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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 $z=0$?

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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!

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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.

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  • $\begingroup$ Ah you made it really easy to see! Thanks a lot!! $\endgroup$ – Evelyn Venne Sep 16 '18 at 10:47

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