Curl of a function and vector field

  • #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. Homework Statement

"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: "

Homework Equations


curlF = ∇XF

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!
 

Answers and Replies

  • #2
vela
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Use the product rule.
 
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  • #3
Use the product rule.
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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