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Directional Derivatives

  1. Jun 5, 2004 #1


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    I need to find the direction in the xy plane in which one should travel, starting from point (1,1), to obtain the most rapid rate of decrease of
    [tex]f(x,y,z) = (x + y - 2)^2 + (3x - y - 6)^2[/tex]

    now, [tex]\nabla f = (2(x+y-2), 2(3x-y-6))[/tex]

    so I'm thinking now I have to find the the unti vector 'u' which would be the direction in question. Unfortunately I do not know how to go on from here.
    Somehow maximise [tex]\nabla f \cdot u[/tex] (where 'u' is a unit vector, dunno how to do vectors properly in latex :frown: )
  2. jcsd
  3. Jun 5, 2004 #2
    \vec{u} (I looked it up myself just a few hours ago).

    First of all, [tex]\nabla f = (2(x + y - 2) + 6(3x - y -6), 2(x + y - 2) - 2(3x - y - 6))[/tex]
    Second, you are only concerned with what is going on at (1,1), where [tex]\nabla f[/tex] = (-24, 8).

    Now, the question is to minimize (-24, 8)[tex]\cdot\vec{u}[/tex]. Since [tex]\vec{v}\cdot\vec{w} = vw\cos \theta[/tex], where [tex]\theta[/tex] is the angle between them, we see that the minimum occurs when [tex]\theta = \pi[/tex]. I.e., when the vectors are pointing in opposite directions.

    So [tex]\vec{u}[/tex] must be the unital vector in the direction of (24, -8).
  4. Jun 5, 2004 #3


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    Silly me, I was straining all that time trying to figure out why my method is wrong and all the time I had been working with the wrong grad f. Thanks.
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