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ID:6169180
大小:2.00 MB
页数:13页
时间:2018-01-05
《微积分 高等数学 英文练习题以及答案(2)》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、Xi’anJiaotong-LiverpoolUniversitySubgroupNameIDNo.21Exercise2-1ConceptofDerivative1.Themotionofanobjectalongthe-axisfollowsthelaw(m).Find:(1)theaveragespeedoftheobjectduringthetimeintervalfromthe1stsecondtothe2ndsecond;解:(m/sec)(2)theInstantaneousvelocityoftheob
2、jectatthe2ndsecond.解:;所以(m/sec)2.Comparethefollowinglimitswiththedefinitionofthederivativeandthenpointouttherelationbetweenand.(Assumethatexists)(1);解:(2)解:(3).(Hint:)解:3.Find,ifand.解:4.Findderivativesofthefollowingfunctionsbyusingthederivativeformulaofpowerfunc
3、tions:(1);解:,所以(2);解:,所以(3)解:,所以5.Taketwopointswiththeabscissaandontheparabolaanddrawasecantlinethroughthetwopoints.Findthepointontheparabolaatwhichthetangentlinetothecurveisparalleltothissecantline.解:当时,;当时,。从而割线的斜率为;抛物线在任意一点处的切线斜率为,若切线与割线平行,则,解得,对应的函数值为,即点为所求。
4、6.Findtheequationofthetangentandthenormalrespectivelytothecurveatthepoint.解:曲线在点的切线斜率为,从而法线斜率为。所以切线方程为:法线方程为:7.Prove:Theareaoftheregionboundedbythetangentatanypointofthehyperbolaandthetwoaxesequals.证明:设为曲线上任意一点,则。由得,切线方程:,切线在两个坐标轴上的截距为,故切线与坐标轴所围三角形的面积为,即是一个常数。Xi
5、’anJiaotong-LiverpoolUniversitySubgroupNameIDNo.21Exercise2-2DifferentiationRules(I)1.Findthederivativesofthefollowingfunctions:(1);解:(2);解:(3);解:(4);解:(5);解:(6);解:(7);解:(8);解:或则(9);解:(10).解:2.Findthederivativesofthefollowingfunctionsatgivenpoints:(1).Findand;解:
6、,所以(2).Find;解:所以3.Writetheequationofthetangenttothecurveatthepointwherethecurvecrossestheaxis.解:,曲线与轴的交点为,从而有,。所以曲线在点处的切线方程为,即;所以曲线在点处的切线方程为,即。Xi’anJiaotong-LiverpoolUniversitySubgroupNameIDNo.21Exercise2-2DifferentiationRules(II)1.Findthederivativesofthefollowi
7、ngfunctions:(1);解:(2);解:(3);解:(4);解:(5);解:(6);解:(7);解:(8);解:(9);解:(10);解:2.Assumethatisdifferentiablein.Prove:Ifisanevenfunction,thenisanoddfunction;ifisanoddfunction,thenisanevenfunction.证明:设是偶函数,即,对等式两边同时求导得,也就是是奇函数;设设是奇函数,即,对等式两边同时求导得,也就是是偶函数;法二:设是偶函数,即,由导数的定
8、义得也就是是奇函数;同理可证题目的下半部分。3.Findtheequationsofthetangentandnormalrespectivelytothecurveatthepointcorrespondingto解:由1(8)的结果知,当时曲线在处的切线方程为:;法线方程为:Xi’anJiaotong-LiverpoolUni
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