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.

$V ' = 4 π V ' = 4 π$

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.

The only critical point(s) are where both $fx=0fx=0$ and $fy=0fy=0$ and that is . However, since $fxxfxy-fxy2=0fxxfxy-fxy2=0$ at the critical point, the Second Derivative Test fails. By graphing, we can see that this point 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.

The point on the surface $x2-2xy+y2-x+y=zx2-2xy+y2-x+y=z$ closest to the point $(1,2,-3)(1,2,-3)$ is $32,32,032,32,0$.

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.

$∂ u ∂ x = 4 x 3 − 3 y , ∂ u ∂ y = −3 x , d x d t = 2 , d y d t = 3 t 2 , d u d t = 40 t 3 ∂ u ∂ x = 4 x 3 − 3 y , ∂ u ∂ y = −3 x , d x d t = 2 , d y d t = 3 t 2 , d u d t = 40 t 3$

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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