VECTOR CALCULUS

DERIVATIVE OF A SCALAR POINT FUNCTION AND A

VECTOR POINT FUNCTION

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Definition of Scalar Function• If every point P(x, y, z) of a region R of

space has associated with a scalar quantity (x, y, z), then (x, y, z) is a scalar function and a scalar field is said to exist in the region R. Example of scalar fields are temperature, pressure, etc.

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A scalar function defines a scalar field in a region (or on a surface or curve) eg.

1. A temperature function given for a plate which is heated by a candle below its midpoint. So the scalar field is all temperature on the plate

2. The pressure field of the earth’s temperature

3. (distance from origin to the point P in space)

222),,()( zyxzyxfPf

z p

f (P)

x

y

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Definition of Vector Function

• If every point P(x, y, z) of a region R has associated with a vector F(x, y, z), then F(x, y, z) is a vector function and a vector field is said to exist in the region R. Examples of vector fields are force, velocity, acceleration, etc.

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A vector function defines a vector field in region (or a surface or a curve)

E.g.(1) Figure below shows the ‘spin’ field of unit vectors

F = (-yi + xj)/(x2 + y2)1/2 in the plane. The field is not defined at the Origin.

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• E.g.(2) Vectors in the gravitational field F = -GM(xi + yj + zk)/(x2 + y2 + z2)3/2

• E.g.(3) Streamlines in a contracting channel. The water speeds up as the channel narrows and the velocity vectors increases in length

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Gradient of a scalar function (Grad)

• If a scalar function (x, y, z) is continuously differentiable with respect to its variables x, y, z throughout the region, then the gradient of , written grad , is defined as the vector

(vector differential operator)

gradx y z

i j k

x y z

i j k

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Directional Derivatives: For example let say (x, y) = 2x + y represents the temperature on a plate. Compute(i) the rate of change of in x-direction.(ii) the rate of change of in y-direction.(iii) the rate of change of in the direction of i + j.(iv) the maximum rate of change of and its direction.Solution

rate of change of gradx y

i j

(2 ) (2 )2

x y x y

x y

i j i j

(i) (2 ) 2 i i j i

(ii) (2 ) 1 j i j j

2 2

3ˆ ˆ(iii) Let (2 )

2 21 1

a i j i ja i j a a i j

a

2 2(iv) maximum rate of change 2 2 1 5i j

direction same or opposite direction as 2 i j8

ExampleFind grad for = 3x2+2y2+z2 at the point (1,2,3). Hence calculate (i) the directional derivative of at (1,2,3) in the direction of vector (2,2,1); (ii) the maximum rate of change of the function at (1,2,3) and its direction.Solution

gradx y z

i j k

2 2 2 2 2 2 2 2 2(3 2 ) (3 2 ) (3 2 )x y z x y z x y z

x y z

i j k

6 4 2x y z i j k

(1, 2,3) 6(1) 4(2) 2(3) 6 8 6 i j k i j k

(i) Let 2 2 a i j k

2 2 2

2 2 1 34ˆ(1,2,3) (6 8 6 ) (12 16 6)

3 32 2 1

i j ka i j k

2 2 2(ii) maximum rate of change (1,2,3) 6 8 6 6 8 6 136 i j k

direction 6 8 6 i j k9

Example

If f = x3y2z, find (i) f

(ii) a unit vector normal to the contour at the point (1,1,1).

(iii) the rate of change of f at (1,1,1) in the direction of i.

(iv) the rate of change of f at (1,1,1) in the direction of the unit vector

Solution2 2 3 3 2(i) 3 2f x y z x yz x y i j k

2 2 2

(1,1,1) 3 2 1(ii) unit vector normal (3 2 )

(1,1,1) 143 2 1

f

f

i j ki j k

(iii) (1,1,1) (3 2 ) 3f i i j k i1

ˆ(iv) (1,1,1) (3 2 ) ( )3

f n i j k i j k

1 6 6 3(3 2 1) 2 3

33 3

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Example A paraboloid has equation 2z = x2 + y2. Find the unit normal vector to the surface at the point (1, 3, 5). Hence obtain the equation of the normal and the tangent plane to the surface at that point.

Solution

2z = x2 + y2 2z x2 y2 = 0

The unit vector in the direction of grad is called the unit normal vector at P.

at point (1,3,5) normal vector 2 6 2 n i j k

2 2 2

2 6 2 2 6 2 3ˆunit normal vector

44 112 6 2

i j k i j k i j kn

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kjinz

yxz

y

yxz

x

yxz

)2()2()2(

vectornormal222222

kjin 222vectornormal yx

The equation of line through (1, 3, 5) in the direction of this normal is

and the equation of the tangent plane is

(-1)(x – 1) + (-3)(y – 3) + (1)(z – 5) = 0

x + 3y – z = 5

3ˆunit normal vector

11

i j kn

1 3 5

1 3 1

x y z

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Divergence of a Vector Field

• The operator applies to a vector function F(x, y, z) will give divergence of F written in short as div F. It is defined as

where F = F1i + F2j + F3k

• Note that the divergence of a vector is a scalar quantity.

31 21 2 3div

FF FF F F

x y z x y z

F i j k i j k

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• The implication of the divergence is easily understood by considering the behaviour of a fluid.

• The divergence (of a the vector field representing velocity) at a point in a fluid (liquid or gas) is a measure of the rate per unit volume at which the fluid is flowing away from a point (source) or flowing towards the point (sink).

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• Imagine that the vector field F below gives the velocity of some fluid flow. It appears that the fluid is exploding outward from the origin.

• This expansion of fluid flowing with velocity field F is captured by the divergence of F, which we denote div F. The divergence of the above vector field is positive since the flow is expanding.

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• In contrast, the below vector field represents fluid flowing so that it compresses as it moves toward the origin. Since this compression of fluid is the opposite of expansion, the divergence of this vector field is negative.

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• When the divergence is everywhere zero, the flow entering any element of the space is exactly balanced by the outflow. This implies that the lines of flow of the field F(r) where div F = 0 must either form closed curves or finished at boundaries or extend to infinity. Vectors satisfying this condition are sometimes termed solenoidal.

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Curl of a vector field

• The curl operator denoted by , acts on a vector and gives another vector as a result,

• The curl of a vector field is slightly more complicated than the divergence. It captures the idea of how a fluid may rotate. We define the curl of F, denoted F, by a vector that points along the axis of the rotation and whose length corresponds to the speed of the rotation.

1 2 3

1 2 3

curl F F Fx y z x y z

F F F

i j k

F F i j k i j k

.

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• For a three dimensional vector field, then you can imagine the curl something like this:

• The curl vector is perpendicular to the vector field. • Note that: to see where curl F should point, we use the

right hand rule.• When the curl of velocity of a fluid’s particle is zero,

there is no rotation of the particle and the motion is said to be curl-free or irrotational. When the curl is non-zero, the motion is rotational.

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Example(a) Find the divergence of the vector field F(x, y, z) = x2yi +

2y3zj + 3zk, in general term and at the point (1, 0, 3).(b) Find the value of a for which F = (2x2y+z2)i + (xy2 x2z)j +

(axyz 2x2y2)k is incompressible.Solution

31 21 2 3(a) div where

FF FF F F

x y z

F F F i j k

2div 2 6 3xy y z F2(1,0,3) 2(1)(0) 6(0 )(3) 3 3 F

(b) div 0 if incompressibleF

31 2 0FF F

x y z

2 2 2 2 2 2(2 ) ( ) ( 2 )

0x y z xy x z axyz x y

x y z

4 2 0

6 6

xy xy axy

axy xy a

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ExampleFind curl v for the following two-dimensional vector fields (i) v = xi + 2j (ii) v = -yi + xjIf v represents the surface velocity of the flow of water, describe the motion of a floating leaf.Solution

(0) ( ) (0) ( ) ( ) ( )(1 ( 1)) 2

x y x y

y z x z x y

i j k k k

(0) (2) (0) ( ) ( ) ( )x z x

y z x z x y

i j k 0

= no rotation.v 0

=+ve

counterclockwise rotation.

v

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0

curl(ii)

xyzyx

kji

vv

02

curl(i)

xzyx

kji

vv

ExampleGiven F = (xy xz)i + 3x2j + yzk, find curl F at the origin and at the point P = (1,2,3).Solution

2

curl

3

x y z

xy xz x yz

i j k

F F

2 2( ) (3 ) ( ) ( ) (3 ) ( )yz x yz xy xz x xy xz

y z x z x y

i j k

22

kjikji xxzxxxz 5)6()0()0(

0kjiF )0(500)0,0,0(

kjiF 53)3,2,1(