最新高等数学在线作业Word文档格式.docx
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最新高等数学在线作业Word文档格式.docx
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32C.
41D.
sinxyz,ye(3)函数的全微分为
sinxy2dz,e[ycosxydx,xycosxydy]A.
sinxy2dz,e[ycosxydx,(1,xycosxy)dy]B.
cosxy2dz,e[ycosxydx,(1,xycosxy)dy]C.
cosxy2dz,e[ycosxydx,(1,xysinxy)dy]D.
2234z,x,y8,(cm)(4)在一个侧面为旋转抛物面的容器内装有的水,现在容器内又注入水
3128(cm)至,则水面比原来升高了
A.6cm
B.2cm
C.8cm;
68D.cm.
1,当,1,x,0,f(x),,1,当0,x,1f(x),(5)周期为2的函数在一个周期内的表达式为若记它的傅
17,,s,,s(x)4,,里叶级数的和函数为,则的值为
A.1
B.-1
C.0
1
2D.
三(解下列各题.
(1)设有两点A(-7,2,-1)和B(3,4,10).若平面过点B且垂直于线段AB,求平
d,,面的方程及点A到平面的距离.
x,2y,1zL:
,M0,21,1
(2)求过点(1,2,-1)且与直线垂直并相交的直线方程.
222x,y,1,0x,y,z,xy,3,0z,0(3)求同时垂直于平面及,且与曲面相切的平
面的方程.
aD(4)设有一个等腰直角三角形的平面薄板,腰长各为,各点处的面密度等于该点到直角顶点的距离的平方,求这平面薄板的重心坐标.
1nx,2s(x)(,1)nn2,(5)求幂级数的和函数.
四(求下列各题的解.
PP(0,2,5),,0
(1)从点(2,-1,-1)向平面引垂线,垂足为点,求平面的方程.
2,f(x,2x),f(x,2x),x,f(x,2x),xf(x,y)yx
(2)设函数可微,且,求.
2222(x,y)dxdydz,,Ω,(x,y,z)3,z,x,y,,,Ω(3)计算,其中.五(求解下列各题.
11211x,y,z,x,yz,:
:
L,,L,,12013122
(1)求异面直线及之间的距离.
f,C(D)D,D,D12
(2)若,,证明:
2aaa,,2f(x)dxf(y)dy,f(x)dx,,,,,x00,,
答案:
一填空题:
(2,,6,,5).1:
,(2,,6,,5)AB,(,1,,2,2),AC,(,2,1,,2)AB,AC,提示因为,故知.
ba,ab0c,ba,ab2:
.
ab00a,b,00aba,bba提示因为、夹角平分线的方向,必为的方向,而,,有
ba,ab00c,a,b,ab,
则
ba,ababba,abc0c,,,,cabba,abba,ab.
22e3:
2y,x2y,x,?
f,2esin(x,2y),2ecos(x,2y),y提示
,,,,222,?
f,2esin,2ecos,2e,y,,0,422.
22,rf(r)4:
2222,f,x,y,zdv,f(r)rsin,drd,d,,,,,,,ΩΩ提示
2,RR222,d,sin,d,rf(r)dr,2,rf(r)dr,,,,0000,
可知
2g(r),2,rf(r).
2e5:
,,7,,,,,,2s,s,,f,,e,,,,,,,222,,,,,,提示.
二、选择题
(答案:
22xy,222axyz,,,.cos,,,cos,,aa,(x,y,z)提示设,则
2222y,zx,z,cos,,222aacos,,cos,,cos,,2.故有.故应选(C).
24(2,1,6),cos()a,b,,c,a,ˆ4118.提示
41362525,,,cos()Pc,cc,a,,rjaˆ411832,.故应选(B).
z,z2sinxyxysinxy?
yecosxy,,e,xyecosxy,x,y提示
u,usinxy2?
dz,dx,dy,e[ycosxydx,(1,xycosxy)dy],x,y.故应选(B).
224z,x,yz,h,0提示由与所围的立体的体积是
2,hhh222,,,2,,V,dv,,d,d,dz,,2,h,d,,2,h2,,,,,,,,,,00044,,Ω.
33V,8,(cm)h,2cmV,128,(cm)h,8cm121当2时,;
时,.知
h,h,6cm21.
故应选(A).
17911,,,,,,,,s,s,s,f,1,,,,,,,,4444,,,,,,,,提示.
三(解下列各题
n,{10,2,11}AB,(10,2,11),,,AB解,因为平面,所以可取,得平面的方程为
10(x,3),2(y,4),11(z,10),0,即
10x,2y,11z,148,0.且有
222d,AB,10,2,11,15.或
10,(,7),2,2,11,(,1),148d,,15.22210,2,11
M(1,2,,1),0L解过点且与直线垂直的平面的方程为
2(x,1),(y,2),(z,1),0,即
2x,y,z,1,0.
212,,M,,,,,1M333,,,L1则直线与平面的交点为:
.由于
5555,,MM,,,,,,(,1,,1,1),,013333,,.s,(,1,,1,1)取则得所求的直线的方程为
x,1y,2z,1,,,1,11.
P(x,y,z),,解设所求的平面与曲面相切于点.则平面的法线向量为n,(2x,y,2y,x,2z)x,y,1,0n,(0,0,1)z,01.而平面的法线向量为,平面的
n,(1,1,0)2法线向量为.
?
n,n,n,n12,从而有方程组
n,n,2z,0,,1,n,n,(2x,y),(2y,x),0.2,
z,0,y,,xP于是有.又因为切点在曲面上.
222x,(,x),0,x(,x),3,0.
x,1,x,,1P(1,,1,0),P(,1,1,0)1212可解出.于是有切点.
n,(3,,3,0)P(1,,1,0),,,1对于切点切平面的法线向量为.可知切平面的方程为
3(x,1),3(y,1),0x,y,2,0,即.
n,(,3,3,0)P(,1,1,0),,2对切点,切平面的法线向量为.可知切平面的方程为
3(x,1),3(y,1),0x,y,2,0,即.
解见图12-5,
22,,x,y且面密度.于是
22y(x,y)dxdy,,Dx,22(x,y)dxdy,,D
aya,22ydy(x,y)dx,,00,,axa22dx(x,y)dy,,00
a1,,23315yay,y,(a,y)dya,,,023,,15,,,a.a115,,4233aax,x,(a,x)dx,,,063,,同理,可求得
2215x(x,y)dxdya,,215Dy,,,a.22154(x,y)dxdy,,aD6
故重心的坐标为
22,,a,a,,55,,.
5.
n,xs(x),,x,[,1,1],2(n,1)n,2解设.
n,1n,,,,xx,[xs(x)],,,,,,(n,1)(n,1)n,1,,n2n2,,
n,1,x,x,xg(x).,n,1n,2n1,,xg(x),.,g(0),0n,1n2,这里,,且有.
n1,,,,,,x1n2n,,g(x),,x,x,,x,(,1,1),,,,,n,11,xn2n2n0,,,,,.
故
xx1,g(x),g(x)dx,dx,,ln(1,x),,001,x,于是
[xs(x)],,xln(1,x),
2xx1x2()ln
(1)dln
(1)dxsx,,x,xx,,x,x,x,,0022(1,x)
22,,x1x1,,,,ln(1,x),,x,ln(1,x),,2222,,
2111,x2,ln(1,x),x,x242有
2,x1,11xxxln(1,),,,,[,1,0),(0,1),,sx(),x242,
x0,,0.,
四(求下列各题的解
PPPP,(2,,3,6)00,,解,取为平面的法线向量,则得平面的方程为
2x,3(y,1),6(z,1),0,
2x,3y,6z,9,0即.
,df(x,y)f(x,y)dxf(x,y)dy,?
,xy解
而
x,f(x,2x),
有
,dxf(x,2x)dxf(x,2x)d(2x),,y,2x)xy(这里
2,,xdx,2f(x,2x)dx.y
即
2,1,x,2f(x,2x),y21,x,(,2)fxx,y2得.
2,33324333dddz2(3)d,,,,,,,,,,,,,,,,000,10解原式.
,,fxx(),sin,,,4,,x(4)把展为的幂级数.
2f(x),(sinx,cosx)2解
nn,,,,2(,1)(,1),2n12n,x,x,,,,2(2n,1)!
(2n)!
,n0n0,,
21111,,245?
1,x,x,,x,x,,,22!
3!
4!
5!
,,
n,nn,(3)2x2,(,1),x,(,,,,,).,2n!
n0
五(求解下列各题.
M(1,0,1)MLLLd,,,22122解过直线作平面,使//则在上选定点则点到平面的距离
d即是所求的两条异面直线之间的距离.
n设平面的法线向量为,则有
n,s,s,(0,1,3),(1,2,2),,(4,,3,1).12
M(,1,1,2)L,11在上取点,则平面的方程为
4(x,1),3(y,1),(z,2),0,即
4x,3y,z,5,0.
M(1,0,,1)d,2而点到平面的距离为
4,1,3,0,1,(,1),5426d,,.222134,(,3),1
L//LM(x,y,z)Ld1211111说明如果则两者的距离的求法更简单,在上取定点,再用求直
LLd12线外一点到直线的距离公式,即可求出.
D,D,D12证(见图12-7),
2aaa,,?
f(x)dx,f(x)dxf(y)dy,,,,,000,,
axaa,dxf(x)f(y)dy,dxf(x)f(y)dy,,,,x000
aaaa,f(y)dyf(x)dy,dxf(x)f(y)dy,,,,xx00
aaaa,f(x)dxf(y)dy,f(x)dxf(y)dy,,,,xx00
aa,2f(x)dxf(y)dy.,,x0
综合习题二
n
n3,1n
(1)数项级数的和为.
1n(2x,1),nn1,
(2)幂级数的收敛区间为.
12321x,y,z,x,y,z:
;
L,,L,,12L101211,1(3)已知两条直线的方程是,则过直线
L2且平行于的平面方程是.
z2,z,1z,f(x,y)x,2y,xy,1,z,2e,0,x(4)设决定了函数,则.
22z,h,(h,0)z,x,yΩ(5)设由曲面与平面围成一个体积为1的立体,则h,.
k,1sin,,,2,,nnn1,,k,
(1)设为常数,则级数(A)绝对收敛
(B)条件收敛
(C)发散
k(D)敛散性与有关
an,12n,1lim,2ax,nn,,ann,1
(2)设,则幂级数的收敛半径为
2(A)
(B)1
2(C)
(D)2
,6,xy,158x,y,z,:
L,:
2L,,12,,3.yz,121,(3)直线与直线的夹角为,
6(A)
4(B)
3(C)
2(D)
23,3,2,10,xy:
Γ,,,M3,1,,1z,0.Oy0,(4)求曲线绕轴旋转一周所得的旋转曲面在点处的
切平面方程为
3x,y,z,3,0(A)
3x,y,z,3,0(B)
3x,y,z,3,0(C)
3x,y,z,3,0(D)
222Ω,{(x,y,z)x,y,(z,1),1,z,1,y,0}Ω(5)设物体在空间的位置为.对任意的点
1u,222x,y,z(x,y,z),Ωm,,物体的体密度为,则物体的质量7,6(A)
2,3(B)
,722,,,,,,63,,(C)
,722,,,,,,63,,(D)
5,n1n,n,s
(1)判定级数的敛散性.
22,,sin,n,a,n1,
(2)判定级数的敛散性.
OzC?
ABCAB(3)已知点(1,0,0)及点(0,2,1),在轴上求一点,使的面积最
小.
222x,y,zx,y,zu,x,2y,3z(4)将数33分成3个正数之和,问各等于多少时,函数取
最小值.
2222f(0),0,Ω,{(x,y,z)x,y,z,t},f,C(5)设类,且.求
1222,,limfx,y,zdv4,,,t,0,tΩ.
,,,,11,,,,ln1,,,nnn1,,,,
(1)判定正项级数的敛散性.,
a?
b?
c,1?
3?
2,,
(2)设平面与原点的距离为6,且在坐标轴上的截距之比为.求平面的
方程.
yyz,(1,e)cosx,ye(3)求的极值.
222x,y,z,3(z,0)ΩΩ(4)设是密度均匀的上半球体:
,求的重心坐标.
22z,x,xy,y,3ax,3by
(1)求的极值.
,,1,,f,1,,,,,,,nf(x)f(0),1,f(0),2,,n1,,,
(2)设偶函数具有二阶连续的导数,且.证明:
绝
对收敛.
答案
3
41.
123n,,,,?
,sn23n3333提示,
11231,nn,,,,,,s?
n234nn,1333333
于是
11,n,11111nn33,,,,,,,,ss?
nn2nn,1n,113333331,3,
3,311,n,,,,,s,,n,n,1n,1,22333,,,,.
从而有
9113,,n,,limlim,,,,,ss,,n,n,1n,n,,n,,434323,,,,,.
(,1,0)2.
n,t
t,2x,1,1nt,2x,1n,1提示令,则原来的幂级数变为,知,即幂级数
1n(2x,1),nn1,的收敛区间为(-1,0).
x,3y,z,2,03.
提示所求的平面的法线向量为
ijk
n,10,1,(1,,3,1)
211.
M(1,2,3)M,L)n001则以为法线向量且过点(点的平面方程为(x,1),3(y,2),(z,3),0.
x,3y,z,2,0.
1,y2z,14ze,14.2z,1F,x,2y,xy,1,z,2e提示令,则2z,1,,F,1,y,F,1,4zexz.
F,z1,yx,,2z,1,,xF4ze,1z.
2
5.
Ω提示在柱面坐标系中可以表示为
2Ω,{(,,,,z),,z,h,0,,,h,0,,,2,}.故
2hhh3,,,,,,,V,dv,dddz,2(h,)d2,,,,,,,000,Ω
2,h,1.2有
2h,,.
1.C
sink11?
,22nnn?
提示.给出的级数是正项级数.
k1sin,2nnlim,1n,,1
n由于,
1
nn,1而正项级数发散.故给出的级数发散.故应选(C).
2.A
1n?
atR,,n1n,12提示的收敛半径,
,1,2n12n?
ax,a(x),,nnx,,n1n1.
112x,t,x,22由于,知.故应选(A).
3.C
s,(1,,2,1)s,(,1,,1,2)LL1212提示直线的方向向量,直线的方向向量.而
s,s112cos()s,s,,12ˆ2ss12.
LL312故知与的夹角为.
故应选(C).
4.C
232?
3x,2y,3z,10提示旋转曲面的方程为.
,,M3,1,,1?
0曲面在点处的法线向量为
,,,,n,63,,6,,6,63,,1,,1,
知切平面方程为
,,3x,3,(y,1),(z,1),0,
3x,y,z,3,0.
5.C
,,2cos1142m,dxdydz,d,sin,d,,rdr1,,,,,,22200rcos,x,y,zΩ提示
,,1142,,,,4cossind,,,,2,,,,02cos,,
,4,,4,,,41,,3,,,cos,,,,,,,,,,23cos,,,,,,00,,
,,,,,,,42722,,,,,,12,.,,,,,,,,,,,,23363,,,,,,,,故应选(C).
nu,n5n,n,2?
解所给的级数为正项级数,且.
,1,v3,,n2,,n1n1n取收敛的正项级数,则有n
u5nn,,2nlim,lim,1vn,,n,,1n32n.
5n1n,n,2,?
正项级数是收敛的.
2,,a,22,,,,?
u,sinn,n,a,n,sinn,,,,,,n22n,a,n,,解
2a,n,,
(1)sin22n,a,n,
n当充分大时,有
2a,,01,,,222n,a,n.
所给的级数为交错级数.并有
22aa,,u,sin,u,sinnn,12222n,a,n(n,1),a,(n,1),
2a,ulim,limsin,0n22n,,n,,nan,,及.由莱布尼茨判别法可知,所给的级数是收敛的.
C(0,0,z)Oz解在轴上任取一点,则
AB,(,1,2,1),AC,(,1,0,z),
AB,AC,(,1,2,1),(,1,0,z),(2z,z,1,2),
?
ABC于是的面积S可以表示为
112S,AB,AC,5z,2z,5.222f,5z,2z,5令,则
,,f,10z,2,f,10.
11,,,,1,,f,0C0,0,,,,,z,,f,055,,,,5令,有驻点,且知其为极小值点,也是最小值点.故有点
ABC可使面积最小.此时
11212030S,,,5,,.min2552,55
解令拉格朗日函数为
222F,x,2y,3z,,(x,y,z,33),有方程组
,F,2x,,0,,x,,,F,4y,,0,,y,,F,6z,,0,,z,
x,y,z,33.,
11z,x,y,xx,18,y,9,z,632由前面3个方程可解得再代入最后的限制条件中,知.故
M(18,9,6)0惟一的驻点为时,函数取最小值为u,u,594minM0.
解用球面坐标系,有t22rf(r)dr,,2t,1,,02limddf(r)rsindrlim,,,,44,,,000t,0t,0tt,,
t224rf(r)dr,4tf(t)f(t)0,limlimlimf(0),,,,43t,0t,0t,0t4tt
2,,1111111,,,,,,,,,,,,,uln10,,,,,,,,n2nnnn2nn,,,,,,,,,,解由
111,,,,,0,,(当n,,时).,,222nn,,有
11111,,,,ln10,,,,,,,,,221nnnn2,,,,limlim,,n,,n,,11222nn.
,,,11,,1?
,ln1?
,,,,,2nnn,,n1n,,,,1收敛,收敛.
设平面的方程为解
xyz,,,1,,,,32
6x,2y,3z,6,,0.
由于
66,,,6,,2227627,,.
,,7,知.故知所求的平面的方程为
6x,2y,3z,42,0.
解由方程组
y,,z,(1,e),(,sinx),0,,x,yyy,z,ecosx,e,ye,0.,y,
可解得
sinx,0,,
cosx,1,y.,
它有无穷多个驻点
x,(2k,1),x,2k,,,,,
,y,,2.y,0.(k,N),,及.
x,(2k,1),,,
,2,2y,,2.A,1,e,B,0,C,,e,对于驻点有.
2,2B,AC,e(1,e),0且.故它们不是极值点;
x,2k,,,
y,0.A,,2,0,B,0,C,,1,对于驻点有.2z,2B,AC,,2,0且A,0max,知其为极大值点.对于这无穷多个极大值点,均有.
x,y,0解由对称性知
,2323zdv,,,,,,,dsincosdrdr,,,000Ω,,z,2,322dvd,sin,d,rdr,,,,,,000Ω故
4119,,,,2,,33244,,,3.318,23,,2,1,3,3
3,,0,0,3,,8,,Ω故的质心坐标为.
z,,2x,y,3a,0,,,x,,,z,,x,2y,3b,0.,,yP(2a,b,2b,a),解由方程组解得驻点.
222zzz,,,ABC,,2,0,,,1,,,2.22xxyy,,,,PPP且
22?
B,AC,1,4,,3,0,A,0.
P(2a,b,2b,a)?
点为极小值点,且极小值为
22z,z,3(ab,a,b).P
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