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Gradient of function of x,y,z is perpendicular to point on surface?

  1. Aug 21, 2012 #1
    http://img580.imageshack.us/img580/8859/proofea.jpg [Broken]

    i am having trouble figuring out how to proceed without knowing anything about the function ψ=ψ(x,y,z). all i know is that it's a function of x,y,z. but how can i figure out how the surface changes w.r.t the tangent vector if i dont even know what the surface is??
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 21, 2012 #2

    Dick

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    Write the curve in the surface as (x(t),y(t),z(t)) where t is a parameter and (x'(t),y'(t),z'(t)) is the tangent vector. See what the chain rule tells you about dψ/dt.
     
  4. Aug 21, 2012 #3

    TSny

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    The surfaces are defined by ψ = constant. So, if you make an infinitesimally small displacement that is tangent to such a surface (thereby keeping you within the surface), what is the value of dψ?

    [Edit: I was not aware that this question was also posted in the introductory physics forum]
     
  5. Aug 21, 2012 #4

    vela

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    I merged the threads and moved it to the calculus forum.
     
  6. Aug 21, 2012 #5
    [tex]\frac{\partialψ}{\partial t} = \frac{\partialψ}{\partial x} \frac{\partial x}{\partial t} + \frac{\partialψ}{\partial y} \frac{\partial y}{\partial t} + \frac{\partialψ}{\partial z} \frac{\partial z}{\partial t} [/tex]

    my guess is that dx/dt, dy/dt and dz/dt are all zero. because in that case you can integrate dψ/dt = 0 and get ψ=constant?? but why would dx, dy and dz be constant? is the problem implying that for any surface described by f(x,y,z), that the slope of the tangent line stays the same or something like that?
     
  7. Aug 21, 2012 #6

    Dick

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    You've got the right expression. I think dψ/dt is the thing that should be zero. Why? Isn't the other side a dot product between grad(ψ) and the tangent vector?
     
  8. Aug 21, 2012 #7
    how can the surfaces be described by ψ=constant? wouldnt that mean x,y,z are constant and you're at one specific point on the surface?
     
  9. Aug 21, 2012 #8
    but how can we show that the dot product of grad(ψ) and the tangent vector is 0? or that grad(ψ) is perpendicular to the tangent vector?
     
  10. Aug 21, 2012 #9

    Dick

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    Nooo! Pick ψ=x^2+y^2+z^2=1. There are LOT of points on a sphere of radius 1!
     
  11. Aug 21, 2012 #10

    Dick

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    You didn't answer my first question about why dψ/dt should be zero. Write me an expression for grad(ψ) in terms of x, y, and z and the tangent vector to (x(t),y(t),z(t)), the curve that's on the surface of ψ=constant.
     
  12. Aug 21, 2012 #11
    oh i see....got it...yeah i was looking at it the wrong way
     
  13. Aug 21, 2012 #12
    so by saying that ψ is constant they're saying every point on this surface is the same distance fromt he origin...does that basically mean it HAS to be a sphere? or could it be some other surface?
     
  14. Aug 21, 2012 #13

    Dick

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    No again. Pick ψ=x+y+z=1. Now the surface is a plane. It doesn't really matter what the surface is.
     
  15. Aug 21, 2012 #14
    like this?
    [tex]grad(ψ) = \frac{\partial ψ}{\partial x} i + \frac{\partial ψ}{\partial y} j+ \frac{\partial ψ}{\partial z} k [/tex]
    dχT= dx i + dy j +dz k

    are you saying grad(ψ) is 0 since ψ is constant?
     
    Last edited: Aug 21, 2012
  16. Aug 21, 2012 #15

    vela

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    Again, consider the equation of the plane x+y+z=1. You have ##\psi(x,y,z) = x+y+z##, and ##\nabla\psi## clearly isn't equal to 0.
     
  17. Aug 21, 2012 #16
    i see..but again how am i supposed to differentiate a function when i dont know what the function is? should i just pick any function of x,y,z?
     
  18. Aug 22, 2012 #17

    vela

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    No, you don't need to calculate the actual derivatives. You just need to show that there is a certain relationship between the derivatives.

    Look at posts 3 and 6 again. What is dψ or dψ/dt equal to if you stay on the surface?
     
  19. Aug 22, 2012 #18
    zero?
     
  20. Aug 22, 2012 #19

    HallsofIvy

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    The whole point is that on a 'level surface', f(x, y, z)= constant, the value of f is a constant! If [itex]\vec{u}[/itex] is a unit tangent vector then [itex]D_{\vec{u}} f= 0[/itex] because it is the rate of change of f in the direction of [itex]\vec{u}[/itex] and f doesn't change in that direction. Therefore, we must have [itex]D_{\vec{u}}f= \nabla f\cdot \vec{u}= 0[/itex]. That is, [itex]\nabla f[/itex] is perpendicular to any tangent vector and so perpendicular to the surface.
     
  21. Aug 22, 2012 #20
    thanks! makes sense...

    im still trying to make sense of the meaning of ψ=constant. does it mean that if you take any small area element of the surface it is the same throughout the surface? is that a good way to interpret it?
     
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