APPLICATION OF VECTOR DIFFERENTIATION

\[If\ \overrightarrow{F}={F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k}\]

is a vector function, defined and differentiable at each point (x, y, z)in a certain region of space [i.e., A defines a vector field], then the divergence of  (abbreviated as 'Div ') is defined as, 

\[Div\ \overrightarrow{F} = \nabla\ . \overrightarrow{F}\]
\[ = (\overrightarrow{i}\frac{\partial}{\partial\ x} + \overrightarrow{j}\frac{\partial}{\partial\ y} + \overrightarrow{k}\frac{\partial}{\partial\ z})\ . \ ({F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k})\]
\[ = (\frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z})\]

Basic properties of Divergence:

If A, B are vector functions and ‘f’ is a scalar function, then

\[1) \nabla\ . ( A + B) = \nabla\ .A + \nabla\ .B\]
\[2) \nabla\ . ( fA) = (\nabla\ f) . A + (f.\nabla\ A)\]
\[3)\overrightarrow{F}\ is\ solenoidal\ if\ \nabla\ . \overrightarrow{F}=0\]

Example:

\[If\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (xy^2 - zy^3)\overrightarrow{k}, then\ find\ div\ \overrightarrow{F}\]

Soln:

\[Let\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (xy^2 - zy^3)\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(xyz)}+ \frac{\partial}{\partial\ y} {(3x^2y)}+ \frac{\partial}{\partial\ z} {(xy^2-xy^3)}\]

=   yz  + 3x2 ( 1 )  + ( 0 - y3 )

=  yz  + 3x2 - y3

Example:

\[If\ \overrightarrow{F}=x^2y\overrightarrow{i} + xy^2z\overrightarrow{j} + xyyz\overrightarrow{k}, then\ find\ div\ \overrightarrow{F}\ at\ the\ point\ (1,-1,2)\]

Soln:

\[Let\ \overrightarrow{F}=x^2y\overrightarrow{i} + xy^2z\overrightarrow{j} + xyyz\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(x^2y)}+ \frac{\partial}{\partial\ y} {(xy^2z)}+ \frac{\partial}{\partial\ z} {(xyyz)}\]
\[\nabla\ . \overrightarrow{F} = 2xy + 2xyz + xy\]

At ( 1, -1, 2)

\[\nabla\ . \overrightarrow{F} = 2(1)(-1) + 2(1)(-1)(2) + (1)(-1)\]

=    -2  - 4  - 1

=   -7

Example:

\[Show\ that\ \overrightarrow{F}=3y^4z^2\overrightarrow{i} + 4x^3z^2\overrightarrow{j} + 6x^2y^3\overrightarrow{k} is\ solenoidal\]

Soln:

\[Let\ \overrightarrow{F}=3y^4z^2\overrightarrow{i} + 4x^3z^2\overrightarrow{j} + 6x^2y^3\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(3y^4z^2)}+ \frac{\partial}{\partial\ y} {(4x^3z^2)}+ \frac{\partial}{\partial\ z} {(6x^2y^3)}\]
\[\nabla\ . \overrightarrow{F} = 0+ 0 + 0 = 0\]
\[\overrightarrow{F}\ is\ solenoidal\]

Example:

\[If\ \overrightarrow{F}=2xy\overrightarrow{i} + 3x^2y\overrightarrow{j} - 3pyz\overrightarrow{k} is\ solenoidal\ at\ (1,1,1)\, find\ 'p'\]

Soln:

\[Let\ \overrightarrow{F}=2xy\overrightarrow{i} + 3x^2y\overrightarrow{j} - 3pyz\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(2xy)}+ \frac{\partial}{\partial\ y} {(3x^2y)}- \frac{\partial}{\partial\ z} {(3pyz)}\]
\[\nabla\ . \overrightarrow{F} = 2y + 3x^2- 3py\]

At ( 1, 1, 1)

\[\nabla\ . \overrightarrow{F} = 2(1) + 3(1)^2- 3p(1)\]
\[\nabla\ . \overrightarrow{F} = 2+ 3 - 3p\]
\[Given\ \overrightarrow{F}\ is\ solenoidal\]
\[i.e\ \nabla\ . \overrightarrow{F} =0\]

2  + 3 – 3p =   0

5 – 3p = 0

3p  = 5

p  =  5/3

Curl of a vector function:

\[If\ \overrightarrow{F}={F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k}\]

is a vector function,

\[then\ curl\ \overrightarrow{F}=\nabla\ × \overrightarrow{F}\]
\[curl\ \overrightarrow{F} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {F_1} & {F_2} & {F_3}\\ \end{vmatrix}\]

Irrotational vector  : 

\[A\ vector\ \overrightarrow{F}\ is\ said\ to\ be\ irrotational\ if\]
\[curl\ \overrightarrow{F} =0\]
\[i.e\ \nabla\ × \overrightarrow{F} =0\]

Example:

\[If\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} +(x y ^2- zy^3)\overrightarrow{k}\ then\ find\ curl\ \overrightarrow{F}\]

Soln:

\[Let\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (x y ^2- zy^3)\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {xyz} & {3x^2y} & {x y ^2- zy^3}\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {(x y ^2- zy^3)} - \frac{\partial}{\partial\ z} {(3x^2y)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(x y ^2- zy^3)} - \frac{\partial}{\partial\ z} {(xyz)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} {(3x^2y)} - \frac{\partial}{\partial\ y} {(xyz)})\]
\[ = \overrightarrow{i}( 2xy - 3zy^2 ) -\overrightarrow{j}((y^2 - 0) - xy)+\overrightarrow{k}(6xy - xz))\]
\[\nabla\ × \overrightarrow{F} = [2xy - 3zy^2] \overrightarrow{i} - [y^2 - xy] \overrightarrow{j} + [ 6xy - xz] \overrightarrow{k}\]

Example:

\[Show\ that\ \overrightarrow{F}= x\overrightarrow{i} +y^2\overrightarrow{j} +z^3\overrightarrow{k}\ is\ irrotational\]

Soln:

\[Let\ \overrightarrow{F}=x\overrightarrow{i} +y^2\overrightarrow{j} +z^3\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {x} & y^2 & z^3\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {(z^3)} - \frac{\partial}{\partial\ z} {(y^2)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(z^3)} - \frac{\partial}{\partial\ z} {(x)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} {(y^2)} - \frac{\partial}{\partial\ y} {(x)})\]
\[ = \overrightarrow{i}[ 0- 0 ] -\overrightarrow{j}[0 - 0]+\overrightarrow{k}[0- 0]\]
\[\nabla\ × \overrightarrow{F} = 0\]
\[ \overrightarrow{F}\ is\ irrotational\]

Example:

\[If\ \overrightarrow{F}=( 2x + 2y + 2z)\overrightarrow{i} - (xy + yz + zx)\overrightarrow{j} + 3xyz\overrightarrow{k}\ then\ find\ \nabla\ × \overrightarrow{F}\ and\ \nabla\ × ( \nabla\ × \overrightarrow{F})\]

Soln:

\[Let\ \overrightarrow{F}=( 2x + 2y + 2z)\overrightarrow{i} - (xy + yz + zx)\overrightarrow{j} + 3xyz\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {2x + 2y + 2z} & - (xy + yz + zx) & 3xyz \\ \end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {( 3xyz )} + \frac{\partial}{\partial\ z} {(xy + yz + zx)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(3xyz)} - \frac{\partial}{\partial\ z} {(2x + 2y + 2z)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} -{(xy + yz + zx )} - \frac{\partial}{\partial\ y} {(2x + 2y + 2z )})\]
\[ = \overrightarrow{i}[ 3xz + (0 + y + x) ] -\overrightarrow{j}[(3yz – ( 0 + 0 + 2)]+\overrightarrow{k}[- (y + 0 + z ] – ( 0 + 2 + 0)]\]
\[\nabla\ × \overrightarrow{F} = [3xz + y + x] \overrightarrow{i} - [3yz - 2] \overrightarrow{j} + [ - y - z - 2] \overrightarrow{k}\]
\[\nabla\ × ( \nabla\ × \overrightarrow{F}) =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {3xz + y + x} & - (3yz - 2) & (- y - z - 2) \\ \end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {( - y- z - 2 )} - \frac{\partial}{\partial\ z} {(-3yz+ 2)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(- y- z - 2 )} - \frac{\partial}{\partial\ z} {(3xz – y - x)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} {(-3yz+ 2)} - \frac{\partial}{\partial\ y} {(3xz+ y+ x)})\]
\[ = \overrightarrow{i}[ (-1 -0 - 0) - (-3y - 0)] -\overrightarrow{j}[(- 0 - 0 - 0) – ( 3x - 0 - 0)]+\overrightarrow{k}[(0 - 0 ) – ( 0 +1 + 0)]\]
\[ = \overrightarrow{i}[ - 1 + 3y] -\overrightarrow{j}[3x]+\overrightarrow{k}[-1]\]
\[\nabla\ × ( \nabla\ × \overrightarrow{F}) = \overrightarrow{i}[ 3y – 1 ] -\overrightarrow{j}[3x]-\overrightarrow{k}\]