Q:

# 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 =

Accepted Solution

A:
Answer:(a) The curl of the vector field curl F is equal to the vector:
$$\displaystyle \text{curl } \textbf{F} = -7e^y \cos z \hat{\i} - 9e^z \cos x \hat{\j} - 7e^x \cos y \hat{\text{k}}$$(b) The divergence of the vector field div F is equal to the expression:
$$\displaystyle \text{div } \textbf{F} = 7e^x \sin y + 7e^y \sin z + 9e^z \sin x$$General Formulas and Concepts:
Pre-CalculusMatrices2x2 Matrix Determinant:
$$\displaystyle \left| \begin{array}{ccc} a & b \\ c & d \end{array} \right| = ad - bc$$3x3 Matrix Determinant:
$$\displaystyle \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right| = a \left| \begin{array}{ccc} e & f \\ h & i \end{array} \right| - b \left| \begin{array}{ccc} d & f \\ g & i \end{array} \right| + c \left| \begin{array}{ccc} d & e \\ g & h \end{array} \right|$$VectorsDot Product:
$$\displaystyle a \cdot b = \sum^{n}_{i = 1} a_ib_i$$CalculusDifferentiationDerivativesDerivative NotationDerivative Property [Multiplied Constant]:
$$\displaystyle (cu)' = cu'$$Derivative Rule [Basic Power Rule]:f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Derivative Rule [Product Rule]:
$$\displaystyle (uv)' = u'v + uv'$$Multivariable CalculusPartial DerivativesVector 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: DefineIdentify given.$$\displaystyle f(x, y, z) = < 7e^x \sin y ,\ 7e^y \sin z ,\ 9e^z \sin x >$$Step 2: Find Curl F[Vector Field] Set up [Curl F]:
$$\displaystyle \text{curl \bf{F}} = \left| \begin{array}{ccc}\hat{\i} & \hat{\j} & \hat{\text{k}} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ 7e^x \sin y & 7e^y \sin z & 9e^z \sin x \end{array} \right|$$[Curl F] Simplify [3x3 Matrix Determinant]:
$$\displaystyle \text{curl \bf{F}} = \left| \begin{array}{ccc} \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ 7e^y \sin z & 9e^z \sin x \end{array} \right| \hat{\i} - \left| \begin{array}{ccc} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial z} \\ 7e^x \sin y & 9e^z \sin x \end{array} \right| \hat{\j} + \left| \begin{array}{ccc} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ 7e^x \sin y & 7e^y \sin z \end{array} \right| \hat{\text{k}}$$[Curl F] Simplify [2x2 Matrix Derterminant]:
$$\displaystyle \text{curl \bf{F}} = \bigg[ \frac{\partial f}{\partial y} \bigg( 9e^z \sin x \bigg) - \frac{\partial f}{\partial z} \bigg( 7e^y \sin z \bigg) \bigg] \hat{\i}} - \bigg[ \frac{\partial f}{\partial x} \bigg( 9e^z \sin x \bigg) - \frac{\partial f}{\partial z} \bigg( 7e^x \sin y \bigg) \bigg] \hat{\j}} + \bigg[ \frac{\partial f}{\partial x} \bigg( 7e^y \sin z \bigg) - \frac{\partial f}{\partial y} \bigg( 7e^x \sin y \bigg) \bigg] \hat{\text{k}}}$$We can differentiate the partial derivatives using basic differentiation techniques listed above under "Calculus":\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}\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}\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}∴ we have found the curl of the given vector field.Step 3: Find Div F[Vector Field] Set up [Div F]:
$$\displaystyle \text{div } \textbf{F} = \frac{\partial}{\partial x} \bigg( 7e^x \sin y \bigg) + \frac{\partial}{\partial y} \bigg( 7e^y \sin z \bigg) + \frac{\partial}{\partial z} \bigg( 9e^z \sin x \bigg)$$We can differentiate the partial derivatives just like when finding the curl:\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}∴ we have found the divergence of the given vector field.---Learn more about div and curl: more about multivariable calculus: : Multivariable CalculusUnit: Stokes' Theorem and Divergence Theorem