Problem 19 - Thomas Calculus

Question

In Exercises \(19-22,\) find the work done by \(F\) over the curve in the direction of increasing \(t .\) \begin{equation} \begin{array}{l}{\mathbf{F}=x y \mathbf{i}+y \mathbf{j}-y z \mathbf{k}} \\\ {\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1}\end{array} \end{equation}

Step-by-Step Solution

Answer

The work done is \( \frac{1}{2} \).

Step 1: Understand the problem

We need to calculate the work done by the force field \( \mathbf{F} = xy \mathbf{i} + y \mathbf{j} - yz \mathbf{k} \) along the curve given by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k} \) for \( t \) in the interval \([0, 1]\). The work done is given by the line integral of \( \mathbf{F} \cdot d\mathbf{r} \) over the curve.

Step 2: Calculate \( \mathbf{F}(\mathbf{r}(t)) \)

Substitute \( x = t \), \( y = t^2 \), \( z = t \) from the parameterization into \( \mathbf{F} \). This gives:\[ \mathbf{F}(t) = t(t^2) \mathbf{i} + t^2 \mathbf{j} - t^2 \cdot t \mathbf{k} = t^3 \mathbf{i} + t^2 \mathbf{j} - t^3 \mathbf{k} \]

Step 3: Calculate \( d\mathbf{r} \)

The derivative of \( \mathbf{r}(t) \) with respect to \( t \) is:\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}) = \mathbf{i} + 2t \mathbf{j} + \mathbf{k} \]

Step 4: Compute the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \)

Perform the dot product between \( \mathbf{F}(t) = t^3 \mathbf{i} + t^2 \mathbf{j} - t^3 \mathbf{k} \) and \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t \mathbf{j} + \mathbf{k} \):\[ \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = t^3 \cdot 1 + t^2 \cdot 2t - t^3 \cdot 1 = t^3 + 2t^3 - t^3 = 2t^3 \]

Step 5: Setup the integral

The work done is given by:\[ W = \int_{0}^{1} 2t^3 \, dt \]

Step 6: Evaluate the integral

Evaluate:\[ \int 2t^3 \, dt = \frac{2}{4} t^4 = \frac{1}{2} t^4 \] from \( t=0 \) to \( t=1 \).Then, compute:\[ \left. \frac{1}{2} t^4 \right|_{0}^{1} = \frac{1}{2}(1^4) - \frac{1}{2}(0^4) = \frac{1}{2} \]

Step 7: Final Step: Conclusion

The work done by the force field \( \mathbf{F} \) over the curve \( \mathbf{r}(t) \) in the direction of increasing \( t \) is \( \frac{1}{2} \).

Key Concepts

Work Done in a Force Field

In physics, work is a measure of energy transfer that occurs when a force moves an object over a distance, particularly within a force field. When calculating the work done by a force field, we often use vector calculus, specifically line integrals. This allows us to compute the work along a specific path or curve, which in mathematical terms is represented as an integral of the dot product of the force vector and the differential displacement.
For instance, consider a vector field \( \mathbf{F} \) and a curve parameterized by a vector function \( \mathbf{r}(t) \). The work done by \( \mathbf{F} \) along this curve is given by the line integral:
\[ W = \int \mathbf{F} \cdot d\mathbf{r} \]
This expression captures how the force interacts with the path, summing the infinitesimal contributions along the trajectory. The work computed provides insights into how the field influences motion, perhaps in transporting a particle or charging an electrical device.

Vector Calculus

Vector calculus is a branch of mathematics focused on differentiation and integration of vector fields, typically in three-dimensional Euclidean space \( \mathbb{R}^3 \). This area of calculus is instrumental in various scientific fields like physics and engineering, where vector fields are used to model things like electromagnetic fields, fluid flow, and more.
Understanding vector operations such as dot products, cross products, and gradients is crucial. The line integral is also a fundamental concept where a function or field's behavior is summed over a curve.
Key elements include

• Operations on vectors: addition, scalar multiplication, and dot products.
• Line and surface integrals: used to calculate physical quantities like work or flux.
• Partial derivatives: which help in understanding how a vector field changes across different directions.

Vector calculus combines these operations to study physical systems more comprehensively and is vital for solving complex real-world problems.

Parametric Equations

Parametric equations describe curves, surfaces, and other geometrical shapes where the coordinates are expressed as functions of one or more parameters, often denoted by \( t \) for curves. Instead of expressing \( y \) as a function of \( x \), a parametric representation might describe both \( x \) and \( y \) as separate functions of \( t \):\[ x = f(t), \, y = g(t) \]
This approach provides flexibility in representing complex paths, such as those encountered in physics and engineering problems. Parametric equations are particularly useful in vector calculus and line integrals as they allow turning an abstract curve into an explicit form that is easier to work with.

• It simplifies calculations of derivatives and integrals.
• It allows representation of curves in different dimensions.
• It provides a straightforward way to evaluate motion along a path.

These equations facilitate precise manipulations of curves and motion paths, essential in calculating quantities like work done along a path in a force field.

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. It computes a single real number from two vectors and provides insights into their directional relationship. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is:
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
Key properties and uses of dot products include:

• The result is a scalar rather than a vector.
• Measures how much one vector extends in the direction of another.
• Helps in determining angles between vectors; for vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cosine of the angle \( \theta \) between them can be calculated as \( \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \).

The dot product is integral to calculating work in physics, reflecting how a force applied along a path contributes to movement. In line integrals, the dot product effectively couples the force with the differential path, capturing energy transfer efficiently.