Is There a Method for Finding A When Given B and ∇XA?

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SUMMARY

The discussion centers on the mathematical relationship B = ∇XA, where B is known but A is not directly obtainable. It is established that A is not a single variable, making it impossible to isolate A directly from the equation. However, if the form of A is specified, such as A = a(r)u, it becomes feasible to derive A. This highlights the importance of understanding the structure of A in order to manipulate the equation effectively.

PREREQUISITES
  • Understanding of vector calculus, specifically the gradient operator (∇).
  • Familiarity with multivariable functions and their representations.
  • Knowledge of differential equations and their applications.
  • Basic proficiency in mathematical modeling techniques.
NEXT STEPS
  • Research methods for expressing multivariable functions in terms of their components.
  • Learn about the implications of the gradient operator in vector calculus.
  • Explore techniques for solving differential equations involving multiple variables.
  • Study mathematical modeling to understand how to define functions like A = a(r)u.
USEFUL FOR

Mathematicians, physics students, and engineers who are dealing with multivariable calculus and need to manipulate complex equations involving gradients.

GarageDweller
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Let's say I have this relation

B=∇XA

I know B, now I want A.
What ever do I do?
Is there some tried and true method out there?
 
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GarageDweller said:
Let's say I have this relation

B=∇XA

I know B, now I want A.
What ever do I do?
Is there some tried and true method out there?


no sir, this relation doesn't allowed to find A directly...A isn't single...but if you have the form of A (for example A=a(r)u) u can find it
 
Darn, makes you miss single variable cal
 

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