .

a. $x+y−z=1,x+y−z=1,$ b. $x−1=y=−zx−1=y=−z$

307 .

a. $323,323,$ b. $〈38,6,12〉,〈38,6,12〉,$ c. $24062406$

309 .

$〈 u , v 〉 = 〈 π cos ( π x ) sin ( 2 π y ) , 2 π sin ( π x ) cos ( 2 π y ) 〉 〈 u , v 〉 = 〈 π cos ( π x ) sin ( 2 π y ) , 2 π sin ( π x ) cos ( 2 π y ) 〉$

### Section 4.7 Exercises

311 .

$( 2 3 , 4 ) ( 2 3 , 4 )$

313 .

$(0,0)(0,0)$ $(115,115)(115,115)$

315 .

Maximum at $(4,−1,8)(4,−1,8)$

317 .

Relative minimum at $(0,0,1)(0,0,1)$

319 .

The second derivative test fails. Since $x2y2>0x2y2>0$ for all x and y different from zero, and $x2y2=0x2y2=0$ when either x or y equals zero (or both), then the absolute minimum occurs at $(0,0).(0,0).$

321 .

$f(−2,−32)=−6f(−2,−32)=−6$ is a saddle point.

323 .

$f(0,0)=0;f(0,0)=0;$ $(0,0,0)(0,0,0)$ is a saddle point.

325 .

$f(0,0)=9f(0,0)=9$ is a local maximum.

327 .

Relative minimum located at $(2,6).(2,6).$

329 .

$(1,−2)(1,−2)$ is a saddle point.

331 .

$(2,1)(2,1)$ and $(−2,1)(−2,1)$ are saddle points; $(0,0)(0,0)$ is a relative minimum.

333 .

$(−1,0)(−1,0)$ is a relative maximum.

335 .

$(0,0)(0,0)$ is a saddle point.

337 .

The relative maximum is at $(40,40).(40,40).$

339 .

$(14,12)(14,12)$ is a saddle point and $(1,1)(1,1)$ is the relative minimum.

341 .

A saddle point is located at $(0,0).(0,0).$ 343 .

There is a saddle point at $(π,π),(π,π),$ local maxima at $(π2,π2)and(3π2,3π2),(π2,π2)and(3π2,3π2),$ and local minima at $(π2,3π2)and(3π2,π2).(π2,3π2)and(3π2,π2).$ 345 .

$(0,1,0)(0,1,0)$ is the absolute minimum and $(0,−2,9)(0,−2,9)$ is the absolute maximum.

347 .

There is an absolute minimum at $(0,1,−1)(0,1,−1)$ and an absolute maximum at $(0,−1,1).(0,−1,1).$

349 .

$( 5 , 0 , 0 ) , ( − 5 , 0 , 0 ) ( 5 , 0 , 0 ) , ( − 5 , 0 , 0 )$

351 .

$18 by 36 by 18 in . 18 by 36 by 18 in .$

353 .

$( 47 24 , 47 12 , 235 24 ) ( 47 24 , 47 12 , 235 24 )$

355 .

$x=3x=3$ and $y=6y=6$

357 .

$V=64,000π≈20,372V=64,000π≈20,372$ cm3

### Section 4.8 Exercises

359 .

maximum: $233,233,$ minimum: $−233−233$

361 .

maximum: $(22,0,2),(22,0,2),$ minimum: $(−22,0,−2)(−22,0,−2)$

363 .

maximum: $32,32,$ minimum = $1212$

365 .

maxima: $f(322,22)=24,f(322,22)=24,$ $f(−322,−22)=24;f(−322,−22)=24;$ minima: $f(−322,22)=−24,f(−322,22)=−24,$ $f(322,−22)=−24f(322,−22)=−24$

367 .

maximum: $211211$ at $f(211,611,−211);f(211,611,−211);$ minimum: $−211−211$ at $f(−211,−611,211)f(−211,−611,211)$

369 .

$2.0 2.0$

371 .

$19 2 19 2$

373 .

$( 1 2 3 , −1 2 3 ) ( 1 2 3 , −1 2 3 )$

375 .

$f ( 1 , 2 ) = 5 f ( 1 , 2 ) = 5$

377 .

$f ( 1 3 , 1 3 , 1 3 ) = 1 3 f ( 1 3 , 1 3 , 1 3 ) = 1 3$

379 .

minimum: $f(2,3,4)=29f(2,3,4)=29$

381 .

The maximum volume is $44$ ft3. The dimensions are $1×2×21×2×2$ ft.

383 .

$( 1 , 1 2 , −3 ) ( 1 , 1 2 , −3 )$

385 .

$1.0 1.0$

387 .

$3 3$

389 .

$( 2 5 , 19 5 ) ( 2 5 , 19 5 )$

391 .

$1212$ 393 .

Roughly 3365 watches at the critical point $(80,60)(80,60)$ ### Review Exercises

395 .

True, by Clairaut’s theorem

397 .

False

399 .

401 .

Does not exist

403 .

Continuous at all points on the $x,y-plane,x,y-plane,$ except where $x2+y2>4.x2+y2>4.$

405 .

$∂u∂x=4x3−3y,∂u∂x=4x3−3y,$ $∂u∂y=−3x,∂u∂y=−3x,$ $∂u∂t=2,∂u∂t=2,$ $∂u∂t=3t2,∂u∂t=3t2,$ $∂u∂t=8x3−6y−9xt2∂u∂t=8x3−6y−9xt2$

407 .

$hxx(x,y,z)=6xe2yz,hxx(x,y,z)=6xe2yz,$ $hxy(x,y,z)=6x2e2yz,hxy(x,y,z)=6x2e2yz,$ $hxz(x,y,z)=−3x2e2yz2,hxz(x,y,z)=−3x2e2yz2,$ $hyx(x,y,z)=6x2e2yz,hyx(x,y,z)=6x2e2yz,$ $hyy(x,y,z)=4x3e2yz,hyy(x,y,z)=4x3e2yz,$ $hyz(x,y,z)=−2x3e2yz2,hyz(x,y,z)=−2x3e2yz2,$ $hzx(x,y,z)=−3x2e2yz2,hzx(x,y,z)=−3x2e2yz2,$ $hzy(x,y,z)=−2x3e2yz2,hzy(x,y,z)=−2x3e2yz2,$ $hzz(x,y,z)=2x3e2yz3hzz(x,y,z)=2x3e2yz3$

409 .

$z = x − 2 y + 5 z = x − 2 y + 5$

411 .

$dz=4dx−dy,dz=4dx−dy,$ $dz(0.1,0.01)=0.39,dz(0.1,0.01)=0.39,$ $Δz=0.432Δz=0.432$

413 .

$3 85 , 〈 27 , 6 〉 3 85 , 〈 27 , 6 〉$

415 .

$∇ f ( x , y ) = − x + 2 y 2 2 x 2 y i + ( 1 x − 1 x y 2 ) j ∇ f ( x , y ) = − x + 2 y 2 2 x 2 y i + ( 1 x − 1 x y 2 ) j$

417 .

maximum: $1633,1633,$ minimum: $−1633−1633$

419 .

$2.32282.3228$ cm3

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