Problem 71 - Calculus Volume 3

Question

Force \(\mathbf{F}(x, y, z)=z y \mathbf{i}+x \mathbf{j}+z^{2} x \mathbf{k}\) acts on a particle that travels from the origin to point \((1,2,3) .\) Calculate the work done if the particle travels: a. along \(\quad\) the \(\quad\) path (0,0,0)\(\rightarrow(1,0,0) \rightarrow(1,2,0) \rightarrow(1,2,3)\) along straight-line segments joining each pair of endpoints; b. along the straight line joining the initial and final points. c. Is the work the same along the two paths?

Step-by-Step Solution

Answer

11 along path (a); \(\frac{43}{4}\) along path (b); work is not the same.

Step 1: Calculate Work for First Path Segment

The work done by a force along a path is given by the integral \( W = \int_{C} \mathbf{F} \cdot d\mathbf{r} \), where \( \mathbf{F} = z y \mathbf{i} + x \mathbf{j} + z^{2} x \mathbf{k} \) and \( d\mathbf{r} = (dx, dy, dz) \). For the first segment from \((0,0,0)\) to \((1,0,0)\), only \( x \) changes, so \( y = 0 \) and \( z = 0 \). The force simplifies to \( \mathbf{F} = 0 \) along this segment. Hence, the work done is \( \int_{0}^{1} 0 \cdot dx = 0 \).

Step 2: Calculate Work for Second Path Segment

From \((1,0,0)\) to \((1,2,0)\), \( x = 1 \), \( y \) changes from 0 to 2, and \( z = 0 \). The force reduces to \( \mathbf{F} = \mathbf{j} \). The differential path is \( d\mathbf{r} = (0, dy, 0) \). Then, work done is \( \int_{0}^{2} 1 \cdot dy = [y]_{0}^{2} = 2 \).

Step 3: Calculate Work for Third Path Segment

For this segment \((1,2,0)\) to \((1,2,3)\), \( x = 1, y = 2, z \) changes from 0 to 3. The force \( \mathbf{F} = 6 \mathbf{i} + \mathbf{k} \), and \( d\mathbf{r} = (0,0,dz) \). Work is \( \int_{0}^{3} z^{2} \cdot dz = \left[ \frac{z^3}{3} \right]_{0}^{3} = 9 \).

Step 4: Total Work for Path (a)

Add the work done on each segment: \[ W_a = 0 + 2 + 9 = 11 \].

Step 5: Parameterize the Straight Path for (b)

For the straight path \((0,0,0) \rightarrow (1,2,3)\), parameterize \( x = t, y = 2t, z = 3t \), where \( t \) goes from 0 to 1. Then, \( \mathbf{F} = 6t^2 \mathbf{i} + t \mathbf{j} + 9t^3 \mathbf{k} \). The differential path is \( d\mathbf{r} = (1, 2, 3)dt \).

Step 6: Calculate Work for the Straight Path

Calculate the dot product \( \mathbf{F} \cdot d\mathbf{r} = 6t^2 \cdot 1 + t \cdot 2 + 9t^3 \cdot 3 = 6t^2 + 2t + 27t^3 \). Then calculate work: \[ W_b = \int_{0}^{1} (6t^2 + 2t + 27t^3) dt = [2t^3 + t^2 + \frac{27}{4}t^4]_{0}^{1} = 2 + 1 + \frac{27}{4} = \frac{43}{4} \].

Step 7: Compare Work Done Along Different Paths

The work done along path (a) is 11, whereas along path (b) is \( \frac{43}{4} \). Since these values are not equal, the work done is different for the two paths.

Key Concepts

Path-Dependent Work

When it comes to doing work along a path in a vector field, one important aspect to consider is whether the work is path-dependent. In simpler terms, path-dependent work means that the amount of work done by a force on an object depends on the specific path or trajectory that the object travels through the field. In the context of vector fields, this often happens if the field is not conservative. Conservative vector fields have the unique property where the work done is independent of the path taken and only depends on the starting and ending points. However, if the vector field is not conservative, as in our original exercise, different paths between the same endpoints result in different amounts of work. To determine if a vector field is path-dependent, look for any closed path calculations within the field that do not result in zero work. Path-dependent work implies that changing the trajectory will affect the energy used or work done, as we can see in the two paths computed in the original exercise.

Line Integrals

Line integrals are a fundamental concept in calculating work done in vector fields. They help us integrate (add up) the components of force along a path. Think of a line integral as a way of summing up tiny pieces of work done along every tiny segment of a path to get the total work. The formula for calculating work using line integrals is: \[ W = \int_C \mathbf{F} \cdot d\mathbf{r} \] where \( \mathbf{F} \) represents the force vector field and \( d\mathbf{r} \) represents a small displacement vector along the path \( C \).Line integrals incorporate both the direction and magnitude of force along the path. This is crucial because only the component of force in the direction of the path contributes to the work done. Through the integral, we account for every little piece of path taken, making it a vital tool when calculating path-dependent work.

Vector Calculus

Vector calculus plays a vital role in determining how forces act in fields and how much work is performed depending on different paths. It's a branch of mathematics that deals with vector fields, differentiation, and integration of vector functions, which are essential for understanding work and energy in physics. In the original exercise, vector calculus is used to decompose the force into its components and integrate them over the path. These components are influenced by variable factors such as position and direction, meaning the force can change magnitude and direction at different points along the path.When working with vector fields in vector calculus:

• The force vector \( \mathbf{F}(x, y, z) \) is expressed in terms of its components \( i, j, k \), which represent directions in three-dimensional space.
• Calculations involve dot products that combine these components with differential paths, thereby computing how much of the vector field pushes or pulls along the path.

Vector calculus thus provides the framework through which the work done by a vector field is quantified and understood.

Parametrization of Paths

Parametrization is the method of expressing a path through a vector function that depends on one or more parameters. In line integrals, parametrizing a path is critical because it enables us to evaluate integrals over paths that travel through vector fields. In simple terms, parametrization transforms complex paths into simple functions that are easier to handle mathematically.For example, in the exercise, the straight path from \((0,0,0) \rightarrow (1,2,3)\) is parametrized as:

• \( x = t \)
• \( y = 2t \)
• \( z = 3t \)

Where \( t \) varies from 0 to 1, denoting a smooth transition from the starting to the endpoint along the straight path. By employing parametrization, we convert a line segment that might change in three-dimensional space into a one-dimensional interval, thus facilitating computation of work done using line integrals in a clear and systematic manner.