醫(yī)科高等數(shù)學(xué)-習(xí)題課件

1、醫(yī)科高等數(shù)學(xué)-習(xí)題課件)2(5:147P22211zxyz11222zyxu改改為為：間斷點：間斷點：1),(222zyxzyx)2(21:149Pcos()Dxxy d(0,0) ( ,0)( , ) 其中其中D是頂點分別為是頂點分別為 , 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dcossincosdsi

2、n1d2sinsincoscos112d)25(2:100Pdxxx211coscos11cos1121dccos1cos1ln21coscos112dx1x=sin21xcxx221111ln21cxx221111ln21cxx2221111ln21cxx211lncxx211lnln)14(3:101Pdxxx)1ln(2)1ln()1ln(22xxxdxxx解解：原原式式dxxxxxx1122122dxxxxxxx111222dxxx121) 1(2122xxdcx12cxxxx1)1ln(22原原式式dxxxxxx1112216:103P求由求由 及其在點及其在點和和 處切線所圍成的圖

3、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計算以計算以xOy面上的圓周面上的圓周圍成的閉區(qū)域為底圍成的閉區(qū)域為底, , 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 22xyax22zxycosaaxyOddVacos02222240cos41ad2244cos4da2244cos4

4、da2044cos2da 202422cos12da20242cos2cos218da20424cos12cos218da20424cos12cos218da20424cos2cos2238da0284sin2sin2384a3234a)4(2:174P221xcey 答案：答案：)7(2:174Pcxxyxy2ln2sin21答答案案：)7(4:174P1 yy)(xPy 令令)2(6:175P)2sin2cos(213xcxceyx答案：答案：)7(6:175Pxxxy3sin532sin572cos答答案案：)5(12:61P) 1ln2(8xx答案：答案：xxPx3tantanlim)

5、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1x6116123xxxx650666655115611612222323xxxxxxxxxxxxxx0611623x

6、xx0)65)(1(2xxx0)3)(2)(1(xxx3, 2, 1321xxx11123)(2 xxxf極極小小值值11) 1 (, 02) 1 ( ff極大值極大值12)2(, 01)2( ff極極小小值值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(3:10122arcsinarcsinxxdxxdxxxxxx221arcsin2arcsin)1 (1

7、arcsinarcsin222xdxxxx221arcsin2arcsinxxdxx221arcsin2arcsinxxdxxxdxxxxxarcsin1arcsin12arcsin222xdxxxx2arcsin12arcsin22cxxxxx2arcsin12arcsin22232)3(5:101xexP答答案案：dxxPee1ln)20(10:102dxxdxxee111lnlncxxxxxdxxxdxlnlnlnln1ln11lnexxxexxxe22不存在不存在證明極限證明極限yxyxPyx00lim4:147kkkxxkxxkxyx11lim,0原原式式設(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二二階階偏偏導(dǎo)導(dǎo)數(shù)數(shù)求求22ln)2(9:148