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

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  • #1
racnna
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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 don't even know what the surface is??
 
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  • #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.
 
  • #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]
 
  • #4
vela
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I merged the threads and moved it to the calculus forum.
 
  • #5
racnna
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[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?
 
  • #6
Dick
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[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?

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?
 
  • #7
racnna
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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]
how can the surfaces be described by ψ=constant? wouldn't that mean x,y,z are constant and you're at one specific point on the surface?
 
  • #8
racnna
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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?

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?
 
  • #9
Dick
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how can the surfaces be described by ψ=constant? wouldn't that mean x,y,z are constant and you're at one specific point on the surface?

Nooo! Pick ψ=x^2+y^2+z^2=1. There are LOT of points on a sphere of radius 1!
 
  • #10
Dick
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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?

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.
 
  • #11
racnna
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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!

oh i see...got it...yeah i was looking at it the wrong way
 
  • #12
racnna
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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?
 
  • #13
Dick
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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?

No again. Pick ψ=x+y+z=1. Now the surface is a plane. It doesn't really matter what the surface is.
 
  • #14
racnna
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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.

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?
 
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  • #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.
 
  • #16
racnna
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i see..but again how am i supposed to differentiate a function when i don't know what the function is? should i just pick any function of x,y,z?
 
  • #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?
 
  • #18
racnna
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zero?
 
  • #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.
 
  • #20
racnna
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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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