Let $F:\mathbb{R}^2\to \mathbb{R}$ be of class $C^2$, with $F(0, 0) = 0$ and $DF(0, 0) =\begin{bmatrix}2 & 3\end{bmatrix}$.

Question

Let $F:\mathbb{R}^2\to \mathbb{R}$ be of class $C^2$, with $F(0, 0) = 0$ and $DF(0, 0) =\begin{bmatrix}2 & 3\end{bmatrix}$. Let $G:\mathbb{R}^3\to \mathbb{R}$ be defined by the equation $G(x, y, z) = F(x + 2y + 3z - 1, x^3 + y^2 - z^2)$.

(a) Note that $G(-2, 3, -1) = F(0, 0) = 0$. Show that one can solve the equation $G(x, y, z) = 0$ for $z$, say $z = g(x, y)$, for $(x, y)$ in a neighborhood $B$ of $( -2, 3)$, such that $g( -2, 3) = -1$.

(b) Find $Dg( -2, 3)$.

(c) If $D_1 D_1F = 3$ and $D_1 D_2F = -1$ and $D_2D_2F = 5$ at $(0,0)$, find $D_2D_1 g( -2, 3)$.

For (a), I am applying the implicit function theorem, I know that $G(-2,3,-1)=0$ with which I want to find out that $\frac{\partial G}{\partial z}(-2,3,-1)$, I know that $DG(-2,3,-1)=\begin{bmatrix}\frac{\partial F}{\partial x}\frac{\partial}{\partial x}(x+2y+3z-1, x^3+y^2-z^2) & \frac{\partial F}{\partial y}\frac{\partial}{\partial y}(x+2y+3z-1, x^3+y^2-z^2) & \frac{\partial F}{\partial z}\frac{\partial}{\partial z}(x+2y+3z-1, x^3+y^2-z^2) \end{bmatrix}|_{(-2,3,-1)}=\begin{bmatrix}\frac{\partial F}{\partial x}(1,12) &\frac{\partial F}{\partial y}(2,6) &\frac{\partial F}{\partial z}(3,2) \end{bmatrix}$, but I do not know how to determine if $\det \frac{\partial G}{\partial z}\neq 0$, I do not know if I have done the right thing, could someone help me by favor?

For (b), I need to know $\frac{\partial G}{\partial z}$, so I could not make this point

For (c), the truth is that I have no idea what to do here, could someone give me an idea of ​​what I can do? Thank you very much.

Answer 1

Here's a sketch. Let $H$ be the 'intermediate' function, $$ H(x,y,z) = (x+2y+3z - 1 ,\quad x^3+y^2 - z^2)$$ and write $X := (-2,3,-1)$. Then $H(X) = (0,0)$ and $G = F\circ H$. We have by chain rule $$ DG(X) = DF(H(X))\,DH(X) = DF(0,0)\, DH(X)$$ We know $DF(0,0) = [\ 2\quad 3\ ]$, so we need to compute $DH(X)$. As a sanity check, since $G:\mathbb R^3\to\mathbb R$, $DG(X)$ should be a $1\times 3$ matrix, so we are looking for a $2\times 3$ matrix. Can you finish?

$$ DH(X)=\begin{bmatrix}1 & 2 & 3 \\ 12 & 6 & 2\end{bmatrix}$$ $$ DG(X)=[\ 38 \quad 22 \quad 12 \ ] $$ If you don't remember the formula for the derivative, then you can derive it- let $h(x,y) = (x,y,g(x,y))$, $X'=(-2,3), h(X') = X. $ $$Dh = \begin{bmatrix}1 & 0 \\ 0 & 1\\ \partial_x g & \partial_y g \end{bmatrix} = \begin{bmatrix} I_2 \\ \hline Dg \end{bmatrix}$$ $$D[G\circ h](X') = DG(X)Dh(X')$$ Since $G\circ h = 0$ everywhere it is defined, $LHS = 0$, and $$ 0 =[\partial_xG(X) \quad \partial_y G(X)]I_2 + [\partial_z G(X)]Dg(X') = [38 \quad 22] + [12]Dg(X') $$ For ($c$), note the above derived formula for any $Y'$, $$ Dg(Y') = -\frac{1}{\partial_z G(Y')}[\partial_x G(Y') \quad \partial_y G(Y')]$$ can be differentiated further. This lets you compute $D^2g(X')$ but life is a little easier because you only want to compute $D_2D_1g$, which you should be able to do.