Question

Given \(\mathbf{F}_{1}=-2 y \mathbf{i}+(z-2 x) \mathbf{j}+(y+z) \mathbf{k}, \mathbf{F}_{2}=y \mathbf{i}+2 x \mathbf{j}\) (a) Is \(\mathbf{F}_{1}\) conservative? Is \(\mathbf{F}_{2}\) conservative? (b) Find the work done by \(\mathbf{F}_{2}\) on a particle that moves around the ellipse \(x=\cos \theta\) \(y=2 \sin \theta\) from \(\theta=0\) to \(\theta=2 \pi\) (c) For any conservative force in this problem find a potential function \(V\) such that \(\mathbf{F}=-\nabla V\) (d) Find the work done by \(\mathbf{F}_{1}\) on a particle that moves along the straight line from (0,1,0) to (0,2,5) (e) Use Green's theorem and the result of Problem 9.7 to do Part (b) above.

Short answer

The force \(\textbf{F}_{1}\) is conservative while \(\textbf{F}_{2}\) is not. The work done by \(\textbf{F}_{2}\) around the ellipse is 0.

Step-by-step solution

Compute the Curl of \(\textbf{F}_{1}\)

A force field \(\textbf{F}\) is conservative if its curl is zero. Compute the curl of \(\textbf{F}_{1} = -2y \mathbf{i} + (z-2x) \mathbf{j} + (y+z) \mathbf{k} \) using \[ abla \times \textbf{F} = \textbf{i} \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) - \textbf{j} \left( \frac{\partial F_z}{\partial x} - \frac{\partial F_x}{\partial z} \right) + \textbf{k} \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \] Calculate and find the components:

Perform the Partial Derivatives

Evaluate the partial derivatives for \(\textbf{F}_{1}\): \[ \frac{\partial F_z}{\partial y} = 1, \frac{\partial F_y}{\partial z} = 1, \frac{\partial F_z}{\partial x} = 0, \frac{\partial F_x}{\partial z} = 0, \frac{\partial F_y}{\partial x} = -2, \frac{\partial F_x}{\partial y} = -2 \] Therefore, \(abla \times \textbf{F}_{1} = 0 \). Hence, \(\textbf{F}_{1}\) is conservative.

Compute the Curl of \(\textbf{F}_{2}\)

Compute the curl of \(\textbf{F}_{2} = y \mathbf{i} + 2x \mathbf{j} \). Since \(\textbf{F}_{2}\) has no \(\textbf{k}\)-component (z-component), calculate: \[ \frac{\partial (2x)}{\partial x} - \frac{\partial y}{\partial y} = 2 - 1 = 1 \] The curl is not zero, indicating \(\textbf{F}_{2}\) is not conservative.

Work Done by \(\textbf{F}_{2}\) Along Ellipse

Parametrize the ellipse with \( x = \cos \theta, y = 2 \sin \theta \) for \(0 \leq \theta \leq 2 \pi \). \[ d\textbf{r} = \left( \frac{\partial x}{\partial \theta} d\theta \right) \textbf{i} + \left( \frac{\partial y}{\partial \theta} d\theta \right) \textbf{j} \] For the given parametrization, we get: \[ d\r = -\sin\theta \mathbf{i} + 2\cos\theta \mathbf{j} \] Evaluate the line integral: \[ \int_{C} \textbf{F}_{2} \cdot d\r = \int_{0}^{2\pi} \left(\sin\theta \cos\theta + 4-x\sin\theta\cos\theta\right) d\theta \]

Compute the Work Integral

Compute the work done by \(\textbf{F}_{2}\) step-by-step: \[ W = \int_{0}^{2\pi} y\cos\theta \left( -\sin\theta \right) d\theta + 2x \cos\theta \left(2\cos\theta \right) d\theta \] Simplifies to \[ W = \int_{0}^{2\pi} -2\sin^2\theta d\theta + 2 \cos^2\theta d\theta = 2 \left[ -\int_{0}^{2\pi}\sin^2-\int_{0}^{2\pi} \cos^2\theta d\theta \right] = 0 \] Therefore, the work done by \(\textbf{F}_{2}\) on the particle around the ellipse is 0.

Key concepts

vector calculus

Vector calculus is a field of mathematics focused on vector fields—functions that assign a vector to each point in space. It's vital for understanding physical phenomena in many domains, such as fluid dynamics and electromagnetism. Vector calculus involves operations like differentiation and integration of vector fields.

To determine if a force field is conservative (one of the goals in our problem), we often use vector calculus operations like the **curl** and the **gradient**.

Key concepts include:

• **Vector Fields** - aspects of spatial phenomena where each point has a vector (e.g., wind speed).
• **Differentiation** - finding rate of change, such as obtaining the curl.
• **Integration** - summing up quantities, such as finding work done along a path.

Mastering these basics will greatly help in solving problems related to force fields and potential functions.

line integrals

A line integral calculates the accumulation of a field along a path. It is a crucial tool when dealing with work done by a force field. In context, we used it to find the work done by \(\textbf{F}_{2}\) as a particle moves along an ellipse.

The general form of a line integral when dealing with a vector field \(\textbf{F}\) and a path \(\textbf{r}(t)\) parameterized by \(\theta\) is:
\[ \int_{C} \textbf{F} \cdot d\textbf{r} \]
Here's a breakdown:

• **Parameterization**: Representing the path using a parameter (like \(\theta\) in our ellipse example).
• **Differential**: Expressing the small segment of the path (\textbf{dr}) in terms of differentials of the parameter.
• **Dot Product**: Multiplying the vector field and the differential path vector.

Finally, evaluating the integral gives us the total work done as the particle moves along the specified path.

green's theorem

Green's Theorem bridges the concepts of line integrals and double integrals, simplifying complex problems involving work and circulation in a plane.

It states:
\[\int_{C} P dx + Q dy = \int\int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA\]
In our problem, Green's Theorem helps in calculating the work done around an ellipse by transforming the line integral into a double integral over the region enclosed by the path.

Key ideas:

• **Closed Curve**: The path must be a closed loop (like our ellipse).
• **Double Integral**: Integrating over an area instead of a path simplifies calculations.
• **Favorable Conditions**: The theorem is especially useful when the region and path are simple to describe.

Using this approach can be much more efficient than directly computing line integrals, making Green's Theorem a powerful tool in vector calculus.

potential functions

A potential function, \(V\), is a scalar field whose gradient yields a given vector field. In contexts of force fields, it's crucial to find potential functions when demonstrating that a field is conservative.

The relationship is given by:
\[\textbf{F} = -\abla V\]
This implies that the force field is the negative gradient of \(V\). In simpler terms, it describes how \(V\) changes in space.

To find a potential function:

• **Integrate Components**: Partial integration of each component of the vector field.
• **Consistency**: Ensure integration yields a single function that aligns across all dimensions.
• **Conservative Forces**: Essentially, only conservative force fields have a corresponding potential function.

In our problem, calculating \(V\) demonstrates the conservativeness of \(\textbf{F}_{1}\), providing a deeper understanding of the force field structure.

force fields

Force fields represent spatial forces affecting particles within a region, often described using vectors. A force field \(\textbf{F} \) exerts force on a particle depending on its position.

In physics and engineering, understanding force fields is vital as they describe real-world phenomena such as gravity, electromagnetism, and more.
Characteristics to understand:

• **Vector Representation**: Each point in space has a force vector assigned.
• **Conservative vs Non-Conservative**: Conservative fields (like \(\textbf{F}_{1}\)) have potential functions and path-independent work.
• **Work Done**: Calculated as the line integral of the field over a path, important in energy and physics problems.

Understanding these properties allows solving complex real-world problems involving forces more effectively.
Master these basics for a solid foundation in dealing with all kinds of force fields and their applications.