Curl of a function and vector field

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The discussion revolves around the calculation of the curl of a product of a differentiable function and a vector field. The problem states that for a function f and a vector field F, the curl of fF can be expressed as fCurl(F) + ∇f x F. The user initially struggled with the application of the curl formula and the product rule but eventually realized the correct approach. By applying the product rule, they successfully derived the solution, confirming that the expression simplifies to the expected result. The conversation highlights the importance of using the product rule in vector calculus.
arduinohero
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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!
 
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Use the product rule.
 
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vela said:
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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