Tangent plane of an implicit system of equations

Question

The next system of equations implicitly define to the equations $u(x,y)$ and $v(x,y)$

$$xe^{2u+3v}-2uv=1$$ $$ye^{u-v}-\frac{u+1}{v+1}=2x$$

Find the tangent plane equation to the graphic of the function $u(x,y)$ in the point where $x=1,y=2,u=v=0$.

I've been trying to do this exercise but I still do not understand how to use the system of equations to calculate that. I appreciate some explanation of how to answer this.

If someone has some example or reference about the tangent plane of an implicit system of equations. It would be a great help.

I made this, firsts that all, I wanted to calculate $u_x$ and $u_y)$.

1. I define $F_1(x,y,u(x,y),v(x,y))$ and $F_2(x,y,u(x,y),v(x,y))$ as.

$$F_1(x,y,u(x,y),v(x,y))=xe^{2u+3v}-2uv-1$$ $$F_2(x,y,u(x,y),v(x,y))=ye^{u-v}-\frac{u+1}{v+1}-2x$$

2. Now I calculated $\frac{\partial F_1}{\partial x}$,$\frac{\partial F_1}{\partial y}$,$\frac{\partial F_1}{\partial u}$,$\frac{\partial F_1}{\partial v}$, $\frac{\partial F_2}{\partial x}$, $\frac{\partial F_2}{\partial y}$, $\frac{\partial F_2}{\partial u}$ and $\frac{\partial F_2}{\partial v}$.

what I get was

$\frac{\partial F_1}{\partial x}=e^{2u+3v}$, $\frac{\partial F_1}{\partial y}=0$, $\frac{\partial F_1}{\partial u}=e^{2u+3v}-2-2v$, $\frac{\partial F_1}{\partial v}=3e^{2u+3v}-2u$, $\frac{\partial F_2}{\partial x}=-2$, $\frac{\partial F_2}{\partial y}=e^{u-v}$, $\frac{\partial F_2}{\partial u}=ye^{u-v}-\frac{1}{v+1}$, $\frac{\partial F_2}{\partial v}=ye^{u-v}+\frac{u+1}{(v+1)^2}$.

3. I evaluated this in the point (1,2,0,0) and answered the following system of equations $$\begin{bmatrix}-1&3\\1&3\end{bmatrix}\begin{bmatrix}u_x\\v_x\end{bmatrix}=\begin{bmatrix}-1\\2\end{bmatrix}$$

and

$$\begin{bmatrix}-1&3\\1&3\end{bmatrix}\begin{bmatrix}u_x\\v_x\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$

so I get $u_x=3/2,u_y=1/4$ and $v_x=1/6,v_y=1/4$

So I think that the tangent plane of $u(x,y)$ is (2/3)(x-1)+(1/4)(y-2)=0.

So far I have been able to resolve this, I would like to know if I am on the right track or should I follow another procedure.

Answer 1

Once you find $u_x(x_0,y_0)$ and $u_y(x_0,y_0)$, the tangent plane should be $$z=u(x_0,y_0)+u_x(x_0,y_0)(x-x_0)+u_y(x_0,y_0)(y-y_0)$$