• Support PF! Buy your school textbooks, materials and every day products Here!

Please help me differentiate this!

  • Thread starter Sorento7
  • Start date
  • #1
16
0

Homework Statement



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

b is not a function of a

Homework Equations


I want to differentiate this, (the jacobian of the vector field)
dot is the Euclidean inner product.


The Attempt at a Solution


[itex]\acute{u}[/itex].v - [itex]\acute{v}[/itex].u / v2 doesn't seem to work
 

Answers and Replies

  • #2
DryRun
Gold Member
838
4
To begin with, i could barely see anything with that font size...
[tex]\frac{∂ \frac{\vec{a}}{ \vec{a} . \vec{b}} }{∂\vec{a}}[/tex]
 
  • #3
16
0
[itex]\frac{∂ \frac{\vec{a}}{ \vec{a} . \vec{b}} }{∂\vec{a}}[/itex]
 
  • #4
16
0
no answer? wow I thought that should have been an easy differentiation?
 
  • #5
I like Serena
Homework Helper
6,575
176
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.
 
  • #6
16
0
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.
 
  • #7
I like Serena
Homework Helper
6,575
176
So is ##\vec a## a function of (x,y) or something?
 
  • #8
16
0
So is ##\vec a## a function of (x,y) or something?
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.)
 
  • #9
I like Serena
Homework Helper
6,575
176
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...
 
  • #10
16
0
Thanks. That was helpful:approve:
 

Related Threads for: Please help me differentiate this!

Replies
3
Views
1K
Replies
2
Views
667
Replies
0
Views
757
Top