2009大专A班数学分析第十章多元函数微分学自测题解答

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数学分析第十章多元函数微分学自测题解答

一、判断题 (对的记“√”,否则记“X”,每小题2分,共12分)

(×)1.若点(x,y)沿着无数多条平面曲线趋向于点(x0,y0)时, 函数f(x,y)都趋向于A, 则

limf(x,y)?A.

(x,y)?(x0,y0)注:

(x,y)?(x0,y0)limf(x,y)?A时,点(x,y)可沿着任意平面曲线趋向于点(x0,y0).

(×)2. 若两个偏导数fx?(x0,y0) , fy?(x0,y0)都存在,则f(x,y)在点( x0 , y0 )连续. (√)3. 函数f(x,y)在点(x0,y0)的邻域G存在两个偏导数,且两个偏导数在点(x0,y0)连续则

f(x,y)在点(x0,y0)可微.

(×)4. 若 limlimf(x,y)? limlimf(x,y)?A ,则

x?x0y?y0y?y0x?x0 lim(x,y)?(x0,y0)f(x,y)?A .

2注: 如 limlimx?0y?0xyx?y422? limlimxyx?y422y?0x?0?0 ,但 lim(x,y)?(0,0)xyx?y42 不存在.

??(x,y) , fyx??(x,y)在区域D内连续是这两(√)5. 函数z?f(x,y) 的两个二阶混合偏导数 fxy个二阶混合偏导数在区域D内相等的充分条件.

??(x0,y0) , (×)6.设函数z?f(x,y) 在点(x0,y0)的邻域G有二阶连续的偏导数,记A?fxx??(x0,y0) ,??B2?AC,则当??0时, z?f(x,y) 在 ??(x0,y0) ,C?fyyB?fxy点(x0,y0)取得极值.

注: 要求(x0,y0)是稳定点. 二、选择题 (每小题2分, 共14分)

1. 若f(x,y)在点(x0,y0)处可微,则f(x,y)在点(x0,y0)处下列结论不一定成立的是( C ). (A)连续; (B)偏导数存在; (C)偏导数连续; (D)切平面存在. ?4xy22,x?y?0?222. 函数f(x,y)??x?y 在原点(0,0)间断的原因是( B ).

?1, x2?y2?0?(A) 在原点无定义; (B) 在原点极限不存在;

(C) 在原点极限存在但无定义; (D) 在原点极限存在但极限值不等于函数值. 注: 由于 lim4xyx?y222x?0y?kx ? lim24kx2x?0x(1?k)?4k1?k2,故在原点极限不存在.

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?x2?2y2,(x,y)?(0,0)?3. 设f(x,y)??x?y,则fy?(0,0)?( C ).

?0, (x,y)?(0,0)?(A)0; (B)1; (C)2; (D)-1. 注: fy?(0,0)?limf(0,?y)?f(0,0)?y?lim2?y?y?2.

?y?0?y?04. 设u?ln(1?x?y2?z3),则(23?u?x??u?y?12?u?z)(1,1,1)?( D ).

32(A)3; (B); (C) ; (D).

5.设z?f(u,v,w), 而f具有一阶连续的偏导数, u??(x,y)的偏导数存在, v??(x),

w?F(y)均为可导函数,则

?z?x? ( A ).

??fv???(x); (B) fu??x??fv???(x)?fw?F?(y); (A) fu??x??fu?????(x)?fw?F?(y). ??fv???(x)?fw?; (D) fu??x?fv?(C) fu??xy注:

?z?x??z?u?u?x??f?v?v?x??fv???(x) ?fu??x6. 函数f(x,y)?2xy?3x2?2y2?10在点(0,0)处( B ).

(A)取得极小值; (B)取得极大值; (C)无极值; (D)不能判定是否取得极值.

??(x,y)??6,fxy??(x,y)?2,fyy??(x,y)??4, 注:fx?(x,y)?2y?6x,fy?(x,y)?2x?4y,fxx有fx?(0,0)?0,fy?(0,0)?0,???22?0,A??6?0,故在点(0,0)处取得极大值. 7. 函数f(x,y)?x?y?3x?3y的极值点为( C ).

3322(A)(0,0),(0,2); (B)(2,0),(2,2); (C)(0,0),(2,2); (D)(0,2),(2,0).

22???6(x?1),fxy???0,fyy???6(y?1),得稳定点(0,0), 注:由fx??3x?6x,fy??3y?6y,fxx(0,2),(2,0),(2,2);又???36(x?1)(y?1),在点(0,0),(2,2),均有??0. 三、填空题 (每小题2分,共16分)

1. 点集E?{(x,y)y?x2}是 无界开 区域(有界或无界;闭或开);

E的边界为 {(x,y)y?x} .

2 2

2. 函数f(x,y)?24x?y222ln(3?x?y)2?arcsin(3?x?y)的定义域是

222 {(x,y)2?x?y?3,4x?y} .

注:arcsin(3?x2?y2)的定义域是{(x,y)2?x2?y2?4};4x?y2的定义域

1ln(3?x?y)22是{(x,y)4x?y2?0};的定义域是{(x,y)x2?y2?3且x2?y2?2}.

3. lim4x?y1222(x,y)?(,0)2ln(1?x?y)xy? 2ln3?ln4 ; lim2?xy?4xyx?0y?0? ?14 .

4. 设f(x,y)?esin? y?(x?1)arctanx?yxy222??,则fx(1,1)?arctan2 ,fy(1,1)? ??e.

注: fx?(x,y)?yexysin? y?arctanx?yxy2?(x?1)[arctanx?yxy222]?x,

fy?(x,y)?xexysin? y??exycos? y?(x?1)[arctanx?yxy2]?y.

??故fx(1,1)?arctan2, fy(1,1)??ecos????e.

5. 设z?f(x,v),v?v(x,y),其中f,v具有二阶连续偏导数,则

?z?y22??2??? fvv(vy)?fvvyy .

注:

?z?y??f?v?v?yx?fv?(x,v)v?(x,y), y2

?z?y2222??(v??(fv?v?)??fvv)?fv?v??. yyyyy6. 设z?yln(xy), 则

?z?x

2

? y[lnyln(xy)?x2lnyx?1x2] ;

?z?y?xx2? yx?1[xlnyln(xy)?ln(xy)?lny?1] .

注: z?yln(xy),

?z?x?ylnyln(xy)?yxx1x,

?z?y?xyx?1ln(xy)?yx1y,

?z?x

2

2

?ylnx2yln(xy)?yxlnyx?yxlnyx?yx1x2?y[lnx2yln(xy)?2lnyx?1x2],

?z?y?x

2?yx?1lny[xln(xy)?1]?yx?1[ln(xy)?1]?y3

x?1[xlnyln(xy)?lny?ln(xy)?1].

7. 抛物面

在点M( x0 , y0 ,z0 )处的切平面方程和法线方程分别是

z?z0?2ax0(x?x0)?2by0(y?y0) z?z0?2ax0(x?x0)?2by0(y?y0) .

1128. 函数f(x,y)?e2x(x?y2?2y)的极小值点为 (,?1) ,极小值为 ?2e . 1注: 由fx?(x,y)?e2x(2x?2y2?4y?1),fy?(x,y)?e(2y?2),得稳定点(,?1).

22x四、解答题(每小题5分,共40分)

1?cos(xy)xsin(xy)1.求极限

(x,y)?(0,4)lim.

解:

(x,y)?(0,4)lim1?cos(xy)xsin(xy)?(x,y)?(0,4)lim1?cos(xy)(xy)2xysin(xy)y?12?1?4?2.

12. 求极限

(x,y)?(0,2)lim(1?x)1x(x?y)

11x?y1解:

(x,y)?(0,2)lim(1?x)2x(x?y)?xy(x,y)?(0,2)lim[(1?x)]x?e2.

3. 求极限lim(x?y).

x?0y?0222解: lim(x?y)x?0y?022xy?limex?0y?0xyln(x?y),

其中limxyln(x2?y2)?limx?0y?0xyx?y22x?0y?0(x?y)ln(x?y)?0,

2222故 lim(x?y)x?0y?022xy?e0?1.

注: limxyx?y2x?0y?022不存在,但

22xyx?y222?12,即

xyx?y22是有界量, 且

lim(x?y)ln(x?y)?limulnu?0.

x?0y?0u?04. 设z?ln(xy),求dz.

解: 由z?ln(xy)?xlnx?ylny,有

xy2xy22?z?x?lnx?1,

?z?y?2ylny?y,

故dz?(lnx?1)dx?y(2lny?1)dy.

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5. 设f(x,y)?x?y? xy,求fx?(0,0).

f(?x,0)?f(0,0)?x?lim?x?x?1.

解:由定义, fx?(0,0)?lim注: f(x,y)?x?y? 6. 求 u??3x2?y2??x?0?x?0xy是分段函数,不可以直接求导,要用定义求.

?u?x4x?2y的偏导数,

?u?y22.

解法一:u?(3x2?y2)4x?2y?e(4x?2y)ln(3x

?u?x?u?y?e(4x?2y)ln(3x?y)22?y),

6x3x?y2y3x?y2222[4ln(3x?y)?(4x?2y)22];

?e(4x?2y)ln(3x?y)22[2ln(3x?y)?(4x?2y)22].

解法二:由复合函数求导法则,令s?3x2?y2,t?4x?2y,有u?st,

?u?x??u?s?s?x2??u?t?t?x2?tst?16x?s(lns)4?s[6xttts?4lns]

?(3x?y)4x?2y[4ln(3x?y)?(4x?2y)226x3x?yts22];

?u?y??u?s?s?y2??u?t?t?y2?tst?12y?s(lns)2?s[2ytt?2lns]

?(3x?y)4x?2y[2ln(3x?y)?(4x?2y)222y3x?y22].

7. 设z?f(x?y,xy),其中f有连续偏导数,求

22?z?x , ?z?y.

解:设u?x?y,v?xy,有z?f(u,v).于是

?z?x?z?u?u?x?z?v?v?x?2xfu??yfv?,

22???z?y??z?u?u?y??z?v?v?y??2yfu??xfv?.

2228. 设z?f?2x?y,ysinx?, 其中f?u,v?有连续的二阶偏导数, 求

?z?x2,,. 2?x?y?y?z?z解: 设u?2x?y,v?ysinx,有z?f(u,v).

???fvu??,于是 由于f?u,v?有连续的二阶偏导数,故有fuv 5

?z?x??z?u?u?x??z?v?v?x?2fu??ycosxfv?,

?z?y??z?u?u?y??z?v?v?y??fu??sinxfv?.

?z?x22???ycosxfuv??)?ysinxfv??ycosx(2fvu???ycosxfvv??) ?[2fu??ycosxfv?]?x?2(2fuu???4ycosxfuv???ysinxfv??y2cos2xfvv??, ?4fuu?z?x?y2???sinxfuv??)?cosxfv??ycosx(?fvu???sinxfvv??) ?[2fu??ycosxfv?]?y?2(?fuu???(2sinx?ycosx)fuv???ycosxsinxfvv???cosxfv?, ??2fuu?z?y22???2sinxfuv???sin2xfvv??. ???sinxfuv??)?sinx(?fvu???sinxfvv??)?fuu?[?fu??sinxfv?]?y??(?fuu五 证明题 (每小题6分,共18分) 1. 证明 limxyx?y623x?0y?0不存在.

证:当动点(x,y)沿着x轴(y?0)趋向于点(0,0)时,有 limx?0y?0xyx?y3362 ?0,

当动点(x,y)沿着曲线y?x趋向于点(0,0)时,有 lim33xyx?y6x?03y?x ? lim2xx63332x?0x?(x)?12,

故limxyx?y62x?0y?0不存在.

xyx?y0,xyx?y22??2. 证明:函数f(x,y)????22,x?yx?y222?0,?0在原点(0,0)连续且偏导数存在,但不可微.

2证:(1) 由f(x,y)?f(0,0)??x, 有

???0,?????0,?(x,y):x??,y??且(x,y)?(0,0),都有

f(x,y)?f(0,0)??,

即函数f(x,y)原点(0,0)连续.

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(2) fx?(0,0)?limf(?x,0)?f(0,0)?x?x?0?lim0?x?x?0?0,

fy?(0,0)?limf(0,?y)?f(0,0)?y?y?0?lim0?y?y?0?0.

(3)假设f(x,y)在原点(0,0)可微, 则有 df?fx?(0,0)?x?fy?(0,0)?y?0, 又?f?f(0??x,0??y)?f(0,0)??x?y(?x)?(?y)22, ??(?x)?(?y),

22特别取?x??y,有lim?f?df?x?0?y??x??lim(?x)22?x?02(?x)22?12?0,与可微定义矛盾.

1?,?xysin223. 证明:函数f(x,y)??x?y?0,?x?yx?y2?0,?0在原点(0,0)可微.

2证: 由?f?f(0??x,0??y)?f(0,0)??x?ysin1(?x)?(?y)22, ??(?x)?(?y), 22有lim?f??0??lim?x?y(?x)?(?y)22?x?0sin1(?x)?(?y)22?0,

即?f?o(?)?0??x?0??y?o(?),故f(x,y)在原点(0,0)可微且df(0,0)?0.

注:函数f(x,y)在原点(0,0)的两个偏导数fx?(0,0),fy?(0,0)存在,不一定有f(x,y)在(0,0)可

微,即全微分df(0,0)不一定存在.故不能由fx?(0,0)?0,fy?(0,0)?0得到df(0,0)?0.

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