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Visualizing the partial derivatives fxx, fyy, fxy and fyx? + Second Derivative Test?

  1. May 23, 2012 #1
    1. The problem statement, all variables and given/known data
    So fx is how much f changes when you change x. Thus fxx is the rate of change of fx, or geometrically how fast the functions slope is changing. The same can be said for fy and fyy. But what about fxy and fyx? Could someone please explain to me what they mean?

    I want to understand what it means to understand the second derivative test:
    http://dl.dropbox.com/u/64325990/MATH%20253/Capture.PNG [Broken]

    I am not sure what the meaning of the term [fxy(a,b)]^2 is. I can't seem to visualize what's going on when you take the second derivative test and the textbook doesn't have a proof.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 23, 2012 #2

    Mark44

    Staff: Mentor

    Re: Visualizing the partial derivatives fxx, fyy, fxy and fyx? + Second Derivative T

    fxy is the rate of change of fx relative to a change in y.
    fxy is also written as
    $$ \frac{\partial}{\partial y} \frac{\partial f}{\partial x}$$

    Note that the two kinds of notation are a little confusing, as the order of x and y is reversed in the two kinds of notation.
    For [fxy(a, b)]2,
    1. take the partial of f with respect to x
    2. take the partial of fx with respect to y
    3. evaluate the result of step 2 at the point (a, b).
    4. square the result of step 3.

    For example, if f(x, y) = 2x3y2, and we need to evaluate it at (1, 1),
    fx = 6x2y2 and fxy = 12x2y.
    fxy(1, 1) = 12*1*1 = 12

    Squaring that result gives you 144.
     
    Last edited by a moderator: May 6, 2017
  4. May 23, 2012 #3
    Re: Visualizing the partial derivatives fxx, fyy, fxy and fyx? + Second Derivative T

    Thank you but I what I really don't understand is what fxy looks like on a graph. What is it's visual representation? If I can understand that then I think I can answer other questions like:

    Why you would subtract [fxy(a, b)]2, why not add it.
     
  5. May 23, 2012 #4
    Re: Visualizing the partial derivatives fxx, fyy, fxy and fyx? + Second Derivative T

    with the partial with respect to x, you are able to extract the flat structure of the function f but only in the x direction. so you're transforming the function f into another function df/dx, and you find its measures the flatness of f but only in the direction of x. now look at df/dx and we want to find the flat structure of this function but now in the y direction. so we create another function from this one that measures the flatness of f in the direction of x and the flatness of df/dx in the direction of y. this function does not necessarily measure flatness in other directions, in fact a derivative may not even exist in other directions.

    sorry my cumbersome attempt to describe it.
     
  6. May 23, 2012 #5

    Mark44

    Staff: Mentor

    Re: Visualizing the partial derivatives fxx, fyy, fxy and fyx? + Second Derivative T

    The graph of z = f(x, y) requires three dimensions: two for the domain, and one for the range.

    Using my example of f(x, y) = 2x3y2, can you get a graph of that function?
    If so, you should also be able to graph fx(x, y) = 6x2y2 and fxy(x, y) = 12x2y.
    If I recall correctly, this is directly related to what is known as the Hessian matrix (http://en.wikipedia.org/wiki/Hessian_matrix), whose deteriminant consists of the second partials. I don't remember why this plays a role in categorizing critical points, but it does.
     
    Last edited: May 24, 2012
  7. May 24, 2012 #6
    Re: Visualizing the partial derivatives fxx, fyy, fxy and fyx? + Second Derivative T

    Alright it is starting to make sense now, thanks everyone.

    Does anyone know how to complete the square for this? It looks so complicated :S
    http://dl.dropbox.com/u/64325990/MATH%20253/Capture1.PNG [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. May 24, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Re: Visualizing the partial derivatives fxx, fyy, fxy and fyx? + Second Derivative T

    We want to know if the Hessian is positive definite, negative definite, or indefinite. In the first case the point is a strict local min, in the second case a strict local max, and in the third case a saddle point. The determinant is the product of the Hessian's eigenvalues, so if it is > 0 both eigenvalues have the same sign. THAT is why it plays a role.

    RGV
     
    Last edited by a moderator: May 24, 2012
  9. May 24, 2012 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Visualizing the partial derivatives fxx, fyy, fxy and fyx? + Second Derivative T

    In a more abstract, but more general, sense, the "derivative" of a function, f, from Rm to Rn is the linear function, Lf, from Rm to Rn that best approximates f ("best approximates" can be made precise through limit calculations). A linear function from Rn to Rm can be written as an n by m matrix.

    So the first derivative of f, from R3 to R is a "3 by 1" matrix or vector- the gradient vector, in fact. And since the first derivative is from R3 to R3, the second derivative is a linear transformation from R3 to R3- which, of course, can be represented by a 3 by 3 matrix- the "Hessian" that Ray Vickerson mentions:
    [tex]\begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}[/tex]

    Since the "mixed" derivatives are equal, that is a symmetric matrix and there exist a "basis" (coordinate system) in which it is diagonal. That is, there exist x', y' such that the second derivative is
    [tex]\begin{bmatrix}\frac{\partial^2 f}{\partial x'^2} & 0 \\ 0 & \frac{\partial^2 f}{\partial y'^2}\end{bmatrix}[/tex]

    And, since the first derivatives are 0 at a critical point, we can write f as
    [tex]f(x,y)= f(x;_0, y'_0)+ \frac{\partial^2 f}{\partial x'^2)(x'- x'_0)^2+ \frac{\partial^2 f}{\partial y'^2}(y- y_0)^2[/tex]
    plus higher power terms. If those two derivatives are both positive (at [itex](x_0, y_0)[/itex]) then it f is increases in a neighborhood and so f has a minimum there. If they are both negative, f has a maximum there. If one is positive and the other negative, f has a saddle point.

    Note that the determinant of that matrix is
    [tex]\frac{\partial^2f}{\partial x'^2}\frac{\partial^2f}{\partial y'^2}[/tex]
    which is positive if they both have the same sign (so increasing or decreasing) and negative if they have opposite signs (so saddle point). Finally, the determinant of a matrix is independent of the change of coordinates so that it is sufficient to check
    [tex]\frac{\partial^2f}{\partial x^2}\frac{\partial^2f}{\partial y^2}[/tex]
    without having to actually make that change of variable.

    Unfortunately, that doesn't work for dimensions higher than 3. The sign of the determinant, the sign of the product of the diagonal elements, does not tell us much about the sign of the individual terms.
     
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