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
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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
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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
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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.
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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
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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 )
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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
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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
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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
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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
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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
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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
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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
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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
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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
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