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

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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