Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 15.3 PARTIAL DERIVATIVES ET SECTION 14.3 ¤ 181
59. u =ln x 2 + y 2 =ln(x 2 + y 2 ) 1/2 = 1 2 ln(x2 + y 2 ) ⇒ u x = 1 1
2 x 2 + y · 2x = x
2 x 2 + y , 2
u xy = x(−1)(x 2 + y 2 ) −2 (2y) =−
2xy
(x 2 + y 2 ) and u 2 y = 1 1
2 x 2 + y · 2y =
2
u yx = y(−1)(x 2 + y 2 ) −2 (2x) =−
2xy
(x 2 + y 2 ) 2 . Thus u xy = u yx .
y
x 2 + y 2 ,
61. f(x, y) =3xy 4 + x 3 y 2 ⇒ f x =3y 4 +3x 2 y 2 , f xx =6xy 2 , f xxy =12xy and
f y =12xy 3 +2x 3 y, f yy =36xy 2 +2x 3 , f yyy =72xy.
63. f(x, y, z) =cos(4x +3y +2z) ⇒
f x = − sin(4x +3y +2z)(4) = −4sin(4x +3y +2z), f xy = −4cos(4x +3y +2z)(3) = −12 cos(4x +3y +2z),
f xyz = −12(− sin(4x +3y +2z))(2) = 24 sin(4x +3y +2z) and
f y = − sin(4x +3y +2z)(3) = −3sin(4x +3y +2z),
f yz = −3cos(4x +3y +2z)(2) = −6cos(4x +3y +2z), f yzz = −6(− sin(4x +3y +2z))(2) = 12 sin(4x +3y +2z).
65. u = e rθ sin θ ⇒ ∂u
∂θ = erθ cos θ +sinθ · e rθ (r) =e rθ (cos θ + r sin θ),
∂ 2 u
∂r ∂θ = erθ (sin θ)+(cosθ + r sin θ) e rθ (θ) =e rθ (sin θ + θ cos θ + rθ sin θ),
∂ 3 u
∂r 2 ∂θ = erθ (θ sin θ)+(sinθ + θ cos θ + rθ sin θ) · e rθ (θ) =θe rθ (2 sin θ + θ cos θ + rθ sin θ).
67. w = x
y +2z = x(y +2z)−1 ⇒ ∂w
∂x =(y +2z)−1 ,
∂ 2 w
∂y ∂x = −(y +2z)−2 (1) = −(y +2z) −2 ,
∂ 3 w
∂z ∂y ∂x = −(−2)(y +2z)−3 (2) = 4(y +2z) −3 4
=
and ∂w
(y +2z) 3 ∂y = x(−1)(y +2z)−2 (1) = −x(y +2z) −2 ,
∂ 2 w
∂x∂y = −(y +2z)−2 ,
∂ 3 w
∂x 2 ∂y =0.
f(3 + h, 2) − f(3, 2)
69. By Definition 4, f x (3, 2) = lim
which we can approximate by considering h =0.5 and h = −0.5:
h→0 h
f x (3, 2) ≈
f(3.5, 2) − f(3, 2)
0.5
=
22.4 − 17.5
0.5
=9.8, f x (3, 2) ≈
f(2.5, 2) − f(3, 2)
−0.5
=
10.2 − 17.5
−0.5
=14.6. Averaging
f(3 + h, 2.2) − f(3, 2.2)
these values, we estimate f x (3, 2) to be approximately 12.2. Similarly, f x (3, 2.2) = lim
which
h→0 h
we can approximate by considering h =0.5 and h = −0.5: f x(3, 2.2) ≈
f x(3, 2.2) ≈
f(2.5, 2.2) − f(3, 2.2)
−0.5
=
9.3 − 15.9
−0.5
To estimate f xy(3, 2), wefirstneedanestimateforf x(3, 1.8):
f x (3, 1.8) ≈
f(3.5, 1.8) − f(3, 1.8)
0.5
=
20.0 − 18.1
0.5
f(3.5, 2.2) − f(3, 2.2)
0.5
=
26.1 − 15.9
0.5
=13.2. Averaging these values, we have f x(3, 2.2) ≈ 16.8.
=3.8, f x (3, 1.8) ≈
f(2.5, 1.8) − f(3, 1.8)
−0.5
=
12.5 − 18.1
−0.5
Averaging these values, we get f x (3, 1.8) ≈ 7.5. Nowf xy (x, y) = ∂ ∂y [f x(x, y)] and f x (x, y) is itself a function of two
=20.4,
=11.2.