VECTOR FUNCTION- GRADIENT OF A SCALAR, DIRECTION DERIVATIVE, DIVERGENCE OF A VECTOR, CURL OF A VECTOR, SCALAR POTENTIAL

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.