Problem 43 - Calculus Early Transcendentals

Question

Find the work done by the force field \(\mathbf{F}\) on a particle that moves along the curve \(C\). $$ \begin{array}{l}{\mathbf{F}(x, y, z)=x y \mathbf{i}+y z \mathbf{j}+x z \mathbf{k}} \\ {C: \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k} \quad(0 \leq t \leq 1)}\end{array} $$

Step-by-Step Solution

Answer

The work done by the force field along the curve is \( \frac{27}{28} \).

Step 1: Vector Field and Parametrization

The force field is given as \( \mathbf{F}(x, y, z) = x y \mathbf{i} + y z \mathbf{j} + x z \mathbf{k} \). The parametrization of the curve \( C \) is given by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \), with \( 0 \leq t \leq 1 \). This means as the particle moves, its coordinates \( (x, y, z) \) can be replaced by \( (t, t^2, t^3) \).

Step 2: Substitute Parametrization into the Force Field

Substitute \( x = t \), \( y = t^2 \), and \( z = t^3 \) into the force field \( \mathbf{F} \). This gives: \[ \mathbf{F}(t) = t \cdot t^2 \mathbf{i} + t^2 \cdot t^3 \mathbf{j} + t \cdot t^3 \mathbf{k} = t^3 \mathbf{i} + t^5 \mathbf{j} + t^4 \mathbf{k}. \]

Step 3: Find the Derivative of the Position Vector

Determine the derivative of the position vector \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \). The derivative \( \mathbf{r}'(t) \) is the velocity vector: \[ \mathbf{r}'(t) = \frac{d}{dt}(t \mathbf{i}) + \frac{d}{dt}(t^2 \mathbf{j}) + \frac{d}{dt}(t^3 \mathbf{k}) = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}. \]

Step 4: Compute the Dot Product

Compute the dot product \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \):\[ \mathbf{F}(t) \cdot \mathbf{r}'(t) = (t^3 \mathbf{i} + t^5 \mathbf{j} + t^4 \mathbf{k}) \cdot (\mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}). \]Calculate this as:\[ = t^3 \cdot 1 + t^5 \cdot 2t + t^4 \cdot 3t^2 = t^3 + 2t^6 + 3t^6. \]Combine terms: \[ = t^3 + 5t^6. \]

Step 5: Integrate the Dot Product Over the Interval

Integrate the result from Step 4 over the interval \( 0 \leq t \leq 1 \) to find the work done:\[\int_{0}^{1} (t^3 + 5t^6) \, dt = \int_{0}^{1} t^3 \, dt + \int_{0}^{1} 5t^6 \, dt.\]Calculate each integral separately:\[ \int_{0}^{1} t^3 \, dt = \left. \frac{t^4}{4} \right|_{0}^{1} = \frac{1}{4},\]\[\int_{0}^{1} 5t^6 \, dt = \left. \frac{5t^7}{7} \right|_{0}^{1} = \frac{5}{7}.\]Add the results:\[\frac{1}{4} + \frac{5}{7} = \frac{7}{28} + \frac{20}{28} = \frac{27}{28}.\]

Step 6: Conclusion: Calculating the Work Done

Thus, the work done by the force field along the curve is \( \frac{27}{28} \).

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and their derivatives and integrals. It extends calculus concepts to multiple dimensions and is foundational for understanding dynamics in physical systems. In vector calculus, the focus is on vectors that have both magnitude and direction, rather than just numerical values. This makes it ideal for representing forces, velocities, and other physical quantities that change over space and time.

Key concepts in vector calculus include:

• Gradient: Measures how much a quantity changes as you move in different directions, represented as a vector showing the direction of greatest increase.
• Divergence: Indicates how much a vector field spreads out from a point. A positive divergence signifies a source, while a negative divergence indicates a sink.
• Curl: Measures the rotation or twisting of a vector field. If the curl is zero, the field is irrotational.

These concepts are used in various applications, such as electromagnetism and fluid dynamics, to analyze how forces interact in a multidimensional space. Understanding these components is essential when dealing with more complex physical systems.

Line Integrals

Line integrals extend the concept of integration to functions over curves. They are particularly useful in calculating work done by force fields, such as when a particle travels along a parametric curve in a field. In essence, a line integral computes a sum of values along a path based on a given function or vector field.

In performing a line integral:

• You parametrize the curve by expressing it in terms of a parameter, often denoted as t.
• Next, you evaluate the vector field along this parametrized path. In our example, \( \mathbf{F}(t) = t^3 \mathbf{i} + t^5 \mathbf{j} + t^4 \mathbf{k} \).
• The subsequent step involves finding the derivative of the position vector to obtain the velocity vector. This demonstrates how the position varies in relation to t.
• Finally, you compute the dot product of the evaluated vector field and the velocity vector to find how much force is exerted along the path, and integrate this from 0 to 1 to determine the work done.

Line integrals are often used in physics to find work done or to calculate the circulation of a field along a closed path, making them a vital concept in vector calculus.

Parametric Curves

Parametric curves offer a way to describe a path in space through parameterization, typically using a variable like t. This approach allows you to express complex curves as functions of this parameter, providing a powerful tool for visualizing and analyzing paths in multidimensional space.

Each parametric curve is defined by a set of equations: one for each coordinate. In the given exercise, the curve is represented by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \). As t varies from 0 to 1, the curve traces a path in three-dimensional space.

Benefits of using parametric curves include:

• Flexibility: They can represent a wide variety of complex shapes and motions, including helices and loops.
• Simplification: Analyzing curves becomes more straightforward using one parameter instead of calculating directly from x, y, z coordinates.
• Connectivity: Useful in calculus for evaluating line integrals and exploring how different quantities change over a path.

Parametric curves are invaluable in analyzing systems where dynamics along a path need to be evaluated, such as tracking particle motion along a curve influenced by a force field.