Conservative and irrotational vectors

  • Thread starter Bevyclare
  • Start date
  • Tags
    Vectors
In summary, you had trouble solving the problem because you were not sure what u1 and u2 were. You were told that the contour runs from x=1 to x=6, so you just need to substitute for y in terms of u in the first term and figure out what u1 and u2 are.
  • #1
Bevyclare
7
0
1. Please can someone assist me in solving the attached vector problem.

I have made several attempt to no avail





3.
 

Attachments

  • Question.pdf
    69.7 KB · Views: 264
Physics news on Phys.org
  • #2
You've almost got part (i) finished. You just have to keep going. You got to this point:

[tex]J=\int_C (ye^{xy}dx+(xe^{xy}+1)dy[/tex]

Then you introduced a parameter u and apparently wrote x=u and y=u3 since C is the curve [itex]y=x^3[/itex]. This gave you dx=du and dy=3u2du, so you got to here:

[tex]J=\int_{u_1}^{u_2} (ye^{u^4}du+3u^3e^{u^4}du+3u^2du)[/tex]

So far, so good. You just need to substitute for y in terms of u in the first term and figure out what u1 and u2 are. You were told the contour runs from x=1 to x=6. Since you know how x and u are related, you can easily see what u1 and u2 equal.

All the terms can easily be integrated either directly or with a simple substitution
 
  • #3
Vela,
Thanks for your response.

I still cannot solve the problem.

Please let someone assist.
 
  • #4
Where are you haveing trouble?
You already had
[tex]J=\int_{u_1}^{u_2} (ye^{u^4}du+3u^3e^{u^4}du+3u^2du)[/tex]
and vela pointed out that, since [itex]y= u^3[/itex], that is the same as
[tex]J=\int_{u_1}^{u_2} (u^3e^{u^4}du+3u^3e^{u^4}du+3u^2du)[/tex]
so it is really just
[tex]J=\int_{u_1}^{u_2} (4u^3e^{u^4}du+3u^2du)[/tex]

You are told that x runs from 1 to 6. When x= 1 what is u? When x= 6, what is u?

If you still can't do it, please show your work so we can see exactly where you have a problem.
 
  • #5
Hello everyone

Please find attached the output of my integration and substitution.

However, I am not convinced that this is correct.

Please assist me.
 

Attachments

  • Question2.doc
    25.5 KB · Views: 184
  • #6
Please let someone assist me by reviewing my attached solution.

I don't think I'm correct.

Please I am counting on you.
 

FAQ: Conservative and irrotational vectors

What is a conservative vector?

A conservative vector is a type of vector field in which the line integral along any closed path is equal to zero. This means that the work done by the vector field on a particle moving along a closed path is independent of the path taken.

What is an irrotational vector?

An irrotational vector is a type of vector field in which the curl, or rotation, of the vector field is equal to zero at every point in space. This means that the vector field is path-independent and has a potential function.

What is the difference between conservative and irrotational vectors?

The main difference between conservative and irrotational vectors is that conservative vectors have a path-independent line integral, while irrotational vectors have a curl equal to zero at every point.

What are some real-world examples of conservative and irrotational vectors?

Conservative vectors can be found in physical systems such as gravitational and electric fields. Irrotational vectors can be seen in fluid flow, as the velocity of a fluid particle is independent of the path taken.

How are conservative and irrotational vectors used in science and engineering?

Conservative and irrotational vectors are used in various scientific and engineering applications, such as in the study of fluid dynamics, electromagnetism, and mechanics. They are also used in fields such as economics and finance to model and analyze systems with path-independent properties.

Similar threads

Replies
2
Views
3K
Replies
1
Views
3K
Replies
24
Views
2K
Replies
6
Views
3K
Replies
2
Views
1K
Replies
10
Views
1K
Replies
2
Views
1K
Replies
3
Views
2K
Back
Top