Question

For the force field \(\mathbf{F}=(y+z) \mathbf{i}-(x+z) \mathbf{j}+(x+y) \mathbf{k},\) find the work done in moving a particle around each of the following closed curves: (a) the circle \(x^{2}+y^{2}=1\) in the \((x, y)\) plane, taken counterclockwise; (b) the circle \(x^{2}+z^{2}=1\) in the \((z, x)\) plane, taken counterclockwise; (c) the curve starting from the origin and going successively along the \(x\) axis to (1,0,0), parallel to the \(z\) axis to ( 1,0,1 ), parallel to the \((y, z)\) plane to (1,1,1) and back to the origin along \(x=y=z\) (d) from the origin to \((0,0,2 \pi)\) on the curve \(x=1-\cos t, y=\sin t, z=t,\) and back to the origin along the \(z\) axis.

Short answer

Work done is zero for curves (a) and (b), -1 for (c), non-zero for parabola (d). Integral basis each transforms curve Sect.

Step-by-step solution

Verify the Force Field is Conservative

The work done by a conservative force field over a closed curve is zero. Calculate the curl of the force field \( \mathbf{F} = (y+z) \mathbf{i} - (x+z) \mathbf{j} + (x+y) \mathbf{k}\) using the formula for the curl, \( abla \times \mathbf{F}\).

Compute the Curl of the Force Field

The curl is given by \[ abla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{k} \] Compute each partial derivative to find \[ abla \times \mathbf{F} = (1 - (-1)) \mathbf{i} + (1 - 1) \mathbf{j} + (-1 - 1) \mathbf{k} = 2 \mathbf{i} + 0 \mathbf{j} - 2 \mathbf{k} \].

Determine the Implication of a Non-zero Curl

Since \( abla \times \mathbf{F} eq 0 \), the force field is not conservative. This means we cannot directly conclude that the work over the closed curves is zero; we will need to compute the line integral for each path.

Curve (a) - Compute Work Around Circle in (x,y) Plane

For the circle \( x^2 + y^2 = 1 \), parameterize using \( x = \cos t \), \( y = \sin t \), \( z = 0 \) for \( 0 \leq t \leq 2\pi \). Then \( d\mathbf{r} = (-\sin t) \mathbf{i} + \cos t \mathbf{j} dt \). Substituting into the force field gives \( \mathbf{F}(\mathbf{r}(t)) = (\sin t) \mathbf{i} - (\cos t) \mathbf{j} \). Thus the line integral \[ W = \int_0^{2\pi} [(\sin t)(-\sin t) + (-\cos t)(\cos t)] dt \]. Simplify and compute the integral to find \( W = 0 \).

Curve (b) - Compute Work Around Circle in (x,z) Plane

For the circle \( x^2 + z^2 = 1 \), parameterize using \( x = \cos t \), \( z = \sin t \), \( y = 0 \) for \( 0 \leq t \leq 2\pi \). Then, \( d\mathbf{r} = (-\sin t) \mathbf{i} + \cos t \mathbf{k} dt \). Substituting into the force field gives \( \mathbf{F}(\mathbf{r}(t)) = (\sin t) \mathbf{i} - (\cos t) \mathbf{j} + (\cos t) \mathbf{k} \). Thus the line integral \[ W = \int_0^{2\pi} [(\sin t)(-\sin t) + (\cos t)(\cos t)] dt \]. Simplify and compute the integral to find \( W = 0 \).

Curve (c) - Compute Work Along Three Segments

Split the curve into three segments: 1. From (0,0,0) to (1,0,0): \( z = 0, y = 0 \), \( d\mathbf{r} = dx \mathbf{i} \). Work done: \( \int_0^1 y+z dx = 0 \). 2. From (1,0,0) to (1,0,1): \( x = 1, y = 0 \), \( d\mathbf{r} = dz \mathbf{k} \). Work done: \( \int_0^1 x+y dz = 1 \). 3. From (1,0,1) to (1,1,1): \( x = 1, z = 1 \), \( d\mathbf{r} = dy \mathbf{j} \). Work done: \( \int_0^1 -(x+z) dy = -2 \). Sum: \( 0 + 1 - 2 = -1 \). 4. From (1,1,1) back to (0,0,0): \( x = y = z t, dt \mathbf{r} \). Integrate: \( \int_0^1 F_x i - \int F_y j + \int d \rightarrow -1, +1, 0 \), sum is 0.

Curve (d) - Compute Work from Parametric Curve

For the curve \( x = 1 - \cos t, y = \sin t, z = t \), use parametric expression \[ (d=F=(dx= find/force \)). Compute: \( \int_0^{2\pi}(f-z).dt \].

Key concepts

Understanding Force Fields

A force field represents a spatial distribution of forces. In vector calculus, a force field can be described with a vector function that assigns a vector (representing force) to each point in a scalar field. Visualize it like wind in the air. At every point in the air, wind has a direction and strength. For example, given a force field \( \mathbf{F} = (y+z) \mathbf{i} - (x+z) \mathbf{j} + (x+y) \mathbf{k} \), each component i, j, k (or x, y, z) depends on the coordinates of the point in space.

In the given problem, we need to understand how this force field interacts with particles moving along specified paths. This involves using the concepts of line integrals and curl to compute the work done by the force along these paths. Understanding force fields forms the foundation for these computations.

Line Integral for Work Calculation

The line integral is a way to compute the work done by a force field in moving a particle along a certain path. To compute the line integral, we parameterize the path, replace parameters in the force field equation, and integrate over the parameter range.

For a vector field \( \mathbf{F} \), and a path \( \mathbf{r}(t) \), the line integral from \( t=a \) to \( t=b \) is:

\[ W = \int_a^b \mathbf{F} \cdot d\mathbf{r} \] where \( d\mathbf{r} \) is derived from the parameterized path.

Imagine you are walking along a marked trail through a field where the wind (force field) constantly changes in strength and direction. The work you do against the wind is summing up all the little pieces of work done at every point of your walk. This is essentially what the line integral calculates. In steps 4 and 5 of the problem, we computed the work around circular paths in different planes by parameterizing the circles and performing the line integral.

Curl of a Vector Field

The curl of a vector field measures the rotation or the 'twisting force' of the field at a point. It helps to determine whether a vector field is conservative (path-independent). Mathematically, the curl of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is given by:

\[ abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \]

In the problem’s step 2, we calculated the curl and found it was non-zero: \( abla \times \mathbf{F} = 2\mathbf{i} - 2\mathbf{k} \). This tells us that the force field is not conservative, meaning the work done by the force field along a closed path is not necessarily zero. When dealing with trajectories that form closed curves, the presence of a non-zero curl indicates that these curves might enclose regions where the field exhibits rotational characteristics, contributing to non-zero work.