UNIT–2
Vector differential calculus
Vector function- A vector function can be defined as below-
If a vector ‘r’ is a function of a scalar variable ‘t’, then-
We define the derivative of a vector function as-
We can denote it as-
Similarly is the second order derivative of
Note- gives the velocity and gives acceleration.
Rules for differentiation-
1.
2.
3.
4.
5.
Example-1: A particle moves along the curve , here ‘t’ is the time. Find its velocity and acceleration at t = 2.
Sol. Here we have-
Then, velocity
Velocity at t = 2,
=
Acceleration =
Acceleration at t = 2,
Example-2: If and then find-
1.
2.
Sol. 1. We know that-
2.
Suppose f(x, y, z) be the scalar function and it is continuously differentiable then the vector-
Is called gradient of f and we can write is as grad f.
So that-
Here is a vector which has three components
Properties of gradient-
Property-1:
Proof:
First we will take left hand side
L.H.S =
=
=
=
Now taking R.H.S,
R.H.S. =
=
=
Here- L.H.S. = R.H.S.
Hence proved.
Property-2: Gradient of a constant (
Proof:
Suppose
Then
We know that the gradient-
= 0
Property-3: Gradient of the sum and difference of two functions-
If f and g are two scalar point functions, then
Proof:
L.H.S
Hence proved
Property-4: Gradient of the product of two functions
If f and g are two scalar point functions, then
Proof:
So that-
Hence proved.
Property-5: Gradient of the quotient of two functions-
If f and g are two scalar point functions, then-
Proof:
So that-
Example-1: If , then show that
1.
2.
Sol.
Suppose and
Now taking L.H.S,
Which is
Hence proved.
2.
So that
Example: If then find grad f at the point (1,-2,-1).
Sol.
Now grad f at (1 , -2, -1) will be-
Example: If then prove that grad u , grad v and grad w are coplanar.
Sol.
Here-
Now-
Apply
Which becomes zero.
So that we can say that grad u, grad v and grad w are coplanar vectors.
Directional derivative-
Let ϕ be a scalar point function and let ϕ(P) and ϕ(Q) be the values of ϕ at two neighbouring 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 =
Example: Find the directional derivative of 1/r in the direction where
Sol. Here
Now,
And
We know that-
So that-
Now,
Directional derivative =
Example: Find the directional derivative of
At the points (3, 1, 2) in the direction of the vector .
Sol. Here it is given that-
Now at the point (3, 1, 2)-
Let be the unit vector in the given direction, then
at (3, 1, 2)
Now,
Example: Find the directional derivatives of at the point P(1, 1, 1) in the direction of the line
Sol. Here
Direction ratio of the line are 2, -2, 1
Now directions cosines of the line are-
Which are
Directional derivative in the direction of the line-
Divergence (Definition)-
Suppose is a given continuous differentiable vector function then the divergence of this function can be defined as-
Curl (Definition)-
Curl of a vector function can be defined as-
Note- Irrotational vector-
If then the vector is said to be irrotational.
Vector identities:
Identity-1: grad uv = u grad v + v grad u
Proof:
So that
Graduv = u grad v + v grad u
Identity-2:
Proof:
Interchanging , we get-
We get by using above equations-
Identity-3
Proof:
So that-
Identity-4
Proof:
So that,
Identity-5 curl (u
Proof:
So that
Curl (u
Identity-6:
Proof:
So that-
Identity-7:
Proof:
So that-
Example-1: Show that-
1.
2.
Sol. We know that-
2. We know that-
= 0
Example-2: If then find the divergence and curl of .
Sol. we know that-
Now-
Example-3: Prove that
Note- here is a constant vector and
Sol. Here and
So that
Now-
So that-
Example-4: Find the curl of F(x,y,z) = 3i+2zj-xk
Ans.
Curl F =
=
= i -
= (0-2)i-(-1-0)j+(0-0)k
= -2i+j
Example-5: What is the curl of the vector field F= ( x +y +z ,x-y-z,)?
Solution:
Curl F =
=
=
= (2y+1)i-(2x-1)j+(1-1)k
= (2y+1)i+(1-2x)j+0k
= (2y+1, 1-2x,0)
Example-6: Find the curl of F = ()i +4zj +
Solution:
Curl F=
=
=(0-4)i-(2x-0)j+(0+1)k
=(-4)i – (2x)j+1k
=(-4,-2x,1)
Solenoidal field-
Ansolenoidal field F is characterised by the following conditions-
1.
2. Flux along every closed surface is zero.
3.
Note- In an solenoidal field for which , the vector F can always be expressed as the curl of a vector function V.
So that-
Irrotational field-
An irrotatonal field F is characterised by the following conditions-
1.
2. Circulation along every closed surface is zero.
3.
Note- In an irrotational field for which , the vector F can always be expressed as the gradient of a scalar function provided the domain is simply connected.
So that-
Here the scalar function is called the potential.
Example-1: Prove that the vector field is irrotational and find its scalar potential.
Sol. As we know that if then field is irrotational.
So that-
So that the field is irrotational and the vector F can be expressed as the gradient of a scalar potential,
That means-
Now-
………………… (1)
……………………. (2)
Integrating (1) with respect to x, keep ‘y’ as constant-
We get-
…………….. (3)
Integrating (1) with respect to y, keep ‘x’ as constant-
We get-
…………….. (4)
Equating (3) and (4)-
and
So that-
Example-2:Prove that the vector field is solenoidal and irrotational.
Sol. We know that if then the vector field will be solenoidal.
So that-
=
So that the vector field is solenoidal.
Now for irrotational field we need prove-
So that-
Thus, the vector field F is irrotational.
Example-3:Show that the vector field is irrotational and find the scalar potential function.
Sol. Now for irrotational field we need prove-
So that-
So that the vector field is irrotational.
Now in order to find the scalar potential function-