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Überprüfen Sie das Vektorfeld $$ \mathbf{w}(x, y, z)=\left(\begin{array}{c} y z \cos (x y z)+2 x z \\ x z \cos (x y z)+2 y z^{2} \\ x y \cos (x y z)+x^{2}+2 y^{2} z \end{array}\right) $$ auf die Potentialfeldeigenschaft und berechnen Sie gegebenenfalls eine Stammfunktion.

Short Answer

Expert verified
Yes, \( \mathbf{w} \) is a potential field with scalar potential \( \phi(x, y, z) = x^2 z + y^2 z^2 + \sin(xyz) \).

Step by step solution

01

Understanding a Potential Field

A vector field \( \mathbf{w}(x, y, z) \) is a potential field if there exists a scalar function \( \phi(x, y, z) \) such that \( abla \phi = \mathbf{w} \). To check if \( \mathbf{w} \) is a potential field, we need to verify if the field is conservative, which means the curl of \( \mathbf{w} \) must be zero.
02

Calculate the Curl

The curl of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by \( abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \). Apply this formula to \( \mathbf{w}(x, y, z) \).
03

Verify the Curl is Zero

After calculating, we find that \( abla \times \mathbf{w} = (0, 0, 0) \), which means the curl of \( \mathbf{w} \) is zero. Thus, \( \mathbf{w} \) is a conservative vector field.
04

Find the Scalar Potential Function

To find the scalar potential function \( \phi \), solve \( abla \phi = \mathbf{w} \). This involves integrating each component of \( \mathbf{w} \):1. Integrate the first component with respect to \( x \), then add a function \( g(y, z) \).2. Integrate the second component with respect to \( y \), include a function \( h(x, z) \).3. Integrate the third component with respect to \( z \), include a function \( k(x, y) \).Ensuring the consistency of these integrations will provide \( \phi(x, y, z) \).
05

Solve for the Scalar Potential

Through consistent integration and simplification, we find the scalar potential function:\[ \phi(x, y, z) = x^2 z + y^2 z^2 + \sin(xyz) \]

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Key Concepts

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

Vector Field
A vector field is a function that assigns a vector to every point in space. In simpler terms, it means that for each location in a region, there is a vector that describes some quantity such as force or velocity. Consider the vector field \( \mathbf{w}(x, y, z) \) from our exercise. It maps a vector to each point defined by the coordinates \( x, y, z \).Understanding vector fields:
  • They have both magnitude and direction.
  • Used to model many physical phenomena, like electromagnetic fields, fluid flow, and gravitational fields.
  • Usually defined in 2D or 3D space.
By analyzing vector fields, we gain insights into how quantities change and behave across space—which can be crucial in fields such as physics and engineering.
Scalar Potential Function
A scalar potential function \( \phi(x, y, z) \) is a scalar value assigned to every point in a field, such that its gradient (or derivative across all dimensions) yields the vector field itself. For a vector to be described as being derived from potential, it must have its components originating from the derivatives of a single scalar function.Key features of a scalar potential function:
  • Represents potential energy in fields like gravity and electrostatics.
  • If a vector field \( \mathbf{F} \) is conservative, the scalar potential \( \phi \) can be found such that \( abla \phi = \mathbf{F} \).
In our exercise, after confirming it's a potential field, the scalar function \( \phi(x, y, z) = x^2 z + y^2 z^2 + \sin(xyz) \) was deduced, underlying the consistency in deriving vector components from a single potential.
Curl of a Vector Field
The curl of a vector field is a measure of the rotation of the field around a point. It's like sensing how and if the field swirls around given locations in space. Mathematically, it's denoted as \( abla \times \mathbf{F} \).Main aspects of the curl:
  • Calculated as a vector, it represents the axis and magnitude of rotation.
  • Used to determine whether a field is conservative.
  • Zero curl means no net rotation, indicating potential conservativeness.
In our exercise, when we calculated \( abla \times \mathbf{w} \) and found it to be zero, it confirmed that the field \( \mathbf{w} \) is conservative, allowing us to search for the scalar potential function.
Conservative Field
A conservative vector field is one where the field is irrotational, meaning curl-free. It implies that the work done around any closed path within the field is zero, just as in gravitational or electrostatic fields.Characteristics of conservative fields:
  • Has a scalar potential function; the field can be expressed as the gradient of this potential.
  • No energy is lost from moving from one point and back to the same point.
  • Typically involves forces like gravity, where potential energy can be clearly described.
In our solution context, because \( abla \times \mathbf{w} = 0 \), \( \mathbf{w} \) is conservative, affirming a potential function exists, aiding in solving physics problems where energy conservation is critical.

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Most popular questions from this chapter

Berechnen Sie das vektorielle Kurvenintegral des Vektorfeldes $$ \mathbf{v}(x, y, z)=\left(\begin{array}{c} y e^{x y} \\ x e^{x y} \\ z^{2} \end{array}\right) \text { entlang der Kurve } \gamma(t)=\left(\begin{array}{c} \cos t \\ \sin ^{2} t \\ t^{2} \sin t \cos t \end{array}\right), t \in[0,2 \pi] $$

Berechnen Sie das vektorielle Kurvenintegral des Vektorfeldes $$ \mathbf{v}(x, y, z)=\left(\begin{array}{c} y e^{x y} \\ x e^{x y} \\ z^{2} \end{array}\right) \text { entlang der Kurve } \gamma(t)=\left(\begin{array}{c} \cos t \\ \sin ^{2} t \\ t^{2} \sin t \cos t \end{array}\right), t \in[0,2 \pi] $$

Zeigen Sie, dass die folgenden Vektorfelder \(\mathbf{v}: G \rightarrow \mathbb{R}^{n}\) eine symmetrische JACOBI-Matrix haben, aber keine Gradientenfelder in \(G\) sind. Woran liegt das? \((a) \mathbf{v}(x, y, z)=\frac{c}{x^{2}+y^{2}}\left(\begin{array}{c}-y \\ x \\\ 0\end{array}\right), \quad G=\left\\{(x, y, z) \in \mathbb{R}^{3} \mid(x, y) \neq(0,0)\right\\}\) (b) \(\quad \mathbf{w}(x, y)=\left(\frac{\frac{y}{x^{2}+y^{2}}+y}{x-\frac{x}{x^{2}+y^{2}}}\right), \quad G=\left\\{(x, y) \in \mathbb{R}^{2} \mid(x, y) \neq(0,0)\right\\}\)

Der Bereich \(B\) aus dem 1. Quadranten \((x \geq 0, y \geq 0)\) des \(\mathbb{R}^{2}\) wird durch die Kurven $$ y=3-x^{2}, \quad y=x^{2}+1 \quad \text { und } \quad x=0 $$ begrenzt. Berechnen Sie das Integral $$ \int_{B} \frac{x}{\sqrt{y}} d F $$

Berechnen Sie das Integral des Vektorfeldes \(\mathbf{v}(x, y, z)=\left(-\frac{y}{x^{2}+y^{2}}, \frac{x}{x^{2}+y^{2}}, 2\right)^{T}\) entlang der Schraubenlinie $$ \gamma(t)=(\cos t, \sin t, t)^{T}, t \in[0,2 \pi] $$

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