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# Gradient , Directional Derivative , Divergence , Curl

In this presentation we will learn Del operator, Gradient of scalar function , Directional Derivative, Divergence of vector function, Curl of a vector function and after that solved some example related to above.

Directional derivative in math
Divergence in math
Curl in math
Gradient , Directional Derivative , Divergence , Curl in mathematics
Gradient , Directional Derivative , Divergence , Curl in math
Gradient , Directional Derivative , Divergence , Curl

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### Gradient , Directional Derivative , Divergence , Curl

1. 1. • Operator is called vector differential operator defined as  ˆˆ ˆi j k x y z             
2. 2.  ( , , )x y z• If is a scalar function of three variables and is differentiable, the gradient of is defined as Where , is a scalar function is a vector function ˆˆ ˆgrad i j k x y z                  
3. 3. If ,determine at point P=(1,3,2). solution 2 3 2 2 x yz xy z   grad       2 3 2 2 3 2 2 2 3 2 2 2 2 3 2 2 2 3 2 2 2 2 2 , 2 , 3 2 , ˆˆ ˆ ˆˆ ˆ2 2 3 2 (1,3,2) ˆˆ ˆ84 32 72 x yz xy z xyz y z x z xyz x yz xy z x y z Therefore i j k x y z xyz y z i x z xyz j x yz xy z k at P i j k                                            
4. 4. If A and B are two scalars ,then 1) 2) ( )A B A B      ( ) ( ) ( )AB A B B A    
5. 5. Directional derivative of in the direction of a is ˆ. d a grad ds   where, Which is a unit vector in the direction of .dr ˆ dr a dr 
6. 6. Compute the directional derivative of at the point (1,2,-1) in the direction of the vector A=2i+3j-4k. Solution 2 2 2 2x z xy yz    Directional derivative of in the direction of a Where, And ,  ˆ. d a grad ds   ˆˆ ˆgrad i j k x y z                 ˆ A a A  2 2 2 2x z xy yz    Hence ,      2 2 2 ˆˆ ˆ2 2 4 2 (1,2, 1), ˆˆ ˆ6 9 3 xz y i xy z j x yz k At i j k              
7. 7. Also given ,thenˆˆ ˆ2 3 4A i j k       22 2 2 3 4 29 , 1 ˆˆ ˆˆ 2 3 4 29 , ˆ. 51 29 A Therefore A a i j k A then d a grad ds             
8. 8. If ,the divergence of A is defined as ˆˆ ˆ x y z A i j ka a a     . ˆ ˆˆ ˆ ˆ ˆ. . x y z yx z divA A i j k i j k x y z x y z a a a aa a                        
9. 9. If , determine at point (1,2,3) . solution 2 2 ˆˆ ˆA x yi xyzj yz k   divA   . ˆ ˆˆ ˆ ˆ ˆ. . x y z yx z divA A i j k i j k x y z x y z a a a aa a                         2 2divA xy xz yz   (1, 2,3) 2(1)(2) (1)(3) 2(2)(3) 13 at divA divA    
10. 10. If ,the curl of A is defined byˆˆ ˆ x y z A i j ka a a    ˆ ˆˆ ˆ ˆ ˆ ˆˆ ˆ x y z x y z curlA A i j k i j k x y z i j k curlA x y z a a a a a a                         
11. 11. If ,determine curl A at (1,3,-2). solution 4 2 2 2 2 2 ˆˆ ˆ( ) ( )A y x z i x y j x yzk     4 2 2 2 2 2 2 2 3 ˆˆ ˆ ˆˆ( 2 2 ) (2 4 ) (1,3, 2), ˆˆ ˆ2 8 106 i j k curlA A x y z y x z x y x yz x zi xyz x z j x y k At curlA i j k                       
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