Help with conservative line integrals.

Click For Summary
SUMMARY

The discussion focuses on evaluating the conservative line integral of the form ydx + xdy + 4dz, confirming its exactness and deriving the potential function. The user successfully identifies the partial derivatives as df/dx = y, df/dy = x, and df/dz = 4, leading to the function f(x,y,z) = xy + 4z + C. The confusion arises in understanding the integration process and the introduction of functions g(y,z) and h(z) during the derivation of the potential function. The final potential function is established as xy + 4z + K, accounting for any constants.

PREREQUISITES
  • Understanding of conservative vector fields
  • Familiarity with partial derivatives
  • Knowledge of integration techniques in multivariable calculus
  • Ability to differentiate functions of multiple variables
NEXT STEPS
  • Study the concept of exact differentials in multivariable calculus
  • Learn about conservative vector fields and their properties
  • Explore integration techniques for functions of several variables
  • Review the method of finding potential functions from vector fields
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus and vector analysis, as well as anyone seeking to deepen their understanding of conservative line integrals and potential functions.

seang
Messages
184
Reaction score
0
I'm really having trouble with this, I can't even understand the example in my textbook. The example problem is:

Show that ydx + xdy + 4dz is exact and then evaluate it over some limits.

I can show that its exact, and once I find the function, I can evaluate it, so I really just need help getting the function. So the book says:

df/dx = y; df/dy = x; df/dz = 4; I understand this part, then the next step is:

f(x,y,z) = xy + g(y,z); Got this too, but now the fun begins. The next line is:

df/dy = x + dg/dy = x or dg/dy = 0; I have no idea what they're talking about here. Where is the dg/dy from? And why is 'x' what the equation is supposed to equal? The next step is:

f(x,y,z) = xy + h(z); followed by:

df/dz = 0 + dh/dz = 4 or h(z) = 4z + C; Again, I have no idea where the dh/dz if from. Also, above df/dz = 4, now it is 0? I understand that if you differentiate 4, you get 0, but what's with the discrepancy?

Any help is appreciated! I know this is long so thanks for your time
 
Physics news on Phys.org
That seems like a really confusing way to explain the procedure, at least if I've gotten what it's trying to get at straight.

If you have the following function:

[tex]xy+4z[/tex]

and find the partial derivatives, you'll get:

[tex]\frac{\partial (xy + 4z)}{\partial x}=y + 0[/tex]

[tex]\frac{\partial xy + 4z}{\partial y}=x + 0[/tex]

[tex]\frac{\partial xy + 4z}{\partial z}=0 + 4[/tex]

What you're doing is the inverse. Integrate the partial derivative with respect to x and you should get the original function -- with the exception of any constants. With respect to x, 4z is just a constant.

You do the same with the partial derivative with respect to y and should also get the original function, with the exception of any constants (4z in this case).

You do the same with the partial derivative with respect to z and get the original function, with the exception of any constants (xy in this case).

Any unique terms are combined into one equation to get the original function. When you integrated with respect to x and with respect to y, you got xy for both (plus some constant). You only include the xy once in your final function, since you didn't get a new unique solution. When you integrated with respect to z, you got a new unique solution - 4z. You combine the unique solutions to get the potential function of xy + 4z.

Of course, since xy + 4z + 6 gets you the same partial derivatives, when you're going in reverse, there will be no way to know that the 6 should have been in the potential function, so your potential function should be xy + 4z + K to cover any constants that had a partial derivative of 0 for all three variables.

In your example, your first integration with respect to x yielded xy + g(y,z), since the partial derivative of any function consisting of just y, z, or yz, yielded a partial derivative with respect to x of 0. And so on.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
3K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K