Line integral of a conservative vector field

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SUMMARY

The discussion focuses on calculating the work done by the force field F(x,y) = (1/2)xy i + (1/4)x^2 j on a particle moving from (0,0) to (1,1) along the curve defined by x = y^2. The parameterization r(t) = t i + sqrt(t) j for 0 ≤ t ≤ 1 is derived, leading to the calculation of work done as the integral from 0 to 1 of (5/8)t^(3/2) dt, resulting in a total work of 1/4. The discussion highlights the infinite ways to parameterize the curve, with alternatives such as x = t^2 and y = t also being valid.

PREREQUISITES
  • Understanding of vector fields and line integrals
  • Familiarity with parametric equations
  • Knowledge of calculus, specifically integration techniques
  • Basic concepts of conservative vector fields
NEXT STEPS
  • Study the properties of conservative vector fields and their implications on line integrals
  • Learn advanced techniques for parameterizing curves in vector calculus
  • Explore the application of Green's Theorem in calculating line integrals
  • Investigate the relationship between work done and potential energy in conservative fields
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Students and educators in calculus, particularly those studying vector calculus, as well as professionals in physics and engineering who require a solid understanding of line integrals and conservative fields.

stevecallaway
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Homework Statement



This is an example in my book, and this is the information in the question.

Find the work done by thr force field F(x,y) = (1/2)xy i + (1/4)x^2 j (with i and j vectors) on a particle that moves from (0,0) to (1,1) along each path (graph shows a x=y^2 curve from (0,0) to (1,1)). This is the information in the question.

Homework Equations


This is the answer given in this example. r(t)=t i + sqrt(t) j for 0<=t<=1, so that dr = (i + 1/(2sqrt(t))dt and F(x,y)=(1/2)t^(3/2) i + (1/4)t^2 j. Then the work done is integral from 0 to 1, (5/8)t^(3/2)dt=(1/4)t^(5/2) = 1/4


The Attempt at a Solution


My question is how do you attain the r(t)=t i + sqrt(t) j equation from the information given?
 
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so
[tex]r(t) = (x(t), y(t)) = (t,\sqrt{t})[/tex]
is a parametric representation of x = y^2, try substituting in and see if it satisfies the equation

there are actually infinite ways to paremeterise the curve, this one was chosen by starting with x(t) = t, and finding y(t)
 
As landance said, there are an infinite number of ways to parameterize a curve. Another perfectly valid one would be [itex]x= t^2[/itex], [itex]y= t[/itex].
 

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