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Suppose $f: \mathbb{R}^n \to \mathbb{R}$ is differentiable at $x$. Let $d_xf$ denote the derivative of $f$ at $x$. Let $L$ be the level set through $x$, $L = \{y \in \mathbb{R}^n: f(y) = f(x)\}$. Suppose $v$ is a tangent vector at $x$ that is tangent to the level set $L$. Then the claim is that $d_xf(v) = 0$. Why is this true? Is there a rigorous justification of this?

I have seen various answers, like Why is the gradient normal? and Why gradient vector is perpendicular to the plane, but I couldn't really find a rigorous justification of that particular fact, that $d_xf(v) = 0$. I can see the intuition but I'd like a proof if possible.

Also, what does it mean precisely when we say "Suppose $v$ is a tangent vector at $x$ that is tangent to the level set $L$"? Can all these facts and notions be defined and proved in the usual multivariable context of Euclidean space $\mathbb{R}^n$ (like in a normal or advanced Calc III course) or do we need an excursion into differential geometry or something? I'd just like to know because some of the multivariable calculus texts/resources I've seen, as well as some answers on this site, mostly seem to gloss over the details and just roughly justify it by appealing to geometric intuition, which I think is useful but I would also like a proof.

twosigma
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  • it is needed that, at $x$, the gradient $d_xf$ is different from the zero vector, so the implicit function theorem can be applied. – janmarqz Jul 29 '20 at 16:00
  • can you see the intuition? how? – C.F.G Jul 29 '20 at 16:05
  • @C.F.G I mean that for the cases like $n = 1, 2, 3$ I have seen several geometric and intuitive answers/arguments, but I can't provide a (analytical) proof (maybe because I don't even know what some of the precise definitions are, like "tangent vector at $x$ that is tangent to $L$") – twosigma Jul 29 '20 at 16:18
  • @twosigma: Consider the 2-sphere $\Bbb S^2$ and a curve $\gamma$ on it. suppose $p\in \Bbb S^2$ and also $p\in \gamma$. Then all vectors belong to tangent space of $ \gamma$ are $T_p\gamma={\lambda \gamma'|\lambda\in \Bbb R}$. this tangent vectors are tangent to sphere too but there are other vectors (like $(T_p\gamma)^\perp$) that are not tangent to $\gamma$. – C.F.G Jul 29 '20 at 16:54
  • Remember that $L$ is a level curve or surface. The function value doesn't change as long as we restrict our movement to $L$. Therefore, if we move in any direction, defined by a unit vector, the directional derivative is zero. Since the direction is a unit vector and the gradient is not zero, the two must be orthogonal.

    That's the intuition.

     – John Douma Jul 29 '20 at 17:33
  • @JohnDouma That was my solution but it has been downvoted and I deleted that. – user Jul 29 '20 at 17:44
  • $\displaystyle \Phi\left(\vec{r}\right) = \mbox{constant} \implies \Phi\left(\vec{r} + \mathrm{d}\vec{r}\right) - \Phi\left(\vec{r}\right) = 0 \implies \nabla \Phi\left(\vec{r}\right)\cdot\mathrm{d}\vec{r} = \color{red}{0}$ – Felix Marin Jul 29 '20 at 18:03

2 Answers2

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If you're granting the fact (given by the implicit function theorem) that the level set actually has a tangent plane at $x$, then any tangent vector is the velocity vector of some curve $\gamma(t)$ contained in the level set. We may assume that $\gamma(0)=x$ and $\gamma'(0)=v$. Then $f(\gamma(t)) = \text{constant}$ (by definition of level set), and so, by the chain rule, $$0=d_xf(\gamma'(0)) = d_xf(v),$$ as you desired.

Ted Shifrin
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  • I do not fully understand the implicit function theorem, and I will revisit it, but does its hypotheses require continuity of the partial derivatives? Do you know if that is necessary for my question to make sense? I only assumed differentiability. – twosigma Aug 01 '20 at 04:17
  • You of course needed to assume $d_x f\ne 0$. But if do not have the hypotheses of the implicit function theorem, what do you mean by the tangent space at $x$ of the level set? Give me a definition, please. Consider, for example, $f(x,y)=y-g(x)=0$, where $g(x) = \begin{cases} x^2, & x \text{ rational} \ 0, & \text{otherwise}\end{cases}$. This function is differentiable only at $x=0$. Does it make sense to talk about the tangent space to the set $y=g(x)$? Yes, there is a tangent line at the origin, but there certainly is not decent structure to this level set. – Ted Shifrin Aug 01 '20 at 04:38
  • Ok, honestly I did not have any definition of tangent space, so that was partly why I asked my question -- out of that confusion. I will revisit this notion when I am more familiar. Thank you. – twosigma Aug 01 '20 at 04:40
  • @Ted Shifrin ,do you mean "some curve" in the sense of "there exists a curve"? Because it seams that "any tangent vector" might not be the velocity vector of just any curve. – Michael Levy Jan 27 '21 at 15:35
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Given the definition of level set, it is a $(n-1)$-dimensional surface in $\mathbb{R}^n$: the set of all points $y$ of $\mathbb{R}^n$ that satisfy a single condition, i.e. $f(y)=f(x),$ with $x$ fixed.

So, the level sets can be represented in parametric form in the neighborhood of each point as $y=y(q),$ where $q\in U\subseteq\mathbb{R}^{n-1}$ are parameters. Given that for each $q\in U$ the corresponding point $y(q)$ belongs to the level set, we have the identity $$ f(y(q))=f(x),\qquad\forall q\in U. $$ Differentiating this relation we have $$ \sum_{i=1}^n\frac{\partial f}{\partial x_i}(y(q))\frac{\partial y_i}{\partial q_k}(q)=0,\qquad \forall k=1,\ldots,n-1, $$ but $\frac{\partial f}{\partial x_i}(y(q))$ are the components of the gradient vector in $y(q)$, while, for fixed $k,$ $\frac{\partial y_i}{\partial q_k}(q)$ are the components of the $k$-th coordinate tangent vector, with respect to the given parametrization, let's call it $t_k(q)$, so the preceding relation can be interpreted as an inner product $$ \nabla f(y(q))\cdot t_k(q) = 0,\qquad\forall k=1,\ldots,n-1. $$ Finally, every tangent vector $v$ in $q$ is a linear combination of the $t_k(q)$, i.e. $v=\sum_{k=1}^{n-1}v_kt_k(q),$ then \begin{align} \nabla f(y(q))\cdot v &= \nabla f(y(q))\cdot\sum_{k=1}^{n-1}v_kt_k(q) = \\ &= \sum_{k=1}^{n-1}v_k[\nabla f(y(q))\cdot t_k(q)] = 0. \end{align}

rschwieb
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Vincenzo Tibullo
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