Vector Calculus: Calculate Work via Line Integral from (-1,2) to (1,3)

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Homework Help Overview

The problem involves calculating the work done by a force field represented by the vector function F(x,y) = (y^2, 2xy) as a particle moves along a straight path from the point (-1,2) to (1,3). The context is vector calculus, specifically focusing on line integrals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the force vector and the appropriate method for calculating work via line integrals. There is confusion about integrating a vector and the correct expression for work in this context. Some participants suggest parameterizing the line and using the dot product in the integral.

Discussion Status

The discussion is ongoing, with participants clarifying the representation of the force vector and exploring the next steps for setting up the line integral. Some guidance has been offered regarding the parameterization of the path and the formulation of the integral.

Contextual Notes

There is mention of difficulties with LaTeX formatting and a lack of clarity about integrating vectors, which may affect the participants' understanding of the problem setup.

mikeyrichster
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Homework Statement



A force is applied to a particle, defined by:

F(x,y)= (y^2, 2xy) << This is a vertical bracket with the y^2 ontop of the 2xy

The path of the particle is straight. The particle moves from (-1,2) to (1,3)

i) Calculate the work that the force F does as the particle moves along the path C by evaluating the appropreate line integral directly


I don't know where to start with this. I know that work: w = fd so to get the distance i must do some integration, but how do you integrate a vector like that?


Can vertical please help!? - would be much appreciated


Many thanks
 
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mikeyrichster said:

Homework Statement



A force is applied to a particle, defined by:

F(x,y)= (y^2, 2xy) << This is a vertical bracket with the y^2 ontop of the 2xy

What do you mean by that? Force is a vector, not a fraction?
The path of the particle is straight. The particle moves from (-1,2) to (1,3)

i) Calculate the work that the force F does as the particle moves along the path C by evaluating the appropreate line integral directly


I don't know where to start with this. I know that work: w = fd so to get the distance i must do some integration, but how do you integrate a vector like that?

w = fd only works for constant forces along straight lines. Perhaps you mean

[tex]W = \int_C \vec F \cdot d\vec R[/tex]
 
Hi there! thanks for your help

sorry about the fraction thing.. what i ment was that it is a vector but written vertically not horizontally... so still a vector but not (y^2, 2xy) .. I am damn useless at LaTeX so i can't make it with a large bracket and the y^2 at the top and the 2xy at the bottom.. hope that makes sense

Also i was only making a guess with the w=fd thing, I am sure your expression is what I am looking for as it uses vectors.

What would be the next logical thing to do with it?

Thanks again
 
Click on the formula below to see the LaTeX for writing the column vector.

[tex]\textbf{F} = \begin{pmatrix} y^2 \\ 2xy \end{pmatrix}[/tex]
 
mikeyrichster said:
What would be the next logical thing to do with it?

Thanks again

Parameterize the line as

[tex]\vec R(t) = \langle x(t), y(t), z(t)\rangle[/tex]

and use

[tex]\int_C \vec F \cdot d\vec R = \int_a^b \vec F(t)\cdot \frac {d\vec R}{dt}\, dt[/tex]
 

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