Curl of a function and vector field

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SUMMARY

The discussion focuses on the calculation of the curl of a product of a differentiable function \( f \) and a differentiable 3-dimensional vector field \( F \). The established solution is \( \text{curl}(fF) = f \text{curl}(F) + \nabla f \times F \). The user initially struggled with applying the product rule correctly but ultimately simplified the expression using vector calculus principles, leading to the correct formulation. This highlights the importance of understanding the product rule in vector calculus.

PREREQUISITES
  • Understanding of vector calculus, specifically curl and divergence.
  • Familiarity with the product rule in differentiation.
  • Knowledge of vector fields and their representations in three dimensions.
  • Proficiency in using the gradient operator \( \nabla \).
NEXT STEPS
  • Study the properties of curl and divergence in vector fields.
  • Learn how to apply the product rule in vector calculus contexts.
  • Explore examples of differentiable functions and vector fields in three dimensions.
  • Investigate the implications of the curl operation in physics, particularly in fluid dynamics.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are studying vector calculus and need to understand the application of the curl operation in differentiable functions and vector fields.

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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