Potential function of a gradient field.

In summary, the conversation discusses finding the line integral of a vector field from the origin along the x-axis to a point and then counterclockwise around a circle to another point. The use of a potential function and the consideration of the vector field's gradient are mentioned, but ultimately the suggestion is made to approach the problem as two separate integrals using different parameterizations.
  • #1
carstensentyl
34
0

Homework Statement


For the vector field = -yi + xj, find the line integral along the curve C from the origin along the x-axis to the point (6, 0) and then counterclockwise around the circumference of the circle x2 + y2 = 36 to the point (6/sqrt(2), 6/sqrt(2)) . Give an exact answer.



Homework Equations





The Attempt at a Solution



I got a potential function f = xy. I tried to consider it as a path independent potential function, and plugged in F(6/sqrt(2), 6/sqrt(2))-F(0,0). The answer I got was 18. Which webassign calls wrong. Then again, webassign incorrect on about 20% of the problems, which makes learning math very frustrating.
 
Physics news on Phys.org
  • #2
grad(f) is yi+xj. That's not what you want. The vector field doesn't have a gradient function, it has nonzero curl. You really have to do the path integral.
 
  • #3
Thanks. I just read a few chapters ahead into the curl stuff.
 
  • #4
sorry to bring this up again but I got the same question with the OP and I am just confused what the real answer should be if the vector field doesn't have a grad function
 
  • #5
Normally one does path integrals before "potential functions" so this is a peculiar question.

from the origin along the x-axis to the point (6, 0) and then counterclockwise around the circumference of the circle [itex]x^2 + y^2 = 36[/itex] to the point (6/sqrt(2), 6/sqrt(2)).
Do it as two separate integrals. On the x-axis use x= t, y= 0. On the circle [itex]x^2+ y^2= 36[/itex] use x= [itex]6cos(\theta)[/itex], [itex]y= 6 sin(\theta)[/itex].
 

1. What is a gradient field?

A gradient field is a mathematical function that describes the rate of change of a physical quantity in a particular direction at every point in space.

2. What is the potential function of a gradient field?

The potential function of a gradient field is a scalar function that is used to describe the potential energy of a system at any given point in space. It is related to the gradient field by the formula: F = -∇V, where F is the gradient field and V is the potential function.

3. How is a gradient field used in physics?

Gradient fields are used in physics to describe the behavior and interactions of physical quantities such as temperature, pressure, and electric or magnetic fields. They are also used in vector calculus to calculate work, potential energy, and other important physical quantities.

4. What is the relationship between a gradient field and a conservative force?

A conservative force is a force that can be described by a gradient field. This means that the work done by the force is independent of the path taken, and only depends on the initial and final positions. In other words, the work done by a conservative force is equal to the negative change in potential energy.

5. How is the potential function of a gradient field calculated?

The potential function of a gradient field can be calculated by using the formula V = -∫F·dr, where V is the potential function, F is the gradient field, and the integral is taken over the path between two points. This integral can also be evaluated using the fundamental theorem of calculus.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
667
  • Calculus and Beyond Homework Help
Replies
8
Views
4K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Replies
4
Views
962
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
8K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
Back
Top