Vector Calculus

1. Differential Operators: If corresponding to each point P in a region R of space there corresponds a scalar denoted by Ф(P) then Фis said to be a scalar point function for the region R. If the coordinates of P be ( x, y, z), then Ф(P) = Ф(x, y, z)
If P0 is the fixed point whose coordinate is ( x0 , y0 , z0 ), then the distance of the point P from the fixed point P0 is given by,
Ф ( P ) =   $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$
2. Vector Fields: (Vector Point function):
(a) If corresponding to each point P of a region R2 of space there corresponds a vector defined by f(P) the f is called a vector point function for the region R2.
If the coordinates of P be ( x, y ), then   f(P) = f( x, y ) = f1 ( x, y ) i + f2 ( x, y ) j
(b) If corresponding to each point P of a region R3 of space there corresponds a vector defined by f(P) the f is called a vector point function for the region R3.
If the coordinates of P be ( x, y, z), then  
  f(P) = f( x, y, z) = f1 ( x, y, z) i + f2 ( x, y, z) j + f3 ( x, y, z) k

Example:  A vector field on R2 is defined by F( x, y) = y i - x j. Describe F by sketching some of the vectors F(x, y).
Solution: We have, F( x, y) = y i - x j
Let us plug in some values for the domain of F to find the corresponding vector in the field.
If ( x,y ) = (1,0) then F( 1,0) = 0 - j = ( 0,-1)
If (x,y) = ( 0,1) then, F( 0,1) = y i  + 0 = ( 1,0)
If (x,y) = ( -1, 0) then, F( -1, 0) = 0 -  j = ( 0,1)
If (x,y) = ( 0,-1) then, F( 0, -1) = - i = ( -1, 0)

If (x,y) = ( 2,1) then F(2,1) = (1,-2)
If (x,y) = (1,-2) then F(1,-2) = ( -2,-1)
If (x,y) = ( -2, -1) then F(-2, -1) =  ( -1, 2)
If (x,y) = ( -1,2) then F( -1, 2) = ( 2,1)

Tangent to Circle

It is obvious that each row is tangent  to a circle with center at the origin.

3. Directional derivatives of scalar point function: Let Ф(P) be a scalar point function and A be a given point in the region of its definition. Through A if AP is drawn in ine side of A and a point P is taken on it, then
 $\lim_{p\to A}$ $\frac{\Phi (P)-\Phi (A)}{AP}$=$\lim_{\delta x\to }$ $\frac{\Phi (x+\delta x,y,z)-\Phi (x,y,z)}{\delta x}$=$\frac{\partial  \Phi }{\partial x}$
$\frac{\partial  \Phi }{\partial y}$, $\frac{\partial  \Phi }{\partial Z}$ are defined in a similar manner.

4. Directional derivatives of vector point  function along coordinate axes: We are aware that            f ( x, y, z) = f1 ( x, y, z) i + f2 ( x, y, z) j + f3 ( x, y, z) k

then  $\frac{\partial f}{\partial x}$ = $\frac{\partial f_1}{\partial x}$i + $\frac{\partial f_z}{\partial x}$j + $\frac{\partial f_3}{\partial x}$k  similarly $\frac{\partial f}{\partial y}$, $\frac{\partial f}{\partial z}$  are defined.

5. Directional derivative of a scalar point function along any line whose direction cosines are l , m, n is  written as, l  $\frac{\partial  \Phi }{\partial x}$ +m $\frac{\partial  \Phi }{\partial y}$+ n $\frac{\partial  \Phi }{\partial z}$

6. Directional derivative of a vector point function along any line with direction cosines,
is   , l  $\frac{\partial  f }{\partial x}$+m $\frac{\partial  f }{\partial y}$+n $\frac{\partial  f }{\partial z}$   , where  $\frac{\partial f}{\partial x}$ = $\frac{\partial f_1}{\partial x}$i + $\frac{\partial f_z}{\partial x}$j + $\frac{\partial f_3}{\partial x}$k, $\frac{\partial f}{\partial y}$ = $\frac{\partial f_1}{\partial y}$i + $\frac{\partial f_2}{\partial y}$j + $\frac{\partial f_3}{\partial y}$k,

and $\frac{\partial f}{\partial z}$ = $\frac{\partial f_1}{\partial z}$i + $\frac{\partial f_2}{\partial z}$j + $\frac{\partial f_3}{\partial z}$k, .

7. The vector differential operator  ( read as “del” or “nabla” ) is defined as below.
 ∇ = i $\frac{\partial }{\partial x}$+j  $\frac{\partial }{\partial y}$+ k $\frac{\partial  }{\partial z}$
If Ф is any scalar function,  ∇  Ф = i $\frac{\partial \Phi }{\partial x}$+j $\frac{\partial\Phi  }{\partial y}$+ k $\frac{\partial \Phi  }{\partial z}$  [ also called as Gradient Ф or∇ Ф ]

8. Divergence  and Curl of a vector. We know that for a scalar point function Ф, grad Ф ( Ф) is a vector point function. Divergence of the vector point function f is a scalar point function where as curl f is a vector point function.
a.   Div f = ∇ . f =(i $\frac{\partial }{\partial x}$+j $\frac{\partial }{\partial y}$+kz) ,( f1  i + f2  j + f3  k )
       =   $\frac{\partial f_1}{\partial x}$+ $\frac{\partial f_2}{\partial y}$+ $\frac{\partial f_3}{\partial z}$

b. Curl f =∇  x f =$\begin{vmatrix}
i &j  &k \\
\frac{\partial }{\partial x} &\frac{\partial }{\partial y}  &\frac{\partial }{\partial z} \\
f_1 &f_2  &f_3
\end{vmatrix}$



9. Laplacian Operator  $\bigtriangledown _2\Phi $
If $\Phi $ is any scalar function, then

 $\bigtriangledown _2\Phi $    =∇ . ∇ Ф
 
                           =(i $\frac{\partial }{\partial x}$+j $\frac{\partial }{\partial y}$+kz). i $\frac{\partial \Phi }{\partial x}$+j $\frac{\partial\Phi  }{\partial y}$+ k $\frac{\partial \Phi  }{\partial z}$

=$\frac{\partial^2 \Phi }{\partial x}$+ $\frac{\partial^2 \Phi  }{\partial y}$+ $\frac{\partial^2 \Phi  }{\partial z}$
If f is any vector function, then
   
   $\bigtriangledown _2\Phi $   = $\frac{\partial^2 \Phi }{\partial x}$+ $\frac{\partial^2 \Phi  }{\partial y}$+ $\frac{\partial^2 \Phi  }{\partial z}$

Vector Calculus Solutions

If f = xz i – 2 y z  j + 3z x k, find div f, curl f at ( 1, -1, 1)
div f$\frac{\partial f_1}{\partial x}$ + $\frac{\partial f_2}{\partial y}$ + $\frac{\partial f_3}{\partial z}$

  = $\frac{\partial f(xz)}{\partial x}$ + $\frac{\partial f(-2z)}{\partial y}$ + $\frac{\partial f(3xz)}{\partial z}$
   = z – 2z + 3x = -z + 3x
 = -1 + 3 = 2

curl f = $\begin{vmatrix}
i & j & k\\
 \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\
 f_1&f_2  &f_3
\end{vmatrix}$


= i [0 –(-2y) ] – j [3z – x ] + k [0 – 0 ]
        = 2 y i – 3z j
        = -2 i – 3 j [ at ( 1, -1, 1) ]

Line Integrals

 $\int _c F.dr$  is called the line or tangent line integral of F(t) along the curve C.

Also,$\int _c F.dr$  =$\int_{a}^{b}$ $(f_{11}$ $\frac{dx}{dt}$ + $f_{12}$ $\frac{dy}{dt}$ + $f_{13}$ $\frac{dz}{dt}$)dt

Stokes Theorem

 The line integral of the tangential component of a vector function F taken around a simple closed curve C is equal to the normal surface integral of curl F taken over any surface S having C as its boundary.
(i.e) IF F is continuously differentiable vector function and S is a surface bounded by a curve C, then
$\int _c F.dr$ =$\int _s curl F . n dS$where, n is the unit normal vector at any point S which is drawn in the sense in which a right handed screw would move when rotated in the sense of description of C.

Green’s Theorem

Let P(x,y) and Q(x,y) be two functions defined in a region A in the xy plane with a simple closed curve C as its boundary. Then,
$\int(Pdx+Qdy)$=$\int \int$ ($\frac{\partial q}{\partial x}$ - $\frac{\partial P}{\partial y}$)dx dy

Divergence Theorem

If F is continuously differentiable vector point function and S is a closed surface enclosing a region V,  then                                                            
$\int _s curl F . n dS$=$\int _v div F dV$


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