Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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

    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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook