Easy proof of Why del(phi) is normal to a surface?

1. Mar 30, 2008

cpfoxhunt

1. The problem statement, all variables and given/known data

Prove that for a 3d space s, defined by a function f(x,y,z) = 0 , a unit vector normal to surface at the point (a,b,c) is given by $$\nabla$$f(a,b,c) / modulus of $$\nabla$$f(a,b,c)

(Apologies for the bad use of latex)

2. Relevant equations

None really

3. The attempt at a solution

I can only seem to gesture at this - it was given us as a definition. I know that d(phi)/ds where phi is a surface and s is a distance = $$\nabla$$(Phi).A , and that surely if a is tangential the LHS is equal to zero, but I'm a bit stuck from there and not sure if i've used things I need to prove. Its only a few marks, but is bugging me.

Any help is greatly appreciated,
Cheers
Cpfoxunt

2. Mar 30, 2008

HallsofIvy

Staff Emeritus
Are you aware that $D_{\vec{v}}\phi= \nabla\phi\cdot\vec{v}$, where $D_{\vec{v}}$ is the derivative in the direction of the unit vector $\vec{v}$? From that, it follows that if $\vec{v}$ is tangent to a "level surface" $\phi$= constant, then $D_{\vec{v}}\phi= \nabla\phi\cdot\vec{v}= 0$. That is, $\nabla\phi$ is perpendicular to any tangent vector and so to the surface itself.

3. Mar 30, 2008

cpfoxhunt

I am aware of that (and now see why its important). That was what i was trying to get at with my attempt. But i'm still not quite happy - it doesn't explain the unit vector evaluated at (a,b,c), and I feel like I should try to gesture at why D(phi) = del(phi).V ?

4. Mar 30, 2008

HallsofIvy

Staff Emeritus
I don't understand what you mean by "explain the unit vector". If you recognize that $\nabla \phi$ is perpendicular to the surface, then of course dividing by its own length will give a unit vector in that direction.

As for why $D_{\vec{v}}\phi= \nabla\phi\cdot\vec{v}$, a unit vector can always be written in the form $a\vec{i}+ b\vec{j}+ c\vec{k}$ where, of course, $\sqrt{a^2+ b^2+ c^2}= 1$. We can write a line in that direction as $x= x_0+ at$, $y= y_0+ bt$, and $z= z_0+ bt$. So $\phi(x,y,z)$ in that direction is $\phi( x_0+ at, y_0+ bt, z_0+ ct)$. Differentiate that with respect to t (using the chain rule) and see what you get.