Chapter 20: Problem 2
Verify that div \(\mathbf{F}=0\) and curl \(\mathbf{F}=\mathbf{0}\) for the given vector field \(\mathbf{F}(x, y)\) by examining the corresponding complex function \(g(z)=P(x, y)-i Q(x, y)\). Find a complex potential for the vector field and sketch the equipotential lines. \(\mathbf{F}(x, y)=-y \mathbf{i}-x \mathbf{j}\)
Short Answer
Step by step solution
Express Vector Field Components
Compute Divergence
Compute Curl
Form Complex Function
Determine Complex Potential
Sketch Equipotential Lines
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Fields
In our exercise, the vector field is given as \( \mathbf{F}(x, y) = -y \mathbf{i} - x \mathbf{j} \). This means that at any point \((x, y)\), the vector has components \(-y\) in the \(x\)-direction and \(-x\) in the \(y\)-direction.
- The component \(-y\) is associated with the unit vector \( \mathbf{i} \), which points along the horizontal axis.
- The component \(-x\) is tied to \( \mathbf{j} \), indicating direction along the vertical axis.
Divergence and Curl
- Divergence measures how much a vector field spreads out from a point. It can be thought of as the net flow of the vector field's magnitude away from that point. A positive divergence indicates a source, while a negative divergence indicates a sink.
- Curl measures the rotation or the swirling of a vector field around a point. A non-zero curl implies the field exhibits circular motion around the point.
Complex Potential
In this context, the exercise identifies the function \( g(z) = -y + ix \), which is derived from the vector field components. Recognizing \( g(z) \) as\(-iz\), the integration yields the complex potential function \( f(z) = -\frac{i}{2}z^2 + C \), where \(C\) is an integration constant.
- This potential function encapsulates the characteristics of the vector field.
- It allows the derivation of the field's components through differentiation.
Equipotential Lines
For the complex potential function \( f(z) = -\frac{i}{2}z^2 \), the equipotential lines are determined using the imaginary part of the potential function. By expressing \( z \) in terms of its real and imaginary parts, \( z = x + iy \), the imaginary part becomes \(-xy\).
For this particular vector field, equipotential lines are hyperbolas defined by \(-xy = \text{constant}\). Understanding and plotting these lines provides insight into how the field behaves, showing the general flow and symmetry of the forces or velocities represented by the vector field.