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

Quote:

Originally Posted by sluggerbroth

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

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

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