Using logarithms in vector Calculus

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SUMMARY

The discussion centers on the derivation of equation (3) from equations (1) and (2) in vector calculus, specifically involving the logarithmic function. The key equations are (1) (∇⋅ωn)/n = (n∇⋅ω + ω⋅∇n)/n, (2) L = log(n), and (3) ∇⋅ω = ∇⋅ω + ω⋅∇L. The transition from (1) to (3) involves recognizing that 1/n can be expressed as ∇L, utilizing the multivariable chain rule to simplify the expression. The discussion clarifies the role of the gradient operator and the cancellation of terms in the derivation process.

PREREQUISITES
  • Understanding of vector calculus concepts, specifically divergence and gradient operators.
  • Familiarity with logarithmic functions and their derivatives.
  • Knowledge of the multivariable chain rule in calculus.
  • Ability to manipulate and simplify mathematical equations involving vectors and scalars.
NEXT STEPS
  • Study the application of the multivariable chain rule in vector calculus.
  • Learn about the properties of logarithmic differentiation in multivariable contexts.
  • Explore advanced topics in vector calculus, such as divergence theorem and Stokes' theorem.
  • Practice derivations involving gradients and divergences in various mathematical contexts.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with vector calculus and need to understand the application of logarithmic functions in derivations.

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Homework Statement


My mentor has run me through the derivation of equation (3) bellow. I am unsure how he went from (1) to (3) by incorporating the log term from eq(2). In eq(3) it seems he just canceled the relevant n terms and then identified 1/n as the derivative of L however if this were the case then I am unsure why only one of the two terms on the RHS of eq(3) contains the L term.

How did he get from (1) to (3) using eq(2)?

Homework Equations


1) (∇⋅ωn)/n = (n∇⋅ω + ω⋅∇n)/n
2) L = log(n)
3) ∇⋅ω = ∇⋅ω + ω⋅∇L
where w is a 3 dimensional vector and n is a scalar.

The Attempt at a Solution


I think he may have just identified 1/n = ∇L but then if this were true, where did ∇n go from the second term in eq(1)? Also if this were true the same ∇L term would then be found in the first term wouldn't it?
 
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Your equation 3 is missing stuff on the left-hand side.

In general for a logarithm you will have ##d\log(f) = df/f##.
 
I finally realized he used the multivariable chain rule to turn ∇n*1/n into ∇L. Thanks for your help though!
 
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