HOMEWORK #3 – SOLUTIONS Page 888 10) 46) 14 − 8
=3
3−1
12 − 8
fy ( 2,1) ≈
= −2
0−2
fx ( 2,1) ≈
yz = ln ( x + z )
y
∂z
1 ⎛
∂z ⎞
∂z
1 ∂z
1
=
⋅ ⎜1 + ⎟ ⇒ y −
=
⇒ ∂x x + z ⎝
∂x ⎠
∂x x + z ∂x x + z
1 ⎞ ∂z
1
∂z
1
⎛
=
⇒
=
⎜⎝ y −
⎟⎠
x + z ∂x x + z
∂x y ( x + z ) − 1
y
∂z
1 ∂z
1 ⎞ ∂z
⎛
+z=
⋅ ⇒⎜y−
= −z ⇒
⎟
∂y
x + z ∂y ⎝
x + z ⎠ ∂y
−z ( x + z )
∂z
=
∂y y ( x + z ) − 1
72) a) u = x 2 + y 2 ⇒ u xx = 2, u yy = 2, 2 + 2 ≠ 0 ⇒ No. b) u = x 2 − y 2 ⇒ u xx = 2, u yy = −2, 2 − 2 = 0 ⇒ Yes. c) u = x 3 + 3xy 2 ⇒ u xx = 6x, u yy = 6x, 6x + 6x ≠ 0 ⇒ No. d) u = ln x 2 + y 2
ux =
u xx
1
1
⋅
x 2 + y2 2 x 2 + y2
(x
=
2
)
+ y 2 ⋅1 − x ⋅ 2x
(x
2
+y
)
2 2
By symmetry, u yy =
=
⋅ 2x =
x
x + y2
2
−x 2 + y 2
(x
+ y2
2
)
2
x 2 − y2
(x
2
+ y2
)
2
u xx + u yy = 0 ⇒ u is a solution.
e) u = sin x cosh y + cos x sinh y
u x = cos x cosh y − sin x sinh y
u xx = − sin x cosh y − cos x sinh y
u y = sin x sinh y + cos x cosh y
u yy = sin x cosh y + cos x sinh y
u xx + u yy = 0 ⇒ u is a solution.
f) u = e− x cos y − e− y cos x
u x = −e− x cos y + e− y sin x
u xx = e− x cos y + e− y cos x
u y = −e− x sin y + e− y cos x
u yy = −e− x cos y − e− y cos x
u xx + u yy = 0 ⇒ u is a solution.
80) T ( x, y ) =
60
1 + x 2 + y2
a) Tx =
−60 ( 2x )
(1 + x
2
+ y2
)
2
−240
20  C
⇒ Tx ( 2,1) =
=−
36
3 m
b) Ty =
−60 ( 2y )
(1 + x
2
+ y2
Page 899 4) z = y ln x, (1, 4, 0 )
)
2
−120
10 C
⇒ Ty ( 2,1) =
=−
36
3 m
z = f ( a,b ) + fx ( a,b ) ( x − a ) + fy ( a,b ) ( y − b )
y
⇒ fx (1, 4 ) = 4
x
fy = ln x ⇒ fy (1, 4 ) = 0
f (1, 4 ) = 0; fx =
z = 0 + 4 ( x − 1) + 0 ( y − 4 ) ⇒ z = 4x − 4
18) f ( 0, 0 ) = 1
y + cos 2 x = f ; (0, 0)
fx =
1
2 y + cos 2 x
⋅ 2 cos x ( − sin x ) ⇒ fx ( 0, 0 ) = 0
1
1
fy =
⋅1 ⇒ fy ( 0, 0 ) =
2
2 y + cos 2 x
y + cos 2 x ≈ 1 + 0 ⋅ x +
20) f ( x, y ) = ln ( x − 3y )
f ( 7, 2 ) = ln1 = 0
1
1
y = 1+ y
2
2
( 7, 2 )
1
⋅1 ⇒ fx ( 7, 2 ) = 1
x − 3y
1
fy =
⋅ ( −3) ⇒ fy ( 7, 2 ) = −3
x − 3y
fx =
ln ( x − 3y ) ≈ 0 + 1( x − 7 ) − 3 ( y − 2 ) = x − 3y − 1
f ( 6.9, 2.06 ) ≈ −.28
Actual value = -.328