Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If \(\mathbf{F}=\langle-y, x\rangle\) and \(C\) is the circle of radius 4 centered at (1,0) oriented counterclockwise, then \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}=0\). b. If \(\mathbf{F}=\langle x,-y\rangle\) and \(C\) is the circle of radius 4 centered at (1,0) oriented counterclockwise, then \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}=0\). c. A constant vector field is conservative on \(\mathbb{R}^{2}\). d. The vector field \(\mathbf{F}=\langle f(x), g(y)\rangle\) is conservative on \(\mathbb{R}^{2}\) (assume \(f\) and \(g\) are defined for all real numbers). e. Gradient fields are conservative.

Short answer

In summary: a. False: The circulation of the given vector field around the curve is not equal to 0. b. True: The circulation of the given vector field around the curve is equal to 0. c. True: Constant vector fields have zero curl, making them conservative in \(\mathbb{R}^{2}\). d. True: Vector fields with the form \(\langle f(x), g(y)\rangle\) have zero curl, making them conservative. e. True: Gradient fields have zero curl, making them conservative vector fields.

Step-by-step solution

Statement a

We need to prove whether the statement is true or false about the field \(\mathbf{F}=\langle -y, x\rangle\) on the given curve (circle of radius 4 centered at (1,0)). To do this, we compute the circulation \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\). First, parameterize the curve: \(r(t) = \langle 1 + 4\cos{t}, 4\sin{t} \rangle\) for \(0\le t\le 2\pi\) Next, we find the derivative of \(r(t)\): \(r'(t) = \langle -4\sin{t}, 4\cos{t} \rangle\) Now we substitute the parameterization into \(\mathbf{F}\): \(\mathbf{F}(r(t)) = \langle -4\sin{t}, 1 + 4\cos{t} \rangle\) The line integral on curve C is: \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} = \int_{0}^{2\pi} \mathbf{F}(r(t)) \cdot r'(t) dt = \int_{0}^{2\pi} (-4\sin t)(-4\sin t) + (1 + 4\cos t)(4\cos t) dt\) Calculating the integral, we get: \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} =\int_{0}^{2\pi} (16\sin^2{t}+4\cos^2{t} + 4\cos{t}) dt\) Notice that \(16\sin^2{t} + 4\cos^2{t} = 4(4\sin^2{t} + \cos^2{t}) \ge 4 \neq 0\) The integral is not equal to 0, which means statement a is false.

Statement b

We need to prove whether the statement is true or false about the field \(\mathbf{F}=\langle x, -y\rangle\) on the given curve (circle of radius 4 centered at (1,0)). To do this, we compute the circulation \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\). Substitute the parameterization into \(\mathbf{F}\): \(\mathbf{F}(r(t)) = \langle 1+4\cos{t},-4\sin{t} \rangle\) The line integral on curve C is: \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} = \int_{0}^{2\pi} \mathbf{F}(r(t)) \cdot r'(t) dt = \int_{0}^{2\pi} (1+4\cos t)(-4\sin t) + (-4\sin t)(4\cos t) dt\) Calculating the integral, we get: \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} = 0\) The integral is equal to 0, which means statement b is true.

Statement c

A constant vector field is conservative if the curl of the vector field is equal to zero since this is equivalent to the line integral being path independent. Curl for vector fields in \(\mathbb{R}^{2}\) is given by \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\). In this case, P and Q are constants, so their partial derivatives are zero, resulting in the curl being equal to zero. Thus, constant vector fields are conservative in \(\mathbb{R}^{2}\), and statement c is true.

Statement d

For a vector field \(\mathbf{F}=\langle f(x), g(y)\rangle\) to be conservative in \(\mathbb{R}^{2}\), its curl must be equal to zero. Curl of \(\mathbf{F}=\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\). Since \(f(x)\) is a function of \(x\) only and \(g(y)\) is a function of \(y\) only, this means that the partial derivatives \(\frac{\partial f}{\partial y}\) and \(\frac{\partial g}{\partial x}\) are zero. Therefore, the curl of \(\mathbf{F}\) is zero, and the vector field is conservative. Statement d is true.

Statement e

A gradient field is the field generated by taking gradients of a scalar function \(\phi\), i.e., \(\mathbf{F} = \nabla \phi\). The curl of a gradient field is \(\nabla \times\nabla \phi\). Using vector calculus identities, the curl of a gradient field is always equal to zero. Since curl is zero, gradient fields are conservative. Thus, statement e is true.

Key concepts

Conservative Vector Field

A conservative vector field is one where the line integral from one point to another is path independent. This means that, no matter what path you take between the two points, the line integral will always have the same value. An intriguing characteristic of such fields is that if you take any closed loop path, the line integral of the field around that loop will be zero. This is deeply connected to physical concepts, for example, in a conservative force field, no energy is lost when moving around a closed path.

In the context of our example, for a vector field to be conservative over \(\mathbb{R}^2\), the curl of the field must be zero everywhere. Mathematically, for a two-dimensional vector field \(\mathbf{F} = \langle P(x, y), Q(x, y) \rangle\), the curl is zero if \(\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}\). Applying this understanding to Statements a, b, and d from the exercise verifies their truthfulness, whereas for Statement a, the vector field in question did not satisfy the properties of a conservative vector field since the line integral did not output zero.

Circulation and Curl

The terms 'circulation' and 'curl' are used to describe the tendency of a vector field to swirl or rotate around a point. Circulation refers to the line integral around a closed curve, essentially measuring the amount of the field that is ''circulating'' around a path. If the circulation is non-zero, it indicates the presence of a ''rotational'' component in the field.

Curl, on the other hand, is a vector operator that describes the infinitesimal rotation at a point in a vector field. In \(\mathbb{R}^2\), the curl of a vector field \(\mathbf{F}\) is given as \(abla \times \mathbf{F}\) which simplifies to the scalar \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\) for a field \(\mathbf{F} = \langle P, Q \rangle\). A zero curl indicates that the field is irrotational, which is a fundamental property of conservative fields, as seen in the statements c, d, and e from the textbook exercise.

Gradient Field

A gradient field is generated by the gradient of a scalar function \(\phi(x, y)\), indicated as \(\mathbf{F} = abla \phi\), where \(abla\) denotes the gradient operator. When you take the gradient of a scalar field, you're essentially calculating the directional rate of change at each point, giving you a vector field where each vector points in the direction of greatest increase of the function.

One of the intrinsic properties of gradient fields is their conservativeness. The curl of a gradient field is always zero, which follows from the fact that mixed partial derivatives of a smooth function commute - a conclusion of Clairaut's theorem. This applies directly to Statement e, where the truth of the statement relies on the property that all gradient fields are conservative because their curl is zero.

Parameterization of Curves

Parameterization is the process of expressing a curve as a function of a variable, typically \(t\), which is called the parameter. Through parameterization, we can describe the position of a point on the curve as it 'travels' over time from \(t = a\) to \(t = b\). This is particularly useful for line integrals, as it allows us to convert them into definite integrals that can be evaluated with standard calculus techniques.

In our example, we parameterized the circle of radius 4 centered at (1,0) for Statements a and b, translating it into a set of equations depending on \(t\). This shows the relevance of proper parameterization; it simplifies the process of calculating line integrals over curves in vector fields, as well as determining the properties of the fields, such as whether the vector field is conservative or if it has a circulation around a closed curve.