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

[cpfoxhunt]

Homework Statement

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 \nablaf(a,b,c) / modulus of \nablaf(a,b,c)

(Apologies for the bad use of latex)

Homework Equations

None really

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

 

[HallsofIvy]

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.

 

[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 ?

 

[HallsofIvy]

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.