Equation of a Tangent Plane Find the equation of the tangent plane to the given surface at the given point. 
$x=u^2, y=v^2, z=uv$ at $u=1, v=1$
How would you find the tangent plane when the surface is in this format?
Thanks.
 A: The tangent vectors to the surface are
$$
\mathbf{t}_u=\left(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}\right)\\
\mathbf{t}_v=\left(\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v}\right)
$$
The normal to the surface is $\mathbf{n}=\mathbf{t}_u\times\mathbf{t}_v$.
The equation of the plane is $\mathbf{n}\cdot(P-P_0)=0$.
Are this elements sufficient to build your answer?
A: An alternative to @enzotib's answer is to see if you can find an implicit equation for the surface and use the gradient. In this case, $z^2=xy$, so with $F(x,y,z)=xy-z^2$, $\nabla F=\langle F_x,F_y,F_z\rangle$ evaluated at $(1,1,1)$ will give a normal vector to the plane. As with @enzotib's answer, $\vec{n}\cdot(P-P_0)=0$ gives the equation of the tangent plane.
Note that the surface $z^2=xy$ is not exactly the same as the original one, since the original surface has only one sheet, and this implicit equation has a second sheet with $x,y<0$. But at $u=1,v=1$, which is at $(1,1,1)$, this is irrelevant.
This approach only generalizes if you are successfully able to represent the parametrized surface with an implicit equation.
