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

  • Thread starter cpfoxhunt
  • Start date
16
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1. 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 [tex]\nabla[/tex]f(a,b,c) / modulus of [tex]\nabla[/tex]f(a,b,c)

(Apologies for the bad use of latex)

2. Homework 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 = [tex]\nabla[/tex](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
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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Are you aware that [itex]D_{\vec{v}}\phi= \nabla\phi\cdot\vec{v}[/itex], where [itex]D_{\vec{v}}[/itex] is the derivative in the direction of the unit vector [itex]\vec{v}[/itex]? From that, it follows that if [itex]\vec{v}[/itex] is tangent to a "level surface" [itex]\phi[/itex]= constant, then [itex]D_{\vec{v}}\phi= \nabla\phi\cdot\vec{v}= 0[/itex]. That is, [itex]\nabla\phi[/itex] is perpendicular to any tangent vector and so to the surface itself.
 
16
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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
Science Advisor
Homework Helper
41,736
894
I don't understand what you mean by "explain the unit vector". If you recognize that [itex]\nabla \phi[/itex] is perpendicular to the surface, then of course dividing by its own length will give a unit vector in that direction.

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

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