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Chapter 3: Problem 11

Berechnen Sie \(^{c} \int \vec{v} \mathrm{~d} \vec{r}\) f眉r \(\vec{v}=\left(2 x-y,-y^{2} z^{2}, x y z\right)\) und \(C \cdot \vec{r}(t)=(\cos t, \sin t, 0), 0 \leqq t \leqq 2 \pi\). Was f眉r eine \(\mathrm{K}\) urve ist \(C\) ? Ist \(\vec{v}\) konservativ?

Short Answer

Expert verified
The line integral is \( \pi \). Curve \( C \) is a circle. \( \vec{v} \) is not conservative.

Step by step solution

01

Parameterization of the Curve

The curve \( C \) is parameterized as \( \vec{r}(t) = (\cos t, \sin t, 0) \) with \( 0 \leq t \leq 2\pi \). This describes a circle in the \( xy \)-plane centered at the origin with a radius of 1.
02

Calculate the Derivative of \( \vec{r}(t) \)

The derivative of \( \vec{r}(t) = (\cos t, \sin t, 0) \) with respect to \( t \) is \( \vec{r}'(t) = (-\sin t, \cos t, 0) \). This derivative represents the tangent vector to the curve \( C \) at any point \( t \).
03

Write the Line Integral

The line integral over the vector field \( \vec{v} \) along the curve \( C \) is given by \[^{c} \int \vec{v} \cdot \mathrm{d} \vec{r} = \int_{0}^{2\pi} \vec{v}(\vec{r}(t)) \cdot \vec{r}'(t) \ dt.\]
04

Substitute \( \vec{v}(\vec{r}(t)) \) and Calculate the Dot Product

Substitute \( \vec{v} = (2x - y, -y^2 z^2, x y z) \) into the parameterization, giving \( \vec{v}(\vec{r}(t)) = (2\cos t - \sin t, 0, 0) \) since \( z = 0 \). Compute\[ \vec{v}(\vec{r}(t)) \cdot \vec{r}'(t) = (2\cos t - \sin t, 0, 0) \cdot (-\sin t, \cos t, 0) = -2\cos t \sin t - \sin^2 t. \]
05

Evaluate the Integral

The integral becomes \[ \int_{0}^{2\pi} (-2\cos t \sin t - \sin^2 t) \ dt. \]Breaking it into two parts \[ \int_{0}^{2\pi} -2\cos t \sin t \ dt \] and \[ \int_{0}^{2\pi} -\sin^2 t \ dt. \]Using the identity \( \cos(2t) = 1 - 2\sin^2 t \), simplify them: \[ \int_{0}^{2\pi} -\sin^2 t \ dt = \frac{1}{2} \int_{0}^{2\pi} (1 - \cos(2t)) \ dt = \pi. \] Thus, \( ^{c} \int \vec{v} \cdot \mathrm{d} \vec{r} = 0 + \pi = \pi. \)
06

Analyze the Curve Type and Vector Field

The parametric equation \( \vec{r}(t) = (\cos t, \sin t, 0) \) reveals that \( C \) is a circle in the \( xy \)-plane. To see if \( \vec{v} \) is conservative, verify if its curl is \( \vec{0} \). Calculate \[ abla \times \vec{v} = \left(\frac{\partial}{\partial y}(xy) - \frac{\partial}{\partial z}(-y^2 z^2), \frac{\partial}{\partial z}(2x - y) - \frac{\partial}{\partial x}(xy), \frac{\partial}{\partial x}(-y^2 z^2) - \frac{\partial}{\partial y}(2x - y) \right) = \vec{0} \] shows \( \vec{v} \) is not conservative as the curl is not zero.

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Key Concepts

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

Parameterization
Parameterization is a method used to describe a curve using a single variable, often called a parameter. In this exercise, the curve \( C \) is parameterized as \( \vec{r}(t) = (\cos t, \sin t, 0) \) where \( 0 \leq t \leq 2\pi \). This parameterization represents a circle in the \( xy \)-plane, centered at the origin with a radius of 1. Here鈥檚 how parameterization helps:
  • It simplifies the representation of geometric objects such as curves and surfaces.
  • Allows us to perform calculus operations, like differentiation and integration, on these objects.
  • Helps to explicitly describe the position of a point on the curve as a function of the parameter.
The choice of parameter, \( t \) in this case, advances from \( 0 \) to \( 2\pi \), guiding the motion along the circle, which is crucial for evaluating line integrals around the closed path.
Vector Field
A vector field is a function that assigns a vector to every point in space. In this problem, the vector field \( \vec{v} = (2x - y, -y^2 z^2, xyz) \) is given. Each component of \( \vec{v} \) is a function of the spatial coordinates \( x, y, \) and \( z \). Understanding vector fields is essential in fields such as physics and engineering because:
  • They visually represent the direction and magnitude of vectors in a given space.
  • Useful for analyzing vector quantities like velocity and force fields.
  • Provide insights into how vectors change across different regions of space.
In practical applications, vector fields can illustrate fluid flow, magnetic fields, or gravitational forces, aiding in various analyses.
Conservative Vector Field
A conservative vector field is a vector field where the line integral is path-independent, depending only on the endpoints. For a vector field \( \vec{v} \) to be conservative, its curl must be zero. This means:
  • The field can be expressed as the gradient of some potential function.
  • Work done in moving along a closed path in the field is zero.
  • Without loss of generality, there is no energy loss or gain when moving around a closed path.
In this exercise, calculating the curl of \( \vec{v} \) reveals that it is not zero, indicating \( \vec{v} \) is not a conservative vector field. This implies that the work done along the path \( C \) may depend on the path taken, not just the start and end points.
Curve Analysis
Curve analysis involves understanding the properties and behaviors of curves, which are often parameterized, as with curve \( C \) in this exercise. It provides insights into:
  • Geometric properties like length, curvature, and path shape.
  • The interaction between curves and vector fields through operations like line integrals.
  • Differential properties, such as tangents and normals, which are analyzed via derivatives.
In this step-by-step solution, the curve \( C \) is identified as a circle in the \( xy \)-plane through parameterization. By analyzing the integral over this curve, insights into how vectors influence this path are gained. The derivative \( \vec{r}'(t) \) provides the tangent vector needed to evaluate changes along the path. This analysis is central for interpreting physical applications, like computing work done by forces along a trajectory.

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Most popular questions from this chapter

Es sei \(C\) cine a) cinmal, b) \(n\)-mal durchlaufene Kreislinie im Raum, die die \(z\)-Achse nicht schneidet und \(\vec{v}(x, y, z)=\frac{1}{x^{2}+y^{2}} \cdot(-y, x, 0)\). Welchen Wert hat \(^{c} \oint \vec{v} \mathrm{~d} \vec{r}^{\prime} ?\)

Man bestimme den h枚chsten und tiefsten Punkt auf der Ellipse mit der Gleichung \(2 x^{2}+6 x y+3 y^{2}+6=0\).

Die Funktionen \(f\) und \(g\) seien auf \(\mathbb{R}\) zweimal stetig differenzierbar und \(c \in \mathbb{R} .\) Zeigen Sie, da脽 f眉r die Funktion \(u\) mit \(u(x, t)=f(x+c t)+g(x-c t)\) die sogenannte Wellengleichung \(u_{n}=c^{2} \cdot u_{x x}\) gilt.

Folgende Funktionen sind auf Stetigkeit zu untersuchen. a) \(f(x, y)=\sin \left(x^{-} y+\sqrt{y}\right)\) b) \(f(x, y)=\ln \left(x^{2}+\sqrt{x y}\right)\) c) \(f(x, y)= \begin{cases}x \cdot \cos \left(x^{2}+y\right), & \text { wenn } x>0 ; \\ x^{2}, & \text { wenn } x \leqq 0\end{cases}\) d) \(f(x, y)= \begin{cases}\frac{\mathrm{e}^{y}-1}{x^{2}+y^{2}}, & \text { wenn }(x, y) \neq(0,0) \\ 0, \quad \text { wenn } x=y=0 .\end{cases}\)

Wie lautet die Gleichung der Fl盲che, die entsteht, wenn die Kurve mit der Gleichung \(z=\ln x\) f眉r \(x>0\) um die \(z\)-Achse rotiert?
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