Question

(a) Check product rule (iv) (by calculating each term separately) for the functions

$\mathit{A}\mathbf{=}\mathit{x}\hat{\mathbf{x}}\mathbf{+}\mathbf{2}\mathit{y}\hat{\mathbf{y}}\mathbf{+}\mathbf{3}\mathit{z}\mathbf{}\hat{\mathbf{z}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{B}\mathbf{=}\mathbf{3}\mathit{x}\hat{\mathbf{x}}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{}\hat{\mathbf{y}}$

(b) Do the same for product rule (ii).

(c) Do the same for rule (vi).

Short answer

a) The product rule (iv) is proven.

b) The product rule (ii) is proven.

c) The product rule (vi) is proven.

Step-by-step solution

prove rule (iv)

To compute an expression substitute the vectors and other required expression and then simplify.

The vector A and B are defined as A=xi+2yj and B=3yi-2xj . Find th cross product of vectors A and B .

$$\begin{gathered} \mathbf{A}\mathbf{\times }\mathbf{B}\mathbf{=}\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ \mathrm{x}& 2\mathrm{y}& 3\mathrm{z}\\ 3\mathrm{y}& -2\mathrm{x}& 0\end{array}\right| \\ =\left(0+6\mathrm{xz}\right)\mathrm{i}=(0+9\mathrm{yz})\mathrm{j}+(-2{\mathrm{x}}^2-6{\mathrm{y}}^2)\mathrm{k} \\ =6\mathrm{xzi}-9\mathrm{yzj}-2\left({\mathrm{x}}^2+3{\mathrm{y}}^2\right)\mathrm{k}........\left(1\right) \end{gathered}$$

The product rule (iv) is expressed as$\nabla .\left(A\times B\right)=B.\left(\nabla \times A\right)-A.\left(\nabla \times B\right)$

Compute left side of product rule (iv), as:

$$\begin{gathered} \nabla .\left(A\times B\right)=\left(\frac{\partial }{\partial X}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(6xzi-\left(9yzj9yzj\right)-2\left(x^2+3y^2\right)k\right) \\ =\frac{\partial }{\partial X}\left(6xzi\right)-\frac{\partial }{\partial y}\left(9yzj\right)-\frac{\partial }{\partial z}2\left(x^2+3y^2\right) \\ =9z+6z \\ =15z \end{gathered}$$

Find the curl of vector A as follows:

$$\begin{gathered} \nabla \times A=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& 2y& 3z\end{array}\right| \\ =\left(\frac{\partial }{\partial y}\left(3z\right)-\frac{\partial }{\partial z}\left(2y\right)\right)i-\left(\frac{\partial }{\partial x}\left(3z\right)-\frac{\partial }{\partial z}\left(x\right)\right)j+\left(\frac{\partial }{\partial z}\left(3z\right)-\frac{\partial }{\partial y}\left(x\right)\right)k \\ =i\left(0-0\right)-j\left(0-0\right)+k\left(0-0\right) \\ =0 \end{gathered}$$

The value of $\nabla \times A$ is 0,, then value of $B.\left(\nabla \times A\right)$ is also zero.

Find the curl of vector $B$as follows:

$$\begin{gathered} \nabla \times B=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 3y& -2x& 0\end{array}\right| \\ =\left(\frac{\partial }{\partial y}\left(0\right)-\frac{\partial }{\partial z}\left(-2x\right)\right)i-\left(\frac{\partial }{\partial x}\left(0\right)-\frac{\partial }{\partial z}\left(-3y\right)\right)j+\left(\frac{\partial }{\partial x}\left(-2x\right)-\frac{\partial }{\partial y}\left(3y\right)\right)k \\ =-5k \end{gathered}$$

Now compute the value of $A.\left(\nabla \times B\right)$, as follows:

$$\begin{gathered} A.\left(\nabla \times B\right)=\left(xi+2yj+3zk\right).\left(-5k\right) \\ =-15k \end{gathered}$$

compute the value of $B.\left(\nabla \times A\right)-A.\left(\nabla \times B\right)$, as follows

$$\begin{gathered} B.\left(\nabla \times A\right)-A\left(\nabla \times B\right)=0-\left(-15k\right) \\ =15k \end{gathered}$$

This concludes that the value of $\nabla .\left(A\times B\right)$is equal to that of

$B.\left(\nabla \times A\right)-A.\left(\nabla \times B\right)$. Thus $\nabla .\left(A\times B\right)=B.\left(\nabla \times A\right)-A.\left(\nabla \times B\right)$. Hence rule (iv) is proven.

 prove product rule (ii)

To compute an expression substitute the vectors and other required expression and then simplify

The product rule (ii) is expressed as

$\nabla \left(A.B\right)A\times \left(\nabla \times B\right)-B\times \left(\nabla \times A\right)+\left(A.\nabla \right)B+\left(B.\nabla \right)A$

Compute left side of product rule (ii), as:

localid="1657524418399" $$\begin{gathered} \nabla \left(A.B\right)=\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(xi+2yj+3zk\right).\left(3yj-2xj\right) \\ =\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(3xy-4xy\right) \\ =\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(-xy\right) \\ =i\frac{\partial }{\partial x}\left(-xy\right)+j\frac{\partial }{\partial y}\left(-xy\right)+k\frac{\partial }{\partial z}\left(-xy\right) \\ =-yi-xj \end{gathered}$$

It is already computed that $\nabla \times A$ is 0 , then value oflocalid="1657525801615" $B\times \left(\nabla \times A\right)$is also zero andlocalid="1657525816634" $\nabla \times Bis-5k$.

Find the value of localid="1657524992996" $A\times \left(\nabla \times B\right)$as follows:

$$\begin{gathered} A\times \left(\nabla \times B\right)=\left|\begin{array}{ccc}i& j& k\\ x& 2y& 3z\\ 0& 0& -5\end{array}\right| \\ =i\left(-10y-0\right)-j\left(-5x-0\right)+z\left(0\right) \\ =-10\mathrm{yi}+5\mathrm{xj} \\ \mathrm{Find}\mathrm{the}\mathrm{curl}\mathrm{of}\mathrm{vector}\mathrm{B}\mathrm{as}\mathrm{follows}: \\ \nabla \times \mathrm{B}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ \frac{\partial }{\partial \mathrm{x}}& \frac{\partial }{\partial \mathrm{y}}& \frac{\partial }{\partial \mathrm{z}}\\ 3\mathrm{y}& -2\mathrm{x}& 0\end{array}\right| \\ =\left(\frac{\partial }{\partial \mathrm{y}}\left(0\right)-\frac{\partial }{\partial \mathrm{z}}\left(-2\mathrm{x}\right)\right)\mathrm{i}-\left(\frac{\partial }{\partial \mathrm{y}}\left(0\right)-\frac{\partial }{\partial \mathrm{z}}\left(3\mathrm{y}\right)\right)\mathrm{j}+\left(\frac{\partial }{\partial \mathrm{x}}\left(-2\mathrm{x}\right)-\frac{\partial }{\partial \mathrm{y}}\left(3\mathrm{y}\right)\right)\mathrm{k} \\ =-5\mathrm{k} \end{gathered}$$

Find the value of $\left(A.\nabla \right)B$.

$$\begin{gathered} \left(A.\nabla \right)B=\left(x\frac{\partial }{\partial x}+2y\frac{\partial }{\partial y}+3z\frac{\partial }{\partial z}\right)\left(3yi-2xj\right) \\ =x\frac{\partial }{\partial x}\left(3yi-2xj\right)+2y\frac{\partial }{\partial y}\left(3yi-2xj\right)+3z\frac{\partial }{\partial z}\left(3yi-2xj\right) \\ =x\left(-2\right)i=2y\left(3\right)j+3z\left(0\right)k \\ =-2xi+6yj \end{gathered}$$

Find the value of $\left(B.\nabla \right)A$.

$$\begin{gathered} \left(B.\nabla \right)A=\left(3y\frac{\partial }{\partial x}-2x\frac{\partial }{\partial y}\right)\left(xi-2yj+3zk\right) \\ =3y\frac{\partial }{\partial x}\left(xi-2xj+3zk\right)-2x\frac{\partial }{\partial y}\left(xi-2yj+3zk\right) \\ =3yi-2x\left(2\right)j \\ =3yi-4xj \end{gathered}$$

 compute right side of rule (ii)

To compute an expression substitute the vectors and other required expression and then simplify.

$$\begin{gathered} \mathrm{Substitiute}3yi-4xj\mathrm{for}\left(\mathrm{B}.\nabla \right)\mathrm{A},-2\mathrm{xi}+6\mathrm{yj}\mathrm{for}\left(\mathrm{A}.\nabla \right)\mathrm{B},\mathrm{for}\mathrm{B}\times \left(\nabla .\mathrm{A}\right) \\ \mathrm{and}-10\mathrm{yi}+5\mathrm{xj}\mathrm{for}\mathrm{A}\times \left(\nabla \times \mathrm{B}\right)\mathrm{into} \end{gathered}$$

$$\begin{gathered} \mathrm{Thus}\mathrm{the}\mathrm{value}\mathrm{of}\nabla \left(\mathrm{A}.\mathrm{B}\right)\mathrm{is}\mathrm{same}\mathrm{as}\mathrm{that}\mathrm{of} \\ \mathrm{A}\times \left(\nabla \times \mathrm{B}\right)-\mathrm{B}\times \left(\nabla \times \mathrm{A}\right)+\left(\mathrm{A}.\nabla \right)\mathrm{B}+\left(\mathrm{B}.\nabla \right)\mathrm{A}.\mathrm{Hence}\mathrm{the}\mathrm{rule}\left(\mathrm{ii}\right)\mathrm{is}\mathrm{proven}. \end{gathered}$$

 Prove product rule (vi)

To compute an expression substitute the vectors and other required expression and then simplify

The product rule (iv) is expressed as

$\nabla \times \left(A\times B\right)=\left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+B\left(\nabla .A\right)$

Compute left side of product rule (iv), as:

$$\begin{gathered} \nabla \times \left(A\times B\right)=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& 2y& 3z\end{array}\right| \\ =\left(\frac{\partial }{\partial y}\left(3z\right)-\frac{\partial }{\partial z}\left(2y\right)\right)i-\left(\frac{\partial }{\partial x}\left(3z\right)-\frac{\partial }{\partial z}\left(x\right)\right)j+\left(\frac{\partial }{\partial z}\left(2y\right)-\frac{\partial }{\partial y}\left(x\right)\right)k \\ =i\left(-12y-19z\right)-j\left(-4x-6x\right)+k\left(0-0\right) \\ =-21yi+10xj \end{gathered}$$

Find the value of $B\left(\nabla .A\right)$.

$$\begin{gathered} B\left(\nabla .A\right)=\left(3yi-2xj\right)\left(\frac{\partial }{\partial x}\left(x\right)+\frac{\partial }{\partial y}\left(2y\right)+\frac{\partial }{\partial z}\left(3z\right)\right) \\ =\left(3yi-2xj\right)\left(1+2+3\right) \\ =6\left(3yi-2xj\right) \\ =18yi-12xj \end{gathered}$$

Find the value oflocalid="1657528608425" $\left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+\left(\nabla .A\right)A\left(\nabla .B\right)$.

$$\begin{gathered} \left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+B\left(\nabla .A\right)=\left[\begin{array}{c}\left(3yi-4xj\right)-\left(-2xj+6yi\right)\\ +0-\left(18yi-12xj\right)\end{array}\right] \\ =-21yi+10xj \end{gathered}$$$$\begin{gathered} \left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+B\left(\nabla .A\right)=\left[\begin{array}{c}\left(3y-4xj\right)\left(-2xj-6yi\right)\\ +0-\left(18yi-12xj\right)\end{array}\right] \\ =-21yi+10xj \end{gathered}$$

Thus the value of $\nabla \times \left(A\times B\right)$ is same as that of

$\left(B.\nabla \right)A-\left(A.\nabla \right)B+A\left(\nabla .B\right)+B\left(\nabla .A\right)$Hence the rule (iv) is proven.