# Differentiation of Vectors : Curl & Divergence of a Vector Function

### Differentiation of Vectors :

Suppose P be the position of a moving particle at any instant of time with respect to origin ‘ O ‘ .

Let $\vec{OP} = \vec{r}$

Then differentiation of $\vec{r}$ with respect to time i.e. $\displaystyle \frac{\vec{dr}}{dt}$ gives the velocity of the particle at P . Similarly , $\frac{\vec{d^2 r}}{dt^2}$ gives the acceleration of the particle at P .

### Rules of Differentiation :

Let U and V be the two functions :

(i) $\displaystyle \frac{\vec{d}}{dt}(\vec{U} + \vec{V}) = \frac{\vec{dU}}{dt} + \frac{\vec{dV}}{dt}$

(ii) $\displaystyle \frac{\vec{d}}{dt}(\vec{U} . \vec{V}) = \vec{U}.\frac{\vec{dV}}{dt} + \frac{\vec{dU}}{dt}.\vec{V}$

(iii) $\displaystyle \frac{\vec{d}}{dt}(\vec{U} \times \vec{V}) = \frac{\vec{dU}}{dt} \times \vec{V} + \vec{U} \times \frac{\vec{dV}}{dt}$
(Note: Order of function should not be changed )

### Scalar and Vector Point functions :

Scalar Point function : If a function φ (x , y, z) defines a scalar at every point of region or space , then φ (x , y, z) is called a Scalar point function .

Vector Point function : If a function F (x , y, z) defines a vector at every point of region or space , then F (x , y, z) is called a Vector point function .

### Vector Differential Operator , Del i.e. ∇

Vector Differential Operator is defined as :

$\displaystyle \nabla = \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}$

Gradient of a scalar point function φ is defined as , ∇φ or grad φ

grad φ = $\displaystyle \nabla \phi = (\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}) \phi$

$\displaystyle = \hat{i}\frac{\partial \phi}{\partial x} + \hat{j}\frac{\partial \phi}{\partial y} + \hat{k}\frac{\partial \phi}{\partial z}$

It is a Vector quantity .

Total differentiation is given as :

$\displaystyle d\phi = \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y}dy + \frac{\partial \phi}{\partial z} dz$

### Directional Derivative

Directional Derivative : Suppose φ (x , y, z) be a function .

The component of ∇φ in the direction of $\vec{r}$ = $\nabla \phi . \hat{r}$ is called Directional Derivative of φ (x , y, z) in the direction of $\vec{r}$ .

Example: For the function $\phi(x,y) = \frac{x}{x^2 + y^2}$ ; Find the magnitude of the directional derivative along a line making an angle 30° with the positive X-axis at (0,2)

Solution : Directional Derivative = ∇φ

$\displaystyle \nabla \phi = (\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}) \phi$

$\displaystyle \nabla \phi = (\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}) (\frac{x}{x^2 + y^2})$

$\displaystyle = \hat{i}(\frac{1}{x^2 + y^2}-\frac{x (2x)}{(x^2 + y^2)^2}) – \hat{j} \frac{x(2y)}{(x^2 + y^2)^2}$

$\displaystyle = \hat{i} \frac{y^2 – x^2}{(x^2 + y^2)^2} – \hat{j} \frac{2 x y}{(x^2 + y^2)^2}$

Now , Directional Derivative at (0,2) will be

$\displaystyle = \hat{i} \frac{2^2 – 0}{(x^2 + y^2)^2} – \hat{j} \frac{0}{(0 + 2^2)^2}$

$\displaystyle \nabla \phi = \frac{\hat{i}}{4}$

Now , Required Directional Derivative at the point (0,2) will be

$\displaystyle = \frac{\hat{i}}{4} (cos30^o \hat{i} + sin30^o \hat{j})$

$\displaystyle = \frac{\hat{i}}{4} ( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j})$

$= \frac{\sqrt{3}}{8}$

### Divergence of a Vector Function

Let $\vec{F} = F_1 \hat{i} + F_2 \hat{j} + F_3 \hat{k}$

Then , $\displaystyle div \vec{F} = \nabla .\vec{F}$

$\displaystyle \nabla .\vec{F} = (\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}) (F_1 \hat{i} + F_2 \hat{j} + F_3 \hat{k})$

$\displaystyle \nabla .\vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}$

Thus , $div \vec{F}$ is a Scalar Function .

Example: If $\displaystyle u = x^2 + y^2 + z^2$ and $\displaystyle \vec{r} = x \hat{i} + y\hat{j} + z \hat{k}$ ; then find $div(u \vec{r})$ in terms of u .

Solution: $\displaystyle div(u \vec{r}) = (\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}) . [( x^2 + y^2 + z^2)(x \hat{i} + y\hat{j} + z \hat{k})]$

$= (\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}) . [( x^2 + y^2 + z^2) x \hat{i} + ( x^2 + y^2 + z^2) y \hat{j} + ( x^2 + y^2 + z^2)z \hat{k} ]$

$\displaystyle = (3 x^2 + y^2 + z^2 ) + ( x^2 + 3 y^2 + z^2) + ( x^2 + y^2 + 3 z^2)$

$\displaystyle = 5 (x^2 + y^2 + z^2 ) = 5 u$

### Curl of a Vector Function

Let $\vec{F} = F_1 \hat{i} + F_2 \hat{j} + F_3 \hat{k}$

Then , $\displaystyle curl \vec{F} = \nabla \times \vec{F}$

$\displaystyle \nabla \times\vec{F} = (\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}) \times (F_1 \hat{i} + F_2 \hat{j} + F_3 \hat{k})$

$\displaystyle = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_1 & F_2 & F_3 \end{array} \right|$

$\displaystyle = \hat{i} (\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z}) – \hat{j} (\frac{\partial F_3}{\partial x} -\frac{\partial F_1}{\partial z}) + \hat{k} (\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y})$

Thus , $curl \vec{F}$ is a Vector Function .

Note : If $curl \vec{F} = 0$ , then field F is said to be irrotational .

If $div .\vec{F} = 0$ , then field F is said to be Solenoidal .

Example : Show that $\displaystyle \vec{F} = (y^2 + 2 x z^2)\hat{i} + (2 x y – z)\hat{j}+ (2 x^2 z – y + 2 z)\hat{k}$ is irrotational and find its scalar potential .

Solution: $curl \vec{F} = \nabla \times\vec{F}$

$\displaystyle \nabla \times\vec{F} = (\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}) \times [(y^2 + 2 x z^2)\hat{i} + (2 x y – z)\hat{j} + (2 x^2 z – y + 2 z)\hat{k}]$

$\displaystyle = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^2 + 2 x z^2 & 2 x y – z & 2 x^2 z – y + 2 z \end{array} \right|$

$= (-1 + 1)\hat{i} – (4 x z – 4 x z)\hat{j} + (2 y – 2y)\hat{k}$

= 0

As $curl \vec{F} = 0$ , hence F is irrotational .

Example: Show that the vector field represented by $\displaystyle \vec{F} = (z^2 + 2 x + 3 y)\hat{i} + (3 x + 2 y + z)\hat{j}+ (y + 2 z x)\hat{k}$ is irrotational but not Solenoidal . Also Obtain a scalar function φ such that $grad \phi = \vec{F}$

Solution: $curl \vec{F} = \nabla \times\vec{F}$

$\displaystyle = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 + 2 x +3 y & 3 x + 2 y + z & y + 2 z x \end{array} \right|$

$= (1-1)\hat{i} – (2z-2z)\hat{j} + (3-3)\hat{k} = 0$

As $curl \vec{F} = 0$ , hence F is irrotational .

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