Line integral with the inverse square field

In summary, the homework statement asks for someone to integrate F(x) = x / |x|^3 along the straight line from (1,0,0) to (2,-2,1). The attempt at a solution does not mention what vector field is being integrated over, how to find the partial derivatives, or what the potential function might be.
  • #1
wumple
60
0

Homework Statement



Integrate F(x) = x / |x|^3 along the straight line from (1,0,0) to (2,-2,1).

Homework Equations



Line integral = int (F dot dx)

The Attempt at a Solution



I don't know where to start. Usually I do line integrals by parameterizing the line and the vector field but I'm not sure what to do with this. Also, is the field centered at (1,0,0)? I don't really understand the question that well. Is this a conservative vector field? How can I check? Can I use the fundamental theorem of calculus? Is there some sort of potential function I can calculate? I just need someone to bump me in the right direction towards those questions.
 
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  • #2
its a conservative vector field if it can be written in terms of some scalar potential, say [itex] \phi (x)[/itex] .

then the vector field can if the gradient of the field [itex] \vec{F}(x) = \nabla \phi (x)[/itex]
 
  • #3
if you can do that, the line intergal will just be the diffenerence in potential
 
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  • #4
How would I go about figuring out what the potential function is?
 
  • #5
well, that depends exactly what your vector field is, as its not very clear, but wirte the partial dereivatives in terms of the functions & see if anything jumps out at you

otherwise, parameterise your line, in term of say t, then find the vector field at each point on teh line given by t and carry out the line intergal. dx will be in a constant dierction along the line

so do you actually mean:

[tex] \vec{F}(x)

= \frac{\vec{x}} {|\vec{x}|^3}

= \frac{1}{ (x^2+y^2+z^2)^{\frac{3}{2}} }

\begin{pmatrix} x \\ y \\ z \end{pmatrix} [/tex]
 
  • #6
Yeah, that's what I meant...I wasn't sure how to write vector hats. What would the partial derivatives be? I don't know how to express that line without parameterizing it in terms of t, but if t is my only variable then how can I write a gradient field?
 
  • #7
click on the tex if you want to see how its written

as for doing the line integral, its fine to parametrise in terms of t. you will end up with something like:
[tex] \vec{p}(t) = \vec{a} + t \vec{u}[/tex]

now evaluate value of the gradient field at each point on the line, [itex] \vec{F}(\vec{p}(t)) [/itex].
 
  • #8
as for finding the potential for, it doesn't always exist for an arbitrary vector field

a vector field derived from a scalar potential always has zero curl, so that could be one check

One way to find it is to equate the form of [tex] \nabla \phi(x) [/tex] with your vector field, and see if the resulting partial differntial equations are eay to solve.

In this case, due to the symmetry of the field I would try first converting ther fiedl to spherical coordinates, then looking at the gradient in those coords.
 
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  • #9
Ok I think I got it. Thanks so much for your help!
 
  • #10
no worries - its worth seeing both methods & the potential will usually be quicker if its easy to find
 
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1. What is a line integral with the inverse square field?

A line integral with the inverse square field is a type of mathematical calculation used in physics and engineering to determine the work done by a force that varies inversely with the square of the distance. This is often seen in fields such as electromagnetism and gravity.

2. How is a line integral with the inverse square field calculated?

To calculate a line integral with the inverse square field, you would need to use the fundamental theorem of calculus, which states that the integral of a function is equal to the difference of its values at the endpoints. In this case, the function would be the inverse square field and the endpoints would be the starting and ending points of the line.

3. What are some real-world applications of line integrals with the inverse square field?

Line integrals with the inverse square field have many practical applications, such as calculating the force of gravity between two objects, determining the electric field around a charged particle, and analyzing the flow of fluids in pipes or channels.

4. Are there any limitations to using line integrals with the inverse square field?

Like any mathematical calculation, there are limitations to using line integrals with the inverse square field. These include the assumption that the field is constant along the line and that the function being integrated is well-behaved. In some cases, other methods may need to be used to accurately calculate the desired value.

5. How does the magnitude of the force affect the line integral with the inverse square field?

The magnitude of the force will impact the value of the line integral with the inverse square field. As the force increases, the work done along the line will also increase. This can be seen in the equation for the line integral, where the force is multiplied by the distance traveled along the line.

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