Calculating a tangent line to an implicitly given function

Question

Let $F:\mathbb{R}^3 \to \mathbb{R}^2 $ be continuously differentiable and let $p\in\mathbb{R}^3 $ for which $F(p)=0$ . Assume $rank DF|_p =2 $ and denote by $E$ the set $E=(x\in \mathbb{R}^3 | F(x)=0) $ .

Implicit function theorem guarantees the existence of a neighberhood $B$ of $p$ such that $B\cap E$ is a smooth curve (/path).

If $DF|_p = \begin{bmatrix} 1 & -1 & 2 \\[0.3em] 3 & 0 & 1 \end{bmatrix} $ how can one calculate a tangent line to the curve $B\cap E$ I just mentioned ? (at $p$ obviously).

Thanks !

Answer 1

If $\gamma(t)$ is some differentiable parametrization of $E \cap B$, then $F(\gamma(t)) = 0$. By the chain rule, $0 = \frac{d}{dt}F(\gamma(t)) = DF(\gamma'(t)) = 0$ for all $t$ for which $\gamma(t)$ is defined. Thus $\gamma'(t) \in \ker DF$. This implies that $\gamma'(t)$ can, up to a scalar multiple, be had by solving the linear system $DF(\gamma'(t)) = 0$. With $DF_p$ as given, we see that with $\gamma'(t_0) = (u, v, w)^T$, where $\gamma(t_0) = p$, we have

$u - v +2w = 0, \tag{1}$

and

$3u + w = 0; \tag{2}$

the solution to (1), (2) is the one-dimensional family of vectors $w(-1/3, 5/3, 1)^T$; this in turn implies the requisite tangent line is $\hat p + w(-1/3, 5/3, 1)$, where we take $\hat p$ to mean the ordered triple $(p_1, p_2, p_3)$ of coordinates of the point $p$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!