# Vector field question + reasoning

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

#### Attachments

• lineintegral.pdf
107.6 KB · Views: 169

SteamKing
Staff Emeritus
Homework Helper
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?

Stephen Tashi
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,

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?

HallsofIvy
Homework Helper
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.

Stephen Tashi
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.

SteamKing
Staff Emeritus