## VECTOR DIFFERENTIATION

#### Vector point function and Vector field:

Let P be any point in a region ‘D’ of space.  Let r be the position vector of P.  If there exists a vector function F corresponding to each P,  then such a function F is called a vector function and the region D is called a vector field.

Example:  consider the vector function

$\overrightarrow{F}= (x-y)\overrightarrow{i}+ xy\overrightarrow{j}+ yz\overrightarrow{k}\ ————(1)$

Let P be a point whose position vector is

$\overrightarrow{r}= 2\overrightarrow{i}+ \overrightarrow{j}+ 3\overrightarrow{k}\ in\ the\ region\ D\ of space.$

At P , the value of F is obtained by putting x = 2, Y = I, z = 3 in F

$i.e\ at\ P,\ \overrightarrow{F}= \overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k}$

Thus, to each point P of the region D, there corresponds a vector F given by the  vector function (I). Hence F is a vector point function (of scalar variables x, y, z) and the region D is a vector field.

#### Scalar point function and scalar field:

If there exists a scalar 'f' given by a scalar function 'f' corresponding to each point P (with position vector r) in a region D of space, ‘f' is called a scalar point function and D is called a scalar field.

Example: let P be a point whose position vector is

$\overrightarrow{r}= 2\overrightarrow{i}+ \overrightarrow{j}+ 3\overrightarrow{k}$

Consider f= xyz + xy + z

Then the value of f at P is obtained by putting x = 2, y = I, z = 3

i.e., At P, f= 2.1.3 + 2.1 + 3 = II

Hence the scalar' II ' is attached to the point P.

The function 'f' is a scalar point function (of scalar variables x, y, z), and D is a

scalar field.

Note : There can be vector and scalar function of one or more scalar variables.

#### Vector differential operator

$The\ vector\ differential\ operator\ 'DEL'\ denoted\ as\ '\nabla'\ is\ defined\ by$
$\nabla = \frac{\partial}{\partial\ x}\overrightarrow{i} + \frac{\partial}{\partial\ y}\overrightarrow{j} + \frac{\partial}{\partial\ z}\overrightarrow{k}$
$\overrightarrow{i},\ \overrightarrow{j},\ \overrightarrow{k}\ are\ unit\ vectors\ in\ x,\ y,\ z\ directions$
$This\ operator\ \nabla\ is\ used\ in\ defining\ the\ gradient,\ divergence\ and\ curl.$
$Properties\ of\ \nabla\ are\ similar\ to\ those\ of\ vectors$

The operator is applied to both vector and scalar functions.

$If\ \phi(x,y,z)\ is\ a\ scalar\ function$

defined at each point (x, y, z) in a certain region of space and is differentiable

$the\ gradient\ of\ \phi\ (shortly\ written\ as\ grad\Phi)\ is\ defined\ as$
$\nabla \phi= (\frac{\partial}{\partial\ x}\overrightarrow{i} + \frac{\partial}{\partial\ y}\overrightarrow{j} + \frac{\partial}{\partial\ z}\overrightarrow{k})\phi$
$= (\frac{\partial\phi}{\partial\ x})\overrightarrow{i} + (\frac{\partial\phi}{\partial\ y})\overrightarrow{j} + (\frac{\partial\phi}{\partial\ z})\overrightarrow{k}$

#### Basic properties of the Gradient

$If\ \phi\ and\ \psi\ are\ two\ scalar\ functions$
$1)\ grad(\phi + \psi) = grad\ \phi + grad\ \psi \ or\ \nabla( \phi + \psi) = \nabla\phi + \nabla\psi$
$2)\ grad(\phi \psi) = \phi\ grad\phi +\psi\ grad\psi \ or\ \nabla( \phi + \psi) = \nabla\phi + \nabla\psi$
$3)\ \phi(x,y,z)\ = c\ (being\ constant)\ represents\ a\ surface$
$then\ unit\ normal\ to\ the\ surface\ \phi\ at\ the\ point\ (x,y,z)\ is\ \frac{\nabla\phi}{|\nabla\phi|}$
$4)\ angle\ between\ surfaces= angle\ between\ normals$
$\theta = \cos ^-1 ( \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}})$

Example: If  f = x2yz,  find grad f at the point (1,-2,1)

Soln: f = x2yz

$\nabla\ f=(\frac{\partial\ f}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ f}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ f}{\partial\ z})\overrightarrow{k}------------(1)$
$\frac{\partial\ f}{\partial\ x} = yz\ \frac{\partial}{\partial\ x}({x^2})$

= yz ( 2x )

$\frac{\partial\ f}{\partial\ x} = 2xyz$
$\frac{\partial\ f}{\partial\ y} = x^2z\ \frac{\partial}{\partial\ y}(y)$

= x2z ( 1 )

$\frac{\partial\ f}{\partial\ y} = x^2z$
$\frac{\partial\ f}{\partial\ z} = x^2y\ \frac{\partial}{\partial\ z}(z)$

= x2y ( 1 )

$\frac{\partial\ f}{\partial\ z} = x^2y$

Equation (1) becomes

$\nabla\ f=(2xyz)\overrightarrow{i} + (x^2z)\overrightarrow{j} + (x^2y)\overrightarrow{k}$

At the point (1,-2, 1),

$\nabla\ f=2(1)(-2)(1)\overrightarrow{i} + (1)^2(1)\overrightarrow{j} + (1)^2(-2))\overrightarrow{k}$
$\nabla\ f=-4\overrightarrow{i} + \overrightarrow{j} - 2\overrightarrow{k}$

Example: Find the unit normal to the surface xy +yz + zx= 3 at the point (1, 1, 1).

Soln:

$\phi(x,y,z)\ = c\ (being\ constant)\ represents\ a\ surface$
$then\ unit\ normal\ to\ the\ surface\ \phi\ at\ the\ point\ (x,y,z)\ is\ \frac{\nabla\phi}{|\nabla\phi|}$
$Here\ \phi = xy +yz + zx$
$\nabla\ \phi=(\frac{\partial\ \phi}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ \phi}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ \phi}{\partial\ z})\overrightarrow{k}------------(1)$
$\frac{\partial\ \phi}{\partial\ x} = y \frac{\partial}{\partial\ x}(x) + 0 + z \frac{\partial}{\partial\ x}(x)$

= y ( 1 )  + z ( 1 )

$\frac{\partial\ \phi}{\partial\ x} = y + z$
$\frac{\partial\ \phi}{\partial\ y} = x \frac{\partial}{\partial\ y}(y) + z \frac{\partial}{\partial\ y}(y) + 0$

= x ( 1 )  + z ( 1 )

$\frac{\partial\ \phi}{\partial\ y} = x + z$
$\frac{\partial\ \phi}{\partial\ z} = 0 + y \frac{\partial}{\partial\ z}(z) + x \frac{\partial}{\partial\ z}(z)$

= y ( 1 )  + x ( 1 )

$\frac{\partial\ \phi}{\partial\ z} = y + x$

Equation (1) becomes

$\nabla\ \phi=(y + z)\overrightarrow{i} + (x + z)\overrightarrow{j} + (y + x)\overrightarrow{k}$

At the point (1,1, 1),

$\nabla\ \phi=(1 + 1)\overrightarrow{i} + (1 + 1)\overrightarrow{j} + (1 + 1)\overrightarrow{k}$
$\nabla\ \phi= 2\overrightarrow{i} + 2\overrightarrow{j} + 2\overrightarrow{k}$
$|\nabla\phi| = \sqrt{(2)^2 + (2)^2 + (2)^2 }$
$= \sqrt{(4 + 4 + 4) }$
$= 2\sqrt{3}$
$unit\ normal\ to\ the\ surface\ \phi\ at\ the\ point\ (1,1,1)\ =\ \frac{\nabla\phi}{|\nabla\phi|}$
$= \frac{2\overrightarrow{i}+ 2\overrightarrow{j}+ 2\overrightarrow{k}}{2\sqrt{3}}$

Example:  Find the acute angle between the surface xy2z = 4 and  x2 + y2 + z2 =6  at the point (2, 1, 1).

Soln:

Let f = xy2z = 4 be the surface  ------ (1)

Normal vector to (1) at (2,1,1)

$\nabla\ f=(\frac{\partial\ f}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ f}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ f}{\partial\ z})\overrightarrow{k}------------(A)$
$\frac{\partial\ f}{\partial\ x} = y^2 z \frac{\partial}{\partial\ x}(x)$

= y2z ( 1)

$\frac{\partial\ f}{\partial\ x} = y^2z$
$\frac{\partial\ f}{\partial\ y} = x z \frac{\partial}{\partial\ y}(y^2)$

= x z  (2y)

$\frac{\partial\ f}{\partial\ y} = 2xyz$
$\frac{\partial\ f}{\partial\ z} = x y^2\ \frac{\partial}{\partial\ z}(z)$

= xy2  (1)

$\frac{\partial\ f}{\partial\ z} = xy^2$
$\nabla\ f = y^2z \overrightarrow{i} + 2xyz\overrightarrow{j} + x y^2\overrightarrow{k}$

At the point (2,1, 1),

$\nabla\ f = (1)^2 (1) \overrightarrow{i} + (2) (1) (1)\overrightarrow{j} + (2)(1)^2\overrightarrow{k}$
$\nabla\ f = \overrightarrow{i} + 4\overrightarrow{j} + 2\overrightarrow{k} = a (say)$

Let g = x2 + y2 + z2 =6  be the surface  ------ (2)

Normal vector to (2) at (2,1,1)

$\nabla\ g = (\frac{\partial\ g}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ g}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ g}{\partial\ z})\overrightarrow{k}------------(B)$
$\frac{\partial\ g}{\partial\ x} = 2x$
$\frac{\partial\ g}{\partial\ y} = 2y$
$\frac{\partial\ g}{\partial\ z} = 2z$
$\nabla\ g = 2x \overrightarrow{i} + 2y\overrightarrow{j} + 2z\overrightarrow{k}$

At the point (2,1, 1),

$\nabla\ g = 2 (2) \overrightarrow{i} + (2) (1) \overrightarrow{j} + (2)(1)\overrightarrow{k}$
$\nabla\ g = 4\overrightarrow{i} + 2\overrightarrow{j} + 2\overrightarrow{k} = b (say)$

Angle between the surfaces = Angle between the normal to them

= Angle between a and b

$= \cos ^-1 ( \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}})$
$= \cos ^-1 ( \frac{(\overrightarrow{i} + 4\overrightarrow{j} + 2\overrightarrow{k}).(4\overrightarrow{i} + 2\overrightarrow{j} + 2\overrightarrow{k}}{\sqrt{(1)^2 + (4)^2 + (2)^2 }\sqrt{(4)^2 + (2)^2 + (2)^2 }})$
$= \cos ^-1 ( \frac{4 + 8 + 4}{\sqrt{1 + 16 + 4 }\sqrt{16 + 4 + 4}})$
$= \cos ^-1 ( \frac{16}{\sqrt{21}\sqrt{24}})$
$= \cos ^-1 ( \frac{16}{\sqrt{(3) (7)}\ 2\sqrt{(3) ( 2 )}})$
$= \cos ^-1 ( \frac{8}{3\sqrt{14}})$