Question

Find the work done by vector field F(x, y, z) = xi + 15xyj − (x + z)k on a particle moving along a line segment that goes from (1, 4, 2) to (0, 5, 1).

Answer 1

The work done by the vector field on a particle moving from point (1, 4, 2) to (0, 5, 1) can be found using a line integral of the field along the straight path connecting these points, which requires further steps to simplify and compute the integral.

Step-by-step explanation:

The work done by a vector field on a particle along a path can be calculated using a line integral of the vector field along the path. Given the vector field F(x, y, z) = xi + 15xyj − (x + z)k, to calculate the work done by this field as a particle moves from point A (1, 4, 2) to point B (0, 5, 1), we need to integrate the field along the straight line segment connecting these two points.

First, we find the parametric equations for the line segment. Let λ be the parameter that changes from 0 at A to 1 at B:

• x = 1 - λ
• y = 4 + λ
• z = 2 - λ

Then, the vector field along this line is F(λ) = (1 - λ)i + 15(1 - λ)(4 + λ)j - (1 - λ + 2 - λ)k.

The differential of displacement along the path is dρ = dx i + dy j + dz k = (-di + dj - dk).

Now compute the work done using the dot product of the vector field and the displacement differentials. We take the integral from λ = 0 to λ = 1:

W = ∫01 F(λ) · dρ dλ.

The exact value of the integral would require further work to solve, which typically involves simplifying the integral and then integrating term by term.

Answer 2

The work done by the vector field F(x, y, z) = xi + 15xyj − (x + z)k on a particle moving along a line segment from (1, 4, 2) to (0, 5, 1) is -16 units of work.

Step-by-step explanation:

The work done by the vector field F(x, y, z) = xi + 15xyj − (x + z)k on a particle moving along a line segment from (1, 4, 2) to (0, 5, 1) can be calculated using the formula:

Work = ∫ F · dr

Here, F represents the vector field and dr represents the infinitesimal displacement vector along the line segment. To find the work, we need to find the dot product between the vector field and the displacement vector.

Let's first calculate the displacement vector:

dr = (0 - 1)i + (5 - 4)j + (1 - 2)k = -i + j - k

Now, let's calculate the dot product:

F · dr = (1)(-1) + (15)(1)(-1) + (0)(-1) = -1 - 15 - 0 = -16

Therefore, the work done by the vector field F on the particle moving along the line segment is -16 units of work.