# Finding a curve, such that a function is $0$.

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?

• It's impossible to find a curve explicitly. This is a blatant application of the Implicit Function Theorem. – Ted Shifrin Jan 19 '20 at 19:20
• Oh, okay - I thought so...! – Analysis Jan 19 '20 at 19:28

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