91Ó°ÊÓ

Since the motion is along a line segment between points A(0, 0, 0) and B(-1, 2, -4), we can represent the line segment parametrically as follows: $$\langle x, y, z\rangle = A + t (B - A) = \langle 0, 0, 0\rangle + t(\langle -1, 2, -4\rangle - \langle 0, 0, 0\rangle)$$ So, $$\langle x, y, z\rangle = \langle -t, 2t, -4t\rangle$$ where \(0 \le t \le 1.\)

To compute the work done or the line integral, we first need to find the derivative of $$\langle x, y, z\rangle$$ with respect to t, which is $$\frac{d\langle x, y, z\rangle}{dt} = \langle -1, 2, -4\rangle$$ Now, we have to substitute $$\langle x, y, z\rangle$$ into the force field to find $$\mathbf{F}(x, y, z):$$ $$\mathbf{F}(\langle -t, 2t, -4t\rangle) = e^{-t+2t}\langle 1, 1, -4t\rangle = e^{t}\langle 1, 1, -4t\rangle$$ Next, we need to compute the dot product of $$\mathbf{F}(x, y, z)$$ and $$\frac{d\langle x, y, z\rangle}{dt}:$$ $$\mathbf{F}(\langle -t, 2t, -4t\rangle) \cdot \frac{d\langle x, y, z\rangle}{dt} = e^{t}\langle 1, 1, -4t\rangle \cdot \langle -1, 2, -4\rangle = -e^{t} + 2e^{t} + 16te^{t}$$ Now, we can compute the line integral: $$W = \int_0^1 (-e^{t} + 2e^{t} + 16te^{t}) \, dt$$

We will check if the force field is conservative by calculating the curl ($$\nabla \times \mathbf{F}$$) and determining if it is zero. If the curl of the force field is zero, then the force field is conservative. First, let's find the curl: $$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{x+y} & e^{x+y} & ze^{x+y} \end{vmatrix}$$ This results in: $$\nabla \times \mathbf{F} = \langle 0, 0, 2e^{x+y} - e^{x+y} \rangle = \langle 0, 0, e^{x+y} \rangle$$ The curl of the force field is not zero, so the force field is not conservative. Since the force field is not conservative, we cannot find a potential function for it. Thus, we can proceed to finish calculating the line integral.

We now need to compute the integral: $$W = \int_0^1 (-e^{t} + 2e^{t} + 16te^{t}) \, dt$$ Let's do this step by step: $$\int (-e^{t} + 2e^{t}) \, dt = -e^{t} + 2e^{t}$$ $$\int 16te^{t} \, dt = 16e^{t}(t-1)$$ Combine the results and apply the bounds from 0 to 1: $$W = \left[-e^{t} + 2e^{t} + 16e^{t}(t-1)\right]_0^1$$ $$W = [-e + 2e + 16e(0-1)] - [0] = e(-1+2-16)$$ So, the work required to move the object from point A to point B is $$W = -13e$$ units of work.