Parameterization of a path to find work

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Homework Help Overview

The discussion revolves around evaluating the work done by a two-dimensional force along different paths from the origin to the point P = (1, 1). The force is defined as F = (x², 2xy), and one of the paths is parameterized as x = t³, y = t².

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss substituting the parameterized path into the force equation and setting up the integral for work. Questions arise about the correctness of the integral setup and how to check the method against other paths.

Discussion Status

Some participants have confirmed the integral setup appears correct, while others are exploring the limits of the parameter t for the parameterized path. There is an ongoing inquiry into how to determine these limits based on the path endpoints.

Contextual Notes

Participants are considering the transition from a two-variable function to a single-variable parameterization and are questioning the appropriateness of their chosen limits for integration.

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Homework Statement


Evaluate the work done by the two-dimensional force F = ( x2, 2xy ) along each of the following three paths joining the origin to the point P = (1, 1) :
The first two are fine
The last path is: the path given parametrically as x = t3, y = t2 with a parameter t

Homework Equations

The Attempt at a Solution


I think you just take x = t3, y = t2 and you plug them into F = ( x2, 2xy ) and then you set up the integral to find work ∫F⋅ds in terms of t. ∫(t6,2t5)⋅(3t2, 2t)dt = ∫(3t8+4t6)dt. Is this correct?
 
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How would you check?

1. what happens when you use the same method/reasoning on the first two paths?
2. how did the work for the first two paths compare with each other? What do you expect for the third one? What do you get?
 
Vitani11 said:
∫(t6,2t5)⋅(3t2, 2t)dt = ∫(3t8+4t6)dt. Is this correct?
Yes. You don't spell it out, but it looks like you have understood that ##\vec{ds}=(dx, dy)=(\dot x.dt, \dot y.dt)=(\dot x, \dot y).dt##.
 
Okay. Now that I understand this part - what about the limits of t? In the original it was a function of two variables going up to the point (1,1). Now that I have paramaterized this (if that is the right wording) what would the limits be in t as a single variable? As in how do I go about figuring it out? I integrated from 0 to 1 because intuitively that makes sense but I'm usually wrong.
 
Vitani11 said:
Okay. Now that I understand this part - what about the limits of t? In the original it was a function of two variables going up to the point (1,1). Now that I have paramaterized this (if that is the right wording) what would the limits be in t as a single variable? As in how do I go about figuring it out? I integrated from 0 to 1 because intuitively that makes sense but I'm usually wrong.
What are the values of the parameter at the path endpoints?
 

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