Problem 72 - Calculus Volume 3

Question

Find the work done by vector field \(\mathbf{F}(x, y, z)=x \mathbf{i}+3 x y \mathbf{j}-(x+z) \mathbf{k}\) on a particle moving along a line segment that goes from (1,4,2) to (0,5,1) .

Step-by-Step Solution

Answer

The work done by the vector field is 9.5 units.

Step 1: Understanding the Problem

We need to calculate the work done by the vector field \( \mathbf{F}(x, y, z) = x \mathbf{i} + 3xy \mathbf{j} - (x + z) \mathbf{k} \) on a particle moving from the point \((1,4,2)\) to \((0,5,1)\). The line segment can be parameterized and the work done can be calculated using the integral of the dot product of \( \mathbf{F} \) and \( d\mathbf{r} \).

Step 2: Parameterize the Path

Parameterize the line segment from \((1,4,2)\) to \((0,5,1)\) as \( \mathbf{r}(t) = (1-t) \mathbf{i} + (4 + t) \mathbf{j} + (2-t) \mathbf{k} \), where \(t\) ranges from 0 to 1.

Step 3: Compute the Path Differential

Calculate the differential \( d\mathbf{r} \) along the path. Since \( \mathbf{r}(t) = (1-t, 4+t, 2-t) \), the differential is \( d\mathbf{r} = ([-1, 1, -1]) dt \).

Step 4: Substitute Parameters into Vector Field

Substitute the parameterized path into the vector field to get \( \mathbf{F}(\mathbf{r}(t)) = (1-t) \mathbf{i} + 3(1-t)(4+t) \mathbf{j} - ((1-t) + (2-t)) \mathbf{k} \). Simplifying, \( \mathbf{F}(\mathbf{r}(t)) = (1-t) \mathbf{i} + (3(4+t)(1-t)) \mathbf{j} - (3 - 2t) \mathbf{k} \).

Step 5: Dot Product of \( \mathbf{F}(\mathbf{r}(t)) \) and \( d\mathbf{r} \)

Calculate the dot product \( \mathbf{F}(\mathbf{r}(t)) \cdot d\mathbf{r} = ((1-t), 3(4+t)(1-t), -(3 - 2t)) \cdot (-1, 1, -1) \), which simplifies to \( -(1-t) + 3(4+t)(1-t) + 3 - 2t \).

Step 6: Simplify and Integrate Over Path

Simplify the expression: \( - (1-t) + 3(4+t)(1-t) + 3 - 2t = 3(4+t)(1-t) - 2 + t \). Integrate this over \(t\) from 0 to 1: \[ \int_0^1 \left[ 12 - 3t + 3t - 3t^2 - 2 + t \right] \, dt \] which simplifies to \[ \int_0^1 \left[ 10 - 3t^2 + t \right] \, dt \].

Step 7: Carry Out the Integration

Evaluate the integral: \[ \int_0^1 (10 - 3t^2 + t) \, dt = \left[ 10t - \frac{3t^3}{3} + \frac{t^2}{2} \right]_0^1 = [10 - 1 + \frac{1}{2} - (0)]_0^1 = 9.5 \].

Key Concepts

Line Integrals

Understanding line integrals is pivotal when working with vector fields. A line integral calculates the aggregate effect of a vector field, in terms of work, along a specific path or curve. In simpler terms, imagine a particle moving through a vector field, such as water currents or magnetic fields. The line integral will help you determine how much total "work" the field does on this particle as it travels along its path.

Think of the path as being cut into infinitesimally small segments, over which we calculate the work done. The choice of path significantly impacts the result of the line integral. Adding up all these tiny contributions gives us the total work done, which can be physically interpreted as the influence the field imposes along that path. In our case, the line integral evaluates the work done by field \( \mathbf{F}(x, y, z) \) from point (1,4,2) to (0,5,1).

The calculation involves parameterizing the path and performing an integration that considers both the vector field and the direction of movement.

Vector Field Parameterization

Parameterization of a path is a crucial step to solve problems involving line integrals in vector fields. By assigning a parameter, typically "\( t \)", to span between two points, we translate the path into a single function that describes every point along the path. In this exercise, the path from \((1,4,2)\) to \((0,5,1)\) was parameterized.

With \( \mathbf{r}(t) = (1-t) \mathbf{i} + (4+t) \mathbf{j} + (2-t) \mathbf{k} \), when \( t \) is 0, it starts at \((1,4,2)\), and when \( t \) is 1, it ends at \((0,5,1)\).

Parameterization simplifies the integration process by aligning the variable vector field components with a single variable path. This transformation facilitates the computation of both differential change \( d\mathbf{r} \) and the insertion of these values back into the vector field \( \mathbf{F} \) for further calculations.

Dot Product Integration

Dot product integration is the stage where we bring together both the parameterized vector field and the path differential. It involves calculating the dot product between the vector field at each point \( \mathbf{F}(\mathbf{r}(t)) \) and the differential path vector \( d\mathbf{r} \).

This dot product is essential as it helps compute the component of the vector field that aligns with the path, which is key to determining work done by the field. In our example, the expression \( \mathbf{F}(\mathbf{r}(t)) \cdot [-1, 1, -1] \) simplifies into a form that could be integrated easily over the defined interval of \( t \).

Following the computation of the dot product, the next step is to integrate this expression. The resulting integral, in the domain defined by \( t \), reflects the total work done by the vector field as the particle traveled along the specified path. This approach effectively intertwines algebraic manipulation and calculus for finding solutions to vector field problems.