1st year maths unit 5(a) Vector calcus

            Mathematics – II (203191152)

Unit – 5(a) Vector Calculus (Lecture Note)

Scalar point function:

• If to each point (x, y, z ) of a region R in space there corresponds a number or

a scalar f = f( x, y, z) then, f is called a scalar point function and R is called a

scalar field.

• For example

(i) the temperature field in a body.

(ii) The pressure field of the air in the earth’s atmosphere.

(iii)The density of a body.

These quantities take different values at different points.

Note: A scalar field which is independent of time is called a stationary or steady-state

scalar field.

Vector point function:

If to each point (x, y, z) of a region R in space there corresponds a vector

v (x, y, z) = v1i + v2 j + v3 k then, v is called a vector point function and R is called

a vector field.

For example

(i) the velocity of a moving fluid at any instant.

(ii) The gravitational force.

(iii)The electric and magnetic field intensity.

Note: A vector field which is independent of time is called a stationary or steady-state

vector field.

Vector differential operator -

The vector differential operator is denoted by ∇ (del or nabla) and is defined as

𝛻= 𝑖̂𝜕𝜕𝑥 + 𝑗̂ 𝜕𝜕𝑦+ 𝑘̂

𝜕𝜕𝑧

Gradient of a scalar field:- For a given scalar function ∅ (𝑥,𝑦,𝑧 ) the gradient of ∅ is denoted by 𝑔𝑟𝑎𝑑 ∅ or ∇∅ is defined as

𝛻∅ = 𝑖̂𝜕𝜙𝜕𝑥 + 𝑗̂ 𝜕𝜙𝜕𝑦+ 𝑘̂

𝜕𝜙𝜕𝑧

Example: Find the gradient of ∅ = 3x2 y − y3z2 at the point (1, −2, 1) .

Sol: ∇∅ = î∂∅∂x + ĵ∂∅∂y + k̂∂∅∂z

= î(6xy) + ĵ(3x2 − 3y2z2) + k̂ (−2y3z)

At the point (1, −2,1 )

∇∅ = −12 î− 9 ĵ− 16 k̂ .

Example: Evaluate ∇er2, where r2 = x2 + y2 + z2

Solution:

r2 = x2 + y2 + z2

Differentiating r partially with respect to x, y, z2r∂r∂x = 2x 

⟹∂r/∂x =xr2r∂r/∂y = 2y 

⟹∂r/∂y =y/r2r=∂r/∂z = 2z ⟹∂r/∂z =z

Example: Find a unit normal vector to the surface 𝒙𝟑 + 𝒚𝟑 + 𝟑𝒙𝒚𝒛 = 𝟑 at the point (𝟏,𝟐,−𝟏)

Sol. ∅(𝑥,𝑦,𝑧)= 𝑥3 + 𝑦3 + 3𝑥𝑦𝑧− 3=0

𝛻∅ = 𝑖̂ 𝜕∅𝜕𝑥+ 𝑗̂ 𝜕∅𝜕𝑦+ 𝑘̂

𝜕∅𝜕𝑧 =𝑖̂ (3𝑥2 + 3𝑦𝑧 )+𝑗̂ (3𝑦2+ 3𝑥𝑧 )+𝑘̂ (3𝑥𝑦)

At the point (1,2,−1) 𝛻∅ = −3𝑖̂+ 9𝑗̂ + 6𝑘̂

𝑛̂ =𝛻∅|𝛻∅|=−3𝑖̂ + 9𝑗̂ +6𝑘̂

√126

Example: Find a unit normal vector to the surface 𝒙𝟐𝒚 + 𝟑𝒙𝒛𝟐 = 𝟖 at the point (𝟏,𝟎,𝟐) 𝛻∅ = 𝑖̂ 𝜕∅𝜕𝑥+ 𝑗̂ 𝜕∅𝜕𝑦+ 𝑘̂

𝜕∅𝜕𝑧𝛻∅=(2𝑥𝑦+3𝑧2)𝑖̂+(𝑥2)𝑗̂+6(𝑥𝑧)𝑘̂

𝛻∅(1,0,2)=12𝑖̂+𝑗̂+12𝑘̂

Directional Derivative:-

The directional derivative of scalar point function ∅ (𝑥, 𝑦, 𝑧) in the direction of vector 𝑎̂ , is the component of ∇∅ in the direction of 𝑎̂ .

If 𝑎̂ is the unit vector in the direction of a, then the directional derivative of ∅ in the direction of a is

𝐷 ∅ = 𝛻∅∗𝑎̂

Examples: Find the directional derivative of ∅ (𝒙,𝒚,𝒛) = 𝒙𝟑 − 𝒙𝒚𝟐 − 𝑧 at point 𝑷( 𝟏,𝟏,𝟎) in the direction of 𝒗= 𝟐𝒊̂−𝟑𝒋̂+𝟔𝒌̂

Sol. Here,∅ (𝑥,𝑦,𝑧) = 𝑥3 − 𝑥𝑦2 − 𝑧 𝛻∅ = 𝑖̂ 𝜕∅𝜕𝑥+ 𝑗̂ 𝜕∅𝜕𝑦+ 𝑘̂

𝜕∅𝜕𝑧 = 𝑖̂(3𝑥2 − 𝑦2) +𝑗̂ (−2𝑥𝑦) +𝑘̂ (−1) 𝐴𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 (1,1,0) 𝛻∅ = 2𝑖̂−2𝑗̂−𝑘̂

The direction derivative of ∅ at point 𝑃 (1,1,0) in the direction of is v is 𝐷𝜙 = 𝛻∅ ∗𝑣̂= 𝛻∅ ∗ 𝑣|𝑣| ( 2𝑖̂−2𝑗̂−𝑘̂) ∗( 2𝑖̂−3𝑗̂+6𝑘̂)/√49 =47.

Example: Find the directional derivative of 𝝓=𝟔𝒙𝟐𝒚+𝟐𝟒𝒚𝟐𝒛−𝟖𝒛𝟐𝒙 at (𝟏,𝟏,𝟏) in the direction of 𝒗= (𝟐𝒊−𝟐𝒋+𝒌) . Hence, find the maximum value.

Solution: 𝑔𝑟𝑎𝑑 𝜙=𝑖𝜕𝜕𝑥(6𝑥2𝑦+24𝑦2𝑧−8𝑧2𝑥 )+𝑗𝜕𝜕𝑦(6𝑥2𝑦+24𝑦2𝑧−8𝑧2𝑥 )+𝑘𝜕𝜕𝑧(6𝑥2𝑦+24𝑦2𝑧−8𝑧2𝑥 )

𝛻𝜙=(12𝑥𝑦−8𝑧2)𝑖+(6𝑥2+48𝑦𝑧)𝑗+(24𝑦2−16𝑧𝑥)𝑘 𝛻𝜙1,1,1=4𝑖+54𝑗+8𝑘

Directional derivative in the direction of 𝑣= (2𝑖−2𝑗+𝑘) at the point (1,1,1)

=𝛻𝜙𝑣|𝑣| =(4𝑖+54𝑗+8𝑘)2𝑖−2𝑗+𝑘|2𝑖−2𝑗+𝑘| =(4𝑖+54𝑗+8𝑘)(2𝑖−2𝑗+𝑘)√4+4+1

=8−108+83=−923

Maximum value of directional derivative =|𝛻𝜙|

Divergence of a vector function:

Let 𝐹 = 𝐹1𝑖̂+𝐹2 𝑗̂ +𝐹3𝑘̂ be a vector function then, divergence of F is

div F OR 𝛻𝐹 =(𝑖̂𝜕𝜕𝑥+𝑗̂ 𝜕𝜕𝑦+𝑘̂ 𝜕𝜕𝑧)(𝐹1𝑖̂+𝐹2 𝑗̂ +𝐹3𝑘̂ )=𝜕𝐹1𝜕𝑥+𝜕𝐹2𝜕𝑦+ 𝜕𝐹3𝜕𝑧

Note: 1) If 𝛻.𝐹 = 0 then, the vector function F is called solenoidal or incompressible

2) In hydrodynamics (the study of fluid motion), a velocity field that is divergence free is called incompressible.

3) In the study of electricity and magnetism, a vector field that is divergence free is called solenoidal.

Example: If 𝑭 = 𝒙𝟐 𝒛 𝒊̂−𝟐𝒚𝟑 𝒛𝟑 𝒋̂+ 𝒙𝒚𝟐𝒛 𝒌̂ then, find divergence of F at( 𝟏,−𝟏,𝟏)

Sol. Here, 𝐹 = 𝑥2 𝑧 𝑖̂−2𝑦3 𝑧3 𝑗̂+ 𝑥𝑦2𝑧 𝑘̂

𝛻𝐹 =𝜕𝐹1𝜕𝑥+𝜕𝐹2𝜕𝑦+ 𝜕𝐹3𝜕𝑧

=𝜕(𝑥2 𝑧)𝜕𝑥 +𝜕(−2𝑦3 𝑧3)𝜕𝑦+𝜕(𝑥𝑦2𝑧)𝜕𝑧 = 2𝑥𝑧−6𝑦2 𝑧3+𝑥𝑦2

At ( 1,−1,1) 𝛻𝐹 = −3

Example: Show that 𝐀̅=𝟑𝒚𝟒𝒛𝟐𝒊+𝟒𝒙𝟑𝒛𝟐𝒋−𝟑𝒙𝟐𝒚𝟐𝒌 is a solenoidal.

Solution: 𝛻A̅ =𝜕𝐴1𝜕𝑥+𝜕𝐴2𝜕𝑦+ 𝜕𝐴3𝜕𝑧 =𝜕(3𝑦4𝑧2)𝜕𝑥+𝜕(4𝑥3𝑧2)𝜕𝑦+ 𝜕(−3𝑥2𝑦2)𝜕𝑧=0

Hence Given function is solenoidal.

Example: Determine the constant a such that 𝑨=(𝒂𝒙𝟐𝒚+𝒚𝒛)𝒊+(𝒙𝒚𝟐−𝒙𝒛𝟐)𝒋+(𝟐𝒙𝒚𝒛−𝟐𝒙𝟐𝒚𝟐)𝒌 is solenoidal

Curl of a vector function:-

Let 𝐹 = 𝐹1𝑖̂+𝐹2 𝑗̂ +𝐹3𝑘̂ be a vector function then, curl of F is

curl F or 𝛻×𝐹 =[𝑖̂𝑗̂𝑘̂

𝜕𝜕𝑥𝜕𝜕𝑦𝜕𝜕𝑧𝐹1𝐹2𝐹3]

Note:- If 𝛻×𝐹 = 0 then, the vector function F is called Irrotational or conservative.

Example: If 𝑭 = 𝒙𝒛𝟑 𝒊̂−𝟐𝒙𝟐𝒚𝒛𝒋̂ +𝟐𝒚𝒛𝟒 𝒌̂then , find curl of F at (𝟏,−𝟏,𝟏)

Sol. Here, 𝐹 = 𝑥𝑧3 𝑖̂−2𝑥2𝑦𝑧𝑗̂ +2𝑦𝑧4 𝑘̂

𝛻×𝐹 = [ 𝑖̂𝑗̂𝑘̂

𝜕𝜕𝑥𝜕𝜕𝑦𝜕𝜕𝑧𝑥𝑧3−2𝑥2𝑦𝑧2𝑦𝑧4] = 𝑖̂(2𝑧4+2𝑥2𝑦)+𝑗̂ (3𝑥𝑧2)+𝑘̂ (−4𝑥𝑦𝑧)

At point (1, −1,1)

𝛻×𝐹 = 3𝑗̂+4𝑘̂.

Example: Show that 𝒓 =𝒙 𝒊̂+𝒚𝒋̂ +𝒛 𝒌̂tis Irrotational.

=𝑖̂(𝑒𝑥𝑦𝑧(𝑥𝑧)−𝑒𝑥𝑦𝑧(𝑥𝑦))−𝑗̂(𝑒𝑥𝑦𝑧(𝑦𝑧)−𝑒𝑥𝑦𝑧(𝑥𝑦))+𝑘̂(𝑒𝑥𝑦𝑧(𝑦𝑧)−𝑒𝑥𝑦𝑧(𝑥𝑧))

 =𝑒𝑥𝑦𝑧(𝑖̂(𝑥𝑧−𝑥𝑦)−𝑗̂(𝑦𝑧−𝑥𝑦)+𝑘̂(𝑦𝑧−𝑥𝑧))

At (1,2,3) =𝑒6(𝑖̂−4𝑗̂+3𝑘̂)

Example: 𝐴=2𝑋𝑌𝑍2𝑖+[𝑥2𝑧2+𝑧𝑐𝑜𝑠 𝑦𝑧 ]𝑗+(2𝑥2𝑦𝑧+𝑦𝑐𝑜𝑠 𝑦𝑧 )𝑘

Find 𝑐𝑢𝑟𝑙 𝐴