Let $f, g : \mathbb{R}^3\to \mathbb{R}$ be functions of class $C^1$ "In general," one expects that each of the equations $f(x, y, z) = 0$ and $g(x , y, z) = 0$ represents a smooth surface in $\mathbb{R}^3$, and that their intersection is a smooth curve. Show that if $(x_0,y_0 , z_0)$ satisfies both equations, and if $\partial(f, g)/\partial(x,y, z)$ has rank $2$ at $(x_0, y_0, z_0)$, then near $(x_0, y_0, z_0)$, one can solve these equations for two of $x, y, z$ in terms of the third, thus representing the solution set locally as a parametrized curve.

I think I should apply the implicit function theorem to $(f,g)$ and I already know that $(f,g)(x_0, y_0, z_0)=0$, I just have to prove that $\det \partial (f,g)/\partial z\neq 0$, but I do not know how to do this, could someone help me please? What does it mean that P and how does this help me get what I want? How do I represent the solution in a parameterized curve? Thank you very much.


1 Answer 1


(I don't know which formulation of the implicit function theorem you are referring to. Given the formulation in Wikipedia the text in bold in your question is exactly the statement of the theorem. In the following I shall use the more basic inverse function theorem. For this I have to set up a map $h:\>{\mathbb R}^3\to{\mathbb R}^3$.)

Put $(x_0,y_0,z_0)=:p_0$, and assume that ${\partial(f,g)\over\partial(x,y)}\biggr|_{p_0}$ has rank $2$. Set up the map $$h:\quad{\mathbb R}^3\to{\mathbb R}^3,\qquad (x,y,z)\mapsto\left\{\eqalign{u&:=f(x,y,z) \cr v&:=g(x,y,z)\cr w&:=z\cr}\right.\quad.$$ Then $h(p_0)=(0,0,z_0)=:q_0$, and ${\rm det}\bigl(dh(p_0)\bigr)\ne0$ by assumption. It follows that $h$ maps a neighborhood $U$ of $p_0$ diffeomorphically onto a neighborhood $V$ of $q_0$. The function $$\gamma:\quad t\mapsto h^{-1}(0,0,t)=:\bigl(x(t),y(t),t\bigr)\in{\mathbb R}^3\tag{1}$$ is then $C^1$ in a neighborhood of $t=z_0$ and produces the curve you are interested in. Note that the $z$-component of $\gamma$ has derivative $\equiv1$; hence the representation $(1)$ of $\gamma$ is regular.

  • $\begingroup$ Could you be clearer please? For what purpose do you define that function? $\endgroup$
    – user482152
    Mar 26, 2018 at 1:37

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