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Why does this work? Directional Derivatives.

  1. Oct 16, 2011 #1
    Find the directional derivative of:
    f(x,y) = (x^2)y-y^3, (2,1,3)


    Can somebody explain to me why we did the step we did?

    First we found the directional derivative of f(x,y,z)
    ▽f(x,y)
    = <2xy, (x^2)-3(y^2), a>
    ▽f(2,1)
    = <2(2)(1), (2^2)-3(1^2)), a>
    = <2(2), 4-3, a> = <4,1, a>

    and then we... did some sorcery mathematics?!
    f(x,y) = (x^2)y-y^3
    z = f(x, y)
    F(x,y,z) = f(x,y) - z
    ▽F(x,y,z) = <2xy, (x^2)-3(y^2), -1>
    ▽F(2,1,3) = <4,1,-1>

    A dot product for some reason...
    <4,1,a><4,1,-1> = 0
    4(4)+1(1)-a = 0
    16+1 = a
    17 = a

    Therefore, the directional derivative is, <4,1,17>.
    __________________________________________

    My question is: if we are given a function f(x,y) and are told to find the directional derivative at the point (xo,yo,zo) is it safe to say that we can always solve it like we did above?

    Du▽f(x,y,z) = <fx(x,y), fy(x,y), <fx(x,y), fy(x,y)><fx(x,y), fy(x,y)>>?

    Can somebody explain to me why the equation above is valid? This isn't homework, I'm just reviewing my notes and couldn't understand what happened here. This is just for my own understanding... and because of that I'm worried I'm posting this in the wrong thread.

    Thank you for taking the time to read my question.
     
  2. jcsd
  3. Oct 16, 2011 #2

    Bacle2

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    Science Advisor

    This is just a variant of the chain rule. The partials are the directional derivatives in the direction of the standard x,y,z axes, so that the chain rule just multiplies by 1 at most.

    Basically, when you do the quotient [f(x+th,y,z)-f(x,y,z)]/h as h->0, the chain rule
    will multiply the partial by the directional vector.
     
  4. Oct 16, 2011 #3
    So that equation, Du▽f(x,y,z) = <fx(x,y), fy(x,y), <fx(x,y), fy(x,y)><fx(x,y), fy(x,y)>>, will always hold (when asked for the directional derivative of f(x,y) at point (x,y,z))?
     
  5. Oct 16, 2011 #4

    Bacle2

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    Science Advisor

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