A particle moves along line segments from the origin to the points (1, 0, 0), (1, 4, 1), (0, 4, 1), and back to the origin under the influence of the force field
F(x, y, z) = z^2i + 5xyj + 3y^2k.
To find the work done by the force field on the particle as it moves along the given path, we can integrate the dot product of the force field and the differential displacement along the path.
The work done, W, is given by the equation:
W = ∫ F · dr
where F is the force field and dr is the differential displacement along the path.
Let's break down the path into segments and calculate the work done on each segment.
Segment 1: From the origin (0, 0, 0) to (1, 0, 0)
The displacement vector dr1 is (dx1, dy1, dz1) = (1, 0, 0) - (0, 0, 0) = (1, 0, 0)
The force vector F1 at (0, 0, 0) is F(0, 0, 0) = (0^2)i + 5(0)(0)j + 3(0^2)k = 0i + 0j + 0k = 0
W1 = F1 · dr1 = (0, 0, 0) · (1, 0, 0) = 0
Segment 2: From (1, 0, 0) to (1, 4, 1)
The displacement vector dr2 is (dx2, dy2, dz2) = (1, 4, 1) - (1, 0, 0) = (0, 4, 1)
The force vector F2 at (1, 0, 0) is F(1, 0, 0) = (0^2)i + 5(1)(0)j + 3(0^2)k = 0i + 0j + 0k = 0
W2 = F2 · dr2 = (0, 0, 0) · (0, 4, 1) = 0
Segment 3: From (1, 4, 1) to (0, 4, 1)
The displacement vector dr3 is (dx3, dy3, dz3) = (0, 4, 1) - (1, 4, 1) = (-1, 0, 0)
The force vector F3 at (1, 4, 1) is F(1, 4, 1) = (1^2)i + 5(1)(4)j + 3(4^2)k = 1i + 20j + 48k
W3 = F3 · dr3 = (1, 20, 48) · (-1, 0, 0)
= -1 + 0 + 0 = -1
Segment 4: From (0, 4, 1) back to the origin (0, 0, 0)
The displacement vector dr4 is (dx4, dy4, dz4) = (0, 0, 0) - (0, 4, 1) = (0, -4, -1)
The force vector F4 at (0, 4, 1) is F(0, 4, 1) = (1^2)i + 5(0)(4)j + 3(4^2)k = 1i + 0j + 48k
W4 = F4 · dr4 = (1, 0, 48) · (0, -4, -1)
= 0 + 0 + (-48) = -48
Now, we can calculate the total work done by summing the work done on each segment:
W = W1 + W2 + W3 + W4
= 0 + 0 + (-1) + (-48)
= -49 units of work
Therefore, the total work done by the force field on the particle as it moves along the given path is -49 units of work.
To find the work done by the force field on the particle as it moves along the given path, we need to calculate the line integral of the force field along that path.
The line integral of a vector field F along a curve C is given by the following formula:
∫C F⋅dr = ∫C (F⋅T) ds
where F is the vector field, dr is the differential displacement vector along the curve C, T is the unit tangent vector pointing in the direction of C, and ds is the differential arc length.
To find the line integral, we need to parameterize the path of the particle. Let's divide the path into three segments:
1. From the origin to (1, 0, 0): This segment can be parameterized as r(t) = ti, where 0 ≤ t ≤ 1.
2. From (1, 0, 0) to (1, 4, 1): This segment can be parameterized as r(t) = (1 + 3t)i + 4tj + tk, where 0 ≤ t ≤ 1.
3. From (1, 4, 1) to the origin: This segment can be parameterized as r(t) = (1 - t)i + (4 - 4t)j + (1 - t)k, where 0 ≤ t ≤ 1.
Now, let's calculate the line integrals for each segment and then add them together to find the total work done:
1. For the segment from the origin to (1, 0, 0):
∫C F⋅dr = ∫(0 to 1) (t^2) dt
Integrating (t^2) with respect to t gives us (1/3)t^3. Evaluating this from 0 to 1 gives us (1/3) - (0/3) = 1/3.
2. For the segment from (1, 0, 0) to (1, 4, 1):
∫C F⋅dr = ∫(0 to 1) [(1 + 3t)(5(1 + 3t)t) + 4(1 + 3t)(4t) + 3(4t)^2] dt
Simplifying and integrating term by term gives us:
∫[(5t + 15t^2 + 15t^3) + (16t + 48t^2) + 48t^2] dt
Integrating each term gives us:
[(5/2)t^2 + (5/4)t^3 + (5/5)t^4] + [8t^2 + (16/3)t^3] + (16/3)t^3
Evaluating this integral from 0 to 1 gives us:
(5/2) + (5/4) + (5/5) + 8 + (16/3) + (16/3) = 81/4.
3. For the segment from (1, 4, 1) to the origin:
∫C F⋅dr = ∫(0 to 1) [(1 - t)^2(2t) + 5(1 - t)(4 - 4t)t + 3(4 - 4t)^2] dt
Simplifying and integrating term by term gives us:
∫[(2t - 2t^2) + (20t - 20t^2) + (48 - 96t + 48t^2)] dt
Integrating each term gives us:
[t^2 - (2/3)t^3] + [10t^2 - (20/3)t^3] + [48t - 48t^2 + (16/3)t^3]
Evaluating this integral from 0 to 1 gives us:
1 - (2/3) + 10 - (20/3) + 48 - 48 + (16/3) = 26/3.
Finally, adding up the three line integrals gives us the total work done:
1/3 + 81/4 + 26/3 = (1 + 81/4 + 26/3)/12 = (201 + 324 + 104)/12 = 629/12.
Therefore, the work done by the force field on the particle as it moves along the given path is 629/12.
so what about it? If you want the work done, then you need to break it up into two parts. The work done along the line from (1,0,0) to (1,4,1) is
∫[1,0,0 .. 1,4,1] F•dr = ∫[1,0,0 .. 1,4,1] z^2 dx + 5xy dy + 3y^2 dz
= (1+20+12)-(0+0+0)
= 33
Now do the other line segment.
Maybe review line integrals.