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Equation of a Tangent Plane

  1. Jun 24, 2005 #1
    I know that a tangent plane is given by:

    [tex]a(x-x_0)+b(y-y_0)+c(z-z_0)=0[/tex]

    Where <a,b,c> is the gradient vector. When I was given the problem of writing an equation (z=...) for it, I replaced <a,b,c> with the partials that compose the gradient vector:

    [tex]\left<a,b,c\right >=\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right >[/tex]

    Now I have:

    [tex]\frac{\partial f}{\partial x}(x-x_0)+\frac{\partial f}{\partial y}(y-y_0)+\frac{\partial f}{\partial z}(z-z_0)=0[/tex]

    This is where I am having trouble solving for z because of the partial in front of it.

    Any suggestions?

    Thanks a lot.
     
  2. jcsd
  3. Jun 24, 2005 #2
    i don't understand whats the problem here
    :: Though partial is infront of z put it is operating on f only ::
     
  4. Jun 24, 2005 #3
    I mean, when I try to solve for z, I come up with:

    [tex]\frac{\partial f}{\partial z}(z-z_0)=-\frac{\partial f}{\partial x}(x-x_0)-\frac{\partial f}{\partial y}(y-y_0)[/tex]

    From here, I want to solve for z, but when I do so I come up with something different than it should be.
     
  5. Jun 24, 2005 #4

    mathwonk

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    its called "division".
     
  6. Jun 24, 2005 #5
    Of course, I know that. I'm just saying that when I solve, I come up with something totally different than what it should be. I was basically wondering whether I set it up wrong, but I guess not.
     
  7. Jun 25, 2005 #6
    ?????? i think u don't want to solve PDE
     
  8. Jun 25, 2005 #7
    I might be mistaken but with your formula, isn't it supposed to be the partial derivatives evaluated at the given point so its not just df/dx but (df/dx)(x,y,z). In that case the partial derivative part is just a constant. In any case I think its better to find the tangent plane from the gradient vector "dotted" with (X-x_0) where X = (x,y,z,) and x_0 is your given point.
     
  9. Jun 25, 2005 #8

    HallsofIvy

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    Then you should tell us exactly what the problem says, what answer you got, and what you were told the correct answer should be.

    Any time your answer does not match the answer given in the book, there are three possibilities:

    1) Your answer might be wrong!

    2) The books answer might be wrong! (Doesn't happen often but it happens)

    3) You might have the same the same answer as the book but written in a different form.
    (A friend and I once came up with "different" answers to a problem. It took us three days to show that they were, in fact, the same.)
     
  10. Jun 25, 2005 #9
    Apparently, the answer is supposed to be:

    [tex]z=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+f(x_0,y_0)[/tex]

    When I try to solve my way, I come up with that PDE (I think) and I don't want to just multiply both sides by [tex]\frac{\partial z}{\partial f}[/tex], so I need to know how the real answer was arrived at.

    Thanks again for all of your help.
     
    Last edited: Jun 26, 2005
  11. Jun 26, 2005 #10

    HallsofIvy

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    We have asked you repeatedly to state exactly what the problem is. In your first post you take the gradient of the function f(x,y,z) but in your last post you have f(x0,y0). How is the surface given? Is it f(x,y,z)= constant or z= f(x,y)?

    If it is the latter, then the answer you give in your last post is correct and follows immediately from the formula you give in your first post by thinking of the surface as F(x,y,z)= f(x,y)-z which has gradient <fx,fy,-1>. The fz you are worrying about is just -1.
    For example, if z= x2+ xy, to find the tangent plane at (1,1) let F(x,y,z)= x2+ xy- z so that the gradient is <2x+ y, x, -1> which is <3, 1, -1> at (1,1). z(1,1)= 2 so the tangent plane there is 3(x-1)+ 1(y-1)- 1(z- 2)= 0 or
    z= 3(x-1)+ (y-1)+ 2.

    If it is the former, then the answer in your last post is impossible. What you would do is use the formula in your first post evaluating the derivatives at the given point so they are just numbers.

    For example, if f(x,y,z)= x2+ xy+ z2= 3, to find the tangent plane at (1, 1, 1), find the gradient of f= <2x+ y, x, 2z> and evaluate it at (1, 1, 1):
    <3, 1, 2>. The equation of the tangent plane is 3(x-1)+ 1(y-1)+ 2(z-1)= 0.
     
  12. Jun 26, 2005 #11
    Alright that answered my question. Thank you.

    I realize I didn't make my question very clear and I appologize about that. :)
     
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