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Proof of ||∇h||^2 when h=fog

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Let f:R2−>R be a differentiable function at any point, and g be the function g:R3−>R2defined by:

    g(u,v,w)=(g1,g2)=(u2+v2+w2,u+v+w)

    consider the function h=fog and prove that

    ||∇h||^2 = 4(∂f/∂x)^2*g1 + 4(∂f/∂x)(∂f/∂y)*g2 + 3(∂f/∂y)^2.


    3. The attempt at a solution

    H=fog=f(g1,g2)

    ∇h=∇(fog)=<∇f(g1,g2),∇g(u,v,w)> (dot product)

    ∇g(u,v,w)=(2u+2v+2w, 3)

    ∇f(x,y)=(∂f/∂x)+(∂f/∂y) evaluated at g1 and g2 respectively.

    <∇f(g1,g2),∇g(u,v,w)>=(∂f/∂x)∇g1+(∂f/∂y)∇g2=∇h

    I stopped here because looking ahead, I can see that I will ultimately be wrong here. I can see that (∂f/∂x)^2 + (∂f/∂x)(∂f/∂y) + (∂f/∂y)^2 will come from squaring my answer for ∇h., the 3 in front of (∂f/∂y)^2 is the same as ∇g2, and the 4 in front of (∂f/∂x)^2 may come from factoring out the 2 from ∇g2 and then squaring it. Any help? I would just like to know where in my process I went wrong and what I should be doing for the following step. Thank you!
     
  2. jcsd
  3. Oct 3, 2014 #2

    BvU

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    Could you proofread your post and place superscripts in the right places ? Good start!

    Then write a dot product where you mean a divergence and nothing where you mean a gradient.

    Perhaps viewing a few TeX examples here can get you a lot more help, a lot faster too!
    Example: ##\nabla \cdot f## is made with ##\#\# ## \nabla \cdot f ##\#\# ##
     
    Last edited: Oct 3, 2014
  4. Oct 3, 2014 #3

    Orodruin

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    Well, to start with, the gradient of ##f\circ g## should be a vector. For each individual gradient ##\nabla f## and ##\nabla g##, the gradient should have more indices than the original function, i.e., ##\nabla f## should be a vector (2D) and ##\nabla g## should be a 3x2 matrix. In your attempt, the gradient of f seems to be a scalar and that of g seems to be a 1x2 matrix. How can you remedy this?
     
  5. Oct 3, 2014 #4
    Thank you! I actually immediately figured it out when you mentioned the matrices.
     
  6. Oct 3, 2014 #5

    LCKurtz

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  7. Oct 3, 2014 #6
    Honestly because I believe I got the answer and I had to turn in the homework the following morning. I've been extremely busy with homework and preparing for my first college midterms.
     
  8. Oct 3, 2014 #7

    LCKurtz

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    Well, it wouldn't have hurt to post a quick "Thanks, I've got it now" or something similar to give the thread closure. It is frustrating to helpers to spend time in a thread only to have the OP never return. Some of us will quit responding to users that do that repeatedly.
     
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