Please answer correctly, its in the attachment.
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Answer:
Work = e+24
F is not conservative.
Step-by-step explanation:
To find the work required to move an object in the force field
along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.
Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with
r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1
so
r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t) with 0≤ t ≤ 1
is a parameterization of the segment.
the work W required to move an object in the force field F along the straight line from A to B is the line integral
where C is the segment that goes from A to B.
Integrating by parts the last integral:
and
To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24
Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.
The segment that goes from A to (1,0,0) can be parameterized as
r(t) = (t,0,0) with 0≤ t ≤ 1
so the work required to move the particle from A to (1,0,0) is
The segment that goes from (1,0,0) to B can be parameterized as
r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1
so the work required to move the particle from (1,0,0) to B is
Hence, the work required to move the particle from A to B along D is
e - 1 + (25e)/2 = (27e)/2 -1
since this result differs from e+24, the force field F is not conservative.