Vector field question + reasoning

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Im doing some revision of vector calculus and came across the following problem

Q: Calculate the work done by the force field F = 3xyi - 2j in moving from A: (1,0,0) to D: (2,0,0) and then from D: (2,0,0) to B: (2,sqrt(3),0)

I got stuck and decided to look at the answers. In the answers (part b of q5 in the document attached), the author assumed that dy=0, and based on this, he assumed that the integral of the vector field was 0. (n.b. r = r(t) = x(t)i + y(t)j + z(t)k )

How he came to this conclusion is beyond me, so could anyone shed some light on what Im misunderstanding?

thanks
 

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  • #2
SteamKing
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Im doing some revision of vector calculus and came across the following problem

Q: Calculate the work done by the force field F = 3xyi - 2j in moving from A: (1,0,0) to D: (2,0,0) and then from D: (2,0,0) to B: (2,sqrt(3),0)

I got stuck and decided to look at the answers. In the answers (part b of q5 in the document attached), the author assumed that dy=0, and based on this, he assumed that the integral of the vector field was 0. (n.b. r = r(t) = x(t)i + y(t)j + z(t)k )

How he came to this conclusion is beyond me, so could anyone shed some light on what Im misunderstanding?

thanks
The force F moves from point A to point D. Notice anything about the coordinates of these two points?
 
  • #3
Stephen Tashi
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Q: Calculate the work done by the force field F = 3xyi - 2j in moving from A: (1,0,0) to D: (2,0,0) and then from D: (2,0,0) to B: (2,sqrt(3),0)

(part b of q5 in the document attached), the author assumed that dy=0,
Along AD there is no change in y coordinate. Is the integration along AD what you are asking about ?
 
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Stephen, yes integration along AD is what Im asking about. SteamKing, I did notice that there was no change in y coordinate. However, why would the curve integral be 0 (as the answers say)? It makes no sense as some force is used to move the particle from A to D so it cannot be 0?
 
  • #5
HallsofIvy
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I notice that, in your pdf attachment, the integral is specifically along the straight line from A to D and then from D to B but you do not say that in your post here. A= (1, 0, 0) and D= (2, 0, 0) so parametric equations for that line are x= t, y= 0, z= 0 with t going from 1 to 2. Alternatively, x= t+ 1, y= 0, z= 0, with t going from 0 to 1. The crucial point is that y and z are equal to 0 for any point on the line from (0, 0, 0) to (1, 0, 0) so we always have y and z constant on that line. The derivative of any constant is, of course, 0. dx= dt, dy= 0, dz= 0.
(I am puzzled why you only asked about "dy" and not about "dz"!)
 
  • #6
Stephen Tashi
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It makes no sense as some force is used to move the particle from A to D so it cannot be 0?
Assuming "i" is the vector in the x-direction, the field exerts no force in the x-direction along a line where y = 0. In practical terms, any force however small and acting for however short a distance is sufficient to give a mass some velocity in the x-direction and cause it to eventually move from A to D. Hence "the work necessary" has lower limit zero, so we say "the work necessary" is zero.
 
  • #7
SteamKing
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Stephen, yes integration along AD is what Im asking about. SteamKing, I did notice that there was no change in y coordinate. However, why would the curve integral be 0 (as the answers say)? It makes no sense as some force is used to move the particle from A to D so it cannot be 0?
The x-component of F = 3xy also depends on y. Since y = 0 all along AD, then the x-component of F = 0 and the work F⋅ds also is zero.
The y-component of F is a constant and = -2, but ds = dy = 0, so the F⋅ds = 0 here as well.
 

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