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Finding potential functions Determine whether the following vector fields are conservative on the specified region. If so, determine a potential function. Let \(R^{*}\) and \(D^{*}\) be open regions of \(\mathbb{R}^{2}\) and \(\mathbb{R}^{3}\), respectively, that do not include the origin. $$\mathbf{F}=\frac{\langle x, y\rangle}{\sqrt{x^{2}+y^{2}}} \text { on } R^{*}$$

Short Answer

Expert verified
If so, find the potential function. Answer: No, the vector field $$\mathbf{F}=\frac{\langle x, y\rangle}{\sqrt{x^{2}+y^{2}}} \text { on } R^{*}$$ is not conservative in the specified region \(R^{*}\), as its curl is not zero. Therefore, there is no potential function associated with it.

Step by step solution

01

Calculate the curl of the vector field

To determine whether the vector field is conservative, we need to calculate its curl. For a two-dimensional vector field \(\mathbf{F} = \langle P, Q\rangle\), the curl is given by: $$\text{curl } \mathbf{F} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$$ Here, the components P and Q of the vector field \(\mathbf{F}\) are: $$P = \frac{x}{\sqrt{x^2+y^2}}, Q = \frac{y}{\sqrt{x^2+y^2}}$$ Now, let's calculate the partial derivatives of P and Q with respect to x and y.
02

Calculate the partial derivatives of P and Q

First, compute the partial derivatives with respect to x and y: $$\frac{\partial P}{\partial x} = \frac{y^2}{(x^2+y^2)^{3/2}}, \quad \frac{\partial Q}{\partial y} = \frac{x^2}{(x^2+y^2)^{3/2}}$$
03

Check if the curl of the vector field is zero

Now, substitute the values of the partial derivatives into the expression of curl: $$\text{curl } \mathbf{F} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{y^2 - x^2}{(x^2+y^2)^{3/2}}$$ We can see that the curl is not zero, so the vector field \(\mathbf{F}=\frac{\langle x, y\rangle}{\sqrt{x^{2}+y^{2}}}\) on \(R^{*}\) is not conservative. Since the vector field is not conservative, there is no potential function associated with it.

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Key Concepts

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

Potential Function
In the context of vector fields, a potential function is quite a significant concept. A potential function for a vector field is a scalar function whose gradient produces the original vector field. Specifically, for a two-dimensional vector field \( \mathbf{F} = \langle P, Q \rangle \), if there exists a scalar function \( \phi \) such that \( abla \phi = \langle P, Q \rangle \) or in other words, \( P = \frac{\partial \phi}{\partial x} \) and \( Q = \frac{\partial \phi}{\partial y} \), then \( \phi \) is called the potential function of \( \mathbf{F} \).

A critical point about potential functions is that they can only be found for conservative vector fields. A vector field is conservative if its line integral between two points is independent of the path taken. This independence from path suggests the existence of a potential function. This means that if a potential function exists, the work done by the vector field along any path between two points only depends on the positions of those points and not on the trajectory taken.
Curl of a Vector Field
The curl of a vector field is a measure that indicates the rotation of the vector field at a point. For two-dimensional vector fields like \( \mathbf{F} = \langle P, Q \rangle \), the curl is a scalar and is found by taking the difference between the partial derivatives of the second component with respect to 'x' and the first component with respect to 'y'. Mathematically, it is expressed as \( \text{curl } \mathbf{F} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \).

The curl gives an idea of the 'twisting' nature at a certain point in the field. If the curl of a vector field is zero everywhere in a region, the field is said to be irrotational in that region, which often indicates the field is conservative. For conservative fields, it's like rolling a ball in a smooth bowl—no matter which way you roll it, it can return to the original point without any swirling motion, hence no 'curl' or rotation.
Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus and play a crucial role in understanding vector fields. They represent the rate at which a function changes as one of its variables is varied while the others are held constant. For a function with two variables, \( f(x, y) \), we can compute the partial derivative with respect to 'x' while keeping 'y' constant, denoted as \( \frac{\partial f}{\partial x} \), and similarly, compute the partial derivative with respect to 'y' while keeping 'x' constant, denoted as \( \frac{\partial f}{\partial y} \).

When dealing with vector fields, we use partial derivatives to compute various quantities like divergence, gradient, and curl. These derivatives provide information about the behavior of the field in different regions and directions. The interplay between the partial derivatives of the vector field's components determines whether the field is conservative and if a potential function can be found. As seen in the exercise, mismatched partial derivatives contributed to a nonzero curl, indicating that the vector field in question is not conservative and lacks a potential function.

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Most popular questions from this chapter

What's wrong? Consider the radial field \(\mathbf{F}=\frac{\langle x, y\rangle}{x^{2}+y^{2}}\) a. Verify that the divergence of \(\mathbf{F}\) is zero, which suggests that the double integral in the flux form of Green's Theorem is zero. b. Use a line integral to verify that the outward flux across the unit circle of the vector field is \(2 \pi\) c. Explain why the results of parts (a) and (b) do not agree.

Mass and center of mass Let \(S\) be a surface that represents a thin shell with density \(\rho .\) The moments about the coordinate planes (see Section 16.6 ) are \(M_{y z}=\iint_{S} x \rho(x, y, z) d S, M_{x z}=\iint_{S} y \rho(x, y, z) d S\) and \(M_{x y}=\iint_{S} z \rho(x, y, z) d S .\) The coordinates of the center of mass of the shell are \(\bar{x}=\frac{M_{y z}}{m}, \bar{y}=\frac{M_{x z}}{m},\) and \(\bar{z}=\frac{M_{x y}}{m},\) where \(m\) is the mass of the shell. Find the mass and center of mass of the following shells. Use symmetry whenever possible. The cylinder \(x^{2}+y^{2}=a^{2}, 0 \leq z \leq 2,\) with density \(\rho(x, y, z)=1+z\)

Miscellaneous surface integrals Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward. $$\begin{aligned} &\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S, \text { where } S \text { is the hemisphere } x^{2}+y^{2}+z^{2}=a^{2}\\\ &\text { for } z \geq 0, \text { and where } \mathbf{r}=\langle x, y, z\rangle \end{aligned}$$

Maximum surface integral Let \(S\) be the paraboloid \(z=a\left(1-x^{2}-y^{2}\right),\) for \(z \geq 0,\) where \(a>0\) is a real number. Let \(\mathbf{F}=\langle x-y, y+z, z-x\rangle .\) For what value(s) of \(a\) (if any) does \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) have its maximum value?

Green's Theorem as a Fundamental Theorem of Calculus Show that if the circulation form of Green's Theorem is applied to the vector field \(\left\langle 0, \frac{f(x)}{c}\right\rangle\), where \(c>0\) and \(R=\\{(x, y): a \leq x \leq b, 0 \leq y \leq c\\},\) then the result is the Fundamental Theorem of Calculus, $$ \int_{a}^{b} \frac{d f}{d x} d x=f(b)-f(a) $$

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