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Directional derivatives

  1. Jun 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f(x, y) = x^2y^3 + xy.
    Is there a direction at (-1; 2) in which the rate of change of f is equal to 18?
    Justify your answer.


    2. Relevant equations



    3. The attempt at a solution
    plugging this into the directional derivative formula, i get
    18 = -v1 + 13v2, where v1 and v2 are the components of the vector in whose direction the rate of change is 18.
    But i do not know where to go from here...
     
  2. jcsd
  3. Jun 11, 2009 #2

    danago

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    What do you know about the magnitude of the gradient vector at a point?
     
  4. Jun 11, 2009 #3

    HallsofIvy

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    I don't see that the "magnitude of the gradient vector" has anything to do with this problem. I do not, however, see how Jennifer1990 got that result. The gradient of [itex]f(x,y)= x^2y^3+ xy[/itex] is [itex]\nabla f(x,y)= (2xy^3+y)\vec{i}+ (3x^2y^2+ x)\vec{j}[/itex] and so [itex]\nabla f(-1, 2)= (2(-1)(2)^3+ 2)\vec{i}+ (3(-1)^2(2)^+ (-1))\vec{j}= -14\vec{i}+ 5\vec{j}[/itex]. So the directional derivative, in the directon of vector [itex]v_1\vec{i}+ v_2\vec{j}[/tex] is [itex]-14v_1+ 5v_2= 18[/itex]. Now remember that you do not want [itex]v_1[/itex] and [itex]v_2[/itex] themselves, you just want the direction, the ratio between them. Another way to do this is to remember that a unit vector, in the direction at angle [itex]\theta[/itex] to the x-axis, is given by [itex]cos(\theta)\vec{i}+ sin(\theta)\vec{j}[/itex]. Solve [itex]-14 cos(\theta)+ 5 sin(\theta)= 18[/itex] for [itex]\theta[/itex].
     
  5. Jun 11, 2009 #4

    danago

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    Isn't the direction of the gradient vector the one in which the function changes most rapidly, where its magnitude is the value of this greatest rate of change? When i made my suggestion, what i was getting at was the fact that the magnitude of del f is less than 18, hence there is no direction in which the rate of change is 18. My apologies if this is not correct.
     
  6. Jun 12, 2009 #5
    Danago's argument is correct, since in general [tex]\nabla[/tex]w is the fastest increase of w and the magnitude of [tex]\nabla[/tex]w is the maximum directional derivative in that direction. HallsofIvy's logic is correct as well but the attempt to solve for theta will come up short. Keep in mind that he did not compute the "j component = 5" correctly if you attempt to go forward with what he did.
     
  7. Jun 12, 2009 #6

    HallsofIvy

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    Sorry Danago, that was just too clever for me!


    Oh! The j component is [itex]3(-1)^2(2)^2+ (-1)= 11[/itex]. Thanks, chriscolose.
     
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