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

  1. Mar 30, 2008 #1
    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 [tex]\nabla[/tex]f(a,b,c) / modulus of [tex]\nabla[/tex]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 = [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
     
  2. jcsd
  3. Mar 30, 2008 #2

    HallsofIvy

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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.
     
  4. Mar 30, 2008 #3
    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 ?
     
  5. Mar 30, 2008 #4

    HallsofIvy

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    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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