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Curl of a function and vector field

  1. Dec 17, 2014 #1
    Hello, I'm having some difficulty with a conceptual question on a practice test I was using to study. I have the answer but not the solution unfortunately.

    1. The problem statement, all variables and given/known data

    "For every differentiable function f = f(x,y,z) and differentiable 3-dimensional vector field F=F(x,y,z), the vector field Curl(fF) equals: "

    2. Relevant equations
    curlF = ∇XF

    3. The attempt at a solution
    The solution is apparently: fCurl(F)+∇f x F
    I am a little lost the process for this question. I attempted to "solve" the problem using a general case. Essentially I let F = <P, Q, R> and multiplied in "f," so fF = <fP, fQ, fR>.
    I then took the curl using the formula. I was left with the following:

    curl(fF) = <d/dy(fR)-d/dz(fQ), d/dz(fP)-d/dx(fR), d/dx(fQ)-d/dy(fP)>

    I am unsure of where to go from here. I originally was going to factor out "f," but then realized that that is not necessarily possible due to it being within the derivative operator. Any suggestions for a next step would be very helpful!
     
  2. jcsd
  3. Dec 17, 2014 #2

    vela

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    Use the product rule.
     
  4. Dec 17, 2014 #3
    I don't know why I didn't think of that... Thank you, that was very helpful!

    For anyone interested, I used product rule to differentiate everything and simplified to the following:

    <f(Ry-Qz)+fyR-fzQ, f(Pz-Rx)+fzP-fxR, f(Qx-Py)+fxQ-fyP>

    I then split the vector into a sum of vectors:

    <f(Ry-Qz), f(Pz-Rx), f(Qx-Py)> + <fyR-fzQ, +fzP-fxR, fxQ-fyP>

    The first vector being added is equal to fcurlF, and the second is quite clearly <fx, fy, fz> crossed with <P, Q, R>. Since <fx, fy, fz> is in fact ∇f, I rewrote the summation as fcurlF + (∇f)XF, which is of course the solution. Thanks again for the help!
     
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