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VECTOR FUNCTION
CONTENT
• INTRODUCTION
• GRADIENT OF A SCALAR
• DIRECTION DERIVATIVE
• DIVERGENCE OF A VECTOR
• CURL OF A VECTOR
• SCALAR POTENTIAL
INTRODUCTION
In this chapter, a vector field or a scalar field
can be differentiated w.r.t. position in three
ways to produce another vector field or scalar
field. The chapter details the three
derivatives, i.e.,
1. gradient of a scalar field
2. the divergence of a vector field
3. the curl of a vector field
VECTOR DIFFERENTIAL
OPERATOR
* The vector differential Hamiltonian operator
DEL(or nabla) is denoted by ∇ and is defined
as:
  

  
= i + j +k
x y z
GRADIENT OF A SCALAR
* Let f(x,y,z) be a scalar point function of
position defined in some region of space. Then
gradient of f is denoted by grad f or ∇ f and is
defined as
grad f = ∇f = i
𝜕f
𝜕x
+ j
𝜕f
𝜕y
+ k
𝜕f
𝜕z
ograd f is a vector quantity.
ograd f or ∇f , which is read “del f ”
*
 Find gradient of F if F = 𝟑x 𝟐y- y 𝟑z 𝟐 at
(1,1,1, )
Solution: by definition,
∇f = 𝑖
𝜕f
𝜕𝑥
+ 𝑗
𝜕f
𝜕𝑦
+ 𝑘
𝜕f
𝜕𝑧
= -(6xy)i + (3 𝑥2
-3 𝑦2
𝑧2
)j – (2 𝑦3
z)k
= -6i+0j-2k ans
DIRECTIONAL DERIVATIVE
* The directional derivative of a scalar point function
f at a point f(x,y,z) in the direction of a vector a , is
the component ∇f in the direction of a.
*If a is the unit vector in the direction of a, then
direction derivative of ∇f in the direction of a of is
defined as
= ∇f .
a
|a|
*
 Find the directional derivative of the function f (x, y) = x2y3
– 4y at the point (2, –1) in the direction of the vector v = 2 i
+ 5 j.
Solution: by definition, ∇f = 𝑖
𝜕f
𝜕𝑥
+ 𝑗
𝜕f
𝜕𝑦
+ 𝑘
𝜕f
𝜕𝑧
∇f = 𝑖 2𝑥y3 + 𝑗 3x2y2 − 4
at (2,-1) = −4𝑖 +8j
Directional derivative in the direction of the vector 2 i + 5 j
= ∇f .
a
|a|
= (−4𝑖 +8j).
3 i + 4 j
| 9+16|
=
𝟑𝟐
𝟓
ans
DIVERGENCE OF A VECTOR
* Let f be any continuously differentiable vector
point function. Then divergence of f and is
written as div f and is defined as
which is a scalar quantity.
 

  
31 2
ff f
div f = •f = + +
x y z
*
Find the divergence of a vector
A= 2xi+3yj+5zk .
Solution: by definition,
=
𝜕
𝜕x
(2x) +
𝜕
𝜕𝑦
(3y) +
𝜕
𝜕𝑧
(5z)
= 2+3+5
= 10 ans
 

  
31 2
ff f
div f = •f = + +
x y z
SOLENOIDAL VECTOR
* A vector point function f is said to be
solenoidal vector if its divergent is equal
to zero i.e., div f=0 at all points of the
function. For such a vector, there is no
loss or gain of fluid.
0321  fff zyxf
*
Show that A= (3y4z 2 i+(4 x3z 2)j-(3x2y2)k
is solenoidal.
Sol: here, A= (3y4z 2)i+(4 x3z 2)j-(3x2y2)k
by definition,
=
𝜕
𝜕x
(3y4z 2) +
𝜕
𝜕𝑦
(4 x3z 2) +
𝜕
𝜕𝑧
(3x2y2)
= 0
Hence, it is a solenoidal.
0321  fff zyxf
CURL OF A VECTOR
* Let f be any continuously differentiable vector
point function. Then the vector function curl of
f(x,y,z) is denoted by curl f and is defined as
 
     zyxfzyxfzyxf
zyx
zyx
,,,,,,
f,,curl
321







kji
f
*
Find the curl of A= (xy)i-(2 xz )j+(2yz)k at the point
(1, 0, 2).
Solution: here, A = (xy)i-(2xz )j+(2yz)k
by definition,
yzxzxy
zyx
22







kji
 
     zyxfzyxfzyxf
zyx
zyx
,,,,,,
f,,curl
321







kji
f
= (2z-2x)i – (0-0)j + (2z-x)k
At (1, 0, 2)
= 2i - 0j + 3k ans
k2j2i22 






























 xy
y
xz
x
xy
z
yz
x
xz
z
yz
y
IRROTATIONAL VECTOR
* Any motion in which curl of the velocity vector
is a null vector i.e., curl v=0 is said to be
irrotational.
*Otherwise it is rotational
  0f zyx ,,curl
*
Show that F=(2x+3y+2z)i + (3x+2y+3z)j +
(2x+3y+3z)k is irrotational.
Solution: F=(2x+3y+2z)i+(3x+2y+3z)j+(2x+3y+3z)k
by definition,
 
3z3y2x3z2y3x2z3y2x
f,,curl








zyx
zyx
kji
f
  0f zyx ,,curl
= (3-3)i – (2-2)j + (3-3)k
=0
hence, it is irrotational.
k2z3y2x3z2y3x 












yx
j2z3y2x3z3y2x 










zx
i
zy 











 3z2y3x3z3y2x
SCALAR POTENTIAL
*If f is irrotational, there will always exist a
scalar function f(x,y,z) such that
f=grad g.
This g is called scalar potential of f.
*
A fluid motion is given by V=(ysinz-sinx)i + (xsinz+2yz)j +
(xycosz+y2) . Find its velocity potential.
Solution: V=(ysinz-sinx)i + (xsinz+2yz)j + (xycosz+y2)
by definition,
∇∅ = 𝑣
i
𝜕∅
𝜕x
+ j
𝜕∅
𝜕y
+ k
𝜕∅
𝜕z
= (ysinz-sinx)i+(xsinz+2yz)j+(xycosz+ y2)
by equating corresponding equation we get,
* 𝜕∅
𝜕x
= ysinz-sinx
integrating w r to x ; ∅= xysinz + 𝑐𝑜𝑠𝑥 + 𝑓1 y, z
* 𝜕∅
𝜕𝑦
= xsinz+2yz
integrating w r to y ; ∅= xysinz + zy2 +𝑓2 x, z
* 𝜕∅
𝜕𝑧
= xycosz+ y2
integrating w r to z ; ∅= xysinz + zy2
+𝑓3 x, 𝑦
Hence, ∅= xysinz + zy2
+ 𝑐𝑜𝑠𝑥 +C
*
DERIVATIVES FORMULA
1 The Del Operator ∇ =
𝜕
𝜕x
i +
𝜕
𝜕y
j +
𝜕
𝜕z
k
2 Gradient of a scalar function is a vector
quantity.
grad f = ∇f = i
𝜕f
𝜕x
+ j
𝜕f
𝜕y
+ k
𝜕f
𝜕z
3 Divergence of a vector is a scalar
quantity.
∇.A
4 Curl of a vector is a vector quantity. ∇*A
0.
0


A

• So, any vector differential equation of the form
B=0 can be solved identically by writing B=.
• We say B is irrotational.
• We refer to  as the scalar potential.
• So, any vector differential equation of the form
.B=0 can be solved identically by writing B=A.
• We say B is solenoidal or incompressible.
• We refer to A as the vector potential.
VECTOR FUNCTION

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VECTOR FUNCTION

  • 2. CONTENT • INTRODUCTION • GRADIENT OF A SCALAR • DIRECTION DERIVATIVE • DIVERGENCE OF A VECTOR • CURL OF A VECTOR • SCALAR POTENTIAL
  • 3. INTRODUCTION In this chapter, a vector field or a scalar field can be differentiated w.r.t. position in three ways to produce another vector field or scalar field. The chapter details the three derivatives, i.e., 1. gradient of a scalar field 2. the divergence of a vector field 3. the curl of a vector field
  • 4. VECTOR DIFFERENTIAL OPERATOR * The vector differential Hamiltonian operator DEL(or nabla) is denoted by ∇ and is defined as:        = i + j +k x y z
  • 5. GRADIENT OF A SCALAR * Let f(x,y,z) be a scalar point function of position defined in some region of space. Then gradient of f is denoted by grad f or ∇ f and is defined as grad f = ∇f = i 𝜕f 𝜕x + j 𝜕f 𝜕y + k 𝜕f 𝜕z ograd f is a vector quantity. ograd f or ∇f , which is read “del f ”
  • 6. *  Find gradient of F if F = 𝟑x 𝟐y- y 𝟑z 𝟐 at (1,1,1, ) Solution: by definition, ∇f = 𝑖 𝜕f 𝜕𝑥 + 𝑗 𝜕f 𝜕𝑦 + 𝑘 𝜕f 𝜕𝑧 = -(6xy)i + (3 𝑥2 -3 𝑦2 𝑧2 )j – (2 𝑦3 z)k = -6i+0j-2k ans
  • 7. DIRECTIONAL DERIVATIVE * The directional derivative of a scalar point function f at a point f(x,y,z) in the direction of a vector a , is the component ∇f in the direction of a. *If a is the unit vector in the direction of a, then direction derivative of ∇f in the direction of a of is defined as = ∇f . a |a|
  • 8. *  Find the directional derivative of the function f (x, y) = x2y3 – 4y at the point (2, –1) in the direction of the vector v = 2 i + 5 j. Solution: by definition, ∇f = 𝑖 𝜕f 𝜕𝑥 + 𝑗 𝜕f 𝜕𝑦 + 𝑘 𝜕f 𝜕𝑧 ∇f = 𝑖 2𝑥y3 + 𝑗 3x2y2 − 4 at (2,-1) = −4𝑖 +8j Directional derivative in the direction of the vector 2 i + 5 j = ∇f . a |a| = (−4𝑖 +8j). 3 i + 4 j | 9+16| = 𝟑𝟐 𝟓 ans
  • 9. DIVERGENCE OF A VECTOR * Let f be any continuously differentiable vector point function. Then divergence of f and is written as div f and is defined as which is a scalar quantity.       31 2 ff f div f = •f = + + x y z
  • 10. * Find the divergence of a vector A= 2xi+3yj+5zk . Solution: by definition, = 𝜕 𝜕x (2x) + 𝜕 𝜕𝑦 (3y) + 𝜕 𝜕𝑧 (5z) = 2+3+5 = 10 ans       31 2 ff f div f = •f = + + x y z
  • 11. SOLENOIDAL VECTOR * A vector point function f is said to be solenoidal vector if its divergent is equal to zero i.e., div f=0 at all points of the function. For such a vector, there is no loss or gain of fluid. 0321  fff zyxf
  • 12. * Show that A= (3y4z 2 i+(4 x3z 2)j-(3x2y2)k is solenoidal. Sol: here, A= (3y4z 2)i+(4 x3z 2)j-(3x2y2)k by definition, = 𝜕 𝜕x (3y4z 2) + 𝜕 𝜕𝑦 (4 x3z 2) + 𝜕 𝜕𝑧 (3x2y2) = 0 Hence, it is a solenoidal. 0321  fff zyxf
  • 13. CURL OF A VECTOR * Let f be any continuously differentiable vector point function. Then the vector function curl of f(x,y,z) is denoted by curl f and is defined as        zyxfzyxfzyxf zyx zyx ,,,,,, f,,curl 321        kji f
  • 14. * Find the curl of A= (xy)i-(2 xz )j+(2yz)k at the point (1, 0, 2). Solution: here, A = (xy)i-(2xz )j+(2yz)k by definition, yzxzxy zyx 22        kji        zyxfzyxfzyxf zyx zyx ,,,,,, f,,curl 321        kji f
  • 15. = (2z-2x)i – (0-0)j + (2z-x)k At (1, 0, 2) = 2i - 0j + 3k ans k2j2i22                                 xy y xz x xy z yz x xz z yz y
  • 16. IRROTATIONAL VECTOR * Any motion in which curl of the velocity vector is a null vector i.e., curl v=0 is said to be irrotational. *Otherwise it is rotational   0f zyx ,,curl
  • 17. * Show that F=(2x+3y+2z)i + (3x+2y+3z)j + (2x+3y+3z)k is irrotational. Solution: F=(2x+3y+2z)i+(3x+2y+3z)j+(2x+3y+3z)k by definition,   3z3y2x3z2y3x2z3y2x f,,curl         zyx zyx kji f   0f zyx ,,curl
  • 18. = (3-3)i – (2-2)j + (3-3)k =0 hence, it is irrotational. k2z3y2x3z2y3x              yx j2z3y2x3z3y2x            zx i zy              3z2y3x3z3y2x
  • 19. SCALAR POTENTIAL *If f is irrotational, there will always exist a scalar function f(x,y,z) such that f=grad g. This g is called scalar potential of f.
  • 20. * A fluid motion is given by V=(ysinz-sinx)i + (xsinz+2yz)j + (xycosz+y2) . Find its velocity potential. Solution: V=(ysinz-sinx)i + (xsinz+2yz)j + (xycosz+y2) by definition, ∇∅ = 𝑣 i 𝜕∅ 𝜕x + j 𝜕∅ 𝜕y + k 𝜕∅ 𝜕z = (ysinz-sinx)i+(xsinz+2yz)j+(xycosz+ y2)
  • 21. by equating corresponding equation we get, * 𝜕∅ 𝜕x = ysinz-sinx integrating w r to x ; ∅= xysinz + 𝑐𝑜𝑠𝑥 + 𝑓1 y, z * 𝜕∅ 𝜕𝑦 = xsinz+2yz integrating w r to y ; ∅= xysinz + zy2 +𝑓2 x, z * 𝜕∅ 𝜕𝑧 = xycosz+ y2 integrating w r to z ; ∅= xysinz + zy2 +𝑓3 x, 𝑦 Hence, ∅= xysinz + zy2 + 𝑐𝑜𝑠𝑥 +C
  • 22. * DERIVATIVES FORMULA 1 The Del Operator ∇ = 𝜕 𝜕x i + 𝜕 𝜕y j + 𝜕 𝜕z k 2 Gradient of a scalar function is a vector quantity. grad f = ∇f = i 𝜕f 𝜕x + j 𝜕f 𝜕y + k 𝜕f 𝜕z 3 Divergence of a vector is a scalar quantity. ∇.A 4 Curl of a vector is a vector quantity. ∇*A
  • 23. 0. 0   A  • So, any vector differential equation of the form B=0 can be solved identically by writing B=. • We say B is irrotational. • We refer to  as the scalar potential. • So, any vector differential equation of the form .B=0 can be solved identically by writing B=A. • We say B is solenoidal or incompressible. • We refer to A as the vector potential.