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I am reviewing a proof in my multivariable calculus textbook. After completing a proof of the "Special" Implicit Function Theorem, the case where $F: \Bbb R^{n+1} \rightarrow \Bbb R$ , I read:

Taking a continuously differentiable function $g: \Bbb R^3 \rightarrow \Bbb R$, Consider a level surface $S$ of $g(x,y,z)$, where $g(x_0,y_0,z_0) = c_0$ and $\frac {\partial f} {\partial z} \neq 0$ , then every vector tangent to $S$ at $(x_0,y_0,z_0)$ is tangent to a curve in $S$. The author then states by the implicit function theorem we need only show this for a graph of the form $z=k(x,y)$. Why is this true?

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That's exactly what the Implicit Function Theorem says. Because $\frac{\partial f}{\partial z}\ne0$ (I assume that at $x_0,y_0,z_0)$), there exists a neighbourhood of $(x_0,y_0,z_0)$ where $z=k(x,y)$, with $k$ continuously differentiable, and $g(x,y,k(x,y))=c_0$.

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  • $\begingroup$ "we need only show this for a graph of the form $z=k(x,y)$", when the author says "graph" does he mean $g(x,y, k(x,y)$ ? Perhaps the wording is confusing me. Thanks for your help. $\endgroup$ – IntegrateThis Jun 8 '17 at 1:37
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    $\begingroup$ You are considering level sets of $g$. Close to $(x_0,y_0,z_0)$, you can represent them as $$\{(x,y,k(x,y)):\ x,y\in V\}$$ where $V$ is some open set that contains $(x_0,y_0)$. $\endgroup$ – Martin Argerami Jun 8 '17 at 1:40

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