化工数学(周爱月)习题解答--第6章
化工数学(周爱月)习题解答——第六章
6-2 解:
(a )2(3i -4j +k ) -(j +2i -3k ) =(2? 32) i -(2? 41) j +(2+3) k (b )(c )(d )(e )(f )由书中 6-3 解:(a )(b )(c ) =4 i -9
j +5 k ( i + j ) (2 i -2 j ) =2-2=0 i j k
2(3 i -4 j +k ) ? ( j
2 i -3k ) =23-41=22( i + j +k )
21-3
grad [xy (y 2
+z 3
)]=grad [xy 3
+xyz 3
)]=(y 3
+yz 3
) i +(3xy 2+xz 3) j +3xyz 2
) k div (yi +xj ) =抖y
(-x ) 0
抖x
+y + z
=0 p187例6-5知 ? (1
r
)
? (r -1)
-r -3r
,所以
? 2(1r
) 蜒 r (-1=) ? (r - 3r =) -r -3? r r ? (-r 3)
=-3r -3+ r (+r -5
r ) =-3-r 3+3-r 3=0
a x
a y a z 21-301-3a (b ? c
)
(abc
) =b x
b y b z =1
-21=0
-21=20 c x
c y
c z
-1
1
-4
-4
1
-4
c (a ? b ) (a ? b ) c (abc ) =20
a 创(b c ) =(a c ) =(-2+1+12)( ? b i -2 (b j +k a ) c ) -(2-2-3)(- i + j -4k ) =8 i -19 j -k
(a 创b ) c =(a c ) ? b (b c ) a (d )
=11(i -2j +k ) -(-1-2-4)(2i +j -3k ) =25i -15j -10k
6-4 解:记 a
, a 1=2i +3j +4k 2=i +2j -2k
i j k
a =a
1? a 2234=-14i +8j +k
12-2
a
a 0=a
=
=
-14 i +8 j +k ) ∵ cos q =a 1 a 2
a a =0? q
90
126-5 解:
(1)证明三个向量共面的充分必要条件是 A ( B ?
C ) 0
① 必要性:已知三个向量共面,设F =B C ,
则F 垂直于B , C 所在的平面,即F ^A
(∵A , B , C 共面)有A F =0薮A (B C ) =0
② 充分性:已知 A (B ? C
)
0,当B 垂C 0时,设F =B 则 A F = ?
c o s q 0
∵ A 构0, F
0? cos q 0? q
90
即 F ^A
所以A , B , C 共面。
1
3
-2
1
3
-2
(2)A (B ? C
)
1-53=0-85=0,三个向量共面。2-210-8
5
6-6 解:
C
(A 创B ) (A C ) =[(A 创B ) A ] C =[(A A ) B -(B A ) A ] C
A A A C =(A A )(B C ) -(B A )(A C ) =
B A B C
6-8 解:参照6-6题,令其中第二个括号中的A =c , C =d 即可。
2 抖R ´6-9 已知 R =x 2yi -2y 2zj +xy 2z 2k ,求在点(2,1,-2) 处的抖x 2
2
R 。 y 2
抖R
解:2=2yi ,
抖x
2
2轾抖R ? 2抖x 臌 2
抖R
? 抖x 2
2
R y 2
(2,1,-2)
R 2
=-4zj +2xz k 2y
i j k
2y 00=-4xyz 2j -8yzk =-32j +16k
(2,1,-2)
2
0-4z 2xz
2
(2,1,-2)
2
R
y 2(2,1,-2)
6-12 在t 时刻,从原点到一动点的向量为r =i cos w t +j sin w t ,其中ω为常数。
(1)求动点速度v ,并证明v 垂直于r ;
(2)求加速度a ,并证明其指向原点,且大小与原点到质点的距离成正比;
(3)证明v ´r 是一常向量,因此动点的轨迹曲线处于某一平面内。
解:
dr
=w (-i sin w t +j cos w t ) , (a )动点速度:v =dt
v r =(-w i s i n w t +w j c o w s t ) w i (c w o +s t w j ∴ v ^r
s w i =n t
dv 2
=-w (i cos w t +j sin w t ) =-w 2r (b )动点加速度:a =dt
矢径r 的方向是由原点指向动点,而动点加速度a 的方向与矢径r 的方向
相反,是由动点指向原点,其大小与矢径r 的大小成正比;
(c )r ? v
i j k
cos w t sin w t -w sin w t w cos w t
0=w k ,为常向量。 0
222222432
6-13 解:j A =xy z (xzi -xy j +yzk ) =x y z i -x y zj +xy z k
3 ¶(j A ) ¶(j A ) 24
=4y zi -2y j +0k , 22
3
=4i -2j
6-14 (a )(b )6-15 6-16 6-17 6-18
6-19 抖x z
抖x z (2-, 1, 1)
蝌
R
(u ) du =[(u -u 2
) i +2u 3
j -3k ]du =(12u 2-13u 3) i +14 2
u j -3uk +c ò2
1R (u ) du =轾犏臌(12u -3u ) i +2u j -3uk 2213 14 =-5 i +15
j -3k 1
62 i
j r k
? r
-A sin q A cos q B =0,∴ 1ò2p (r ? r ) d q 0
-A sin q A cos q B
20
d 骣ç d 2A 2dt ççç桫A ? dA ÷dt ÷÷dA ÷dt ? dA
dt A ? A d A
dt 2
dt 2
d 骣 ∴蝌A ? d 2A
珑 dA dt 2dt
dt 珑珑珑桫A ? 鼢
dt
鼢鼢鼢dt d 骣桫A
? dA dt
A ?
dA dt
c
? j
? (3x 2
y
y 3z 2
) =6xyi +(3x 2-3y 2z 2) j -2y 3
zk
? j
(1,-2, -1)
-12 i -9
j -16k
? j
[(2xyz +4z 2
) i +x 2zj +(x 2
y +8xz ) k M 0
]
M =8 i -
j -10k
l =10=
3(2 i - j -2k )
? j
¶l
=逊j
M l (8 i -4
(2 i j -2k M 0
0=j -10k ) ?
) 370
3
3
x 2+y 2+z 2=9, z =x 2+y 2-3在点(2,-1,2)处的夹角。
解:
解:∵解:∵ 解: 解:求曲面
22
解:令 j 1=x 2+y 2+z -9, j 2=x +y -23-z , 则法向量
n 1=? j
1
2(x i +y +j z ) k , 2=n ? j
2
j 2+x i 2-y
k
在点(2,-1,2)处
n 1(2,-1,2) =4i -2j +4k , n 2(2,-1,2) =4i -2j -k
n 1 n 2
cos q ==
n 1n 2
q =54.42?
=
=0.5819
0.302p
U ? 2v v 2U 。
6-20 证明蜒 (U v -v ? U )
证明:
蜒 (U v -v ? U ) 蜒 (U v ) -蜒 (v U )
=蜒U v +U ? 2v 蜒v U -v 2U =U ? 2v v 2U
6-21 解:若v 为管形场,则押 v 0
? v ? [(x 3y ) i +(y -2z ) j + ∴ a =-2
6-22 解:由式(6-66)得
(x +
a z ) k =]+1+1a =2+ a =
2
汛(汛A ) =蜒( A ) -(蜒 ) A =蜒( A ) - A
222
=蜒[ (x yi -2xzj +2yzk )]-? (x yi 2xzj +2yzk )
22
=? (2xy 2y ) -? (x yi 2xzj +2yzk )
=2yi +(2x +2) j -2yi =2(x +1) j
6-24 解:一次作用:蜒 A , 二次作用:蜒 (? A )
A 0
¶A y
¶A
z ) =
y z
¶A
蜒( A ) =? x
抖x
汛(汛A ) =蜒( A -) 蜒(
) A
6-25 求证:
(1)? (xv ) x ? v i v ;
(2)汛(xv ) =x 汛v +i v ;
(3)蜒 (j ) =(蜒 ) j 貉
2
j .
证明:
(1)由式(6-45):? (xv )
x ? v
v ? x
x ? v
v i =x ? v
i v
(2)由式(6-63):汛(xv ) =x 汛v +汛x v =x 汛v +i v (3)设为直角坐标系 ?
抖 抖
i +j +, 抖x y 抖z
2
?
2
22
抖2 2
x 抖2y z
抖 j j 抖 抖j
蜒 (j ) =(i +j +k ) (i +j +k )
抖x y 抖z x 抖y z
2
抖j =+抖x 2
j 2j
+= 2j 22y z
抖 抖 抖
(蜒 ) j =[(i +j +k ) (i +j +k )]j
抖x y 抖z x 抖y z
22+) j = j 22y z
2
2
2
2
抖=(2+抖x
∴蜒 (j ) =(蜒 ) j 貉
6-27 解:(a ) (A ? ) j A ? j
24
4
3
j
223322
(2yzi -x yj +xz k ) (4xyz i +2xz j +6x yz k )
3
4
3
3
2
=8xy z -2x yz +6x yz =2xyz (4yz -x +3x z ) (b )
抖 (B x +B y +B z ) A
抖x y z
2抖22
=(x +yz +xy )(2yzi -x yj +xz k )
抖x y z
222
=2y (xy +z ) i +2x y (x -z ) j +x z (z +2y ) k
(B ? ) A
(c )
(d )6-28
i
j
k
(A
囱) j =2yz -x 2y
xz 2j 抖 抖x
y
z
=[(-x 2
y 抖2抖z -xz
y ) i +(xz 2抖抖x -2yz z ) j +(2yz 抖抖y +x 2
y ) k ](2x 2yz 3) =-2x 3z 2(3xy +z 3) i +4xyz 3(z 4-3xyz ) j +4x 2yz 3
(z +xy ) k
x
A 囱j =A
i
j k
囱(2x 2yz 3) =2yz -x 2y xz 2
4xyz 32x 2z 36x 2yz 2
=-2x 3z 2(3xy +z 3) i +4xyz 3(z 4-3xyz ) j +4x 2yz 3
(z +xy ) k
验证平面格林定理:
骣抖Q 蝌 l
Pdx +Qdy
=
P D
÷
桫
抖x -y ÷÷
闭曲线C 的图形见右图。
解:
蝌 Pdx +Qdy =
l 22
(xy +y ) dx +x dy c
轾122
=犏(xy +y ) dx +x dy 蝌0犏臌==
y =x 2
01
轾0
+犏(xy +y 2) dx +x 2dy 犏臌1(x 2+x 2+x 2) dx
1
y =x
蝌(x
1
3
+x 4+2x 3) dx +
ò
1
骣[1**********](x +3x -3x ) dx =çx +x -x =+-1=-çç0
桫540
54蝌骣çç抖Q P ÷
蝌(x -2y ]dxdy
D
ç桫
抖x -y ? ÷÷dxdy =蝌[2x (x +2y )]dxdy =D
D
=蝌1
轾0犏x
犏臌x
2(x -2y ) dy dx =
1
轾犏臌
(xy -y 2
) x 0
x 2dx =
蝌1
2
-x 2) -(x 3-x 4) dx =
1
犏轾0
臌
(x 0
(x 4-x 3) dx
=骣1桫5x 5-14x 4=1-1=-105420
左边=右边,证毕。
6-29 求椭圆x =a cos q ,
y =b sin q 的面积。
解:矢径 r
=x i + y ,由散度定理的二维形式(j
6-53)得 2蝌dxdy =
蝌骣抖x +y ÷? dxdy xdy -ydx
S
S
桫抖x y ÷÷
=
ò
2p 0
(a cos q ? b cos q
b sin q a sin q ) d q
=ab ò2p 0
d q =2ab p
∴
蝌d x d =
y a p b
S
6-30 解:F =4xzi -y 2j +yzk ,逊F =4z -2y +y =4z -y ,由散度定理得20
? Fd t 蝌 F nd s =蝌蝌
s
D
蝌(4z -D
y ) d t
10
=
蝌蝌(4z -0
1
1
111
y ) dxdydz =
10
1
(2-y ) dydx
1
=蝌(2y -y 2) dx =
020
33
dx =22
F t d
6-31 解:验证散度定理
s =蝌蝌 F n d
s
D
(1) 左边,对x =1的曲面,n =i ,蝌F nd s =
s x =1
1
蝌F idydz =蝌
s x =1
-1
1
-1
1dydz =4
对x =-1的曲面,n =-i ,蝌F nd s =
s x =-1
蝌F idydz =-s x =-1
蝌
-1
11-1
(-1) dydz =4
∴蝌F nd s =4+4=8, 同理,对y =? 1, z
s x =
1
1曲面,也有
蝌F nd s =
s y =? 1
F nd s =3? 8蝌F nd s =8, 即 蝌
s z
1
24
s
(2) 右边
? Fd t 蝌 F ? nd s 蝌蝌
s
D
1-1
蝌
-1
11-1
3dxdydz =3蝌 d t =3? 8
D
24
左边=右边,证毕。
6-40 解:
(3x -3y ) i -6xyj
i j k 抖
=0 汛v =
抖x y z 3x 2-3y 2-6xy 0
(a ) v =? j
2
2
∴ j (x , y ) =x 3-3xy 2是速度势。
(b
)v =
3(x 2+y 2)
(c )由P207的式(6-128)知
? y -v y i +v x j =6xyi +(3x 2-3y 2) j
d y =-v y dx +v x dy =6xydx +(3x 2-3y 2) dy y (x , y ) =3x 2y -y 3+c y (0,0)=0\c =0
y (x , y ) =3x 2y -y 3y (x , y ) =3x 2y -y 3=c
(d )