According to my lecture notes, the normal vector to a surface $f$ is given by $\text{grad}(f) = \underline{\nabla} f$. However, surely the normal vector to the surface $f$ should be perpendicular to the vector $\underline{\nabla}f$?
Asked
Active
Viewed 125 times
0
-
4Duplicate of Find normal vector for the surface F(x,y)=0 and Why is the gradient of implicit surface the normal vector (i.e. parallel to the normal line)? – Jan 01 '16 at 17:49
1 Answers
0
If the surface, $M$, say, is given by $f^{-1}(c)$ for a regular value $c$ of $f$ then $f$ is constant along $M$. So for any curve $\gamma(t)$ in $M$, $f\circ \gamma= c$ and consequently $$\frac{d}{dt}f\circ \gamma = 0 =\langle \gamma^\prime, (\nabla f)\circ\gamma\rangle$$ Since any tangent vector to $M$ arises as a tangent to a curve in $M$, $\nabla f$ is perpendicular to $M$.
Thomas
- 22,361
-
What are the domains and codomains of $\gamma$ and $f$ you had in mind here? – Eli Rose Jan 01 '16 at 17:59
-
@EliRose It does not really matter, but the standard case is, for sure, that $f:\mathbb{R}^n\rightarrow \mathbb{R}$ and $\gamma:(-\varepsilon, \varepsilon ) \rightarrow M := f^{-1}(c)$ – Thomas Jan 01 '16 at 19:44