91Ó°ÊÓ

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

The first quiotient rule is$∇\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\frac{\mathrm{g}\left(∇\mathrm{f}\right)-\mathrm{f}\left(∇\mathrm{g}\right)}{{\mathrm{g}}^2},$The$∇$operator is defined aslocalid="1657516180871" $$\begin{gathered} ∇=\left(\frac{∂}{∂\mathrm{x}}\mathrm{i}+\frac{∂}{∂\mathrm{y}}\mathrm{j}+\frac{∂}{∂\mathrm{z}}\mathrm{k}\right) \\ \mathrm{Compute}\mathrm{the}\mathrm{value}\mathrm{of}∇\left(\frac{\mathrm{f}}{\mathrm{g}}\right),\mathrm{as}: \\ ∇\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\left(\frac{∂}{∂\mathrm{x}}\mathrm{i}+\frac{∂}{∂\mathrm{y}}\mathrm{j}+\frac{∂}{∂\mathrm{z}}\mathrm{k}\right)\left(\frac{\mathrm{f}}{\mathrm{g}}\right) \\ =\mathrm{i}\frac{∂}{∂\mathrm{x}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)+\mathrm{j}\frac{∂}{∂\mathrm{y}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)\mathrm{k}+\frac{∂}{∂\mathrm{z}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right) \\ =\left(\frac{\mathrm{g}{\displaystyle \frac{∂\mathrm{f}}{∂\mathrm{x}}}-\mathrm{f}\left({\displaystyle \frac{∂\mathrm{g}}{∂\mathrm{x}}}\right)}{{\mathrm{g}}^2}\right)+\mathrm{j}\left(\frac{\mathrm{g}{\displaystyle \frac{∂\mathrm{f}}{∂\mathrm{y}}}-\mathrm{f}\left({\displaystyle \frac{∂\mathrm{g}}{∂\mathrm{y}}}\right)}{{\mathrm{g}}^2}\right)+\mathrm{k}\left(\frac{\mathrm{g}{\displaystyle \frac{∂\mathrm{f}}{∂\mathrm{z}}}-\mathrm{f}\left({\displaystyle \frac{∂\mathrm{g}}{∂\mathrm{z}}}\right)}{{\mathrm{g}}^2}\right) \\ =\frac{1}{\mathrm{g}}\left(\frac{∂\mathrm{f}}{∂\mathrm{x}}\mathrm{i}+\frac{∂\mathrm{f}}{∂\mathrm{y}}\mathrm{j}+\frac{∂\mathrm{f}}{∂\mathrm{z}}\mathrm{k}\right)-\frac{\mathrm{f}}{{\mathrm{g}}^2}\left(\frac{∂\mathrm{g}}{∂\mathrm{x}}\mathrm{i}+\frac{∂\mathrm{g}}{∂\mathrm{y}}\mathrm{j}+\frac{∂\mathrm{g}}{∂\mathrm{z}}\mathrm{k}\right) \end{gathered}$$

$\mathrm{Substitiute}∇\mathrm{for}\frac{∂}{∂\mathrm{x}}\mathrm{i}+\frac{∂}{∂\mathrm{y}}\mathrm{j}+\frac{∂}{∂\mathrm{z}}\mathrm{k}\mathrm{into}\mathrm{equation}\left(1\right),\mathrm{as}:$

$$\begin{gathered} ∇\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\frac{1}{g}\left(\frac{∂f}{∂\mathrm{x}}\mathrm{i}+\frac{∂f}{∂\mathrm{y}}\mathrm{j}+\frac{∂f}{∂\mathrm{z}}\mathrm{k}\right)-\frac{\mathrm{f}}{{\mathrm{g}}^2}\left(\frac{∂g}{∂\mathrm{x}}\mathrm{i}+\frac{∂g}{∂\mathrm{y}}\mathrm{j}+\frac{∂g}{∂\mathrm{z}}\mathrm{k}\right) \\ =\frac{1}{g}\left(∇f\right)-\frac{1}{g^2}\left(∇g\right) \\ =\frac{g\left(∇f\right)-f\left(∇g\right)}{g^2} \end{gathered}$$

Thus the first quoitient rule is proven.

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

The second quiotient rule is $∇\left(\frac{\stackrel{â‡\P Ä}{A}}{g}\right)=\frac{g\left(∇\stackrel{â‡\P Ä}{A}\right)-\stackrel{â‡\P Ä}{A}\left(∇g\right)}{g^2}$. Let the vector $\stackrel{â‡\P Ä}{A}$is defined as The operator is defined as $$\begin{gathered} \stackrel{â‡\P Ä}{A}=A_{x}i+A_{y}j+A_{z}k\mathrm{The}∇\mathrm{operator}\mathrm{is}\mathrm{defined}\mathrm{as} \\ ∇=\frac{∂}{∂\mathrm{x}}\mathrm{i}+\frac{∂}{∂\mathrm{y}}\mathrm{j}+\frac{∂}{∂\mathrm{z}}\mathrm{k}. \end{gathered}$$

$$\begin{gathered} \mathrm{Compute}\mathrm{the}\mathrm{value}\mathrm{of}∇\left(\frac{\stackrel{â‡\P Ä}{\mathrm{A}}}{\mathrm{g}}\right),\mathrm{as}: \\ ∇\left(\frac{\stackrel{â‡\P Ä}{A}}{g}\right)=\left(\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k\right)\left(\frac{A_{x}i+A_{y}i+A_{z}k}{g}\right) \\ =i\frac{∂}{∂x}\left(\frac{A_x}{g}\right)+j\frac{∂}{∂y}\left(\frac{A_y}{g}\right)+j\frac{∂}{∂z}\left(\frac{A_z}{g}\right) \\ =\frac{g{\displaystyle \frac{∂A_x}{∂x}}-A_x\left({\displaystyle \frac{∂g}{∂x}}\right)}{g^2}+\frac{g\frac{∂A_y}{∂y}-A_y\left(\frac{∂g}{∂y}\right)}{g^2}+\frac{g\frac{∂A_z}{∂z}-A_z\left(\frac{∂g}{∂z}\right)}{g^2} \\ =\frac{1}{g}\left(\frac{∂A_x}{∂x}+\frac{∂A_y}{∂y}+\frac{∂A_z}{∂z}\right)-\frac{f}{g^2}\left(\frac{∂g}{∂x}+A_y\frac{∂g}{∂y}+A_Z\frac{∂g}{∂z}\right)......\left(1\right) \end{gathered}$$

The dot product of $∇$and vector $\stackrel{â‡\P Ä}{A}$is obtained as ,

$$\begin{gathered} ∇.\stackrel{â‡\P Ä}{A}=\left(\frac{∂}{∂X}i+\frac{∂}{∂y}j+\frac{∂}{∂Z}k\right)\left(A_{x}i+A_{y}j+A_{z}k\right) \\ =\frac{A_x}{∂X}i+\frac{∂A_y}{∂y}j+\frac{∂A_z}{∂Z}k \\ \mathrm{The}\mathrm{dot}\mathrm{product}\mathrm{of}∇\mathrm{g}\mathrm{and}\mathrm{vector}\stackrel{â‡\P Ä}{\mathrm{A}}\mathrm{is}\mathrm{obtained}\mathrm{as}, \\ \stackrel{â‡\P Ä}{\mathrm{A}}.\left(∇\mathrm{g}\right)=\left({\mathrm{A}}_{\mathrm{x}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{j}+{\mathrm{A}}_{\mathrm{z}}\mathrm{k}\right)\left(\frac{∂}{∂\mathrm{X}}\mathrm{i}+\frac{∂}{∂\mathrm{y}}\mathrm{j}+\frac{∂}{∂\mathrm{Z}}\mathrm{k}\right)\mathrm{g} \\ =\left({\mathrm{A}}_{\mathrm{x}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{j}+{\mathrm{A}}_{\mathrm{z}}\mathrm{k}\right)\left(\frac{∂\mathrm{g}}{∂\mathrm{x}}\mathrm{i}+\frac{∂\mathrm{g}}{∂\mathrm{y}}\mathrm{j}+\frac{∂\mathrm{g}}{∂\mathrm{z}}\mathrm{k}\right) \\ ={\mathrm{A}}_{\mathrm{x}}\frac{∂\mathrm{g}}{∂\mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{∂\mathrm{g}}{∂\mathrm{y}}+{\mathrm{A}}_{\mathrm{x}}\frac{∂\mathrm{g}}{∂\mathrm{z}} \end{gathered}$$

$$\begin{gathered} \mathrm{Substitiute}\stackrel{â‡\P Ä}{\mathrm{A}}-\left(∇\mathrm{g}\right)\mathrm{for}{\mathrm{A}}_{\mathrm{x}}\frac{∂\mathrm{g}}{∂\mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{∂\mathrm{g}}{∂\mathrm{y}}+{\mathrm{A}}_{\mathrm{y}}\frac{∂\mathrm{g}}{∂\mathrm{y}}\mathrm{and}∇-\stackrel{â‡\P Ä}{\mathrm{A}}\mathrm{for}\frac{{\mathrm{A}}_{\mathrm{x}}}{∂\mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{∂{\mathrm{A}}_{\mathrm{y}}}{∂\mathrm{y}}+{\mathrm{A}}_{\mathrm{y}}\frac{∂{\mathrm{A}}_{\mathrm{z}}}{∂\mathrm{z}} \\ \mathrm{into}\mathrm{equation}\left(1\right),\mathrm{as}: \\ ∇\left(\frac{\stackrel{â‡\P Ä}{\mathrm{A}}}{\mathrm{g}}\right)=\frac{1}{\mathrm{g}}\left(\frac{∂{\mathrm{A}}_{\mathrm{x}}}{∂\mathrm{x}}\mathrm{i}+\frac{∂{\mathrm{A}}_{\mathrm{y}}}{∂\mathrm{y}}\mathrm{j}+\frac{∂{\mathrm{A}}_{\mathrm{z}}}{∂\mathrm{z}}\mathrm{k}\right)-\frac{\mathrm{f}}{{\mathrm{g}}^2}\left({\mathrm{A}}_{\mathrm{x}}\frac{∂\mathrm{g}}{∂\mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{∂\mathrm{g}}{∂\mathrm{y}}+{\mathrm{A}}_{\mathrm{z}}\frac{∂\mathrm{g}}{∂\mathrm{z}}\right) \\ =\frac{1}{\mathrm{g}}\left(∇.\stackrel{â‡\P Ä}{\mathrm{A}}\right)-\frac{\mathrm{f}}{{\mathrm{g}}^2}\left(\stackrel{â‡\P Ä}{\mathrm{A}}.∇\mathrm{g}\right) \\ =\frac{\mathrm{g}\left(∇.\stackrel{â‡\P Ä}{\mathrm{A}}\right)-\stackrel{â‡\P Ä}{\mathrm{A}}\left(∇\mathrm{g}\right)}{{\mathrm{g}}^2} \end{gathered}$$

Thus the second quoitient rule is proven.

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

The second quiotient rule is $∇×\left(\frac{\stackrel{â‡\P Ä}{A}}{g}\right)=\frac{g\left(∇×\stackrel{â‡\P Ä}{A}\right)-\stackrel{â‡\P Ä}{A}\left(∇g\right)}{g^2}$. Let the vector $\stackrel{â‡\P Ä}{A}$is defined as $\stackrel{â‡\P Ä}{A}=A_{x}i+A_{x}j+A_{x}k$The operator is defined as $∇=\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k$.

Compute the value of $∇×\left(\frac{\stackrel{â‡\P Ä}{A}}{g}\right)$, as:

$$\begin{gathered} ∇\left(\frac{\stackrel{â‡\P Ä}{A}}{g}\right)=\left|\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ \frac{A_x}{g}& \frac{A_y}{g}& \frac{A_z}{g}\end{array}\right| \\ =i\left(\frac{∂}{∂y}\left(\frac{A_z}{g}\right)-\frac{∂}{∂z}\left(\frac{A_y}{g}\right)\right)+j\left(\frac{∂}{∂z}\left(\frac{A_x}{g}\right)-\frac{∂}{∂x}\left(\frac{A_z}{g}\right)\right)+k\left(\frac{∂}{∂x}\left(\frac{A_y}{g}\right)-\frac{∂}{∂y}\left(\frac{A_x}{g}\right)\right) \\ =\frac{1}{g}\left(i\left(\frac{∂A_z}{∂y}-\frac{∂A_y}{∂z}\right)+j\left(\frac{∂A_x}{∂z}-\frac{∂A_z}{∂x}\right)+k\left(\frac{∂A_y}{∂x}-\frac{∂A_x}{∂y}\right)\right) \\ =\frac{g\left(∇×\stackrel{â‡\P Ä}{A}\right)-\stackrel{â‡\P Ä}{A}×\left(∇g\right)}{g^2} \end{gathered}$$

Thus the third quoitient rule is proven.