Advanced level vector
Vector triple product:
1. 
(
) = (.) – (.)
2. ( ×) × = – ×( ×) = – [(.) – (.)] = (.) – (.)
3. (×).(×)×(×) = (.×)²
Prove: Let, × = 
so, (×)×(×) = ×(×) = (.) – (.)
= (×.) – (×.)
= (.×) – (.×)
= ( .×) – 0 [as × = 0]
(×).(×)×(×) = (×).(.×) = (.×
Reciprocal sets of vectors: Let us consider, 
, 
and 
are three sets of vectors with their reciprocal sets of vectors 
, 
and 
respectively then, 
.
.
.
. 
. .
0
and 
[the volume of the body is v = 
]
Differential Geometry:
Let us consider, C is the space curve is defined by the function 
. The vector in the direction of tangent to C is 
.
If s is the arc length measured from a fixed point on C represents the scalar value of u, then 
= 
is unit tangent vector to C.
If changes with respect to s, then 
represents the value of curvature of C. The direction of is the normal at that point on curve C. So, = 
where 
is the unit normal vector to the curve and K = curvature .The radius of curvature is 
= 
.
A unit vector 
perpendicular to the plane of 
and is = × is called binormal to the curve. , and represent trihedral or triad ordinate system at any point of curve C.
Example: A space curve is represented by x = 4Cost, y = 4Sint and z = 3t. Calculate (i) unit tangent vector and (ii) unit normal vector.
The position vector for any point on the curve is 
= 4cost
+ 4sin t
+ 3t
= – 4sint + 4cost + 3
= 
= 
= 
= 5
The unit tangent vector = = .
= – 
sint + cost + 
= – cost – sin t
= 
= – 
cost – sin t
= 
= K.1
[where is the unit normal vector of the curve]
The curvature is K = +
= 
= – cost – sin t.
Normal and tangent vector at any point on the surface:
Let us consider, a surface is represented by the equation 
= (u, v) . A curve can be represented for a fixed value of u = u0as =(u0, v . Similarly for u = u1, the curve is represented =
. Therefore, =(u, v) represents a curve which moves in space and generates a surface s.
(Keeping v = v0 constant) represents a tangent vector at point P to the curve v = v0 at point P. similarly,
(Keeping u = u0 constant) at point P represents a tangent vector to the curve u = u0 at point P.
So, 
× 
is the vector normal to the surface at P.
Example: The equation of a surface is represented by z = x2 + y2. Find the equation for the tangent plane at pointP (2, -1, 2).
Let us consider, x = u, y = v and z = u2 + v2 where (x, y, z) represents a point P on the surface. The position vector of a point P (x, y, z)on the surface is represented by 
= u
+ v
+ (u² + v²)
.
Example: The equation of a surface is represented by z = x2+ y2. Find the equation for the tangent plane at pointP (2,-1,2).
Let us consider, x =u, y = v and z = u2+ v2where (x,y,z) represents a point P on the surface. The position vector of a point P (x,y,z)on the surface is represented by u+v+(u²+v²).
Therefore, 
+ 2u =4(where u = 2)and 
=+2v = -2 (where v = -1)
The normal vector to the surface at point P is 
= × =(+4)×(-2) = -4 + 2+ .
= × = 
= -4 +2+
The position vector of point P(2, -1, 2) is 
= 2– +2 . Let us consider, another point Q(x,y,z) is on a plane which is the tangent plane at point P (2, -1, 2).
The position vector of point Q on the tangent plane is = x+ y +z . The vector
= (– ) = [(x+y+z) – (2 – + 2)] = (x – 2) + (y+1) + (z-2) represents a vector on the tangent plane. As is the normal to the surface at point P(2, -1, 2), then is also perpendicular to . Therefore, . = 0.
So, [(x-2) + (y+1) +(z-2) ].(-4+2+) = 0
Or, -4(x-2) + 2(y+1) + (z-2) = 0
Or, -4x +8 + 2y + 2 + z – 2 = 0
-4x + 2y + z = – 8
This represents the equation of the tangent plane at point P on the surface.
Vector differential operator: This is called as del and written as ∇.∇ is also called as nabla.
Where ∇ = 
+ 
+ 
. This vector operator possesses the properties analogous to that of vector.
The gradient: ∇
= ( + + ) = 
+ 
+ 
( where is the differentiable scaler field.
The component of ∇, in the direction of is ∇. is called the directional derivative of in the direction of .
The divergence: Let us consider, 
= 
+ 
+ 
is the differentiable vector field. The divergence of is ∇.=( + + ). = ( + + ) = 
+ 
+ 
.
The curl: Let us consider, = + is the differentiable vector field. Then the curl or rotation of is ∇×. Here, ∇× = 
= ( 
– 
) + ( 
– 
) + ( 
– 
).
Prove that ∇ is a vector perpendicular to the surface (x, y, z) = c where c is a constant.
Let us consider, =x +y + z be the position vector to any point P (x, y, z) on the surface. Then, 
=dx + dy + dz lies in the tangent plane to the surface at point P.
But d = dx + dy + dz = 0 [as (x, y, z) = constant then, d = 0]
Or, ( + + )
Or, ∇. = 0
Hence, ∇ is perpendicular to dr or to the surface.
Example: 1.Find unit normal vector to the surface x2y + 2xz = 4 at point (1, 1, 2)
The normal vector to the surface x2y + 2xz = 4 is ∇ = ∇( x2y + 2xz) = (2xy+2z) + 
+ 2x +6 + + 2 at (1, 1, 2)
Unit normal to the surface = 
= 
+ 
+
Another unit normal is – – – opposite direction of the above vector.
Angular speed: A rigid body rotates about an axis through a point O with angular speed 
. The linear velocity of a point P of the body with position vector is 
= × . [direction of is that in which a right-hand screw would advance under the given rotation].
The point P rotates in a circle of radius r sin 
. The magnitude of linear velocity is v = 
r sin
∴ = ×
Conservative vector field: If , ∇ × = 0 then is conservative vector field.
If = ∇ 
then is known as scalar potential.
Example: 1. Prove 
= (2xy + 
) + + 3x
is conservative.
= =[ – ] + [ – ] + [ – ] = [0 – 0] + [3z2 – 3z2] + [2x – 2x] = 0
∇ × = 
= [
] + [
– 
] + [ 
] = [0-0] + [
] + [2x-2x] = 0
- Find work done to move a particle from (2, -1, 1) to (1, 1,2) in this field. As W =
. = [(2xy + ) = + 3x].[dx + dy + dz]
W = (2xy +)dx + dy +3xdz = d(y + x) = 
= [1×1 + 8 ×1] – [4(-1) + 1×2] = 9 + 2 = 11
Find scaler potential in previous problem.
3) Find scaler potential in previous problem.
= ∇ = + + = (2xy + ) + + 3x
Then, = 2xy + so, = ∫( 2xy + )dx = y + x + f (y, z)
And = so, = ∫ dy = y + g(x, z)
And = 3x so, = ∫3x dz = x+ h(x, y)
As is a particular function of x, y, z so f(y,z) = 0, g(x, z) = x and h(x, y) = y
∴ = y + x
= Work = (1,1,2) – (2, -1, 1)
- Find the work done in moving a particle in the force field given by = 3zy – 5z + 10x along the curve x = t2+ 1, y = 2t2and z = t3 from t = 1 to t = 2.
The work done W = ∫dw = ∫. = ∫( 3xy – 5z + 10x ). ( dx + dy + dz )
= ∫ 3xydx – 5zdy + 10xdz)
= 
[3(
+1). 2d( +1 ) – 5
d(2) + 10 (+1)d ()
= [3(+1). 2.2tdt – 5 . 4tdt + 10(+1) . 3dt]
= [12
dt +12dt – 20
dt +30dt]
= ( 12 + 10 + 12 + 30)dt
= [2
+ 2 + 3 + 10]
= [2(
-1) + 2(
-1) + 3(
-1) + 10(
-1)
= 126 + 62 + 45 + 70 = 303.
- Calculate the work done to move a particle through a curve y = 2x2 from (0, 0) to (1, 2) in xy plane by the force = 3xy–
.
The work done is W= 
.where = dx +dy [z = 0 here]
W =
dw = (3xy – ). (dx+ dy)
= 
3x (2)dx –
d (2)
= 6
dx – . 4xdx
= 6dx – 16
dx
= – 
- If
and 
are two vectors, then findthe component of perpendicular to .
Let us consider, the vector perpendicular to and is × = AB sin 
where is unit vector perpendicular to the plane containing and i.e. – Z axis here. So,
= –
∴ ×(×) = -AB sin B × = A
sin
∴ Asin = ×
where Asin is the component of perpendicular to .
Gauss’s divergence theorem:
.ds = 
(div) dv
Volume integral of divergence of a vector in a vector field through a volume is equal to the surface integral of the normal component of over the surface which enclose the volume.
Stoke’s law:
. d
= 
curt.
Line integral of the tangential comp. of a vector round any closed path is equal to the normal surface integral of the vector curl over the surface having the path as its boundary.
A force is defined as = 
. Calculate the work done along a semi-circular path with the origin as centre connecting the points (-1, 0) and (1, 0).
(ii) along a straight line joining (-1, 0), (0, 1) and (0, 1), (1, 0). Comparing the two values, comment on the nature of the force.
Here .
= 
Let us consider a = a cos and y = a sin . So, dx = – a sin d and dy = a cos d
Therefore, work done W . = 
.(dx + dy ) = 
= 
=
sin 2d = 0
(ii) (a) Along a straight line joining (-1, 0) (0, 1): The equation of the straight line joining (-1, 0), (0, 1) is x + 1 = y or, dx = dy
Therefore, work done W = .= . (dx + dy ) = = 
= 
∴ .= 
= . 2 [
= 
(b) Along the straight line joining (0, 1), (1, 0):The equation of the straight line is given by y = 1 – x or, dy = – dx
Therefore, work done W = .= . (dx + dy ) = = 
=
∴ .= – 
= – .2
= – 
∴ . joining the two points (-1, 0) and (1, 0) via the above straight lines = – = 0
The force is conservative.
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