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Gradient Question

  1. Jun 25, 2008 #1
    I am given z = 32 - x[tex]^{2}[/tex] - 4y[tex]^{2}[/tex]
    Starting at the point (3,2) in i + j direction,
    find if you are going up or down the hill and how fast.

    The way I thought to proceed was that the gradient would tell me if I was going down or up hill and that [tex]\left|\nabla z \right|[/tex] would give me "how fast". My answer of [tex]\sqrt{292}[/tex] isn't correct however, so I'm obviously doing something wrong. Can anyone point me in the right direction of how to proceed?
     
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  3. Jun 25, 2008 #2

    Hootenanny

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    The magnitude of the gradient effectively gives you the magnitude of the greatest increase/decrease of the function, rather than the rate of change of the function in a given direction, which is what you were asked. Instead of simply calculating the gradient of the function, you need to evaluate the directional derivative of the function at the given point, in the given direction.
     
  4. Jun 25, 2008 #3

    Ben Niehoff

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    What gradient did you compute?

    A few notes:

    1. "The (i + j) direction" is misleading, because (i + j) is not a unit vector!

    2. The directional derivative in the direction of a unit vector [itex]\mathbf{\hat u}[/itex] is given by:

    [tex]\mathbf{\hat u} \cdot \nabla f[/tex]
     
  5. Jun 25, 2008 #4

    The gradient I computed was:
    -2xi - 8yj

    If I am supposed to calculate [tex]\mathbf{\hat u} \cdot \nabla f[/tex], what unit vector am I supposed to use? As you said, i + j isn't a unit vector...
     
  6. Jun 25, 2008 #5

    HallsofIvy

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    Do you not know how to construct a unit vector in the direction of, say, vector u? Just divide by its length.
     
  7. Jun 25, 2008 #6
    I just realized that was what I was over looking. Thanks, if I use

    i + j / |i + j|

    i get the correct answer.
     
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