Question

Prove product rules (i), (iv), and (v)

Short answer

The product rules (i), (iv), and (v) are proved

Step-by-step solution

Find the curl of vector v

To prove any rule, simplify its left and right side, and comare them with each other.

Let the vector v be defined as ${\mathrm{v}}_{x}i+v_{y}j+v_{z}k$and thelocalid="1657359311673" $\nabla$operator is defined as $\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$. The gradient of vector v is obtaind as

localid="1657346308106" $$\begin{gathered} \nabla .v=\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right).\left(v_{x}i+v_{y}j+v_{z}k\right) \\ =\frac{\partial v_x}{\partial x}i+\frac{\partial v_y}{\partial y}j+\frac{\partial v_z}{\partial z} \\ \mathrm{The}\mathrm{curl}\mathrm{of}\mathrm{vector}\mathrm{v}\mathrm{is}\mathrm{obtaind}\mathrm{as} \\ \nabla .\mathrm{v}=\left(\frac{\partial }{\partial \mathrm{x}}\mathrm{i}+\frac{\partial }{\partial \mathrm{y}}\mathrm{j}+\frac{\partial }{\partial \mathrm{z}}\mathrm{k}\right).\left({\mathrm{v}}_{\mathrm{x}}\mathrm{i}+{\mathrm{v}}_{\mathrm{y}}\mathrm{j}+{\mathrm{v}}_{\mathrm{z}}\mathrm{k}\right) \\ =\left(\frac{\partial {\mathrm{v}}_{\mathrm{x}}}{\partial \mathrm{y}}-\frac{\partial {\mathrm{v}}_{\mathrm{x}}}{\partial \mathrm{z}}\mathrm{k}\right)\mathrm{i}+\left(\frac{\partial {\mathrm{v}}_{\mathrm{y}}}{\partial \mathrm{z}}-\frac{\partial {\mathrm{v}}_{\mathrm{y}}}{\partial \mathrm{x}}\mathrm{k}\right)\mathrm{j}+\left(\frac{\partial {\mathrm{v}}_{\mathrm{z}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{v}}_{\mathrm{z}}}{\partial \mathrm{y}}\right)\mathrm{k} \end{gathered}$$

 Prove∇(fg)=∇g+g∇f 

In the expression$\nabla \left(fg\right)=f\nabla g+g\nabla f$, where f,g are two dimensional vectors..

Find the gradient of vector g.

$$\begin{gathered} \nabla \left(\mathrm{g}\right)=\left(\frac{\partial }{\partial \mathrm{x}}\mathrm{i}+\frac{\partial }{\partial \mathrm{x}}\mathrm{j}\right)\left(\mathrm{g}\right) \\ =\frac{\partial }{\partial \mathrm{x}}\mathrm{i}+\frac{\partial }{\partial \mathrm{x}} \end{gathered}$$

Find the gradient of vector f.

$$\begin{gathered} \nabla \left(\mathrm{f}\right)=\left(\frac{\partial }{\partial \mathrm{x}}\mathrm{i}+\frac{\partial }{\partial \mathrm{x}}\mathrm{j}\right)\left(\mathrm{f}\right) \\ =\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\mathrm{j} \\ \mathrm{Now}\mathrm{applying}\mathrm{the}\nabla \mathrm{perator}\mathrm{on}\mathrm{the}\mathrm{product}\mathrm{of}f\mathrm{and}g. \\ \nabla \left(\mathrm{fg}\right)=\frac{\partial \left(\mathrm{fg}\right)}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial \left(\mathrm{fg}\right)}{\partial \mathrm{y}}\mathrm{j} \\ =\mathrm{f}\frac{\partial \mathrm{g}}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial \mathrm{g}}{\partial \mathrm{y}}\mathrm{j}+\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\mathrm{j} \\ =\mathrm{f}\left(\nabla \mathrm{g}\right)+\mathrm{g}\left(\nabla \mathrm{f}\right) \\ \mathrm{Thus},\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{that}\nabla \left(\mathrm{fg}\right)=\mathrm{f}\left(\nabla \mathrm{g}\right)+\mathrm{g}\left(\nabla \mathrm{f}\right) \end{gathered}$$

Find∇×(A×B) .

$$\begin{gathered} Intheexpression\nabla \times \left(A\times B\right)=B\left(\nabla \times A\right)-A\left(\nabla \times B\right),whereA,Baredefined \\ asA=A_{X}I+A_{y}j+A_{z}kandB=B_{X}I+B_{y}j+B_{z}k. \\ FindthecrossproductofthevectorsAandB. \\ A\times B=\left|\begin{array}{ccc}i& j& k\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right| \\ =i\left(A_y{B}_z-A_z{B}_y\right)-j\left(A_x{B}_z-A_z{B}_x\right)+k\left(A_x{B}_y-A_y{B}_x\right) \\ =i\left(A_y{B}_z-A_z{B}_y\right)-j\left(A_z{B}_x-A_z{B}_x\right)+k\left(A_x{B}_y-A_y{B}_x\right) \\ ObtainthedivergenceofvectorA\times B \\ \nabla .\left(A\times B\right)=\frac{\partial }{\partial x}\left(A_y{B}_z-A_z{B}_y\right)+\frac{\partial }{\partial y}\left(A_z{B}_x-A_x{B}_z\right)+\frac{\partial }{\partial x}\left(A_x{B}_y-A_y{B}_x\right).....\left(1\right) \end{gathered}$$

Find B(∇×A)-A(∇×B)  

Find curl of vector A.

$$\begin{gathered} \nabla \times A=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)i-\left(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z}\right)j+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)k \\ =\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)i+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)j+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)k \\ NowevaluateB\left(\nabla \times A\right) \\ B\left(\nabla \times A\right)=\left(B_{X}i+B_{y}j+B_{z}k\right)\left[\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)i-\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)j+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)k\right] \\ =B_x\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)i+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)j+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right) \end{gathered}$$

$$\begin{gathered} FindcurlofvectorB. \\ \nabla \times B=\left(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}\right)i-\left(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}\right)j+\left(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}\right)k \\ =\left(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}\right)i+\left(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}\right)j+\left(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}\right)k \\ \mathrm{Now}\mathrm{evaluate}\mathrm{A}\left(\nabla \times \mathrm{B}\right) \\ \mathrm{A}\left(\nabla \times \mathrm{B}\right)\left({\mathrm{A}}_{\mathrm{X}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{i}+{\mathrm{A}}_{\mathrm{z}}\mathrm{i}\right)\left[=\left(\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{y}}-\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{z}}\right)\mathrm{i}-\left(\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{z}}-\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)\mathrm{j}+\left(\frac{\partial {\mathrm{B}}_{\mathrm{y}}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{B}}_{\mathrm{x}}}{\partial \mathrm{y}}\right)\right] \end{gathered}$$

$$\begin{gathered} A_x\left(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}\right)+A_y\left(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}\right)+A_z\left(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}\right) \\ FindB\left(\nabla \times A\right)-A\left(\nabla \times B\right) \\ BB\left(\nabla \times A\right)-A\left(\nabla \times B\right)=B_x\left(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}\right)+B_y\left(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}\right)+B_z\left(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}\right) \\ -A_x\left(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}\right)+A_y\left(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}\right)+A_z\left(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}\right) \\ =\frac{\partial }{\partial x}\left(A_y{B}_z-A_y{B}_z\right)+\frac{\partial }{\partial y}\left(A_z{B}_x-A_x{B}_z\right)+\frac{\partial }{\partial z}\left(A_x{B}_y-A_y{B}_x\right) \end{gathered}$$

….(2)

From equations (1) and (2) , it can be concluded that

$\nabla \times \left(\mathrm{A}\times \mathrm{B}\right)=\mathrm{B}\left(\nabla \times \mathrm{A}\right)-\mathrm{A}\left(\nabla \times \mathrm{B}\right)$

Find   

Find the curl of fA,

$$\begin{gathered} \nabla \times \left(fA\right)=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ f{A}_x& f{A}_y& f{A}_z\end{array}\right| \\ =\left(\frac{\partial }{\partial y}\left(f{A}_z\right)-\frac{\partial }{\partial z}\left(f{A}_y\right)\right)i-\left(\frac{\partial }{\partial x}\left(f{A}_z\right)-\frac{\partial }{\partial z}\left(f{A}_x\right)\right)+k\left(\frac{\partial }{\partial x}\left(f{A}_y\right)-\frac{\partial }{\partial y}\left(f{A}_x\right)\right) \\ =\left[\begin{array}{c}\left(f\frac{\partial }{\partial y}\left(A_z\right)+A_z\frac{\partial }{\partial z}\left(f\right)-f\frac{\partial }{\partial y}\left(A_y\right)-A_y\frac{\partial }{\partial z}\left(f\right)\right)i-\left(\begin{array}{c}f\frac{\partial }{\partial x}\left(A_z\right)+A_z\frac{\partial }{\partial x}\left(f\right)\\ -f\frac{\partial }{\partial z}\left(A_z\right)+A_z\frac{\partial }{\partial z}\left(f\right)\end{array}\right)j\\ +\left(f\frac{\partial }{\partial y}\left(A_y\right)+A_y\frac{\partial }{\partial z}\left(f\right)-f\frac{\partial }{\partial y}\left(A_x\right)-A_x\frac{\partial }{\partial z}\left(f\right)\right)\end{array}\right] \end{gathered}$$

$$\begin{gathered} =\left[\begin{array}{c}f\left(\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_z}{\partial y}\right)i+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)j+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)k\right)\\ -\left(\left(A_y\frac{\partial f}{\partial z}-A_z\frac{\partial f}{\partial y}\right)i+\left(A_z\frac{\partial f}{\partial x}-A_z\frac{\partial f}{\partial z}\right)i\left(A_x\frac{\partial f}{\partial y}-A_y\frac{\partial f}{\partial x}\right)k\right)\end{array}\right] \\ \mathrm{Thus}\mathrm{using}\mathrm{above}\mathrm{calculations}\mathrm{we}\mathrm{can}\mathrm{write}\nabla \times \left(\mathrm{fA}\right)=\mathrm{f}\left(\nabla \times \mathrm{A}\right)-\mathrm{A}\times \left(\nabla \mathrm{f}\right) \\ ,\mathrm{where}\mathrm{f}\mathrm{is}\mathrm{a}\mathrm{constant}\mathrm{vector}. \end{gathered}$$