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Calculating the directional derivative of a function of two variables

  1. Nov 10, 2009 #1
    The problem statement, all variables and given/known data

    Consider the function:
    z=f(x,y)= log(x^2 + y^2) (x,y)=/=(0,0)

    Calculate the directional derivative of f(x,y) at (x,y)=(1,1) in the direction of the vector (1,2)


    The attempt at a solution

    When i tried to work out the unit vector from the point (1,1) to (1,2) i got (0,1).

    I got partial derivative df/dx = 2x/(x^2 + y^2)
    and partial derivative df/dy = 2y/(x^2 + y^2)

    Then, for gradf at (1,1) i got (1,1)..

    so, for directional derivative i got:

    (gradf at (1,1)) x unit vector = (1,1).(0,1) = 1

    But the answer is 3/root5

    Does anyone know what i have done wrong? Thankyou
     
  2. jcsd
  3. Nov 10, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi wshfulthinker! Welcome to PF! :smile:

    (have a curly d: ∂ and a square-root; √ and a grad: ∇ and try using the X2 tag just above the Reply box :wink:)
    Nooo … you're msunderstanding "the direction of the vector (1,2)" …

    it's not the point (1,2) (which is in direction (0,1) from (1,1), as you say) …

    (1,2) is the actual direction that you're taking the derivative along.

    (otherwise, your method is ok :wink:)
     
  4. Nov 10, 2009 #3
    Hi, thanks for the welcome and showing me the symbols!

    I don't really get it though! where do i use the point (1,2). I'm not even sure what i worked out, i followed the method that were in my lecture notes which were worded almost the exact same way as my actual question (except it said find instead of calculate - i don't know if that means it's different)
     
  5. Nov 10, 2009 #4

    tiny-tim

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    grrr! it's not a point!! :rolleyes:

    it's a direction … use it instead of your (0,1). :smile:
     
  6. Nov 10, 2009 #5
    Okay okay... so i am pretty crap with vectors! But i got the answer finally! :D I found a book which wrote the direction in the i + j form which made more sense to me and didn't make me think it was a point. And yes, i got the answer so i think i kind of understand it now... Thankyou! :)
     
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