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Directional derivative: radial direction

  1. Aug 27, 2006 #1
    (hmm. wasnt to sure about where to post this)

    im given an equation to a 3d surface and asked to find the gradient at a certain point, in the radial direction.

    my question is, what is the radial direction?

    [ the equation is f(x,y) = 3*(x^2)*y + 2*y if its needed ]
     
  2. jcsd
  3. Aug 27, 2006 #2

    HallsofIvy

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    The radial direction, at the point (x,y,z) is the direction of the ray from (0,0,0) through (x,y,z). In particular, a unit vector in that direction is
    [tex]\frac{x}{\sqrt{x^2+ y^2+ z^2}}\vec i+ \frac{y}{\sqrt{x^2+ y^2+ z^2}}\vec j+ \frac{z}{\sqrt{x^2+ y^2+ z^2}}\vec k[/tex].

    The gradient of the function f(x,y)= 3x2y+ 2y is, of course,
    [tex]\nabla f= 6xy\vec i+ (3x^2+ 2)\vec j[/tex]
    and the derivative "in the radial direction" is the dot product
    [tex](6xy\vec i+ (3x^2+ 2)\vec j) \cdot (\frac{x}{\sqrt{x^2+ y^2+ z^2}}\vec i+ \frac{y}{\sqrt{x^2+ y^2+ z^2}}\vec j+ \frac{z}{\sqrt{x^2+ y^2+ z^2}}\vec k)[/tex]
    [tex]= \frac{6x^2y}{\sqrt{x^2+ y^2+ z^2}}\vec i+ \frac{y(2x^2+ 2)}{\sqrt{x^2+ y^2+ z^2}}\vec j[/tex]
     
  4. Aug 27, 2006 #3
    Do you mean

    [tex]= \frac{6x^2y}{\sqrt{x^2+ y^2+ z^2}}+ \frac{y(2x^2+ 2)}{\sqrt{x^2+ y^2+ z^2}}[/tex]

    i.e. sans the i-hat and j-hat? The dot product is a scalar.
     
  5. Aug 27, 2006 #4

    HallsofIvy

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    Oops! Yes, of course.
     
  6. Aug 28, 2006 #5
    thankyou! i was stuck on that one for ages ><

    thanks again for the help.
     
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