Gradient & Smooth Surfaces: Implicit Function Theorem

[Davidllerenav]

Homework Statement

Prove that if $\mathbf{q}$ is a point of a smooth surface $S$ in $\mathbb{R}^m$, then the normal line to $S$ at $\mathbf{q}$ is independent of which smooth function $f:D\subset\mathbb{R}^m\rightarrow\mathbb{R}$ is used to define $S$.

Hint: consider local graph like behaviour and follow the argument of Section$3.8$.

Relevant Equations

None.

Section $3.8$ talks about the gradient and smooth surfaces, defining when the directional derivative $(\partial f/\partial\mathbf{u})(\mathbf{p})$ takes maximum value and that when it equals $0$, then $\mathbf{u}$ is a unit vector orthogonal to $(grad\ f)(\mathbf{p})$.It also says that if $S$ is a level set of a function $f$ $(grad\ f)(\mathbf{p})$ is normal to the level set $S$. I don't know how to use this, since this chapter is about the Implicit Function Theorem.

I've been trying to solve the problem, but I don't know how to begin, can you help me?

 

[fresh_42]

By a surface $S$ defined by $f$ is presumably meant the graph of $f$, i.e. $S=S_f=\{\,(x,f(x))\,|\,x\in D\, \stackrel{f}{\longrightarrow} \mathbb{R}\,\}$. So what has to be proven is, that given $S=S_f=S_g=\{\,(x,g(x))\,|\,x\in D\, \stackrel{g}{\longrightarrow} \mathbb{R}\,\}$ the normal vector at $\mathbf{q}$ does not change. This means if we change the description of $S$ without changing $S$, nothing happens to $\vec{n}_\mathbf{q}$.

Say $S=\{\,(x,y)\,|\,x\in D\,\}$. Now calculate $\vec{n}_\mathbf{q}$ if $y=f(x)$ and if $y=g(x)$. It means that the geometry is the same, even if we change the description of the second coordinate.

At least this is as I understand the question. It is a bit strange since nothing really happens to $S$. I'm not sure I understand the phrase 'independent of which function describes $S$' correctly. I would have expected something else:

If $f,g\, : \,\mathbb{R} \longrightarrow S$ describe two paths in $S$ through $\mathbf{q}$, show that both normal vectors at $\mathbf{q}$ are the same, i.e. different tangents, but the same normal vector.

 

[Davidllerenav]

By a surface $S$ defined by $f$ is presumably meant the graph of $f$, i.e. $S=S_f=\{\,(x,f(x))\,|\,x\in D\, \stackrel{f}{\longrightarrow} \mathbb{R}\,\}$. So what has to be proven is, that given $S=S_f=S_g=\{\,(x,g(x))\,|\,x\in D\, \stackrel{g}{\longrightarrow} \mathbb{R}\,\}$ the normal vector at $\mathbf{q}$ does not change. This means if we change the description of $S$ without changing $S$, nothing happens to $\vec{n}_\mathbf{q}$.

Say $S=\{\,(x,y)\,|\,x\in D\,\}$. Now calculate $\vec{n}_\mathbf{q}$ if $y=f(x)$ and if $y=g(x)$. It means that the geometry is the same, even if we change the description of the second coordinate.

At least this is as I understand the question. It is a bit strange since nothing really happens to $S$. I'm not sure I understand the phrase 'independent of which function describes $S$' correctly. I would have expected something else:

If $f,g\, : \,\mathbb{R} \longrightarrow S$ describe two paths in $S$ through $\mathbf{q}$, show that both normal vectors at $\mathbf{q}$ are the same, i.e. different tangents, but the same normal vector.

I think that I need to say that $S$ is a level set of a continuously differentiable function $f:D\subset\mathbb{R}^m\rightarrow\mathbb{R}$ such that $(grad\ f)(\mathbf{x})\neq\mathbf{0}$. Is that what you mean by $S=S_f=\{\,(x,f(x))\,|\,x\in D\, \stackrel{f}{\longrightarrow} \mathbb{R}\,\}$?

 

[fresh_42]

I meant that $S$ is the graph of $f$, but in this case we have $S=S_f=S_g=\{\,x\in D\,|\,f(x)=g(x)=c\,\}$ which is a cut through the graph. You have to combine the two functions somehow, so that the normal vectors can be compared.

 

[Davidllerenav]

I meant that $S$ is the graph of $f$, but in this case we have $S=S_f=S_g=\{\,x\in D\,|\,f(x)=g(x)=c\,\}$ which is a cut through the graph. You have to combine the two functions somehow, so that the normal vectors can be compared.

Is there any Hint you can give me on how to relate both functions? Also, do I need the Implicit Function Theorem in this problem? Because this chapter is about it.

 

[fresh_42]

We have a graph, or a level set within the graph. This is already the setup for the theorem of implicit functions: $(x,f(x)) = (x,c) = (x,g(x))$. Are you allowed to use the theorem about inverse functions? It is a corollary to the implicit function theorem and says when a function is locally invertible. Then you could consider $g\circ f^{-1}$.

But I haven't done your homework, it is just an idea.

 

[Davidllerenav]

We have a graph, or a level set within the graph. This is already the setup for the theorem of implicit functions: $(x,f(x)) = (x,c) = (x,g(x))$. Are you allowed to use the theorem about inverse functions? It is a corollary to the implicit function theorem and says when a function is locally invertible. Then you could consider $g\circ f^{-1}$.

But I haven't done your homework, it is just an idea.

Yes, I think I can use the inverse function theorem. But can you explain the inverse function part, I didn't understand what you meant.

 

[fresh_42]

Yes, I think I can use the inverse function theorem. But can you explain the inverse function part, I didn't understand what you meant.

You are right, it doesn't work for $f$ and $g$ as given. That's where the hint probably comes into play. Use the graph to make the dimensions match. I just gave you hints what to test, I didn't solve the question; e.g. we have $(x,f(x))=(x,g(x))=(x,c)$ and thus a function $F\, : \,\mathbb{R}^{n+1} \longrightarrow \mathbb{R}^{n+1}$ given as $F(x,y)=(x,f(x))$ and the same for $g$. And how is the normal vector defined?

 

[Davidllerenav]

You are right, it doesn't work for $f$ and $g$ as given. That's where the hint probably comes into play. Use the graph to make the dimensions match. I just gave you hints what to test, I didn't solve the question; e.g. we have $(x,f(x))=(x,g(x))=(x,c)$ and thus a function $F\, : \,\mathbb{R}^{n+1} \longrightarrow \mathbb{R}^{n+1}$ given as $F(x,y)=(x,f(x))$ and the same for $g$. And how is the normal vector defined?

The normal vector is given by the gradient of $F$, right?

 

[fresh_42]

Yes. And here we have to show, that $f$ and $g$ give the same gradient. Therefore we need to translate "used to define $S$" into an equation of $f$, resp. $g$, and that this doesn't change the gradient.
What does it mean if both functions describe $S$?

Have a look on the Wikipedia pages under "normal vector". I found the english and german pages best for our context. The other languages are a bit too short.

 

[Davidllerenav]

Yes. And here we have to show, that $f$ and $g$ give the same gradient. Therefore we need to translate "used to define $S$" into an equation of $f$, resp. $g$, and that this doesn't change the gradient.
What does it mean if both functions describe $S$?

Have a look on the Wikipedia pages under "normal vector". I found the english and german pages best for our context. The other languages are a bit too short.

If both functions describe $S$ they are equivalent.
I found on Wikipedia something about differential variety, but I didn't understand it to be honest.

 

[fresh_42]

If both functions describe $S$ they are equivalent.
I found on Wikipedia something about differential variety, but I didn't understand it to be honest.

A trick I often use is the following: Call the english page, switch to german, which is more detailed on the case we need (and in general, too. They often have real calculations when the english page has only abstract descriptions) , and either try to read it, as formulas are formulas, or let chrome translate the page. It might result in a terrible english, but we do not want to write an essay, so who cares.

 

[WWGD]

I am not sure everyone Reads German , Freshmeister.

 

[WWGD]

David: tangent vectors in a tangent space are equivalence classes of curves in the same direction. I assume f,g may just be different parametrizations of the same surface, and the normal does not depend on the choice of parametrization.

 

[fresh_42]

I am not sure everyone Reads German , Freshmeister.

let chrome translate the page

... and believe it or not, the formulas are actually the same!

 

[WWGD]

... and believe it or not, the formulas are actually the same!

Last I checkd those translations were not always great.

 

[fresh_42]

Last I checkd those translations were not always great.

Yes, not for a test in English class, but enough to understand the content, especially as many words can easily be guessed. One doesn't need a translator to understand "implizite Funktion". It is not optimal but can be done real quick and often gives additional information. Especially Wikipedia pages are not translated, but written by different people in different ways. Me, too, checks sometimes language versions I do not speak well, e.g. Spanish or French.