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Calculate the directional derivative

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the directional derivative, direction, and rate of maximum increase in the direction of v at the given point.
    f(x,y) = x^2 + y^3
    v = <4,3>
    P = (1,2)

    2. Relevant equations
    I know how to calculate directional derivative but I don't know how to calculate rate of maximum increase and direction


    3. The attempt at a solution
     
    Last edited: Nov 8, 2011
  2. jcsd
  3. Nov 8, 2011 #2

    Dembadon

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    Re: Gradient

    A function increases most rapidly at P0 in the direction of grad ƒ evaluated at P0.

    Given the above fact, when would you say that it deceases most rapidly? Hint: you don't need to do any other computations.
     
  4. Nov 8, 2011 #3
    Re: Gradient

    oh wait... is direction the gradient?
     
  5. Nov 8, 2011 #4

    Dembadon

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    Re: Gradient

    The gradient is the vector that is orthogonal (perpendicular) to the level curve at the point you gave. So yes, the gradient has direction, if that's what you meant. :smile:

    For the direction: Do you know how to find a unit vector in the direction of your gradient? You can use it to find the direction in which your function is increasing and decreasing most rapidly. There should be a formula in your book somewhere, perhaps involving cosine? :smile:
     
  6. Nov 8, 2011 #5
    Re: Gradient

    Duf(P) = gradient * cos theta??
    uhh-- so theta is the direction... yes?
     
  7. Nov 8, 2011 #6

    Dembadon

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    Re: Gradient

    Yes. If you think about it, when θ=0, your function will be increasing most rapidly. It will be decreasing most rapidly when θ=[itex]\pi[/itex] (the opposite direction).

    Edit: Keep in mind that your final answer should be a vector. :smile:

    2nd Edit: I need to go to class, so I'll check back in a little over an hour.
     
  8. Nov 8, 2011 #7
    Re: Gradient

    my directional derivative was 44/5 and my gradient was <2,12> so ||gradient|| is sqrt(148). So the direction at the given point is cos-((44/5)/sqrt(148))?

    Is rate of maximum increase as same as maximum rate of increase?
     
  9. Nov 8, 2011 #8

    Dembadon

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    Re: Gradient

    Sorry if I confused you. :redface: I didn't see v at first.

    Since the rate of increase is your directional derivative, the maximum rate of increase will occur when the angle between vectors v and [itex]\nabla[/itex]ƒ is 0. Thus, the maximum rate of increase is [itex]\frac{44}{5}[/itex]cos 0.

    The direction of the maximum rate of increase is when [itex]\nabla[/itex]ƒ is in the direction of v. This follows from the previous step because we need cos θ = 1 in order to obtain the maximum rate of increase, and since the angle between v and [itex]\nabla[/itex]ƒ is 0, the vectors are pointing in the same direction.
     
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