Line Integral and Vector Field Problem

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SUMMARY

The discussion focuses on calculating the work done by the force field F(x,y) = x sin(y)i + yj on a particle moving along the parabola y = x^2 from the point (-1,1) to (2,4). The key method involves using the line integral of the dot product of the force field vector and the differential path vector d\vec{r}. The correct path vector is identified as \vec{r}(x) = x\vec{i} + x^2\vec{j}, leading to d\vec{r} = (\vec{i} + 2x\vec{j})dx. The discussion emphasizes that the Pythagorean theorem is not applicable in this context due to the non-conservative nature of the force field.

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  • Understanding of line integrals in vector calculus
  • Familiarity with vector fields and conservative forces
  • Knowledge of parametric equations for curves
  • Proficiency in calculus, specifically integration techniques
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  • Study the concept of line integrals in vector fields
  • Learn how to determine if a force field is conservative
  • Explore parametric equations and their applications in physics
  • Practice solving work problems involving non-conservative forces
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Homework Statement


Find the work done by the force field F(x,y) = x sin(y)i + yj on a particle that moves along on the parabola y = x^2 from (-1,1) to (2,4).

Homework Equations


Work = line integral of the dot product of Field vector and change in the path
The path is parabola equation.

The Attempt at a Solution


I tried to integrate with respect to x and y instead of t, because I don't know how to find the path vector r(t).
So, i got two integrals for X-direction and y - direction.
Then, I can just use the Pythagorean theorem to find the total work.

Srry, I didn't write any mathematical equations.
 
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The "path vector" is always a position vector:
\vec{r}(x)= x\vec{i}+ y\vec{j}= x\vec{i}+ x^2\vec{j}
so that
d\vec{r}= (\vec{i}+ 2x\vec{j})dx

If the force field is not conservative, the work done to move along two legs of a right triangle might have nothing to do with the work required to move along the hypotenuse. I don't believe that using the "Pythagorean Theorem" will work here.
 
Last edited by a moderator:
HallsofIvy said:
The "path vector" is always a position vector:
\vec{r}(x)= x\vec{i}+ y\vec{j}= x\vec{i}+ x^2\vec{j}
so that
d\vec{r}= (\vec{i}+ 2x\vec{j})dx

If the force field is not conservative, the work done to move along two legs of a right triangle might have nothing to do with the work required to move along the hypotenuse. I don't believe that using the "Pythagorean Theorem" will work here.

Thanks!
 

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