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If we have an implicit surface $g(x, y, z)=...$, the gradient is a normal vector. But why? Do I have to start with tangent vectors and then show the gradient is perpendicular? Not sure how to approach this.

UXkQEZ7
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1 Answers1

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Fix a point $p$ on the surface and take a curve $\gamma=\gamma(t)$ lying on the surface and passing through $p$, say $\gamma(0)=p$. That implies $$ g(\gamma(t))=0 $$ for all $t$. Differentiate this relation termwise.

This argument shows that $\nabla g(p)$ is orthogonal to the velocity vector $\dot{\gamma}(0)$ for all curves on the surface that pass through $p$. By definition, the plane tangent to the surface at $p$ is the one that is composed of those vectors. Therefore, $\nabla g(p)$ is orthogonal to the tangent plane.

Giuseppe Negro
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  • The only question that I always have in this proof: what is the simplest way to show that the set of all tangent vectors at "good point" is a hyperplane? :) – Evgeny Sep 25 '15 at 17:32
  • Thanks! I couldn't really understand the part between termwise differentiation and showing orthogonality, but this answer helped me find the following link which had a more verbose version. http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-36-proof/MIT18_02SC_notes_19.pdf – UXkQEZ7 Sep 25 '15 at 17:42
  • @Evgeny: Here's how I like to see it. Applying an isometry of $\mathbb R^3$ you can make so that the surface is given locally as a graph: $z=f(x, y)$ (with $p\equiv (0,0,0)$). The set of tangent vectors at $(0,0,0)$ is contained in the hyperplane $z=0$ and it coincides with it because it contains the velocities to the curves $\gamma_x(t)=(t, 0, f(t, 0))$ and $\gamma_y(t)=(0, t, f(0, t))$. – Giuseppe Negro Sep 25 '15 at 17:58
  • Any proof will involve the choice of a coordinate system in a way or another. – Giuseppe Negro Sep 25 '15 at 17:59
  • @GiuseppeNegro Perfect, thank you :) I had similar idea in mind. So, in some sense it's a 'cheap version' of implicit function theorem, right? Or even just a fragment of it without naming it so? :) – Evgeny Sep 25 '15 at 18:35
  • It depends on your definition of "surface". Some define a surface as something that is locally a graph. With this definition, we have done nothing more than translating $p$ to the origin and rotating the axes. There are other definitions in use, and to prove that they are equivalent you need the implicit function theorem. I have posed this question myself some years ago, here. – Giuseppe Negro Sep 25 '15 at 20:13
  • @Evgeny: Previous comment was addressed to you. (I noticed I had forgot the @... construct) – Giuseppe Negro Sep 25 '15 at 23:36
  • @GiuseppeNegro Thank you again! :) – Evgeny Sep 26 '15 at 04:37