Theory question about Exact Eqtns and Calculus

  • Thread starter Thread starter flyingpig
  • Start date Start date
  • Tags Tags
    Calculus Theory
Click For Summary
SUMMARY

This discussion centers on the relationship between exact equations and vector fields in calculus. It establishes that if a differential form dQ = A(x,y)dx + B(x,y)dy satisfies the condition \(\frac{A(x,y)}{dy} = \frac{B(x,y)}{dx}\), then a scalar function Q(x,y) exists. Furthermore, it highlights that the line integral of the vector field F = (A(x,y), B(x,y)) is path-independent and depends solely on the endpoints, confirming the connection between exact equations and vector field theory.

PREREQUISITES
  • Understanding of exact differential equations
  • Familiarity with vector fields and line integrals
  • Knowledge of scalar functions in multivariable calculus
  • Basic proficiency in calculus notation and operations
NEXT STEPS
  • Study the properties of exact differential equations in detail
  • Learn about vector fields and their applications in physics
  • Explore the concept of path independence in line integrals
  • Investigate scalar potential functions and their significance in calculus
USEFUL FOR

Students and educators in calculus, particularly those focusing on differential equations and vector calculus, will benefit from this discussion.

flyingpig
Messages
2,574
Reaction score
1

Homework Statement



I've noticed that exact equations take the same form as vector field dot <dx,dy> form, are there any relationships? Because I ended up writing an arrow on top of my F(x,y) everytime I do these exact equation problems

Is there a secret my book is not telling me?
 
Physics news on Phys.org
not sure if i understand the question, but if you have a differential
dQ = A(x,y)dx + B(x,y)dy

and they satisfy
\frac{A(x,y)}{dy} = \frac{B(x,y)}{dx}

then there exists a scalar function Q(x,y), such that dQ
dQ = A(x,y)dx + B(x,y)dy

similarly if you consider the vector field
F = \begin{pmatrix} A(x,y) \\ B(x,y)\end{pmatrix}
a line integral will independent of path, depending only on endpoints and equal to Q(\vec{x}_f) - Q(\vec{x}_i), and you could write:
dQ = F \bullet \vec{dx}
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
1K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
20
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K