I have the following question:
Consider the set $\Gamma \subseteq \mathbb{R}^3$ of solutions of the system
\begin{equation*}
\begin{cases}
x+\ln{y}+2z-2=0 \\
2x+y^2+e^z-1-e=0
\end{cases}
\end{equation*}
Describe $\Gamma$ in a neighborhood of $(0,1,1)$ using the Implicit Function Theorem.
I saw this and it helped me to start doing something…. but I couldn't conclude it.
My attempt: define $f:U\subseteq\mathbb{R}^3\to\mathbb{R}^2$ by $f(x,y,z)=(x+\ln y+2z-2,2x+y^2+e^z-1)$ where $U$ could be $\mathbb{R}\times\mathbb{R}^+\times \mathbb{R}$. Then $\frac{\partial f}{\partial(y,z)}=\begin{matrix}1/y & 2y\\2 & e^z\end{matrix}$ which has non-zero determinant at the mentioned point $(0,1,1)$. So I can use the implicit function theorem to say that $y$ and $z$ can be described as a function of $x$ (in a open neighbourhood of $(0,1,1)$). Following the hint of the link, as I have by implicit function theorem $f(x,y(x),z(x))=0$ in an open neighbourhood I can differentiate both side with respect to $x$. But then I get
$\begin{cases}1+y'/y+2zz'=0\\2+2yy'+e^zz'=0\end{cases}$
which I couldn't solve for $y,z$ or $y',z'$. Also I can't use ODE methods because I didn't learned it in my course.
Could you help me? Is this the correct way I should follow?
Best Answer
The IFT gives you: $\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx} + \frac{\partial f}{\partial z} \frac{dz}{dx} = 0$, that is in matrix form:
$\begin{pmatrix}\frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{pmatrix}% \begin{pmatrix} \frac{dy}{dx} \\ \frac{dz}{dx} \end{pmatrix} = -\frac{\partial f}{\partial x} $.
Solving this linear system at the point $(0, 1, 1)$ will give you the total derivatives $\frac{dy}{dx}$ and $\frac{dz}{dx}$.
Note: you get a system with 2 equations and 2 unknowns - the one your wrote. That shouldn't be too difficult ;)