高等数学 课后习题答案第八章

习题八1.判断下列平面点集哪些是开集、闭集、区域、有界集、无界集？并分别指出它们的聚点集和边界：(1){(x,y)|x≠0};(2){(x,y)|1≤x2+y24};(3){(x,y)|yx2};(4){(x,y)|(x-1)2+y2≤1}∪{(x,y)|(x+1)2+y2≤1}.解：(1)开集、无界集，聚点集：R2，边界：{(x,y)|x=0}.(2)既非开集又非闭集，有界集,聚点集：{(x,y)|1≤x2+y2≤4}，边界：{(x,y)|x2+y2=1}∪{(x,y)|x2+y2=4}.(3)开集、区域、无界集，聚点集：{(x,y)|y≤x2}，边界：{(x,y)|y=x2}.(4)闭集、有界集，聚点集即是其本身，边界：{(x,y)|(x-1)2+y2=1}∪{(x,y)|(x+1)2+y2=1}.2.已知f(x,y)=x2+y2-xytanxy,试求(,)ftxty.解：222(,)()()tan(,).txftxtytxtytxtytfxyty3.已知(,,)wuvfuvwuw,试求(,,).fxyxyxy解：f(x+y,x-y,xy)=(x+y)xy+(xy)x+y+x-y=(x+y)xy+(xy)2x.4.求下列各函数的定义域：2(1)ln(21);zyx11(2);zxyxy2224(3);ln(1)xyzxy111(4);uxyz(5);zxy22(6)ln();1xzyxxy22(7)arccos.zuxy解：2(1){(,)|210}.Dxyyx(2){(,)|0,0}.Dxyxyxy22222(3){(,)|40,10,0}.Dxyxyxyxy(4){(,,)|0,0,0}.Dxyzxyz2(5){(,)|0,0,}.Dxyxyxy22(6){(,)|0,0,1}.Dxyyxxxy22222(7){(,,)|0,0}.Dxyzxyxyz5.求下列各极限：2210ln(e)(1)lim;yxyxxy22001(2)lim;xyxy0024(3)lim;xyxyxy00(4)lim;11xyxyxy00sin(5)lim;xyxyx222222001cos()(6)lim.()exyxyxyxy解：(1)原式=022ln(1e)ln2.10(2)原式=+∞.(3)原式=00441lim.4(24)xyxyxyxy(4)原式=00(11)lim2.11xyxyxyxy(5)原式=00sinlim100.xyxyyxy(6)原式=22222222222()00001()2limlim0.()e2exyxyxxyyxyxyxy6.判断下列函数在原点O(0，0)处是否连续：33222222sin(),0,(1)0,0;xyxyzxyxy33333333sin(),0,(2)0,0;xyxyzxyxy(3)222222222,0,(2)()0,0;xyxyzxyxyxy解：(1)由于3333333322223333sin()sin()sin()0()xyxyxyxyyxxyxyxyxy又00lim()0xyyx，且3333000sin()sinlimlim1xuyxyuxyu,故00lim0(0,0)xyzz.故函数在O(0,0)处连续.(2)000sinlimlim1(0,0)0xuyuzzu故O(0,0)是z的间断点.(3)若P(x,y)沿直线y=x趋于(0，0)点，则2222000limlim10xxyxxxzxx,若点P(x,y)沿直线y=-x趋于(0，0)点，则22222220000()limlimlim0()44xxxyxxxxzxxxx故00limxyz不存在.故函数z在O(0，0)处不连续.7.指出下列函数在向外间断：(1)f(x,y)=233xyxy;(2)f(x,y)=2222yxyx;(3)f(x,y)=ln(1－x2－y2);(4)f(x,y)=222e,0,0,0.xyxyyy解：(1)因为当y=-x时，函数无定义，所以函数在直线y=-x上的所有点处间断，而在其余点处均连续.(2)因为当y2=2x时，函数无定义，所以函数在抛物线y2=2x上的所有点处间断.而在其余各点处均连续.(3)因为当x2+y2=1时，函数无定义，所以函数在圆周x2+y2=1上所有点处间断.而在其余各点处均连续.(4)因为点P(x,y)沿直线y=x趋于O(0,0)时.12000lim(,)limexxyxxfxyx.故(0，0)是函数的间断点，而在其余各点处均连续.8.求下列函数的偏导数：(1)z=x2y+2xy;(2)s=22uvuv;(3)z=xln22xy;(4)z=lntanxy;(5)z=(1+xy)y;(6)u=zxy;(7)u=arctan(x-y)z;(8)yzux.解：(1)223122,.zzxxyxxyyy(2)uvsvu2211,.svsuuvuvvu(3)22222222222111ln2ln(),22zxxyxxxyxxyxyxy222222112.2zxyxyyxyxyxy(4)21122seccsc,tanzxxxxyyyyy222122sec()csc.tanzxxxxxyyyyyy(5)两边取对数得lnln(1)zyxy故221(1)(1)(1).ln(1)1yyyxzyxyxyyxyyxyxxyln(1)(1)(1)ln(1)1ln(1)(1).1yyyyxzxyyxyxyyxyxyyxyxyxyxy(6)1lnlnxyxyxyuuuzzyzzxxyzxyz(7)11221()().1[()]1()zzzzuzxyzxyxxyxy112222()(1)().1[()]1()()ln()()ln().1[()]1()zzzzzzzzuzxyzxyyxyxyuxyxyxyxyzxyxy(8)1.yzuyxxz2211lnln.lnln.yyzzyyzzuxxxxyzzuyyxxxxzzz9.已知22xyuxy，求证：3uuxyuxy.证明：222223222()2()()uxyxyxyxyxyxxyxy.由对称性知22322()uxyyxyxy.于是2223()3()uuxyxyxyuxyxy.10.设11exyz，求证：222zzxyzxy.证明：11112211eexyxyzxxx,由z关于x,y的对称性得1121exyzyy故11111122222211ee2e2.xyxyxyzzxyxyzxyxy11.设f(x,y)=x+(y-1)arcsinxy，求fx(x,1).解：2111(,)1(1)21xfxyyyxxyy则(,1)101xfx.12.求曲线2244xyzy在点（2，4，5）处的切线与正向x轴所成的倾角.解：(2,4,5)1,1,2zzxxx设切线与正向x轴的倾角为α,则tanα=1.故α=π4.13.求下列函数的二阶偏导数：(1)z=x4+y4-4x2y2;(2)z=arctanyx;(3)z=yx;(4)z=2exy.解：(1)2322224812816zzzxxyxyxyxxxy，，由x,y的对称性知22222128.16.zzyxxyyyx(2)222211zyyxxyxyx，2222222222222222222222222222222222222222()022,()()11,12,()()2,()()2.()()zxyyxxyxxyxyzxyxxyyxzxyyxyzxyyyyxxyxyxyzxyxxyxyxxyxy(3)222ln,ln,xxzzyyyyxx21222112111,(1),1ln(1ln),ln(1ln).xxxxxxxxzzxyxxyyyzyxyyyxyxyyzyxyyyxyyx(4)22e2,e,xyxyzzxxy222222222e22e22e(21),e,2e,2e.xyxyxyxyxyxyzxxxxzzzxxyxyyx14.设f(x,y,z)=xy2+yz2+zx2,求(0,0,1),(0,1,0),(2,0,1).xxyzzzxfff解：2(,,)2xfxyzyzx22(,,)2,(0,0,1)2,(,,)2(,,)2,(0,1,0)0,(,,)2(,,)2(,,)0,(2,0,1)0.xxxxyyzyzzzzzzxzzxfxyzzffxyzxyzfxyzzffxyzyzxfxyzyfxyzf15.设z=xln(xy)，求32zxy及32zxy.解：ln()1ln(),zyxxyxyxxy232223221,0,11,.zyzxxyxxyzxzxyxyyxyy16.求下列函数的全微分：(1)22exyz;(2)22yzxy;(3)zyux;(4)yzux.解：(1)∵2222e2,e2xyxyzzxyxy∴222222d2ed2ed2e(dd)xyxyxyzxxyyxxyy(2)∵22223/22212()2zxxyyxyxxyxy2222222223/2()yxyyxyzxyxyxy∴223/2d(dd).()xzyxxyxy(3)∵11,lnzzzyyzuuyxxxzyxy2lnlnyzuxxyyz∴211ddlndlnlnd.zzzyyzyzuyxxxxzyyxxyyz(4)∵1yzuyxxz1lnyzuxxyzlnyzuyxxzz∴121ddlndlnd.yyyzzzyyuxxxxyxxzzzz17.求下列函数在给定点和自变量增量的条件下的全增量和全微分：(1)222,2,1,0.2,0.1;zxxyyxyxy(2)e,1,1,0.15,0.1.xyzxyxy解：(1)22()()()2()9.6881.68zxxxxyyyyzd(2)(4)1.6zxyxxyy(2)()()0.265eee(e1)0.30e.xxyyx