Unit – 4
Vector Differential Calculus
In ordinary differentiate, the derivative 
(where y=f(x)) is defined as 
=
Consider vector 
which may depend for its value on scular variable t. corresponding to change 
in t, let there be a change 
in 
i.e
The vector derivative can now be defined as
& we know that ,
now 
From the definition of vector derivative, following results can be easily established
For vector 
8. 
9. If S is constant 
Application of mechanics in vector differentiate
Consider a particle P moving along a circle of radius r with constant angular. Speed w.
Differentiating 
T
Again diff. we get ,
b. Radial & transverse component of velocity & acc
c
tangential & normal component of Accn’s
Que 1:- a curve is given by the equation 
. Find the angle between tangents
.
Solution :- 
Where 
Now 
Que 2 if 
show that 
has a constant dim
Now 
i.e 
Both cant be true simultaneously
has constant direction.
If cure 
i.e a vector field 
for which the curl vanishes.
For a real-valued function f(x,y,z)f(x,y,z) on R3R3, the gradient ∇f(x,y,z)∇f(x,y,z) is a vector-valued function on R3R3 , that is, its value at a point (x,y,z)(x,y,z) is the vector
∇f(x,y,z)=(∂f/∂x,∂f/∂y,∂f/∂z)=∂f/∂xi+∂f/∂yj+∂f/∂zk
For example, it is often convenient to write the divergence div f as ∇⋅f∇⋅f, since for avectorfield f(x,y,z)=f1(x,y,z)i+f2(x,y,z)j+f3(x,y,z)kf(x,y,z)=f1(x,y,z)i+f2(x,y,z)j+f3(x,y,z)k, the dot product of f with ∇∇ (thought of as a vector) makes sense:
∇⋅f=(∂/∂xi+∂/∂yj+∂/∂zk)⋅(f1(x,y,z)i+f2(x,y,z)j+f3(x,y,z)k)
=(∂/∂x)(f1)+(∂/∂y)(f2)+(∂/∂z)(f3)
=∂f1/∂x+∂f2/∂y+∂f3/ ∂z
=div f
Let ϕ be a scalar point function and let ϕ(P) and ϕ(Q) be the values of ϕ at two neighboring points P and Q in the field. Then,
,
are the directional derivative of ϕ in the direction of the coordinate axes at P.
The directional derivative of ϕ in the direction l, m, n =l 
+ m
+ 
The directional derivative of ϕ in the direction of 
= 
A solenoidal vector field satisfies
∇.B=0 |
|
for every vector 
, where 
is the divergence. If this condition is satisfied, there exists a vector 
, known as the vector potential, such that
B=∇ X A |
|
where ∇ X A is the curl. This follows from the vector identity
∇. B=∇.(∇ X A) =0 |
|
If A is an irrotational field, then
A x r |
|
is solenoidal. If 
and 
are irrotational, then
U x v |
|
is solenoidal. The quantity
|
|
Where 
is the gradient, is always solenoidal.For a function 
satisfying Laplace's equation
∇2ϕ=0 |
|
it follows that ∇ϕ is solenoidal (and also irrotational).
A vector field 
for which the curl vanishes
A conservative vector field (for which the curl 
) may be assigned a scalar potential
\ |
where 
is a line integral.
Q1. Calculate the curl for the following vector field.
F⃗ =x3y2 i⃗ +x2y3z4 j⃗ +x2z2 k⃗
Solution: In order to calculate the curl, we need to recall the formula.
where P, Q, and R correspond to the components of a given vector field: F⃗ =Pi⃗ +Qj⃗ +Rk⃗
=(
(x2z2)−
(x2y3z4) )i⃗ +(
(x3y2)−
(xz2) )j⃗ +(
(x2y3z4)−
(x3y2) )k⃗
=(0−4x2y3z3)i⃗ +(0−2xz2)j⃗ +(2xy3z4−2x3y)k⃗
Thus the curl is
=(−4x2y3z3)i⃗ +(−2xz2)j⃗ +(2xy3z4−2x3y)k⃗
Q2. Find the directional derivative of Θ=x2y cos z at (1,2,π/2) in the direction of a = 2i+3j+2k.
Solution : ∇ϕ = i
+ k 
= 2xy cos zi+ x2 cos zj -x2y sin zk
At (1,2,π/2) ∇ϕ = 0i +0j-2k
Directional directive in the direction of 2i+3j+2k.
=(0i+0j-2k).
=- 
Q3. In what direction from the point (2,1,-1) is the directional derivative of ϕ=x2yz3 maximum? What is its magnitude?
Solution :∇ϕ= i
+ k 
= -4i-4j+12k
Directional derivative is maximum in the direction of ∇Θ. Hence, directional derivative is maximum in the direction of -4i-4j+12k
Its magnitude =
=4
Example 4: Prove that ͞͞͞F = [y2 cos x +z3] i +(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for͞͞͞F (ii) the work done in moving an object in this field from (0, 1, -1) to (
/ 2,-1, 2)
Sol. : (a) The field is conservative if cur͞͞͞͞͞͞F = 0.
Now, cur͞͞͞F = 
̷̷
X 
/ 
y 
/ 
z
i j k
Y2COS X +Z3 2y sin x-4 3xz2 +2
; Cur 
= (0-0) – (3z2 – 3z2) j + (2y cos x- 2y cos x) k = 0
; F is conservative.
(b) Since F is conservative there exists a scalar potential ȸ such that
F = ȸ
(y2 cos x +z3) i + (2y sin x-4) j + (3xz2 + 2) k
= 
i + 
j + 
k
= y2 cos x + z3, 
= 2y sin x – 4, 
= 3xz2 + 2
Now, 
= 
dx + 
dy + 
dz
= (y2 cos x + z3) dx +(2y sin x – 4)dy + (3xz2 + 2)dz
= (y2 cos x dx + 2y sin x dy) +(z3dx +3xz2dz) +(- 4 dy) + (2 dz)
=d(y2 sin x + z3x – 4y -2z)
ȸ = y2 sin x +z3x – 4y -2z
(c) now, work done =
.d ͞r
= 
dx + (2y sin x – 4) dy + ( 3xz2 + 2) dz
= 
(y2 sin x + z3x – 4y + 2z) (as shown above)
= [ y2 sin x + z3x – 4y + 2z ]( 
/2, -1, 2)
= [ 1 +8 
+ 4 + 4 ] – { - 4 – 2} =4
+ 15
Q5. If 
x2zi – 2y3z3j + xy2z2k find dvi 
and curl 
at (1,-1, 1)
Solution : div 
= ∇.
= 
=
= 2xz – 6y2z3 + 2xy2z =(2-6+2) = -2
Curl 
= 
=i(2xyz2 + 6y3z2) – j(y2z2- x2) + k(0-0)
=-8 at (1,-1,1)
Q6. Find the angle between the normal to the surface xy = z2 at the points (1,4,2) and (-3,-3,3)
Solution: let ϕ = xy-z2
∇ϕ= i
=yi + xj -2zk =4i + j -4k
∇ϕ = 3i – 3j-6k
But these are the normal to the surface at given points. Angle between two vectors is given by (4i + j -4k).( 4i + j -4k)= |4i + j- 4k|.|-3i-3j -6K|cos θ.
If θ is the angle between then cos θ= 
= 
.
Questions
1. Find the directional derivative of 
at point A(1,-2,1) in the direction of AB where B is
(2,6,-1).
= 
- 
= (2-1)i + (6+2)j + (-1-1)k
=I + 8j – 2k
Hence, Directional Derivative at A is the direction of 
.
=4(i-8j+4k). 
= 
2. Find the directional derivative of 
at (0,1) in direction making an angle of 
with positive
x axis.
Soln. Directional Derivative
=
=0i-j at (0,1)
Unit vector making angle of 
with axis
=
=
Hence, required directional derivative
=(0i-j). [
] =
3. Prove that 
=0 where 
Soln. Let 
Then, 
= 
=
).[
= 0
4. A vector field is given by 
=
i + (
Show that 
is irrotational and find its scalar potential.
Soln. Curl 
= 
=
=0i+0j+(2xy-2xy)k
=0i+0j+0k
Hence, 
is irrotational.
If 
is the scalar potential, then 
Therefore, 
i + (
= 
Comparing the coefficients of i , j , k
From above eq. common terms are:
Q.5. Prove that 
and hence find f if 
.
Soln. We have 
= 
Here, 
=f(r) and f is a function of r and r is a function of (x,y,z)
+ j
But 
2r
=2x
, 
=
Comparing with 
; f(r)=
+C =
(2,6,-1).
= 
- 
= (2-1)i + (6+2)j + (-1-1)k
=I + 8j – 2k
Hence, Directional Derivative at A is the direction of 
.
=4(i-8j+4k). 
= 
2. Find the directional derivative of 
at (0,1) in direction making an angle of 
with positive
x axis.
Soln. Directional Derivative
=
=0i-j at (0,1)
Unit vector making angle of 
with axis
=
=
Hence, required directional derivative
=(0i-j). [
] =
3. Prove that 
=0 where 
Soln. Let 
Then, 
= 
=
).[
= 0
4. A vector field is given by 
=
i + (
Show that 
is irrotational and find its scalar potential.
Soln. Curl 
= 
=
=0i+0j+(2xy-2xy)k
=0i+0j+0k
Hence, 
is irrotational.
If 
is the scalar potential, then 
Therefore, 
i + (
= 
Comparing the coefficients of i , j , k
From above eq. common terms are:
Q.5. Prove that 
and hence find f if 
.
Soln. We have 
= 
Here, 
=f(r) and f is a function of r and r is a function of (x,y,z)
+ j
But 
2r
=2x
, 
=
Comparing with 
; f(r)=
+C =
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