精品文档---下载后可任意编辑(一)1.解:两点间直线段的方程为:,故所以。2.解:的参数方程为,则所以3.解:故4.解:如图:,,:,,:,,∴5.解:∴6.解:∴xy110xdxdxdxyds21112222110dxxxdxyxLsin2121cos21ayaax20cos12||21sin2121cos21222aaaayx2cos||12cos212||212aa||21cos2sin22222aaadyxds 202222cos21dadsyxL0222cos2cos21dda220222sin22sin221aaatdtdttattatdtyxds222sincos2022222cossinsincosatdtttttttadsyxL20232204233321242attadttta32222212222LyxLyxLyxLyxdsedsedsedse0yxxax 0dxdxds201xyxxax220dxdxds2112taytaxsincos40xadtdttatadtyxdx2222cossin4022020222adtedxedxedseaaxaxLyx2424|22020aeaeeeaaaxaxttayx44343434sincos222222cossin3sincos3ttattadtyxdstdttadtttacossin3cossin92222044373434cossinsincos3tdttttadsyxL37206374sin616cos613attaadtdtatatadtzyxds2cossin22222220202222222222sincosdttaadttatatadsyxzL精品文档---下载后可任意编辑。7.解:8.解:直线段的方程为,化成参数方程为,,,从 1 变到 0 故9.解:直线的参数方程为,,()10.解:11.解:1)原式2)原式12.解:1)的方向余弦,2),故3),故13.解:因为 故原积分与路径无关,于是3203238|312ata 1111511422545122ydyydyyyyxydxLAB123zyxtx3ty2tz ydzxdyxydxxL2233 dtttttt01223232233348787013 dtttx1ty21tz3110tLdzyxydyxdx1...