文档介绍：§:设000(,)Pxy是平面曲线C上一点,C的方程为:(,)0(1)Fxy?.求::切线方程为000000(,)()(,)()0xyFxyxxFxyyy????法线方程为000000(,)()(,)()0yxFxyxxFxyyy????C0P?yxo解题思路:(,)Fxy在0P点满足隐函数定理条件时,方程(1)将存在隐函数:()()yfx??(1)与方程(*),C在0P点的切线与法线方程为:000()()(2)yyfxxx????与0001()(3)()yyxxfx?????,00000(,)()(,)xyFxyfxFxy???代入(2)、(3),即得切线和法线方程。()90xyxy???在点(2,1)处的切线和法线。:33(,)2()9,Fxyxyxy???记xF269,xy??yF269,yx??,,,xyFFF?皆在全平面连续:且(2,1)xF15?0,?(2,1)yF12??0,?:所求切线方程为15(2)12(1)0xy????:所求法线方程为12(2)15(1)0xy???????(0)a?上任一点处的切线方程,并证明这些切线被坐标轴所截取的线段等长。:解0000(,),0,0,xyxy??在此曲线上任取一点且222333(,)Fxyxya???记,00(,)0Fxy?则113322,,33xyFxFy????1**********(,)0,(,)0,33xyFxyxFxyy??????00,,(,),:xyFFFxy?皆在某邻域内连续且00(,)Fxy在点满足隐函数定理存在条件;00(,)xy于是,此曲线在点切线方程为:1133000022()()033xxxyyy??????233300xyaxy??即??230,0xxa此切线与轴的交点为,??2300,yya与轴的交点为;它们之间的距离平方为:????22222330000dxaya????42233300axy???????00(,)(),(),(),(4)xxtyytzztt???????:0(),(),()xtytztt设在上可导,222000()()()0xtytzt??????且000(),(),()xtytzt???(即不同时为零),0[,]t???取,C对应于上一点0PP在点邻近取一点0P??P0t?[]??t00000000()()()(,,)(,,)tttPxyzPxyz?ozyx0P?(,,)Pxxyyzz??????000()()()(,,)ttttttPxyz???????00()(),tttxxx?????其中00()(),tttyyy?????00()().tttzzz?????0PP于是,割线方程为:000xxyyzzxyz????????000xxyyzzxtytzt???????????即0,t??令0PP?则,00;PPPT?割线切线0CP于是,上点切线方程为:000000(5)()()()xxyyzzxtytzt?????????P0t?[]??t0tt???Tozyx0P?此切线的方向向量为:??000(),(),()Txtytzt????(,,)CPxyz:通过曲线上一点定义,00PCP且与点的切线垂直的平面,()()()()()()0(6)xtxxytyyztzz?????????T例3求曲线:?0costuxeudu??,tysin2?tcos?,tez31??在0??t时,,2,1,0???zyx,costext??,sincos2tty???,33tez???,1)0(??x,2)0(??y,3)0(??z切线方程,322110?????zyx法平面方程,0)2(3)1(2?????????zyx即