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(a) find the curl of \(\mathbf{F}\) and (b) determine if \(\mathbf{F}\) is conservative. If it is conservative, find a potential function for it, and, if not, explain why not. $$ \mathbf{F}(x, y, z)=(y-z, z-x, y-x) $$

Short Answer

Expert verified
(a) The curl of \( \mathbf{F} \) is \((0, 0, -2)\). (b) \( \mathbf{F} \) is not conservative.

Step by step solution

01

Understand Curl Operation

The curl of a vector field \( \mathbf{F} = (P, Q, R) \) in three dimensions is given by the formula: \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \). We will apply this to the given vector field.
02

Compute Partial Derivatives

Identify the components of \( \mathbf{F}(x, y, z) = (y-z, z-x, y-x) \) where \( P = y-z \), \( Q = z-x \), and \( R = y-x \). Then calculate the required partial derivatives: \( \frac{\partial R}{\partial y} = 1 \), \( \frac{\partial Q}{\partial z} = 1 \), \( \frac{\partial P}{\partial z} = -1 \), \( \frac{\partial R}{\partial x} = -1 \), \( \frac{\partial Q}{\partial x} = -1 \), and \( \frac{\partial P}{\partial y} = 1 \).
03

Calculate the Curl

Using the partial derivatives calculated, substitute them into the curl formula: \( abla \times \mathbf{F} = (1 - 1, -1 + 1, -1 - 1) = (0, 0, -2) \). So, the curl of \( \mathbf{F} \) is \( (0, 0, -2) \).
04

Determine Conservativeness

A vector field is conservative if its curl is zero, i.e., \( abla \times \mathbf{F} = \mathbf{0} \). Since the curl of \( \mathbf{F} \) is \( (0, 0, -2) \), which is not zero, \( \mathbf{F} \) is not conservative.
05

Conclusion: No Potential Function

Since \( \mathbf{F} \) is not a conservative vector field, it does not have a potential function. The non-zero curl indicates that \( \mathbf{F} \) has rotational effects and cannot be derived from a scalar potential.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Curl of a Vector Field
In vector calculus, the *curl* is an operator that describes the rotation of a vector field. Imagine a vector field as a vast field full of tiny arrows, with each arrow pointing in the direction of the vector at that point. The curl helps us determine how much these arrows "twist" around each other. To compute the curl of a vector field \( \mathbf{F} = (P, Q, R) \), you use the formula: \[abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\]This calculation often involves *partial derivatives*, which measure how a function changes as one variable changes while others stay constant.
  • The first component, \( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \), is the curl along the x-direction.
  • The second component, \( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \), is the curl along the y-direction.
  • The third component, \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \), is the curl along the z-direction.
These individual curls reveal the underlying rotational effects present in the vector field. For example, a curl result of \((0, 0, -2)\) indicates rotation primarily along the negative z-axis. A non-zero curl means the vectors in the field are swirling around, not just pointing from high to low values.
Conservative Vector Field
A *conservative* vector field is a special kind of field where all the curls are zero, symbolically represented as \( abla \times \mathbf{F} = \mathbf{0} \). This means there's no rotational swirling, only potential differences, kind of like elevation changes in a landscape without any corkscrew motion.If a vector field is conservative, it comes from a scalar potential function (think of it as a height map, where each point has a height or potential associated with it). This potential is a scalar field \( f \), where \( \mathbf{F} = abla f \). Why does any of this matter? Knowing whether a vector field is conservative lets you predict movement within the field efficiently. It's like figuring out that uphill and downhill paths can determine your route:
  • Path Independence: If \( \mathbf{F} \) is conservative, the work done moving between two points is path-independent, relying solely on the points themselves, not how you traverse between them.
  • Existence of a potential function: Conservative fields always have a potential function, providing an easier way to compute integrals and understand field behavior.
Since the curl of \( \mathbf{F} \) in our original exercise is \((0, 0, -2)\), which isn't zero, the field \( \mathbf{F} \) is not conservative. This indicates some rotational energy, preventing the existence of a potential function.
Partial Derivatives
Partial derivatives are essential tools in vector calculus, enabling us to measure the rate of change in multi-variable functions. When we take a partial derivative, we treat all other variables as constants and focus on how the function changes with respect to one specific variable.Consider a function \( f(x, y, z) \). Its partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), measures how \( f \) changes when \( x \) changes slightly, while \( y \) and \( z \) remain the same.In the context of computing the curl of a vector field, partial derivatives reveal how each component of the vector field changes relative to the spatial variables:
  • \( \frac{\partial P}{\partial y} \) looks at how the x-component \( P \) shifts as \( y \) changes.
  • Other derivatives like \( \frac{\partial Q}{\partial x} \) or \( \frac{\partial R}{\partial z} \) give clues for how other directional vectors adjust with movement in space.
These derivatives are foundational for applying the curl operation, as seen in the earlier example. Without calculating them accurately, we can't determine if the vector field swirls or documents potential fluid-like flow properties.

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Most popular questions from this chapter

(a) sketch the curve \(C\) over which the integral is taken and (b) evaluate the integral. (Hint: Look for ways to reduce, or eliminate, messy calculation.) \(\int_{C} e^{8 x y} d x-\ln \left(\cos ^{2}(x+y)+\pi^{x} y^{1,000,000}\right) d y,\) where \(C\) is parametrized by: $$ \alpha(t)=\left\\{\begin{array}{ll} (\cos t, \sin t) & \text { if } 0 \leq t \leq \pi \\ (\cos t,-\sin t) & \text { if } \pi \leq t \leq 2 \pi \end{array}\right. $$

Find \(\int_{C} \mathbf{F} \cdot d \mathbf{s}\) if \(\mathbf{F}(x, y)=(-y, x)\) and \(C\) is the circle \(x^{2}+y^{2}=a^{2}\) of radius \(a,\) traversed counterclockwise.

Let \(C\) be a piecewise smooth simple closed curve in \(\mathbb{R}^{2}\), oriented counterclockwise, such that the origin lies in the exterior of \(C\). What are the possible values of the winding number of \(C ?\)

In this exercise, we return to the geometry of curves in \(\mathbb{R}^{3}\) and study the role of the parametrization in defining curvature. Recall that, if \(C\) is a curve in \(\mathbb{R}^{3}\) with a smooth parametrization \(\alpha: I \rightarrow \mathbb{R}^{3},\) then the curvature is defined by the equation: $$\kappa_{\alpha}(t)=\frac{\left\|\mathbf{T}_{\alpha}^{\prime}(t)\right\|}{v_{\alpha}(t)}$$ where \(\mathbf{T}_{\alpha}(t)\) is the unit tangent vector and \(v_{\alpha}(t)\) is the speed. We have embellished our original notation with the subscript \(\alpha\) to emphasize our interest in the effect of the parametrization. We assume that \(v_{\alpha}(t) \neq 0\) for all \(t\) so that \(\kappa_{\alpha}(t)\) is defined. Let \(\beta: J \rightarrow \mathbb{R}^{3}\) be another such parametrization of \(C .\) As in the text, we assume that \(\alpha\) and \(\beta\) are one-to-one, except possibly at the endpoints of their respective domains. As in equation (9.3), there is a function \(g: J \rightarrow I\), assumed to be smooth, such that \(\beta(u)=\alpha(g(u))\) for all \(u\) in \(J .\) In other words, \(g(u)\) is the value of \(t\) such that \(\beta(u)=\alpha(t)\) (a) Show that the velocities of \(\alpha\) and \(\beta\) are related by \(\mathbf{v}_{\beta}(u)=\mathbf{v}_{\alpha}(g(u)) g^{\prime}(u)\). (b) Show that the unit tangents are related by \(\mathbf{T}_{\beta}(u)=\pm \mathbf{T}_{\alpha}(g(u))\). (c) Show that \(\kappa_{\beta}(u)=\kappa_{\alpha}(g(u))\). In other words, if \(\mathbf{x}=\beta(u)=\alpha(t)\) is any point of \(C\) other than an endpoint, then \(\kappa_{\beta}(u)=\kappa_{\alpha}(t) .\) We denote this common value by \(\kappa(\mathbf{x})\), i.e., \(\kappa(\mathbf{x})\) is defined to be \(\kappa_{\alpha}(t)\) for any smooth parametrization \(\alpha\) of \(C,\) where \(t\) is the value of the parameter such that \(\alpha(t)=\mathbf{x} .\) In other words, the curvature at \(\mathbf{x}\) depends only on the point \(\mathbf{x}\) and not on how \(C\) is parametrized.

Find \(\int_{C} e^{y} d x+x e^{y} d y\) if \(C\) is the curve parametrized by \(\alpha(t)=\left(e^{t^{2}}, t^{3}\right), 0 \leq t \leq 1\).

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