joriarty
- 61
- 0
Conservative vector fields and line integrals
A particle is subject to a force F defined by [tex]F\left( x,y \right)=\left(\begin{array}{c} y^{2} \\ 2xy \end{array}\right)[/tex]. The particle moves in a straight line C from (-1,2) to (1,3).[a] Calculate the work done by the force F as the particle moves along the path C by evaluating the appropriate line integral directly.
Show that F is a conservative field and find a function φ such that ∇φ = F.
[c] Use your answer to part (b) to calculate the work done by an alternative method.
-
I have, or at least thought I had, completed this problem. Until the answers I got using each method (parts a and c) were completely different. I have no idea what I have done wrong :(. Probably a really silly mistake but any help is appreciated!
[a] I worked out the equation of the line as being y = 0.5x + 5/2. Thus [tex]W=\int_{\left( -1,2 \right)}^{\left( 1,3 \right)}{\left(\begin{array}{c} y^{2} \\ 2xy \end{array}\right)}\cdot \left(\begin{array}{c} dx \\ dy \end{array}\right)=\int_{\left( -1,2 \right)}^{\left( 1,3 \right)}{y^{2}dx+2xydy}[/tex]
This needs to be written parametrically. Let x=x(t) and y=y(t). Choose t=2y. Thus using the equation of the line between the two points as above, x = 2y - 5= t - 5 and y = t/2, and the limits of the integral become 4 ≤ t ≤ 6.
So dx becomes (dx/dt)dt = tdt and dy becomes (dy/dt)dt = (1/2)dt.
The integral is now [tex]W=\int_{t=4}^{6}{\frac{t^{2}}{4}t\cdot dt\; +2\left( t-5 \right)\left( \frac{t}{2} \right)\frac{1}{2}dt}=\; \int_{t=4}^{6}{\frac{t^{3}}{4}+\frac{t^{2}}{2}-\frac{5t}{2}dt}=\; \frac{196}{3}[/tex]
We want a function φ such that ∇φ = F.
(1) φx = y2
(2) φy = 2xy
Integrating equation (1) with respect to x, I get (3) φ = xy2 + G(y) where G(y) is an arbitrary function of y.
Differentiating (3) with respect to y and setting it equal with (2), I get φy = 2xy + dG/dy = 2xy so dG/dy = 0 and this means that G = c, an arbitrary constant. So φ = xy2 + c.
[c] [tex]\int_{c}^{}{F\cdot dr=\phi \left( 1,3 \right)-\phi \left( -1,2 \right)=13}[/tex]
I am inclined to believe my answer for part [c] more than the one I got for part [a] simply due to the fact that 196/3 seems to be an unusually large number. But I really am stuck with where I have gone wrong, these answers should match!
Thank you :)
FYI this is part of a 10% assignment, but my lecturer does allow us to discuss problems with others, it's just what we write down on the paper that has to be our own understanding and working. Unfortunately no one else I know has even started it yet...
Homework Statement
A particle is subject to a force F defined by [tex]F\left( x,y \right)=\left(\begin{array}{c} y^{2} \\ 2xy \end{array}\right)[/tex]. The particle moves in a straight line C from (-1,2) to (1,3).[a] Calculate the work done by the force F as the particle moves along the path C by evaluating the appropriate line integral directly.
Show that F is a conservative field and find a function φ such that ∇φ = F.
[c] Use your answer to part (b) to calculate the work done by an alternative method.
Homework Equations
-
The Attempt at a Solution
I have, or at least thought I had, completed this problem. Until the answers I got using each method (parts a and c) were completely different. I have no idea what I have done wrong :(. Probably a really silly mistake but any help is appreciated!
[a] I worked out the equation of the line as being y = 0.5x + 5/2. Thus [tex]W=\int_{\left( -1,2 \right)}^{\left( 1,3 \right)}{\left(\begin{array}{c} y^{2} \\ 2xy \end{array}\right)}\cdot \left(\begin{array}{c} dx \\ dy \end{array}\right)=\int_{\left( -1,2 \right)}^{\left( 1,3 \right)}{y^{2}dx+2xydy}[/tex]
This needs to be written parametrically. Let x=x(t) and y=y(t). Choose t=2y. Thus using the equation of the line between the two points as above, x = 2y - 5= t - 5 and y = t/2, and the limits of the integral become 4 ≤ t ≤ 6.
So dx becomes (dx/dt)dt = tdt and dy becomes (dy/dt)dt = (1/2)dt.
The integral is now [tex]W=\int_{t=4}^{6}{\frac{t^{2}}{4}t\cdot dt\; +2\left( t-5 \right)\left( \frac{t}{2} \right)\frac{1}{2}dt}=\; \int_{t=4}^{6}{\frac{t^{3}}{4}+\frac{t^{2}}{2}-\frac{5t}{2}dt}=\; \frac{196}{3}[/tex]
We want a function φ such that ∇φ = F.
(1) φx = y2
(2) φy = 2xy
Integrating equation (1) with respect to x, I get (3) φ = xy2 + G(y) where G(y) is an arbitrary function of y.
Differentiating (3) with respect to y and setting it equal with (2), I get φy = 2xy + dG/dy = 2xy so dG/dy = 0 and this means that G = c, an arbitrary constant. So φ = xy2 + c.
[c] [tex]\int_{c}^{}{F\cdot dr=\phi \left( 1,3 \right)-\phi \left( -1,2 \right)=13}[/tex]
I am inclined to believe my answer for part [c] more than the one I got for part [a] simply due to the fact that 196/3 seems to be an unusually large number. But I really am stuck with where I have gone wrong, these answers should match!
Thank you :)
FYI this is part of a 10% assignment, but my lecturer does allow us to discuss problems with others, it's just what we write down on the paper that has to be our own understanding and working. Unfortunately no one else I know has even started it yet...
Last edited:
