1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please help me differentiate this!

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\frac{∂ \frac{\vec{a}}{ \vec{a} . \vec{b}} }{∂\vec{a}}[/itex]

    b is not a function of a
    2. Relevant equations
    I want to differentiate this, (the jacobian of the vector field)
    dot is the Euclidean inner product.


    3. The attempt at a solution
    [itex]\acute{u}[/itex].v - [itex]\acute{v}[/itex].u / v2 doesn't seem to work
     
  2. jcsd
  3. Apr 17, 2012 #2

    sharks

    User Avatar
    Gold Member

    To begin with, i could barely see anything with that font size...
    [tex]\frac{∂ \frac{\vec{a}}{ \vec{a} . \vec{b}} }{∂\vec{a}}[/tex]
     
  4. Apr 17, 2012 #3
    [itex]\frac{∂ \frac{\vec{a}}{ \vec{a} . \vec{b}} }{∂\vec{a}}[/itex]
     
  5. Apr 18, 2012 #4
    no answer? wow I thought that should have been an easy differentiation?
     
  6. Apr 18, 2012 #5

    I like Serena

    User Avatar
    Homework Helper

    It's not a proper differentiation.
    I can't make out what your formula is supposed to represent.
    It doesn't look like a Jacobian.
     
  7. Apr 18, 2012 #6
    Maybe I phrased it wrong. Its the result of multiplying inversion of an inner product to one of its vector components. (1/[itex]\vec{a}[/itex].[itex]\vec{b}[/itex]) * [itex]\vec{a}[/itex]

    edit: [itex]\vec{b}[/itex] is a constant vector.
     
  8. Apr 18, 2012 #7

    I like Serena

    User Avatar
    Homework Helper

    So is ##\vec a## a function of (x,y) or something?
     
  9. Apr 18, 2012 #8
    Suppose:

    F([itex]\vec{a}[/itex]) = (1/[itex]\vec{a}[/itex].[itex]\vec{b}[/itex]) [itex]\vec{a}[/itex]

    dot is the Euclidean inner product and F is defined as a vector space(R3 → R3)

    I need ∂[itex]\vec{F}[/itex]/∂[itex]\vec{a}[/itex] (given that [itex]\vec{b}[/itex] is an arbitrary constant vector.)
     
  10. Apr 18, 2012 #9

    I like Serena

    User Avatar
    Homework Helper

    Ok, so suppose ##\vec a## is ##[^x_y]##.

    Then ##\vec F(\vec a) = \vec F(x, y) = [^{F_x(x,y)}_{F_y(x,y)}]##.

    In that case the Jacobian is the 2x2 matrix of the partial derivatives of F.

    With your ##\vec F(\vec a) = {\vec a \over \vec a \cdot \vec b}##, you get:
    $$\vec F(\vec a) = {[^x_y] \over x b_x + y b_y}$$
    From this you can calculate the partial derivatives.

    For instance:
    $${\partial F_x \over \partial x} = {\partial\over \partial x}({x \over x b_x + y b_y})$$
    With an application of the quotient rule the result follows...
     
  11. Apr 18, 2012 #10
    Thanks. That was helpful:approve:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook