Given the system of equations
$f_1(x,y,z) = x+z-e^{(x+2y)} + e^{(x+y+z)} = 0$
$f_2(x,y,z) = x+y+z + 2 \sin(x+y+z) = 0$
I have shown that the system can be solved for $x$ and $y$ near the origin ($x=y=z=0$ is a solution). How can I determine the tangent vector to the solution curve $(x(z), y(z))$ at $z=0$?
#
For the matrix $M=\begin{bmatrix}
\frac {\partial f_1}{\partial x} & \frac {\partial f_1}{\partial y} \\
\frac {\partial f_2}{\partial x} & \frac {\partial f_1}{\partial y}
\end{bmatrix}$
at $x=y=z=0$,
$ M= \begin{bmatrix}
1 & -1 \\
3 & 3
\end{bmatrix}$, where $\det M=6\neq 0$. Hence $M$ is regular.
Then there exists an interval $U$ around $z=0$, an open set $V$ around $(x_0,y_0)=(0,0)$, and a function $\phi:U\to V，\phi(z)=(\phi_1(z), \phi_2(z)) = (x,y)$, where $\phi$ gives the parametrisation of the intersection curve.
The point I'm stuck at is, I can get $z=y-\ln(1-y)$ and $x=\ln(1-y)-2y$. When I set $z=0$, it leads to all $x=y=z=0$. What can I do next?
Thanks heaps in advance!
Best Answer
Note that the equations $f_1(x,y,z)=0$ and $f_2(x,y,z)=0$ define two surfaces in $\mathbb{R}^3$. The gradients $\nabla f_1 (0,0,0)$ and $\nabla f_2(0,0,0)$ are vectors at $(0,0,0)$ that are normal to the first and second surface respectively. This means that $\nabla f_1 (0,0,0)$ and $\nabla f_2(0,0,0)$ both are normal to the intersection curve of these surface. Hence the tangent to the curve is: $$\nabla f_1(0,0,0) \times \nabla f_2 (0,0,0).$$
Note that this approach doesn't work if $\nabla f_1(0,0,0)$ and $\nabla f_2 (0,0,0)$ point in the same direction.