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Directional derivative

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data

    (Q) The derivative of f(x,y) at Po(1,2) in the direction i + j is 2sqrt(2) and in the direction of -2j is -3. What is the derivative of f in the direction of -i - 2j? Give reasons for your answers.

    2. Relevant equations

    The directional derivative is given by the formula:

    ∂f/∂x i+∂f/∂y j

    3. The attempt at a solution

    You get simultaneous equations when you apply the above equation and you find that

    ∂f/∂y = 3/2.
    And ∂f/∂x = [4sqrt(2) - 3] / 2.

    Then applying the dot product of this and -i - 2j, you get [-3-4sqrt(2)] / 2 but the answer is supposed to be -7/sqrt(5). How did they get that??:confused:
     
  2. jcsd
  3. Oct 23, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that's true.

    No, that's not true. "The derivative of f(x,y) at Po(1,2) in the direction i + j is 2sqrt(2)" tells you that [itex]f_x/\sqrt{2}+ f_y/\sqrt{2}= 2\sqrt{2}[/itex] (dividing by the length of i+ j) or that [itex]f_x+ f_y=4[/itex]. Since [itex]f_y= 3/2[/itex], that gives [itex]f_x= 5/2[/itex]

    No, take the dot product of [itex](5/2)i+ (3/2)j[/itex] with the unit vector in the direction of -i- 2j.

    Remember that the derivative in the direction of vector v is [itex]\nabla f \cdot v/||v||[/itex].

    You keep forgetting to divide by the length of v.
     
  4. Oct 23, 2007 #3
    Eye opener!!

    Thank-you very much for explicitly exposing my weakness!! I really mean it. Now, I'll never forget to divide by the length! :smile:
     
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