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:(a) The curl of the vector field curl F is equal to the vector:
[tex]\displaystyle \boxed{ \text{curl } \textbf{F} = ze^x \hat{\i} - (yze^x - 7xye^z) \hat{\j} - 7xe^z \hat{\text{k}} }[/tex](b) The divergence of the vector field div F is equal to the expression:
[tex]\displaystyle \boxed{ \text{div } \textbf{F} = 7ye^z + ye^x }[/tex]General Formulas and Concepts:
Pre-CalculusMatrices2x2 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]VectorsDot Product:
[tex]\displaystyle a \cdot b = \sum^{n}_{i = 1} a_ib_i[/tex]CalculusDifferentiationDerivativesDerivative NotationDerivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]Derivative Rule [Basic Power Rule]:f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Derivative Rule [Product Rule]:
[tex]\displaystyle (uv)' = u'v + uv'[/tex]Multivariable CalculusPartial DerivativesVector 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: DefineIdentify given.[tex]\displaystyle f(x, y, z) = 7xye^z \hat{\i} + yze^x \hat{\text{k}}[/tex]Step 2: Find curl F[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}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 7xye^z & 0 & yze^x \end{array} \right|[/tex][curl F] Simplify [3x3 Matrix Determinant]:
[tex]\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}}[/tex][curl F] Simplify [2x2 Matrix Determinant]:
[tex]\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}}[/tex]We can differentiate the partial derivatives using basic differentiation techniques listed above under "Calculus":[tex]\displaystyle \frac{\partial}{\partial y} yze^x = ze^x[/tex][tex]\displaystyle \frac{\partial}{\partial z} 0 = 0[/tex][tex]\displaystyle \frac{\partial}{\partial x} yze^x = yze^x[/tex][tex]\displaystyle \frac{\partial}{\partial z} 7xye^z = 7xye^z[/tex][tex]\displaystyle \frac{\partial}{\partial x} 0 = 0[/tex][tex]\displaystyle \frac{\partial}{\partial y} 7xye^z = 7xe^z[/tex]Substituting in our partial derivative values, we get:[tex]\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}[/tex]∴ we have found the curl of the given vector field.Step 3: Find div F[Vector Field] Set up [div F]:
[tex]\displaystyle \text{div } \textbf{F} = \frac{\partial}{\partial x} 7xye^z + \frac{\partial}{\partial y}0 + \frac{\partial}{\partial z} yze^x[/tex]We can differentiate the partial derivatives and use the same method just like when finding the curl: [tex]\displaystyle \frac{\partial}{\partial x} 7xye^z = 7ye^z[/tex][tex]\displaystyle \frac{\partial}{\partial y} 0 = 0[/tex][tex]\displaystyle \frac{\partial}{\partial z} yze^x = ye^x[/tex][tex]\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}[/tex]∴ 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