求曲线x^(2/3)+y^(2/3)=a^(2/3)在点( √2/4a,√2/4a)处的切线方程和法线方程曲线应该是x^(2/3)+y^(2/3)=a^(2/3)求导,得(2/3)x^(-1/3)+(2/3)y^(-1/3)*y’=0,切线斜率y’=-x^(-1/3)/y^(-1/3)=-1,切线方程:x+y=(√2/2)a,法线斜率
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![求曲线x^(2/3)+y^(2/3)=a^(2/3)在点( √2/4a,√2/4a)处的切线方程和法线方程曲线应该是x^(2/3)+y^(2/3)=a^(2/3)求导,得(2/3)x^(-1/3)+(2/3)y^(-1/3)*y’=0,切线斜率y’=-x^(-1/3)/y^(-1/3)=-1,切线方程:x+y=(√2/2)a,法线斜率](/uploads/image/z/8439034-58-4.jpg?t=%E6%B1%82%E6%9B%B2%E7%BA%BFx%5E%282%2F3%29%2By%5E%282%2F3%29%3Da%5E%282%2F3%29%E5%9C%A8%E7%82%B9%28+%E2%88%9A2%2F4a%2C%E2%88%9A2%2F4a%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E5%92%8C%E6%B3%95%E7%BA%BF%E6%96%B9%E7%A8%8B%E6%9B%B2%E7%BA%BF%E5%BA%94%E8%AF%A5%E6%98%AFx%5E%282%2F3%29%2By%5E%282%2F3%29%3Da%5E%282%2F3%29%E6%B1%82%E5%AF%BC%2C%E5%BE%97%282%2F3%29x%5E%28-1%2F3%29%2B%282%2F3%29y%5E%28-1%2F3%29%2Ay%E2%80%99%3D0%2C%E5%88%87%E7%BA%BF%E6%96%9C%E7%8E%87y%E2%80%99%3D-x%5E%28-1%2F3%29%2Fy%5E%28-1%2F3%29%3D-1%2C%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%3Ax%2By%3D%28%E2%88%9A2%2F2%29a%2C%E6%B3%95%E7%BA%BF%E6%96%9C%E7%8E%87)
求曲线x^(2/3)+y^(2/3)=a^(2/3)在点( √2/4a,√2/4a)处的切线方程和法线方程曲线应该是x^(2/3)+y^(2/3)=a^(2/3)求导,得(2/3)x^(-1/3)+(2/3)y^(-1/3)*y’=0,切线斜率y’=-x^(-1/3)/y^(-1/3)=-1,切线方程:x+y=(√2/2)a,法线斜率
求曲线x^(2/3)+y^(2/3)=a^(2/3)在点( √2/4a,√2/4a)处的切线方程和法线方程
曲线应该是x^(2/3)+y^(2/3)=a^(2/3)
求导,得(2/3)x^(-1/3)+(2/3)y^(-1/3)*y’=0,
切线斜率y’=-x^(-1/3)/y^(-1/3)=-1,
切线方程:x+y=(√2/2)a,
法线斜率=1,
法线方程:y=x.
我看的迷迷糊糊的...
"求导,得(2/3)x^(-1/3)+(2/3)y^(-1/3)*y’=0"
为什么会有y’呢?
"切线斜率y’=-x^(-1/3)/y^(-1/3)=-1"
x与y怎么可以相消呢?
求曲线x^(2/3)+y^(2/3)=a^(2/3)在点( √2/4a,√2/4a)处的切线方程和法线方程曲线应该是x^(2/3)+y^(2/3)=a^(2/3)求导,得(2/3)x^(-1/3)+(2/3)y^(-1/3)*y’=0,切线斜率y’=-x^(-1/3)/y^(-1/3)=-1,切线方程:x+y=(√2/2)a,法线斜率
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