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Partial derivatives.

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    First problem: Let [tex]f(x,y) = x-y[/tex] and u = vi+wj. In which direction does the function decrease and increase the most? And what u (all of them) satisfies Duf = 0

    Second problem: Let [tex]z = f(x,y)[/tex], where [tex]x = 2s+3t[/tex] and [tex]y = 3s-2t[/tex]. Determine [tex]\partial{z^2}/\partial{s^2}[/tex]

    2. Relevant equations

    Gradient and the chain rule

    3. The attempt at a solution

    For the first question in the first problem I've gotten using gradient: increase i-j and decrease -i+j. Am I correct? For the second question all I've gotten so far is (nabla)f(dot)u = 0 = (1-y)v+(x-1)w. Where do I get the second equation to solve both v and w?

    Second problem gives me 4zxx+12zxy+9zyy. Is that completely wrong?
    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 15, 2009 #2


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    What you have done so far is correct. For the second part, here's any easy hint: in 2 dimensions, a perpendicular to vector <a, b> is <-b, a>.

    For the second problem, [tex]f_s= f_x x+_s+ f_y y_s= 2f_x+ 3f_y[/tex] .
    That applies to any function of x,y so it applies to [tex]f_s[/tex] itself.

    [tex]f_{ss}= (f_s)_s= 2(2f_x+ 3f_y)_x+ 3(2f_x+ 3f_y)_y= 4f_{xx}+ 6f_{yx}+ 6f_{xy}+ 9f_yy[/tex]
    and since, for this function, with continuous second derivatives, the mixed derivatives are the same,
    [tex]f_ss= 4f_{xx}+ 12f_{xy}+ 9f_{yy}[/tex]
    exactly as you say.
  4. Feb 15, 2009 #3
    So I just mark (1-y)w-(x-1)v = 0?

    Thank you, I never thought of that and it will make things much easier.
  5. Feb 15, 2009 #4


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    ?? You had already found that the gradient was i- j and you appeared to know that the directional derivative, in the direction of unit vector [itex]\vec{i}[/itex] is the dot product [itex]\nabla f\cdot \vec{v}[/itex]. That will be 0 if and only if the two vectors are orthogonal: the direction in which the derivative is 0 is perpendicular to the gradient vector, i+ j.

  6. Feb 16, 2009 #5
    Thank you. This is a new subject for me and my thinking was too complicated. I found that the vector is either [itex]\vec{i}+\vec{j}[/itex] or [itex]-\vec{i}-\vec{j}[/itex].

    And I just realized that this function is so simple that I could have found the vectors by giving the function a set value and draw it in 2-dimensions. I know there is a fancy name for that, but I've no idea what it is in English.
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