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Math Help - Every tangent plane to graph z^(3)=x^(3)+y^(3) passes thru origin???

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    Member sluggerbroth's Avatar
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    Every tangent plane to graph z^(3)=x^(3)+y^(3) passes thru origin???

    True or false and show me why you say so.
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    Re: Every tangent plane to graph z^(3)=x^(3)+y^(3) passes thru origin???

    Quote Originally Posted by sluggerbroth View Post
    True or false and show me why you say so.
    True. You can explicitly solve for z

    This gives

    z=f(x,y)=\left( x^3+y^3\right)^\frac{1}{3}

    f_x(x,y)=\frac{x^2}{\left( x^3+y^3\right)^\frac{2}{3}}

    f_y(x,y)=\frac{y^2}{\left( x^3+y^3\right)^\frac{2}{3}}

    This gives the equation of the tangent plane at (x_0,y_0)

    z=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+f(x_0,y_0)

    z=\frac{x_0^2}{(x_0^3+y_0^3)^{\frac{2}{3}}}(x-x_0)+\frac{y_0^2}{(x_0^3+y_0^3)^{\frac{2}{3}}}(y-y_0)+\sqrt{x_0^3+y_0^3}

    If you evaluate at (0,0) it gives

    z=\frac{x_0^2}{(x_0^3+y_0^3)^{\frac{2}{3}}}(0-x_0)+\frac{y_0^2}{(x_0^3+y_0^3)^{\frac{2}{3}}}(0-y_0)+\sqrt{x_0^3+y_0^3}

    z=-\frac{x_0^3+y_0^3}{(x_0^3+y_0^3)^{\frac{2}{3}}} + \sqrt{ x_0^3 + y_0^3 }=-\sqrt{x_0^3+y_0^3}+\sqrt{x_0^3+y_0^3}=0
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