HOMEWORK #3 – SOLUTIONS Page 899 4) z = y ln x, (1, 4, 0 )
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 =
fy =
1
2 y + cos x
2
1
2 y + cos 2 x
⋅ 2 cos x ( − sin x ) ⇒ fx ( 0, 0 ) = 0
⋅1 ⇒ fy ( 0, 0 ) =
y + cos 2 x ≈ 1 + 0 ⋅ x +
20) f ( x, y ) = ln ( x − 3y )
f ( 7, 2 ) = ln1 = 0
1
2
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
ln ( x − 3y ) ≈ 0 + 1( x − 7 ) − 3 ( y − 2 ) = x − 3y − 1
fx =
f ( 6.9, 2.06 ) ≈ −.28
Actual value = -.328
26) v = y cos xy
dv = vx dx + vy dy = −y 2 sin xy dx + ( −xy sin xy + cos xy ) dy
30) w = xye xz ⇒ dw = wx dx + wy dy + wz dz =
32) 34) (
)
⎡ x yze xz + ye xz ⎤ dx + xe xz dy + x 2 ye xz dz = ⎣
⎦
ye xz ( xz + 1) dx + xe xz dy + x 2 ye xz dz
( 3, −1) → ( 2.96, −0.95 )
2
2
Δz = ( 2.96 ) − ( 2.96 ) ( −0.95 ) + 3 ( −0.95 ) − ( 32 − 3 ( −1) + 3) = −.7189
dz = zx dx + zy dy = ( 2x − y ) dx + ( −x + 6y ) dy =
( 2 ⋅ 3 − ( −1)) ( −.04 ) + ( −3 − 6 ) (.05 ) = −0.73
z = x 2 − xy + 3y 2
S = 2 ( xy + xz + yz )
x = 80, y = 60, z = 50
dS = 2 ( y + z ) dx + 2 ( x + z ) dy + 2 ( x + y ) dz =
2 (110 ) (.2 ) + 2 ⋅130 (.2 ) + 2 ⋅140 (.2 ) = 152cm 2 = max error
40) 0 < x, y, z, w < 50
P = xyzw
dP = Px dx + Py dy + Pz dz + Pw dw =
yzw dx + xzw dy + xyw dz + xyz dw =
50 3 (.05 ) + 50 3 (.05 ) + 50 3 (.05 ) + 50 3 (.05 ) =
4 ⋅ 50 3 (.05 ) = 25, 000
Page 907 4) ⎛ y⎞
z = tan −1 ⎜ ⎟ , x = et , y = 1 − e−t
⎝ x⎠
dz ∂z dx ∂z dy
=
⋅ + ⋅ =
dt ∂x dt ∂y dt
1
−y
1
1
⋅ 2 ⋅ et +
⋅ ⋅ e−t = 2
2
⎛ y⎞ x
⎛ y⎞ x
1+ ⎜ ⎟
1+ ⎜ ⎟
⎝ x⎠
⎝ x⎠
xe−t − yet
⋅e = 2
=
y2
x + y2
x+
x
t −t
−t
t
e e − 1− e e
1 − et + 1
2 − et
=
=
2t
−t 2
2t
−t 2
e + 1− e
e + 1− e
e2t + 1 − e−t
−y
⋅ et +
2
2
x +y
(
(
6) 1
)
)
−t
(
)
(
)
2
=
dz
dt
w = ln x 2 + y 2 + z 2 , x = sin t, y = cos t, z = tan t
dw ∂w dx ∂w dy ∂w dz
=
⋅ +
⋅ +
⋅ =
dt
∂x dt ∂y dt ∂z dt
1
1
⋅ 2x ⋅ cos t +
⋅ 2y ⋅ ( − sin t ) +
2
2
2
2
2
2
2 x +y +z
2 x +y +z
(
)
(
)
1
x cos t − y sin t + z sec 2 t
2
⋅
2z
⋅
sec
t
=
=
x 2 + y2 + z 2
2 x 2 + y2 + z 2
(
38) )
sin t cos t − cos t sin t + tan t sec 2 t
=
sin 2 t + cos 2 t + tan 2 t
tan t sec 2 t tan t sec 2 t
dw
=
= tan t =
1 + tan 2 t
sec 2 t
dt
40) 1
V = π r2h
r = +1.8 in sec , h = −2.5 in sec
3
dV ∂V dr ∂V dh 2
1
=
⋅ +
⋅
= π rh (1.8 ) + π r 2 ( −2.5 ) ⇒
dt
∂r dt ∂h dt 3
3
dV ⎤
2
1
2
= π (120 ) (140 ) (1.8 ) + π (120 ) ( −2.5 ) =
dt ⎥⎦(120,140 ) 3
3
in 3
in 3
8160π
≈ 25, 635
sec
sec
V = IR ⇒ I =
V
R
V = −.01, R = +.03 dI ∂I dR ∂I dV
=
⋅
+
⋅
=
dt ∂R dt ∂V dt
−V
1
−I
1
⋅ (.03) + ( −.01) =
(.03) + ( −.01)
R2
R
R
R
dI ⎤
−.08
1
−5 A
=
.03
+
−.01
=
−3.1
×
10
(
)
(
)
dt ⎥⎦( 400,.08 ) 400
400
sec
44) ⎛ c + v0 ⎞
f0 = ⎜
fs
⎝ c − vs ⎟⎠
c = 332
m
m
m
, v0 = 34
, v0 = 1.2 2
sec
sec
sec
m
m
, vs = 1.4
sec
sec 2
⎛ 332 + 34 ⎞
f0 = ⎜
( 460 ) = 576.6Hz ( perceived frequency ) ⎝ 332 − 40 ⎟⎠
df0 ∂f0 dv0 ∂f0 dvs
=
⋅
+
⋅
=
dt ∂v0 dt ∂vs dt
⎛ 1 ⎞
dv0
c + v0
dvs
f
⋅
+
f
=
s
⎜⎝ c − v ⎟⎠ s dt
( c − vs )2 dt
s
vs = 40
1
332 + 34
Hz
⎛
⎞
⋅ 460 (1.4 ) = 4.65
⎜⎝
⎟⎠ ⋅ 460 (1.2 ) +
2
332 − 40
sec
( 332 − 40 )