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Consider the vector field. f(x, y, z) = 7ex sin(y), 7ey sin(z), 9ez sin(x) (a) find the curl of the vector field. curl f = (b) find the divergence of the vector field. div f =

Respuesta :

Space

Answer:

(a) The curl of the vector field curl F is equal to the vector:
[tex]\displaystyle \text{curl } \textbf{F} = -7e^y \cos z \hat{\i} - 9e^z \cos x \hat{\j} - 7e^x \cos y \hat{\text{k}}[/tex]

(b) The divergence of the vector field div F is equal to the expression:
[tex]\displaystyle \text{div } \textbf{F} = 7e^x \sin y + 7e^y \sin z + 9e^z \sin x[/tex]

General Formulas and Concepts:
Pre-Calculus

Matrices

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

3x3 Matrix Determinant:
[tex]\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|[/tex]

Vectors

Dot Product:
[tex]\displaystyle a \cdot b = \sum^{n}_{i = 1} a_ib_i[/tex]

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

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

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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

Multivariable Calculus

Partial Derivatives

Vector Calculus (Line Integrals)

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

Div and Curl:

  • [tex]\displaystyle \text{div \bf{F}} = \nabla \cdot \textbf{F}[/tex]
  • [tex]\displaystyle \text{curl \bf{F}} = \nabla \times \textbf{F}[/tex]

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle f(x, y, z) = < 7e^x \sin y ,\ 7e^y \sin z ,\ 9e^z \sin x >[/tex]

Step 2: Find Curl F

  1. [Vector Field] Set up [Curl F]:
    [tex]\displaystyle \text{curl \bf{F}} = \left| \begin{array}{ccc}\hat{\i} & \hat{\j} & \hat{\text{k}} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ 7e^x \sin y & 7e^y \sin z & 9e^z \sin x \end{array} \right|[/tex]
  2. [Curl F] Simplify [3x3 Matrix Determinant]:
    [tex]\displaystyle \text{curl \bf{F}} = \left| \begin{array}{ccc} \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ 7e^y \sin z & 9e^z \sin x \end{array} \right| \hat{\i} - \left| \begin{array}{ccc} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial z} \\ 7e^x \sin y & 9e^z \sin x \end{array} \right| \hat{\j} + \left| \begin{array}{ccc} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ 7e^x \sin y & 7e^y \sin z \end{array} \right| \hat{\text{k}}[/tex]
  3. [Curl F] Simplify [2x2 Matrix Derterminant]:
    [tex]\displaystyle \text{curl \bf{F}} = \bigg[ \frac{\partial f}{\partial y} \bigg( 9e^z \sin x \bigg) - \frac{\partial f}{\partial z} \bigg( 7e^y \sin z \bigg) \bigg] \hat{\i}} - \bigg[ \frac{\partial f}{\partial x} \bigg( 9e^z \sin x \bigg) - \frac{\partial f}{\partial z} \bigg( 7e^x \sin y \bigg) \bigg] \hat{\j}} + \bigg[ \frac{\partial f}{\partial x} \bigg( 7e^y \sin z \bigg) - \frac{\partial f}{\partial y} \bigg( 7e^x \sin y \bigg) \bigg] \hat{\text{k}}}[/tex]

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

[tex]\displaystyle\begin{aligned}\frac{\partial f}{\partial y} \bigg( 9e^z \sin x \bigg) & = 0 \\\frac{\partial f}{\partial z} \bigg( 7e^y \sin z \bigg) & = 7e^y \cos z \\\frac{\partial f}{\partial x} \bigg( 9e^z \sin x \bigg) & = 9e^z \cos x \\\frac{\partial f}{\partial z} \bigg( 7e^x \sin y \bigg) & = 0 \\\end{aligned}[/tex]

[tex]\displaystyle\begin{aligned}\frac{\partial f}{\partial x} \bigg( 7e^y \sin z \bigg) & = 0 \\\frac{\partial f}{\partial y} \bigg( 7e^x \sin y \bigg) & = 7e^x \cos y \\\end{aligned}[/tex]

[tex]\displaystyle\begin{aligned}\text{curl } \textbf{F} & = (0 - 7e^y \cos z) \hat{\i} - (9e^z \cos x - 0) \hat{\j} + (0 - 7e^x\cos y) \hat{\text{k}} \\\text{curl } \textbf{F} & = \boxed{-7e^y \cos z \hat{\i} - 9e^z \cos x \hat{\j} - 7e^x \cos y \hat{\text{k}}} \\\end{aligned}[/tex]

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

Step 3: Find Div F

  1. [Vector Field] Set up [Div F]:
    [tex]\displaystyle \text{div } \textbf{F} = \frac{\partial}{\partial x} \bigg( 7e^x \sin y \bigg) + \frac{\partial}{\partial y} \bigg( 7e^y \sin z \bigg) + \frac{\partial}{\partial z} \bigg( 9e^z \sin x \bigg)[/tex]

We can differentiate the partial derivatives just like when finding the curl:

[tex]\displaystyle\begin{aligned}\text{div } \textbf{F} & = \frac{\partial}{\partial x} \bigg( 7e^z \sin y \bigg) + \frac{\partial}{\partial y} \bigg( 7e^y \sin z \bigg) + \frac{\partial}{\partial z} \bigg( 9e^z \sin x \bigg) \\& = \boxed{7e^x \sin y + 7e^y \sin z + 9e^z \sin x}\end{aligned}[/tex]

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

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Learn more about div and curl: https://brainly.com/question/21029519

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

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Topic: Multivariable Calculus

Unit: Stokes' Theorem and Divergence Theorem

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