人民高等數(shù)學(xué)一第8章課后習(xí)題詳解

88-★1.

f(x,y)

x2y22

，求f )x解f(1,yx

x(yx

x2y2★2.已知函數(shù)f(uvwuwwuv，試求f(xyxyxy解：f(xyxyxyxy)xy★★3.zxyf(xy)y0zx2fx解y0

x2x0f(x0)

fx2★（1）zln(y22x解：要使表達(dá)式有意義，必須y22x1所求定義域?yàn)镈{(xy|y22x1x ★（2）x y解：要使表達(dá)式有意義，必須x 0，D{(x,y)|x yz★★（3）u zx2解

1

x2x2D{(x2x2

z

x2y2★★★（4）z

44xln(1x2y2解

4xy2 1xyln(1x2y2)0D{(x,y)|0x2y21,y2x1x2★★（5）zx1x2yx解：要使表達(dá)式有意義，必須x 1x2y2

D{(x,y)|x2y21,0xln(xey★（1）

x2知識(shí)點(diǎn))ln(xey 解： ln

x2 ★★（2）lim2

xy知識(shí)點(diǎn)思路xy(2 xyxy(2 xy

22 xy★★★（3）lim(x2y2)e(x (xy)2

(x

e解：e

x

lim

x

y)lim2x0,limy

eexxee

xy22

(xy)2ux

lim

lim2u

20

ex

u

u

u lim(x2y2)e(xy)★★（4）

x2x2解：方法一：（x2x2x22x212x2

0x2x2x2x2x2x2x2xx20 |y||y

，又lim|y|x2x2方法三：（極坐標(biāo)代換令xrcos,yrsin，則當(dāng)(xy00)時(shí)r002

limrcosr

limrcossinx2x2

x2x2y2 x2 (x2y2)3知識(shí)點(diǎn)思路00x2x2

u，則(x,y)00)u0limusinu洛必達(dá)lim1cosu

1

★★★（6）

1cos(x2y2)(x2y2)ex2y21cos(x2y2 1cos(x2y2

x2

1cos(x2y2解：

知識(shí)點(diǎn)

x2y2

1cosu

1 u思路：若(x,y沿不同曲線趨于(x0,y0x

(x,y)(0,0)x證：取ykx x

1

k(x,y)(0,0)x

x0(1

11★★★★（2）lim(1xy)xy證：方法一 1

lim(1xy)xylim(1xy)xyxylim[(1xy)xy]xy

，

x0(x若(x,yx軸趨于(00，則上式lim

0

lim(1xy)xye01111

若(x,yy

趨于(0,0)，則

1x

x0(x

y1從而lim(1xyxy方法二

1,

1n1lim(1xy)xy

n)2lim

1)n22ne0

1,

n

n

1 lim(1xy)xylim xyxy1★★★（3）

n

n(n1)解

xxy1xy1(xy)(xy1

x

若(x,yx軸趨于(00，則上式lim0若(x,yy

xx

趨于(0,0)，則上式 11)11)y注：若(x,yyx趨于(00，則lim(x

xy11)lim0xyxy1

x

(xy(xy)(xy1

yy2★（1）f(x,y)

y2解y22x0時(shí)函數(shù)無(wú)定義，故函數(shù)的間斷點(diǎn)集為{(xy|y2★★★（2）f(x,y)xyln(x2y2解：函數(shù)間斷點(diǎn)為(0,0) 由0|xyln(x2y2)| 2y2)ln(x2y2) 21 lim(x2y2)ln(x2y2

ux2

limulnu

lnu洛必

limu

u0u

u0故由定

limxyln(x2y20，故(00f(xy

ye

x0y任，

f(x,y在(00

y2ex2。知識(shí)點(diǎn)。

,x0y任xy

f(xyf(x0y0，則函數(shù)z

f(x,y在(x0,y0

(x0y0的存在性，若(x,y沿不同曲線趨于(x0,y0解：若(x,yx軸趨于(00

lim0x0y0y2ex2

若(x,yy

x2軸趨于(00

111 y0y2ex2

ye故limf(x,y不存在，從而函數(shù)f(x,y在(00偏偏 limf(x0x,y0)f(x0,y0 y f(x0,y0 (x,y 幾何意義：zf(x,y)的偏導(dǎo)數(shù)fx(x0,y0zf(x, 在點(diǎn)(x0y0z0y處的切線Txx limf(x0,y0y)f(x0,y0yx 0yz(x,y),f(x,y),f(x0,y0),f(x,y （1（2）zf(x,y的偏導(dǎo)數(shù)fx(xy),fy(xzf(x,y2 2xyyxDD在區(qū)域D8-★（1）zx3y3x2y2xy3知識(shí)點(diǎn)思路：函數(shù)對(duì)自變量x（y）求導(dǎo)時(shí)將另一自變量y（x）解：z3x2y6xy2

zx36x2y

x2z 解：z

x2y2x

1y

z1

z ；xx2xx2x22

x2

2x2

解：x

x2

(x2y2

；y

x2

(x2y2

z

1

1 解： (ln(xy))2 y

2 x

2x

2y

zsin(xy)cos2(xy)zcos(xy)y2cos(xy)(sin(xy))解y[cos(xy)

zcos(xy)x2x[cos(xy)

z(1xy)y知識(shí)點(diǎn)思路x（y）y（x）看為常量，按一元函數(shù)求導(dǎo)法則求導(dǎo)。在本題中對(duì)自變量x求偏導(dǎo)時(shí)，函數(shù)為xy求偏導(dǎo)時(shí)，函數(shù)為y的冪指函數(shù)。解：zy(1xy)y1(1xy)y(1xy)y1yy2(1xy) z(eln(1xy)y)(eyln(1xy)

eyln(1xy)(ln(1xy)

1(1xy)yln(1xy) xy 1xy （z5在方程兩邊同時(shí)取自然對(duì)數(shù)得lnzyln(1yzxy1zln(1xy)yzy

1z y

xy

zln

ln(1xy)1xyx；y解：z

tany

2csc2xy z

y

( (

)

(x)zuu知識(shí)點(diǎn)思路：函數(shù)對(duì)自變量x（yz）y，z（x，z或x，y）看為常量，按一元函數(shù)求導(dǎo)解：u

)z11z

z1 y

( yx z( x

x

zx；

x

(y)yz(y

) ( y

x(y

yxy★★2.設(shè)f(x,y)x(y1) ，求fx(x,1)xy1解：法一：1

x

fx(x,1)法二

fx(x,y)1(y

1x1xy x21

★★★3.

(x2y)f(x,y)

,x2y2

，fx(xy),fy(x ,x2y2知識(shí)點(diǎn)思路：分段函數(shù)分段點(diǎn)處偏導(dǎo)數(shù)用定義求；段點(diǎn)處應(yīng)用法則求導(dǎo)解：當(dāng)(x,y)(00)

(x)2 f(0x,0)f(0, m |x| f(0,0)

f(0,0y)f(0,0)

|y|lim

|y當(dāng)(xy00)時(shí)

x x2fx(x,y)2x x2

y) x2

x2 x(x2 2x x2

(x2y22

x21

yx2fy(x,y) x2

y) x2

x2 y(x2 x2 x2

(x2y2

x2z★★4.曲線 y

在點(diǎn)(245x知識(shí)點(diǎn)思路z

f(x,y的偏導(dǎo)數(shù)

zf(x,(x,y)表示空間曲 在點(diǎn)(x,y,z)處的切線T

y

于x軸的斜率，解：z2x

21tan

4

2

2

25.x2y2xy★（1）zx2yey解：z2xyey

zx2eyx22

y

2z y

)x2

)y

2x(12z

2y

y

y

2y

y (x xye

x x x x

zarctanyx解：z

y)

z

1 x2

(y

x2 2z

)

； xx2 (x2y22

(x2y2)y2 y2 )

yx2

(x2y2

(x2y2 2 (

) 2)yx (xy★★★（3）zyx解：zyxln

z2

2z

(

lny)

(lny)

y

)x(x1)2

(11

y

(xlny★★6.f(xyzxy2yz2zx2，求

解：

fxz2x

f2xyz2

fyz2zy f2yzx2, y

fzxx所以

z(1,0,2)2,fyz(0,1,0)0，fzzx(2,0,1)2 2 2★★★7.設(shè)z (xy)，其中(u)可導(dǎo)，證明

x

xyy2證 2左邊x22

2

22

(xy)

x3

右邊xy2y(xy)x)2y2x2y(xy3 注：本題中對(duì)抽象函數(shù)(xy)★★8.zxln(xy

3

3 解zln(xyx

yln(xy) 2z

1y

3 0

2z1

3z

x y

如果函zf(xy在點(diǎn)(xy的全增量zf(xxyyf(xy為zAxByo()，其中A,B與x,y無(wú)關(guān)， (x)2(y)2，則稱(chēng)函數(shù)在點(diǎn)(x,y)可微，全微分dzAxBy。（1）zf(x,y在(x,yzf(x,y在(x,y（2）zf(xy在(xy可微，則limzdz0；從而若limzdz0 zf(x,y在(x,y（3）zf(x,y在(x,yzf(x,y在(x,ydzzdxz （4）zf(x,y在(x,yzz在(x,y (xydzzdxz zf(x,y在(x,y的某鄰域內(nèi)偏導(dǎo)數(shù)fx，fy在(x,y連續(xù)，且|x|,|y|都比較小時(shí)，有全增量近似公式zdzfx(x,y)xfy(x,y)y函數(shù)值近似公式f(xxyyf(xyfx(xy)xfy(x習(xí)題8- ★（1）z3xy y知識(shí)點(diǎn)思路

dzzdxzdy解：z6xy1 z3x2

所以dz6xy1)dx3x2x

zsin(xcosy)解：zcos(xcosycos

zcos(xcosy)(xsin所以dzcos(xcosycosydxxsinycos(xcos★★★（3）uxyz解

yxyzln所 ★★2.zln(2x2y2x2,y1解

x22y2x22y2x22所以dz4dx2 xzy★★★3.設(shè)f(x,y,z) ，求dfxzyf11 f

x解：fx

) ) ()2z yz z yz21122f ln(y故fx(1,1,1

從而dzdx★★4.zyx2y1x0.1y0.2時(shí)的全增量zdzx解：zyyy dzyx1 將x2y1x0.1y0.2代入得全增量z10.2)1

dz10.110.2)

2

知識(shí)點(diǎn)x3x3

f(xx,yy)f(x,y)fx(x,y)xfy(x,解

f(x,y)

x1.02,y1.97似值。

x1,y2,x0.02,y fx(x,

3y2fy(x, 2x3

2x3(x(xx)3(yx3

3y2

x 2x3 2x3所 0.02 ★★6.計(jì)算(1.007)298知識(shí)點(diǎn)

213

213思路

f(xx,yy)f(x,y)fx(x,y)xfy(x,解：設(shè)f(xyxyx1.007,y2.98取x1y3x0.007y 又f(xyyxy1,f(xyxyln

f(1,31fx(1,33fy(1,3所以(1.007)29810.007)300213(0.0070(0.02)★★7.x6my8mx2cmy減少5cm知識(shí)點(diǎn)思路

zdzzxz x2x2 則有zdzxxyy其中z

zx2x2

，x6,y8,x0.02,yx2所以zdz60.02)80.05x2 2.8cm★★8.5m4m，高3m20cm，求所需材料的近解xyz，則長(zhǎng)方體體積為Vxyz，從而所需材料的精確值為由題意可知x5y4z3x0.4y0.4z精確值V5434.63.62.813.632m2★★9.IVRRI

I=20A，0.5AR解 dRRdVRdI1dVV I其中V110I20,12,20.51,2 |V |R||dR||IdV||I2dI||I|1

I2121100.50.2375又RI

1105.5

0.240.044RR從而R0.244.4復(fù)合函數(shù)的中間變量均為一元函數(shù)如果函數(shù)uu(tvv(t)在點(diǎn)tzf(uv在對(duì)應(yīng)點(diǎn)(uvzf(u(tv(t))tdzzduz u v復(fù)合函數(shù)中間變量為多如果函數(shù)uu(x,yvv(x,y在點(diǎn)(x,yzf(uv(uvzf[u(x,yv(x,y在對(duì)應(yīng)點(diǎn)(x,y z z z zxuxvxyuyv復(fù)合函數(shù)中間變量既有一元函數(shù)又有多元函數(shù)如果函數(shù)uu(x,y及在點(diǎn)(x,yvvyy點(diǎn)可導(dǎo)，zf(uv)在對(duì)應(yīng)點(diǎn)(uv)出具有連續(xù)偏導(dǎo)數(shù)，則復(fù)合函數(shù)zf[u(x,yvy在對(duì)應(yīng)點(diǎn)(x,y z z zxuxyuyv注：若zf(xyuuu(xy，則zf(xyu(xzffu；zff u ux為fxzf(x,yuy,uxzf(xyu(xyxy 區(qū)別同上 習(xí)題8-★★1.zyxety1e2tdz dt解 dzzdxz x yy

1

1)

1(2e2t(etet

★★2.zex2yxsintyt3dzdt解：dzzdxz x yex2ycostex2y(2)3tesint2t3(cost6t2★★3.zu2v2，而uxy,vxy，求zx解：zzuz zzuz u

v

u

v2u12v2(xy)2(xy)★★4.zx2y2xyz

2u12v2(xy)2(xy)4x解：令ux2y2,vxyzuv,ux2y2vxyzzuz zzuz u

v

u

vvuv12xuvlnu vuv12yuvlnu

2

)xy

2x2x2

y

2y2

2

)xy

x2

x

2y2elnux)vx)evelnux)vx)evxlnux)計(jì)算或用對(duì)數(shù)★★5.zarctan(xyyexdzdx解：dzzz

（zzxzarctan(xyy看為常量 y 1

1

ex(1x)16.（f具有一階連續(xù)偏導(dǎo)數(shù)

uf(x2y2,解sx2y2,txy，則原函數(shù)為uf(s,tsx2y2,txy復(fù)合而成的函數(shù)，按多u

s

2xfyf s t u

s

2yfxf s t

xuf uy解sxty，則原函數(shù)為ufx,ysxty y u

s

t

u

s

f1f s

t

fs0y

s

t zufs

t0yfyf s

t z2

z2

uf(x,xy,知識(shí)點(diǎn)思路x解sxy,txyz，則函數(shù)為uf(xstsxytxyz復(fù)合而成，按復(fù)合函數(shù)求導(dǎo)u

fs

（其中ffx的導(dǎo)數(shù)

s

t fyf u

0

s

xfxzf

s

t u

0

0

txyf t ★★★7.z

f1z1zzf(x2y2 x y 知識(shí)點(diǎn)

u uv思路：本題為抽象函數(shù)f

y的復(fù)合函數(shù)，故要用商式求導(dǎo)法則()

證：令ux2y2

zyf(u)2x2xyf(u) f2 f2zf(u)yf(u)(2y)

f(u)2y2f f2

f2

1z1z12xyf(u)1f(u)2y2f(u)2xy2f(u)xf(u)2xy2fx y f2

f2

xyf2 yf

yf(x2y2

zy2f(x2y2

★★★8.設(shè)uf(xyzx

z，其中f有二階連續(xù)偏導(dǎo)數(shù)，求ux2y2解：令sxyz,tx2y2z2uf(s,tsxyz,tx2y2z2復(fù)合而成的函數(shù)，由求導(dǎo)法則有u

s

t

f2xf

s

f2yf s

t

s

t u

s

t

f2zf s t 函數(shù)f仍為f(s,tsxyz,tx2y2z2st為中間變量以xyz為自變量，且由f有二階連續(xù)偏導(dǎo)數(shù) t fss2xfst2ft2x(fts2xftt) f4xf4x2f 2u

4yfst

4y2

2

，

4zfst

4z2

2

u

3

4(xyz)fst4(xyz)ftt6★★9.z

2f(2xy,ysinx，其中f解：令u2xyvysinx則函數(shù)為uf(uvu2xyvysinx復(fù)合而成，按z

u

v2fycosxf u v 由f(uv為u,v的函數(shù)，所以fu,fv仍為以u(píng),vxy 2(u u2z(2f 2(u u

fufv

fuf

u v

u v)f ycosx(v v))f ycosx(v v)

（2 2 210.x2xyy2（其中f具有二階連續(xù)偏導(dǎo)數(shù)

zf(xy,解：令uxy，則函數(shù)為uf(uyuxyyz

u

0yf

u

xff

（

u

u y的偏導(dǎo)數(shù)，求解時(shí)將中間變量u看作常量f(uy為uyfu,fy仍為以u(píng)yxy 2z(yf

fuf

y2yy(u uy(u y(u uy(u u)

(yf)

fuf

u

2

(xfuf

f

f

fy

f x(u u)

u

u x2fxfxffx2f ★★（2）z

f( ,xy)uyvx2yux

f(uvuyvx2yxz

u

vyf2xyf

u

v1fx2f u

v

u

v x 由f(uv為u,v的函數(shù)，所以fu,fv仍為以u(píng),vxy (yf2xyf

z

y

fuu

fuv)2

f

fvu

fvv)2yf u

v

u

v

y 2

f2xy(fyf2xy)2yf

2yf4x2y2f2yf（f具連續(xù)二階偏導(dǎo)數(shù)f

(yf2xyf

z

y

fuu

fuv)

f

fvu

fvv)2xf

u

v x2

u

v y(f1fx2)1f2xy(f1fx2)2xfuu

x2 vu

yfyf1f2x3yf2xf（fff x2

1 x2f1z

(x

1

x2

xu

v

u

v fx2)x2(f1fx2 vu 1f2xfx4f（fffx2

★★★11.zf(x,yxeucosvyeusinv2z2z

2(

2)

知識(shí)點(diǎn)思路：在本題中將函數(shù)看為u,vxy是中間變量的角色。按鏈?zhǔn)椒▌t對(duì)自變量u,v求導(dǎo)即得證：zfeucosvfeusin zfeusinvfeucosv 2z

(

(fyesineucosv( e2ucos2

eusinv)feucosveusinv(feucosvfeusinv)feusin e2usin2vffeucosv x2zx

f

(f

yeusinv(ye2usin2

eucosv)feucosveucosv(f(eusinv)feucosv)feusin e2ucos2vffeucosv （fff）

2z2z

f

2

2

2u(e2uf

2

2故左邊

)

fyyx2y2右邊，得證★★★12.設(shè)ux(xyy(xy)，其中函數(shù),

0 證vxy，u(v)x(v)y(v)

ux(v)(v)y(v)2u

2

2ux

2

0 一個(gè)方程 若二元方程F(x,y)0確定一元隱函數(shù)yf(x)， F(x,yz)0zf(x,yzFx z 方程組情F(x,y,u,v)若方程組 確定二元函數(shù)uu(x,y),vv(x,G(x,y,u,v)FxFuFyFu則GxGvFuGuFu，GyFu,GuFuGuGuGuGu8-★★1.已知

arctanydyx2x2知識(shí)點(diǎn)思路F(xyFFdyFx解

F(x,y)

x2x2x

F(x,y) 2x2

(1y(yxy

x ) x2F(x,y) 2

1

yy所 dy

2x2yy

1(x

x2注★★2.設(shè)x2yz

0zx解方法一;（zFx

zyFz設(shè)F(x,y,z)x2yz F

F

F 故zFx

xyz

yz xyz；zyz xyz

xz xyzxz xyzx求偏導(dǎo)，得： xy 1z

x)0，整理可得

)z z

xyzyz xyzyz xyzyxy2z

y0，整理可得

)z 故z

xyzxz xyzxz xyz★★★3.z

f(x,y)由方程F(xz,yz)0所確定，證 xzyzzxy 設(shè)G(xyz)F(x

z,yz

ux

z,vyz 則GF1F

z)F

zF；

F

z)F1

FF；

GF1F11F1F y x x2yFzyF

xzFxy2F x v；

vvvv

xFuyF

xFuyFx

y

x2yFzyFxzFxy2F xFyF v z(xFuyF)xy(xFyF)zxyxFuv 方法二xyzxy令uxzvyzx 1

1

2yFzyFyy

F x

vFuyFv xy

y

xzFxy2F vx

y

xFyFx2yFzyFxzFxy2F xFyF 故 z(xFuyF)xy(xFyF)z vxFuyFv方法三：（dzzdxzdy及全微分形式的不變性

dF(x

z,yz

)d故Fd(xzFdyz F(dxydzzdy)Fd(yxdzzdx)

(FuFv)dz(FzFv)dx

zFu

x2yF xy2Fdz ux2F xyF x2yF

xy2F

v

u x2F xyF x

y

x2yFzyFxzFxy2F xFyF 故 z(xFuyF)xy(xFyF)z vxFuyFv★★★4.x2y2z2

zyf()，其中f可導(dǎo)， x解：方法一

F(x,y,z)x2y2z2yf

其中uy則F F2y(f(u)yf(u)(z))2yf(u)

f F2zyf(u)12zf 故zFx

2z f(y2y

f()

( 2y

yf()zf( (y (

2zf(z 2yz 方法二xyzxy的函數(shù)。x求偏導(dǎo)得：2x2zz

z1

z yf(y

y

()2zy2y2zz

z)1zf(

yf(y

y(2z

f( f() f()2 z

f()

()2y z

yf()zf( 2zf( 2yzyf( ★★★5.設(shè)(uv(cxazcybz)0z

f(x, 滿足axbyc證：在方程(cxazcybz)0x(caz)(bz)

z

aby(az)(cbz)

vz cv

ab azbz

bc

c(a c(a v abv

ab

ab 2z2★★6.

2xzy0，求x2解：方法一：（用隱函數(shù)求偏導(dǎo)公式設(shè)F(xyzz32xzy

F F F3z2 3z2

3z2 2

（z

的函數(shù)

(3z22x

2

2

z

2 z(3z22(3z22x) 4z 3z2

(3z2(3z2 (3z22z

(3z22x)y16z y(3z2

（zxy的函數(shù)(3z2x

z

3z2方程（1）x222z2

4z6z(z

3z2

(3z2y3z2z2xz1

（2）

z 3z2方程（2）x6z(z

22 3z2 6z(z

2 2故

3z2

(3z2

2★★7.設(shè)

xz

1

解：設(shè)F(xyzz5xz4yz31，則有Fz4 Fz3, F5z44xz3 故zFx

4 5z44xz3 4

5z44xz3 2z

(5z44xz33yz2)4z3z(5z44xz33yz2)z4(20z3z12xz2z3z26yzz

(5z44xz33yz2 5

6yz)y

(4xz6

)5z44xz33yz2 (5z44xz33yz2

(5z44xz33yz215z1012xz99(5z44xz33yz2

2 又x0,y0時(shí)z1xyz dx,

★★8.設(shè)x2y2z21dz解：方法一dxdy1

xx(z),yy(z)zdxdy

dx2ydy

2z

dx2ydy

2z2當(dāng) 12(y 2z2

dx

zy，dy

x 2x 2x

2(y

y

2(y

y

xx(z),yy(z)dxdydz 2xdx2ydy2zdz dxdy(1y2)dydxzyy

即dxzy y

dyxz y

dz★★9.設(shè)

2z

， dx1dydz2zdz

zz(x),yy(x)x

dx(12z)dx 3z2 0

2

1當(dāng)2y1

0 d

1111z2y1 2y 12y1dy

2y13z24yz2y注8

uuv

yeuucosv

x,y,x,解：方法一

uu(x,yvv(x,y)x1euu

(eusinv)uucosvv

0eu

(eucos usin 故u

1ucos1ucos usineusinv eucos usinv

cosveusin eucos eusin eucos eusinv eucos usin

u

cos

;v

sinv。 eu(sinvcosv)1 ueu(sinvcosv)方法二：（duudxudydudvdx 表示，則表示系數(shù)即為所求

uu(x,yvv(x,y)dyeuducosvduusin把dudv(1sinv2)cos

dvdu

sinvdxcosvdyeu(sinvcosv)1

u

sin

cos eu(sinvcosv) eu(sinvcosv)v

cosv

v

sinv同理可得： ueu(sinvcosv)u； ueu(sinvcosv)x

d2yy[(x1)2(y1)2★★★11.

xy

x2(y證：方程兩邊同時(shí)取對(duì) xylnxln設(shè)F(xyxylnxlny

F11,F1 dyFxy(1

x(y (dy(1x)y)x(y1)y(1x)(y1xdydy(y(1x))

x(y1)

x2(yx(x1)dyy(y

x(x1)y(1x)y(y x(y x2(yy[(x1)2(y1)2x2(y

x2(y注§8.6微分法在幾何上的應(yīng)用空間曲線的切線與（1）xx(t),yy(tzz(t)零。則曲線在某點(diǎn)t0處的切向量為T(mén)x(t0),y(t0z(t0x y z記x(t)x,y(t)y,z(t)z，則切線方程 0 0 x(t y(t z(t y（2）zz(x)，yy(xzz(x)x0可導(dǎo)，則曲線在某點(diǎn)(x0y0z0處的切向量為T(mén)1,y(x0z(x0x y z則切線方 0 0 y(x0 z(x0F(x,y,z) ，F(xiàn),G具連續(xù)偏導(dǎo)數(shù)，則曲線在點(diǎn)M0(x0,y0,z0G(x,y,z) 處切向量為T(mén) }， x y z則切線方程 法平面方程 Gx (xx) (yy) (zz) z x y（1）F(x,yz)0M0(x0y0z0nFx(x0,y0z0Fy(x0,y0z0Fz(x0,y0z0)}Fx(x0,y0,z0)(xx0)Fy(x0,y0,z0)(yy0)Fz(x0,y0,z0)(zz0)x y z法線方程 Fx(x0,y0,z0 Fy(x0,y0,z0 Fz(x0,y0,z0空間曲面（2）zf(xyP0(x0y0n{fx(x0,y0),fy(x0,y0切平面方程 fx(x0,y0)(xx0)fy(x0,y0)(yy0)(zz0)0fx(x0,y0)(xx0)fy(x0,y0)(yy0)z（上式表明函數(shù)zf(xy)在點(diǎn)(x0y0)處的全微分，在幾何上表示曲面zf(x,y在點(diǎn)(x0y0x y z法線方程 fx(x0,y0 fy(x0,y0 的切平面8-★1.x

y1tzt2在t21 解 x(t)

(1

,y(t)1,z(t)2tt2又t2時(shí)x2y3z4x(2)1,y(2)1,z(t 故曲線在t2

{ ,T{x(2),y(2),z(t)}{ ,9x y

3 2z 法平面方程為1(x21y34(z4) ★★2.y22mxz2mx在點(diǎn)(xy 解：由題設(shè)可 2yy(x)2m,2zz(x)1，

y(x)m,z(x) 故曲線在點(diǎn)(xyz

T{1,y(x),z(x)}{1,m,1

xx0yy0z

(xx)m(yy)1(zz) x2y2z2★★★3.求曲線x2y2 在點(diǎn)M0(0,0,a)處的切線方程與法平面方程解：設(shè)F(xyzx2y2z2a2G(xyzx2y2 Fx|M2x|M0Fy|M2y|M0Fz|M2 Gx|M(2xa)|Ma,Gy|M2y|M0,Gz|M

02a0,

2a

2a2,

0zy 0zy

M0M

0

ayxyxM0(00a處的切向量為T(mén){02a2z故所求切線方程為x0，法平面方程為2a2y00yz★★4．xtyt2zt3x2yz4知識(shí)點(diǎn)：空間曲線的切線Tx(t),y(tz(t)}解M(x0y0z0，其對(duì)應(yīng)參數(shù)t x(t1y(t2tz(t3t2，故該點(diǎn)的切線向量為T(mén)12t,3t nx2yz4的法向量為n1n故nT

即14t3t2

解得：t1或t 從而所求點(diǎn)為M(1,11或M(1113 ★★5．x2y2z21xy2z0思路解：設(shè)F(xyzx2y2z21，則F2xF2yF M(x0y0z0，曲面在點(diǎn)Mn2x02y0故2x02y02z0

x2y2z2 11,

6所以切點(diǎn)為 6

)1(x

1)1(y1)2 2)

xy2z6 ★6.zx2y2在點(diǎn)(1,12)解

f(xyx2y2

n{fx,fy,1}{2x,2y,

n{2,2,故切平面方程為2(x12y11(z20即2x2yz4

x1y1z ★★★7.F(nxlznymz)0x1y2z

其中F知識(shí)點(diǎn)證：設(shè)G(xyzF(nxlznymzunxlzvnymz，則GxnFuGynFvGzlFumFv

故nsF(nxlznymz)0★★★8.xyza3（a0，常數(shù)）上任意點(diǎn)處的切平面與三個(gè)坐標(biāo)面所形成的四面體的體知識(shí)點(diǎn)思路證：設(shè)F(xyzxyza3，則FyzFxzF M(x0y0z0，則曲面在該點(diǎn)的法向量為n{y0z0,x0z0,x0

x0)x0z0(yy0)x0y0(zz0)

1故四面體體積為V1|3x3y3z|27

xy

a3

00§8.7方向?qū)?shù)與梯度zzf(x,y)P(x,y)某領(lǐng)域有定義l為自點(diǎn)P出發(fā)的射線，記(x)2y)2，函數(shù)f(x,y)Pl函數(shù)函數(shù)uf(x,yz在(x,yz)zf(x,y在(x,y其中coscoscos為方向lfcos cos l2.函數(shù)uf(x,yzM(x,yzffcosfsinx 向l1.zf(x,yP(x,y)flimf(xx,yy)方向xxy gradf(x,y,z)fifjfk{f,f,f8-0★★1.求函數(shù)uln(xy2z2M(0,12l2110 666：向量l的方向余弦為cos ,cos ,cos666又u ,u

2 ,u x gradf(x,yx gradf(x,y)fifj{f,f

xy2z2 xy2z2u

1,u

2,u x

5y

5z 26 265故l|M05 55

) ( )

（與答案不同★★2.zln(xyy24x(12x解y24xx2ydy4(12y24xkdy

dy ，從而方向的傾角 dx 1x1x 11x1x

z[zcoszsin

1

21 2

★★3.求函數(shù)uxyyzxzP(123P 解：向徑OP{1,2,3}，其方向余弦為cos ,cos ,cosuyz,uxz,uy

123u 4 123★★★4ux2y2z2xtyt2zt3上點(diǎn)(1,1,1)沿曲線在該點(diǎn)的切線正方向解：當(dāng)t1時(shí)，曲線上點(diǎn)為xtyt2zt3的切向量為T(mén){x(ty(tz(t1

1cos1

,cos

,cos2u 2x 2,2x 12 u 12★5.設(shè)f(xyzx23y25z22xy4y

解 f2x f6y2x f(0,0,0) f(0,0, ★★★6.x0A(xy){2xy(x4y2為某二元函數(shù)u(x,y的梯度，其中u(x,y解：

gradu{u,u}x故 又u(x,yxy即2x(x4y24xy2

4x(x4y21)0x0

★★7.求函數(shù)ux2y2z2M(10,1M(0,10 解 u

u2

ugradu(M1{202},gradu(M1)gradu(M2，從而兩梯度的夾角為2(xa)2(y(xa)2(yb)2(zr

，試在空間哪些點(diǎn)處等式|gradu|1(xa)2(y(xa)2(yb)2(z 故u1 2(x

x，

2(xa)2(yb)2(z

uyb，uz r2 ruu x y z所以gradu{ }

, , r2(xa)2(yb)2(z 若|gradu|1r1，即在空間球面(xa)2yb)2(zc)21上|gradu|1§8.8值函數(shù)zf(xy)在點(diǎn)(x0y0某領(lǐng)域內(nèi)有定義，對(duì)領(lǐng)域內(nèi)任一異于(x0,y0的點(diǎn)(x,y)，如f(x,y)f(x0,y0)(f(x,y)f(x0,則稱(chēng)函數(shù)在點(diǎn)(x0y01.（必要條件）zf(x,y在點(diǎn)(x0y0處具有連續(xù)偏導(dǎo)數(shù)，且在點(diǎn)(x0,y0有極值，則必有fx(x0y00,fy(x0y00（可推廣至多元函數(shù)2.（充分條件）zf(x,y在點(diǎn)(x0y0處具有二階連續(xù)fx(x0,y0)0,fy(x0,y0)0令fxx(x0y0A,fxy(x0y0B,fyy(x0y0C，則（1）ACB20(xyA A0ACB20時(shí)，函數(shù)在(xy ACB20求函數(shù)uf(x,yz在條件(x,yz)0方法一：化為無(wú)條件極值。即在方程(x,yz)0zz(x,y，代入目標(biāo)函數(shù)，按L(x,yz)f(x,yz(x,y (x,y,z)Lf(x,y,z)(x,y,z) 由Lf(xyz(xyz)0解出可能極值點(diǎn)(x0y0z0 如：函數(shù)uf(x,yz在條件(x,yz)0(x,yz)0L(x,yz)f(x,yz(x,yz(x,y8-★★1.求函數(shù)f(xyx3y33xy知識(shí)點(diǎn)思路：解方程組fx(x,y)0,fy(x,y)0A,B,CACB2解

3x23y fy3y23x 由(1yx2，代入(2

x(x31)0，故

x0,x故有兩駐點(diǎn)(0又 y3,fyy6y駐點(diǎn)(00A0B3C0ACB290，故(00不是極值點(diǎn)(1,1，A6B3C6ACB2270A60，所以函數(shù)在點(diǎn)(1,1處取得極大值1（與習(xí)題答案不同★★2.求函數(shù)f(xy)x2y222(x2y2x x2y2)2x4x4x(x2y21) x解：解方程組fy2(x2y2)2y4y4y(x2y21) 由（2）

y0，代入（1）

x0,或x1，故有駐點(diǎn)(10),(00),(1

8xy, 4(x23y2對(duì)(10)A8B0C8ACB2640，且得極小值1，

A80，所以函數(shù)在點(diǎn)(10)(10也取得極小值1（xy均為偶函數(shù)對(duì)(00A4B0C4ACB2160，所以(00★★3.ye2x(xy22y解

(2x2y24y1) fye(2y2) 2由（2） y1，代入（1）得2x2410，x1，故駐點(diǎn)為(1, 2y又2y故A4eB0C2eACB28e20A4e0，所以函數(shù)在點(diǎn)112極小值12★★4.求函數(shù)f(xysinxcosycos(xy0xy的極值2解

fxcosxsin(xy) ff

sinysin(xy) +（2）并代入（1）得cosxsinysin(xy0x,y2

3f

sin(xy),fyycosycos(x33233對(duì) )，A3 3

3,B 32

,C ，AC2

30，且A 0，所以函數(shù)★★★5．x2y2z22x2y4z100z

f(x,y知識(shí)點(diǎn)思路：先按隱函數(shù)求導(dǎo)法則求出函數(shù)偏導(dǎo)數(shù)，然后解方程組fx(x,y)0,fy(x,y)0得出函數(shù)的駐A,B,CACB2符號(hào)判定是否為極值點(diǎn)。解：在方程兩邊同時(shí)對(duì)x2x2zz24z y2y2zz24z

z1 zz1 zz1x z解方程組

1

得駐點(diǎn)(11x1y1z2z z(z2)2(1

(1x)(1

(z2)2(1又

(z

;zyy

(z故z2

A1B0C1 ACB210A10(11取得極小值

z6時(shí)A1,B0,C1 ACB210，又A10，故函數(shù)在點(diǎn)(1, （

(x1)2y1)2z2)216（以(1124為半徑的球面16(x1)2(16(x1)2(y16(x1)2(y16(x1)2(y

z246z242★★6.60105方法一知識(shí)點(diǎn)面積解xy米，則問(wèn)題歸結(jié)為在約束條件xy60f(xy)15x10y

L(x,y,)15x10y(xyLx15y0L 10x0L

y32xy 代入（3）

x

，y

，寬為

方法二：（一元函數(shù)最值x60f

15x10x

(x0)f(x15600

x又f(2101200

x

★★72p的矩形繞它的一邊旋轉(zhuǎn)構(gòu)成一個(gè)圓柱體，問(wèn)矩形的邊長(zhǎng)各為多少時(shí)，才能使圓柱體的解xpxx高為px，從而體積為Vx2(p (xxV(2px3x2x

，令

(2px3x20

x23又

(2p6x)| x

x2p2p1p 4最大體 p3注：本題也可采用二元函數(shù)條件極值解決，其中目標(biāo)函數(shù)為圓柱體體積，約束條件為周長(zhǎng)(2p)★★★8.zx2y2xyz1知識(shí)點(diǎn)解：設(shè)橢圓上點(diǎn)的坐標(biāo)為(x,yz)d

x2x2y2d2x2y2z2，其中zx2y2xyz1（約束條件

L(x,y,z,,)x2y2z2(zx2y2)(xyzLx2x2x0Ly 2y2y0LyLz2z

（1）-（2）得(xy)(1)

zx2

xyz 若1，帶回（1）得0，由（3）可得z10，這與 2故yx，由（4）

z2x2，代入（5）

2x22x1

x2

y2

3,z3由問(wèn)題本身的意義知3

3,

3,2

3

3,

3,2

3 95大值點(diǎn)。因?yàn)閐2 595

，最長(zhǎng)距離 95★★9.AB，10995品B

4002x3y0.01(3x2xy3y2)（元解：利潤(rùn)函數(shù)L(xy10x9y[4002x3y0.01(3x2xy

(2)61y80xLL

60.01x0.06y0 又 A0.060x120y80為極大值點(diǎn)，由問(wèn)題的實(shí)際意義知，也為最大值點(diǎn)。120件產(chǎn)品A，80件產(chǎn)品B★★10.p，其溶解溫度(C為p與(p與之間的經(jīng)驗(yàn)公式apb apibpii解：方程組

api6b

6

pi

28365.28a396.6b代入方程組得pp

396.6a6b

i 66解得：a

b1230057.66 所以經(jīng)驗(yàn)公式為2.234p95.35★★★11.已知一組實(shí)驗(yàn)數(shù)據(jù)為(x,y),(x,y (x,y)。現(xiàn)若假定經(jīng)驗(yàn)公式是yax2bxc abc解：設(shè)MnM[y(ax2bxc)]2M(a,b,n

[y(ax2

c)]2M

nnn n

[yi

bxic)]

[y(ax2

c)]2

[yx2ax4bx3cx2]i i

[yixi

[

ax2

c]

x4

x3

x2

yxnnn innn

2 axi xicxiyi

x bxnc

★★12.25A1000B32000A解xyR2x5yx4,3y9,1000x2000y

MaxR2x5xyx2yx0,yyyDCxBx2yyO0Axx4y3x2y8x0y0O(00A(40B(42C(23D(03，且R(000R(40)8,R(42)R(2,3)19，R(0,3)由線性規(guī)劃問(wèn)題最優(yōu)解在區(qū)域邊界處取得，R(231923時(shí)獲(x2y2a2)(2a(x2y2a2)(2a2x2y2解

(x2y2a22a2x2y2

a2x2y2D{(x,y)|a2x2y2★★（1）知識(shí)點(diǎn)

1)xx思路：由題目知該函數(shù)為1型的冪指函數(shù)未定型，可根據(jù)重要極限lim(11)xe 解：原式

)xy

)x]x

lim(11)x

xx

2x1)xye1ex

xx2xyx解 0

x ||xy

1x2xy

|x |y

x又

0,|x|y|x|yx2

0， 定理

lim 2xxxxxy★★★3.

x4x知識(shí)點(diǎn)思路：二重極限中(xyx0y0是指以任意方式趨于該點(diǎn)，若(x,y沿不同曲線趨于(x0y0時(shí)， x2 x2 解

y

，則

k

1

2 k(xy)cos1,(x,y)(0, f(xy

,(x,y)(0,

在點(diǎn)(00知識(shí)點(diǎn)思路x解 lim(xy) cosx

xlim(xy)cos10f(0,x

函數(shù)在點(diǎn)(005.

zxyet20知識(shí)點(diǎn)思路f思路f f( (x)aa

解：ze(xy)2yyex2y2

ze(xy)2xxex2

u

z(x

z(x

(xy)zln|xy解：

1(xy)2z

1(x

z 1(x★★★6.r

x2x2y2

2r2r xx2y2z2xxxyz

(x

z2r x2

x2， (x2y2z2 (x2y2z22r2r2r

y2

x2 x2

(x2y2z22(x2y2z2

(x2y2z2

(x2y2z2 (x2y2z2z

x2y2

★★7.

u 3x231( x21( x2z(x2(x2y2)x2y2

(x2y2xy

u(x2(x2y2)x2y2

11 x211 x2x2x2y2dzudxudyu (x2y2)x2y2

dx

((x2y2)x2y2

x2x2y2★★★8.求uxyyzzx解：uyxy1yzzxxyyzzxlnzxyyzzx(yln xyzuxyyzzx(zln

uxyyzzx(xln dzudxudyudzxyyzzx[(ylnz)dx(zlnx)dy(xln

方法二lnuylnxzlnyxlnx求偏導(dǎo)（注意uxyz的函數(shù)）uyxyyzzx(yuyu yzx yzx

yxy

lnx),

zxy

lny)dzudxudyudzxyyzzx[(ylnz)dx(zlnx)dy(xln

★★9.

z(x2y2

arctanx

2y y 解 x(x2y2 x 1

arctan) x(2x2 arctan

(x1yarctany

arctan2 x(x2y2

x( )

x(2ydz

arctanx[(2xy)dx(2y

1(x12

arctan

yx2

x2

)(1y(xx2 ,xy★★★10.設(shè)f(xyx2

，求

x(x,y),

(x,y解：當(dāng)(x,y)(00)

x2y2 f(0x,0)f(0,

00f(0,0)

f(0,0y)f(0,0)

0 y0當(dāng)(xy00)時(shí)fx(x,y)

2xy(x2y2)x2y2x(x2y2)2

(x2y2 x2(x2y2)x2y2 x2(x2y2fy(x,y)

(x2y2

(x2y2 x2(x2y2 f(x,y)(x2y2)2,xy0 f(x,y)(x2y2)2,xy 0 x2y20

0 x2y20 |xy|sin(x2y2),x2y2★★★11.f(xyx2

,x2y2

，f(x,y)在(0,0)處的可微性知識(shí)點(diǎn)思路：可微的必要條件zdzo()，若limzdz0解：

f(0x,0)f(0, m00f(0,0)

f(0,0y)f(0,0)

0 y0f(x,y在(00dzfx(00)xfy(00)y sin[(x)2(y)2考慮

z

(x)2((x)2 (x (x)2(y)2|xysin[(x)2(y)2(x)22 |k(x)2(1k2|k1k f(x,y在(00(x2y2)

,x2y2★★★★12.f(xy

x2 ,x2y2

，問(wèn)在點(diǎn)(00偏導(dǎo)數(shù)是否存在？（2）偏導(dǎo)數(shù)是否連續(xù)？（3）知識(shí)點(diǎn)思路(x)2

f(0x,0)f(0, (y)2 fy(0,0)

f(0,0y)f(0, 函數(shù)在點(diǎn)(00當(dāng)(xy00)時(shí)fx(x,y)2x

x2

(x2y2)

x2

(x2y2

x2

x2

x2(x,yx軸趨于(00fx(xy在點(diǎn)(00不連續(xù)。x,y的對(duì)稱(chēng)性可知

1212

fy(x,y)2y

x2

2yx2y2

x2

limf(x,y)lim(2y

2

f(xyx0y

x