东北大学大一公共课专业高等数学统考试卷及答案 (5).pdf

该【东北大学大一公共课专业高等数学统考试卷及答案 (5) 】是由【小屁孩】上传分享，文档一共【6】页，该文档可以免费在线阅读，需要了解更多关于【东北大学大一公共课专业高等数学统考试卷及答案 (5) 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:学院与专业:学号:一、填空题[共24分]1、[4分]设z?x4y3?2x,则dz?34dx?12dy?1,2??x?acost?x?ayz2、[4分]曲线?:?y?asint在点?a,0,0?的切线方程为??.?0ac?z?ct??x2?y2?xy?,?x,y???0,0?3、[4分]f(x,y)??x2?y2,则f?0,y??0?x?0,?x,y???0,0?4、[4分]函数z?x2?y2在点P?1,2?处沿从点P?1,2?到点P?2,2?3?方向的方001向导数是1?235、[4分]设L为取逆时针方向的圆周x2?y2?9,则曲线积分??2xy?2y?dx??x2?4x?dy??18?L6、设L为直线y?x上由点A?0,0?到点B?1,1?之间的一段,则曲线积分2?xy2ds?.4L二、[7分]计算二重积分??2xy2ex2yd?,?1,x?y,x?0所围成的闭区D域解:作图知D:0?y?1,0?x?yy11y1e2xy2ex2yd??dy2xy2ex2ydx?y?ex2y?dx?y?ey2?1?dy??1????????02D0000??x2?y2?z2?2三、[7分]计算三重积分???zdv,其中?.由?所确定?z?x2?y2????x2?y2?z2?2解:由交线??z2?z?2?0,z?1,z??2〔舍去〕??z?x2?y212于是投影区域为D:x2?y2?1,?柱坐标下为0???2?,0?r?1,r2?z?2?r22?12?r22?11?1416?7????zdv??d??rdr?zdz??d??r?2?r2?r4?dr???12????2?46?12?00r200四、[7分]计算??xz2dydz??x2y?z2?dzdx??2xy?y2z?dxdy,其中?为半球?z?a2?x2?y2的上侧解:令?:z?0,x2?y2?a2取下侧。则???为半球体?的外侧,由高斯公式11原式?????x2?y2?z2?dv???xz2dydz??x2y?z2?dzdx??2xy?y2z?dxdy??1?2?aa2?a2???5???d??d???2?2sin?d????2xydxdy?2???cos??2????d??2rco??rsin??rdr0?5?000D0004a2??r2??a5?sin2?2?a5〔用对称性可以简化计算〕504501五、[7分]计算???x?y?1?dS,其中?为抛物面z??x2?y2??0?z?1?2?解:z?x,z?y,dS?1?x2?y2dxdy,投影区域为D:x2?y2?2xy2?22322??由对称性,原式???1dS??d??1?r2rdr???1?r2?2?33?1?33?000六、[7分]求u?x?2y?2z在约束条件x2?y2?z2?1下的最大值和最小值解:令L?x?2y?2z???x2?y2?z2?1???1??1?1?x?x?x??L?1?2x?0?2??3?3?x????L??2?2?y?0?13?2??2则?y,??y?,????,?y?or?y?L?2?2?z?0?233?z???122?x2?y2?z2?1???????z??z??z????3?3?122?144?122?144u??,,????????3,u?,?,?????3?333?333?333?333由于最值一定存在,所以最大值为3,最小值为?3?x??z?2z七、[7分]设z?f?x,?,其中f具有二阶连续偏导数,求,.?y??x?x?y?z1?2z??1?x1x解:?f?1?f?,??f?f???f?f?f?x12y?x?y?y?1y2?y212y22y322八、[7分]求微分方程?y?x2e?x?dx?xdy?:原式可以化为一阶线性微分方程y??y?xe?xx由公式?1?11??dx??dx??ln?yexxe?xexdxcelnxxe?xexdxcx?e?xdxc?x?ce?x??????????????????????九、[7分]设f?x?具有二阶连续导数,f?0??0,f??0??1,且??????2????xyx?y?fxy??dx???f?x?xy??dy?0是全微分方程,求fx其此全微分方程的通解。解:由全微分方程的条件知x2?2xy?f?x??f???x??2xy,f???x??f?x??x2,r2?1?0,r??i1,2有特解有形式f*?x??ax2?bx?c,代入原方程得f*?x??x2?2从而通解f?x??ccosx?csinx?x2?2,f??x???osx?2x1212由初值条件c?2?0,c?1,c?2,c?1,1212因此f?x??2cosx?sinx?x2?2原方程即为????2???2??xyx?y?2cosx?sinx?x?2y?dx???2sinx?cosx?2x?xy?dy?0?x2y2?即d???yd??2sinx?cosx?2x????2sinx?cosx?2x?dy?0?2??x2y2?x2y2d????2sinx?cosx?2x?y??0,???2sinx?cosx?2x?y?c?2?2?xn十、[7分]〔非化工类做〕求幂级数?的收敛域及其和函数n?1n?01an?2解:由limn?1?lim?1,从而R?1,??1,1?为收敛区间n??an??1nn?1又x?1时级数发散〔调和级数去掉第一项〕,x??1时级数由莱布尼茨判别法知道其收敛,从而收敛域为[?1,1)?xn?xn?1设??S?x?,则S?0??1,xS?x???n?1n?1n?0n?0??1x1?xS?x????xn?,xS?x??xS?x??0??dx??ln?1?x???1?x1?xn?00?1,x?0?因此S?x???ln?1?x???,?1?x?0,0?x?1?x1?x十一、[6分]〔非化工类做〕将函数f?x??ln展开成x的幂级数1?x解:f?x?的定义域为??1,1?,f?x??ln?1?x??ln?1?x?11??f??x????????1?n?1?xn??2x2n,f?0??ln1?01?x1?x??n?0n?0?2从而f?x???x2n?1,x???1,1?2n?1n?0十二、[6分]〔非化工类做〕证明:n?1???1??2x2在区间???,??上等式?cosnx??成立n2124n?0?2x2证明:对???,??上的偶函数f?x???作周期为2?的周期延拓,再作出其124傅立叶级数由收敛定理即可推出。?2???2x2?2??2x3?由公式a?????dx??x???00??124???1212?002???2x2?2???2x2?sinnx1?cosnx??1?n?1n?0,a?????cosnxdx?????d?0??xd?n??124???124?nn?nn2000n?1?2x2???1?b?0,从而由收敛定理知道???cosnx在???,??上一定成立n124n2n?01十、[7分]〔化工类做〕在曲面z?2x2?y2上求出切平面,使所得的切平面与2平面4x?2y?2z?1?0平行。解:曲面的法向量n??4x,y,?1?应与平面平面4x?2y?2z?1?0的法向量平行,4xy?11?1?21从而有??,?y??1,x?,由于切点在曲面上z?2?????1?2?14?2?22?2?2?1?因此切平面为4?x???2?y?1??2?z?1??0,2x?y?z?1?0?2?十一、[6分]〔化工类做〕设z?z?x,y?是由方程x2?y2?z???x?y?z?所确定的函数,其中??x?可导,求dz解:对方程两边取微分得2xdx?2ydy?dz?????dx?dy?dz?即?1????dz?dz????dz?2xdx?2ydy?????dx?dy???2x????dx??2y????dy?2x????dx??2y????dydz?1??????1?x2?y2sin,x2?y2?0十二、[6分]〔化工类做〕证明函数f?x,y???x2?y2??0,x2?y2?0在原点?0,0?处可微,但f?x,y?在点?0,0?处不连续x1?x2?02?sin?0f?x,0??f?0,0?解:由定义f??0,0??lim?limx2?02?0xx?0x?0x?0x?0同理f??0,0??0y由于11?x2?y2?sin?f??0,0?x?f??0,0?y?x2?y2?sinx2?y2xyx2?y2lim?lim?0x?0x2?y2x?0x2?y2y?0y?0从而函数f?x,y?在原点?0,0?处可微。当x2?y2?01??1?1???3?f??x,y??2xsin?x2?y2cos???x2?y22?2x?xx2?y2x2?y2?2?1x1f??x,y??2xsin??cosxx2?y2x2?y2x2?y21x1由于f??x,0??2xsin??cos,limf??x,0?不存在,因此f?x,y?在点xxxx2x2x2x?0?0,0?处由于limf??x,y?不存在而不连续。xx?0y?0