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Let $F\in C^1(\mathbb{R}^3, \mathbb{R}^2)$ be defined by $$F(x,y,z)=\begin{pmatrix} x^2+2x\cos y+ \sin z \\ x^5+\sin y + 2x\cos z\end{pmatrix}$$

Show: There is a $\varepsilon >0$ and $C^1$-Curve $g: (-\varepsilon , \varepsilon ) \to \mathbb{R}^2$ with $g(0)=0$ and $F(t,g_1(t),g_2(t))=0$ for all $t\in (-\varepsilon , \varepsilon )$. Compute $g'(0)\in \mathbb{R}^2$

My Idea:

The monomials in the front remain the same, so I have to find a vector-valued function with two components, such that $t^2 +2t\cos g_1(t)+ g_2(t) = 0$ and $t^5+\sin g_1(t) + 2t\cos g_2(t)=0$ for all $t\in (-\varepsilon , \varepsilon)$. Do I have to give an explicit curve or is there another way to atleast show the existence?

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  • $\begingroup$ It's impossible to find a curve explicitly. This is a blatant application of the Implicit Function Theorem. $\endgroup$ – Ted Shifrin Jan 19 '20 at 19:20
  • $\begingroup$ Oh, okay - I thought so...! $\endgroup$ – Analysis Jan 19 '20 at 19:28
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Let $F(x,y,z)=0.$ In order to find functions $g_1$ and $g_2$ such that $F(x,g_1(x),g_2(x))=0,$ you need to apply the implicit function theorem. Show that the determinate of the Jacobian $$ J=\det\begin{pmatrix} \frac{\partial F_1}{\partial y} & \frac{\partial F_1}{\partial z} \\ \frac{\partial F_2}{\partial y}&\frac{\partial F_2}{\partial z} \end{pmatrix} \neq 0 $$ when $(x,y,z)=(0,0,0)$. The existence of $g(x)=(g_1(x),g_2(x))$ then follows, defined in a neighborhood $x\in(-\epsilon,\epsilon)$.

For finding $g'(0),$ differentiate $F(x,g_1(x),g_2(x))=F(x,g(x))=0$ with respect to $x$. Then $$ 0=\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}g_1'(x) + \frac{\partial F}{\partial z}g_2'(x) $$ All partials are $2\times 1$ vectors. You should find $\begin{pmatrix}g_1'(x)\\g_2'(x)\end{pmatrix}=-\begin{pmatrix} \frac{\partial F_1}{\partial y} & \frac{\partial F_1}{\partial z} \\ \frac{\partial F_2}{\partial y}&\frac{\partial F_2}{\partial z} \end{pmatrix}^{-1}\begin{pmatrix}\frac{\partial F_1}{\partial x} \\\frac{\partial F_2}{\partial x} \end{pmatrix}$.

You just have to evaluate $(x,y,z)=(0,0,0)$.

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  • $\begingroup$ $\begin{pmatrix}g_1'(0)\\g_2'(0)\end{pmatrix} = (-2,-2)^T$, right? $\endgroup$ – Analysis Jan 19 '20 at 20:27

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