醫(yī)科高等數(shù)學(xué)-習(xí)題ppt課件

1、醫(yī)科高等數(shù)學(xué)醫(yī)科高等數(shù)學(xué)習(xí)題習(xí)題)2(5:147p22211zxyz11222zyxu改為：改為：間斷點(diǎn)：間斷點(diǎn)：1),(222zyxzyx)2(21:149pcos()dxxy d(0,0) ( ,0)( , ) 其中其中d是頂點(diǎn)分別為是頂點(diǎn)分別為 , 和和 的三的三角形閉區(qū)域角形閉區(qū)域 32答案是：答案是：)3(23:149p其中其中d是是由圓周由圓周 及坐標(biāo)軸所圍成及坐標(biāo)軸所圍成的在第一象限內(nèi)的閉區(qū)域；的在第一象限內(nèi)的閉區(qū)域； dyxyxd222211221xyddd2211d22112221121ddtttt11212)28(2:100pddxxcos,sin解：令解：令dcossin

2、cosdsin1d2sinsincoscos112d)25(2:100pdxxx211coscos11cos1121dccos1cos1ln21coscos112dx1x=sin21xcxx221111ln21cxx221111ln21cxx2221111ln21cxx211lncxx211lnln)14(3:101pdxxx)1ln(2)1ln()1ln(22xxxdxxx解：原式解：原式dxxxxxx1122122dxxxxxxx111222dxxx121) 1(2122xxdcx12cxxxx1)1ln(22原式原式dxxxxxx1112216:103p求由求由 及其在點(diǎn)及其在點(diǎn)和和 處

3、切線所圍成的圖形面積處切線所圍成的圖形面積 243yxx (0, 3)(3,0)xyo3335 . 134 xy62 xy42 xy(0, 3)4 y(3,0)2 y5 . 102)34()34(dxxxx35 . 12)34()62(dxxxx5 . 102dxx35 . 1296dxxx5 . 13933105 . 131233xxxx4924:149p計(jì)算以計(jì)算以xoy面上的圓周面上的圓周圍成的閉區(qū)域?yàn)榈讎傻拈]區(qū)域?yàn)榈? , 而以曲面而以曲面為頂?shù)那斨w的體積為頂?shù)那斨w的體積 22xyax22zxycosaaxyoddvacos02222240cos41ad2244cos4da2

4、244cos4da2044cos2da 202422cos12da20242cos2cos218da20424cos12cos218da20424cos12cos218da20424cos2cos2238da0284sin2sin2384a3234a)4(2:174p221xcey 答案：答案：)7(2:174pcxxyxy2ln2sin21答案：答案：)7(4:174p1 yy)(xpy 令令)2(6:175p)2sin2cos(213xcxceyx答案：答案：)7(6:175pxxxy3sin532sin572cos答案：答案：)5(12:61p) 1ln2(8xx答案：答案：xxpx3t

5、antanlim)2(25:632xxxl3sec3seclim222xxxxxlsincos63sin3cos6lim2xxxcos3coslim2xxxlsin3sin3lim23)0()1 (lim)4(25:63axapxx)1ln(limxaxxexxaxe1)1ln(lim221)(11limxxaxalxexaaxe1limae極值極值求求2024228)()4(27:63234xxxxxfp02444244)(23xxxxf0611623xxx是方程的解是方程的解經(jīng)觀察知經(jīng)觀察知1x6116123xxxx650666655115611612222323xxxxxxxxxxxxx

6、x0611623xxx0)65)(1(2xxx0)3)(2)(1(xxx3, 2, 1321xxx11123)(2 xxxf極小值極小值11) 1 (, 02) 1 ( ff極大值極大值12)2(, 01)2( ff極小值極小值11)3(, 02)3( ffdxxxxp)1 (arctan)15(2:100tdtdxtxtx2,2令令dttttt)1 (arctan22原式原式dttt)1 (arctan22td t arctanarctan2ct2arctancx2arctandxxp2arcsin)15(3:10122arcsinarcsinxxdxxdxxxxxx221arcsin2ar

7、csin)1 (1arcsinarcsin222xdxxxx221arcsin2arcsinxxdxx221arcsin2arcsinxxdxxxdxxxxxarcsin1arcsin12arcsin222xdxxxx2arcsin12arcsin22cxxxxx2arcsin12arcsin22232)3(5:101xexp答案：答案：dxxpee1ln)20(10:102dxxdxxee111lnlncxxxxxdxxxdxlnlnlnln1ln11lnexxxexxxe22不存在不存在證明極限證明極限yxyxpyx00lim4:147kkkxxkxxkxyx11lim,0原式原式設(shè)設(shè)極限不存在極限不存在偏導(dǎo)數(shù)偏導(dǎo)數(shù)求求xxyzp)1 ()4(6:147)1ln(xyxez)1ln(xyxexz)1)1(ln(xyxyxy)1ln(xyxeyzxyx12xxyxz)1 ( )1)1(ln(xyxyxy12)1 (xxyx二階偏導(dǎo)數(shù)二階偏導(dǎo)數(shù)求求22ln)2(9:148yxzp2222221yxxyxxz22yxx22222222222222yxxyyxxyxxz22222yxxyyxz2222222yxyxyz有一階連續(xù)偏導(dǎo)有一階連續(xù)偏導(dǎo)fxyyxxfzp),()5(12:148的偏導(dǎo)的偏導(dǎo)對(duì)對(duì)求求yxz,2211x