Unit 6
Vector Differential Calculus
Vector function-
Suppose be a function of a scalar variable t, then-
Here vector varies corresponding to the variation of a scalar variable t that its length and direction be known as value of t is given.
Any vector can be expressed as-
Here , , are the scalar functions of t.
Differentiation of a vector-
We can denote it as-
Similarly is the second order derivative of
Note- gives the velocity and gives acceleration.
Note-
1. Velocity =
2. Acceleration =
The derivative of the velocity vector V(t) is called acceleration A(t), and it is given as-
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.
Example-3: A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.
And find the magnitudes of the velocity and acceleration at time t.
Sol. Suppose
Now,
At t = 0 | 4 |
At t = π/2 | -4 |
At t = 0 | |v|= |
At t = π/2 | |v|= |
Again acceleration-
Now-
At t = 0 | |
At t = π/2 | |
At t = 0 | |a|= |
At t = π/2 | |a|= |
Scalar point function-
If for each point P of a region R, there corresponds a scalar denoted by f(P), in that case f is called scalar point function of the region R.
Note-
Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).
Vector point function-
If for each point P of a region R, then there corresponds a vector then is called a vector point function for the region R.
Vector field-
Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).
Note-
Del operator-
The del operated is defined as-
Example: show that where
Sol. Here it is given-
=
Therefore-
Note-
Hence proved
Key takeaways-
- Any vector can be expressed as-
Here , , are the scalar functions of t.
1. Velocity =
2. Acceleration =
3. Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).
4. Vector field-
Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).
Tangential and normal accelerations-
It is very important to understand that the magnitude of acceleration is not always the rate of change of |V|.
Suppose be the vector-valued function which denotes the position of any object as a function of time.
Then
Then the tangential and normal component of acceleration are given as below-
Example: A object move in the path where t is the time in seconds and distance is measured in feets.
Then find and as functions of t.
Sol.
We know that-
And
Now we will use-
And now-
Key takeaways-
The vector differential operator is written as-
Gradient
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.
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
Grad(uv) = 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-
Key takeaways-
- Gradient of a constant (
If f and g are two scalar point functions, then
5. If f and g are two scalar point functions, then
6. If f and g are two scalar point functions, then-
7.
8.
9. If then the vector is said to be irrotational.
10. grad uv = u grad v + v grad u
11.
12. curl (u
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-
Key takeaways-
- The directional derivative of ϕ in the direction l, m, n= l + m+
- The directional derivative of ϕ in the direction of =
Solenoidal field-
A solenoidal 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-
Key takeaways-
- In an solenoidal field for which , the vector F can always be expressed as the curl of a vector function V.
So that-
2. An irrotatonal field F is characterised by the following conditions-
1.
2. Circulation along every closed surface is zero.
3.
3. 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.
References
1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.
2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.
3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers
4. HK dass, engineering mathematics.