Question

Consider the given vector field. f(x, y, z) = 7xyezi + yzexk (a) find the curl of the vector field. curl f = $7yez+0+yex (b) find the divergence of the vector field. div f = (no response)

Answer 1

Answer:

(a) The curl of the vector field curl F is equal to the vector:
$\displaystyle \boxed{ \text{curl } \textbf{F} = ze^x \hat{\i} - (yze^x - 7xye^z) \hat{\j} - 7xe^z \hat{\text{k}} }$

(b) The divergence of the vector field div F is equal to the expression:
$\displaystyle \boxed{ \text{div } \textbf{F} = 7ye^z + ye^x }$

General Formulas and Concepts:
Pre-Calculus

Matrices

2x2 Matrix Determinant:
$\displaystyle \left| \begin{array}{ccc} a & b \\ c & d \end{array} \right| = ad - bc$

3x3 Matrix Determinant:
$\displaystyle \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right| = a \left| \begin{array}{ccc} e & f \\ h & i \end{array} \right| - b \left| \begin{array}{ccc} d & f \\ g & i \end{array} \right| + c \left| \begin{array}{ccc} d & e \\ g & h \end{array} \right|$

Vectors

Dot Product:
$\displaystyle a \cdot b = \sum^{n}_{i = 1} a_ib_i$

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle (cu)' = cu'$

Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
$\displaystyle (uv)' = u'v + uv'$

Multivariable Calculus

Partial Derivatives

Vector Calculus (Line Integrals)

Del (Operator):
$\displaystyle \nabla = \hat{\i} \frac{\partial}{\partial x} + \hat{\j} \frac{\partial}{\partial y} + \hat{\text{k}} \frac{\partial}{\partial z}$

Div and Curl:

• $\displaystyle \text{div \bf{F}} = \nabla \cdot \textbf{F}$
• $\displaystyle \text{curl \bf{F}} = \nabla \times \textbf{F}$

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle f(x, y, z) = 7xye^z \hat{\i} + yze^x \hat{\text{k}}$

Step 2: Find curl F

1. [Vector Field] Set up [curl F]:
   $\displaystyle \text{curl \bf{F}} = \left| \begin{array}{ccc} \hat{\i} & \hat{\j} & \hat{\text{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 7xye^z & 0 & yze^x \end{array} \right|$
2. [curl F] Simplify [3x3 Matrix Determinant]:
   $\displaystyle \text{curl \bf{F}} = \left| \begin{array}{ccc} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & yze^x \\ \end{array} \right| \hat{\i} - \left| \begin{array}{ccc} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ 7xye^z & yze^x \\ \end{array} \right| \hat{\j} + \left| \begin{array}{ccc} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ 7xye^z & 0 \\ \end{array} \right| \hat{\text{k}}$
3. [curl F] Simplify [2x2 Matrix Determinant]:
   $\displaystyle \text{curl \bf{F}} = \bigg[ \frac{\partial}{\partial y} yze^x - \frac{\partial}{\partial z} 0 \bigg] \hat{\i} - \bigg[ \frac{\partial}{\partial x} yze^x - \frac{\partial}{\partial z} 7xye^z \bigg] \hat{\j} + \bigg[ \frac{\partial}{\partial x} 0 - \frac{\partial}{\partial y} 7xye^z \bigg] \hat{\text{k}}$

We can differentiate the partial derivatives using basic differentiation techniques listed above under "Calculus":

$\displaystyle \frac{\partial}{\partial y} yze^x = ze^x$

$\displaystyle \frac{\partial}{\partial z} 0 = 0$

$\displaystyle \frac{\partial}{\partial x} yze^x = yze^x$

$\displaystyle \frac{\partial}{\partial z} 7xye^z = 7xye^z$

$\displaystyle \frac{\partial}{\partial x} 0 = 0$

$\displaystyle \frac{\partial}{\partial y} 7xye^z = 7xe^z$

Substituting in our partial derivative values, we get:

$\displaystyle \begin{aligned}\text{curl \bf{F}} & = (ze^x - 0) \hat{\i} - (yze^x - 7xye^z) \hat{\j} + (0 - 7xe^z) \hat{\text{k}} \\& = \boxed{ze^x \hat{\i} - (yze^x - 7xye^z) \hat{\j} - 7xe^z \hat{\text{k}}} \\\end{aligned}$

∴ we have found the curl of the given vector field.

Step 3: Find div F

1. [Vector Field] Set up [div F]:
   $\displaystyle \text{div } \textbf{F} = \frac{\partial}{\partial x} 7xye^z + \frac{\partial}{\partial y}0 + \frac{\partial}{\partial z} yze^x$

We can differentiate the partial derivatives and use the same method just like when finding the curl:

$\displaystyle \frac{\partial}{\partial x} 7xye^z = 7ye^z$

$\displaystyle \frac{\partial}{\partial y} 0 = 0$

$\displaystyle \frac{\partial}{\partial z} yze^x = ye^x$

$\displaystyle \begin{aligned}\text{div } \textbf{F} & = \frac{\partial}{\partial x} 7xye^z + \frac{\partial}{\partial y}0 + \frac{\partial}{\partial z} yze^x \\& = 7ye^z + 0 + ye^x \\& = \boxed{7ye^z + ye^x} \\\end{aligned}$

∴ we have found the divergence of the given vector field.

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Learn more about div and curl: https://brainly.com/question/7886206

Learn more about multivariable calculus: https://brainly.com/question/2407209

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Topic: Multivariable Calculus

Unit: Stokes' Theorem and Divergence Theorem