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Magazine Chapter 18: Problem 18
Find \(\nabla \times \mathbf{F}\) and \(\nabla \cdot \mathbf{F}\). $$ \mathbf{F}(x, y, z)=x^{3} \ln z \mathbf{i}+x e^{-y} \mathbf{j}-\left(y^{2}+2 z\right) \mathbf{k} $$
Short Answer
Expert verified
\(\nabla \times \mathbf{F} = (-2y, \frac{x^3}{z}, e^{-y})\), \(\nabla \cdot \mathbf{F} = 3x^2 \ln z - x e^{-y} - 2\).
Step by step solution
01
Understand the Problem
The problem asks for the curl \(abla \times \mathbf{F}\) and the divergence \(abla \cdot \mathbf{F}\) of the vector field \(\mathbf{F}(x, y, z)=x^{3} \ln z \mathbf{i}+x e^{-y} \mathbf{j}-(y^{2}+2z) \mathbf{k}\). The vector field \(\mathbf{F}\) consists of components \(F_1 = x^3 \ln z\), \(F_2 = x e^{-y}\), and \(F_3 = -(y^2 + 2z)\).
02
Find the Curl (\(\nabla \times \mathbf{F}\))
To find the curl, use the formula for the curl of a vector field \(\mathbf{F}=(F_1, F_2, F_3)\):\[abla \times \mathbf{F} = \left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right)\]Compute each component:- \(\frac{\partial F_3}{\partial y} = \frac{\partial}{\partial y}(-y^2 - 2z) = -2y\)- \(\frac{\partial F_2}{\partial z} = \frac{\partial}{\partial z}(x e^{-y}) = 0\)- \(\frac{\partial F_1}{\partial z} = \frac{\partial}{\partial z}(x^3 \ln z) = \frac{x^3}{z}\)- \(\frac{\partial F_3}{\partial x} = \frac{\partial}{\partial x}(-y^2 - 2z) = 0\)- \(\frac{\partial F_2}{\partial x} = \frac{\partial}{\partial x}(x e^{-y}) = e^{-y}\)- \(\frac{\partial F_1}{\partial y} = \frac{\partial}{\partial y}(x^3 \ln z) = 0\)Thus, \[abla \times \mathbf{F} = (-2y - 0, \frac{x^3}{z} - 0, e^{-y} - 0) = (-2y, \frac{x^3}{z}, e^{-y})\]
03
Find the Divergence (\(\nabla \cdot \mathbf{F}\))
To find the divergence, use the formula for the divergence of a vector field \(\mathbf{F}=(F_1, F_2, F_3)\):\[abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\]Compute each partial derivative:- \(\frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}(x^3 \ln z) = 3x^2 \ln z\)- \(\frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y}(x e^{-y}) = -x e^{-y}\)- \(\frac{\partial F_3}{\partial z} = \frac{\partial}{\partial z}(-(y^2 + 2z)) = -2\)Thus, \[abla \cdot \mathbf{F} = 3x^2 \ln z - x e^{-y} - 2\]
04
Present the Solutions
The curl of the vector field is \(abla \times \mathbf{F} = (-2y, \frac{x^3}{z}, e^{-y})\) and the divergence of the vector field is \(abla \cdot \mathbf{F} = 3x^2 \ln z - x e^{-y} - 2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Curl of a Vector Field
The curl of a vector field is an important concept in vector calculus, especially when dealing with vector fields in three dimensions. In the context of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \), the curl gives us a vector that represents the rotation or the twisting of the field around a point. The formula for calculating the curl is given by:\[abla \times \mathbf{F} = \left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right)\]To find the curl of the vector field \( \mathbf{F}(x, y, z) = x^3 \ln z \mathbf{i} + x e^{-y} \mathbf{j} - (y^2 + 2z) \mathbf{k} \), you follow three main steps:
- Calculate the partial derivative of \( F_3 \) with respect to \( y \) and \( F_2 \) with respect to \( z \).
- Calculate the partial derivative of \( F_1 \) with respect to \( z \) and \( F_3 \) with respect to \( x \).
- Calculate the partial derivative of \( F_2 \) with respect to \( x \) and \( F_1 \) with respect to \( y \).
Divergence of a Vector Field
Divergence measures how much a vector field spreads out or converges at a given point. It tells us about the "source" or "sink" behavior of the field within a region. For a vector field \( \mathbf{F} = (F_1, F_2, F_3) \), the divergence is calculated using the formula:\[abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\]Applying this to the vector field \( \mathbf{F}(x, y, z) = x^3 \ln z \mathbf{i} + x e^{-y} \mathbf{j} - (y^2 + 2z) \mathbf{k} \), involves:
- Calculating the partial derivative of \( F_1 \) concerning \( x \).
- Calculating the partial derivative of \( F_2 \) concerning \( y \).
- Calculating the partial derivative of \( F_3 \) concerning \( z \).
Partial Derivatives
Partial derivatives play a crucial role in vector calculus by providing the rate of change of a function concerning one variable while holding other variables constant. They are essential for calculating both curl and divergence, allowing the breakdown of complex, multi-variable expressions into simpler parts.To better understand their application, let's consider our vector field \( \mathbf{F}(x, y, z) = x^3 \ln z \mathbf{i} + x e^{-y} \mathbf{j} - (y^2 + 2z) \mathbf{k} \).
- The partial derivative of \( F_1 = x^3 \ln z \) with respect to \( z \) is \( \frac{x^3}{z} \).
- The partial derivative of \( F_2 = x e^{-y} \) with respect to \( x \) is \( e^{-y} \).
- The partial derivative of \( F_3 = -(y^2 + 2z) \) with respect to \( y \) is \( -2y \).
