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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
B. Tech. (I Sem.), 2020- 21
MATHEMATICS-I (KAS-103T)
MODULE 5 (Vector Calculus)
SYLLABUS: Vector differentiation: Gradient, Curl and Divergence and their Physical
interpretation, Directional derivatives, Tangent and Normal planes. Vector Integration: Line
integral, Surface integral, Volume integral, Gauss’s Divergence theorem, Green’s theorem,
Stoke’s theorem (without proof) and their applications.
Course Outcome:
Remember the concept of vector and apply for directional derivatives, tangent and
normal planes. Also for solving line, surface and volume integrals.
Application in Engineering:
Vector calculus plays an important role in computational fluid dynamics or
computational electrodynamics simulation, differential geometry and in the study
of partial differential equations also useful for dealing with large-scale behavior such as
atmospheric storms or deep-sea ocean currents. It is used extensively in physics
and engineering, especially in the description of electromagnetic fields, gravitational
fields and fluid flow.
CONTENTS
Topic Page No.
- 1.1 Gradient 1. Vector differentiation (3-22)
- 1.1.1 Geometrical Interpretation of Gradient
- 1.1.2 Properties of Gradient
- 1.2 Directional derivative
- 1.3 Divergence of a vector point function
- 1.3.1 Physical Interpretation of Divergence
- 1.4 Curl of a vector point function
- 1.4.1 Physical Interpretation of Curl
- 1.5 Tangent and Normal Planes
- 2.1 Line Integral 2. Vector Integration (23- 69 )
- 2.1.1 Work done by a force
- 2.1.2 Circulation
- 2.2 Surface integral
- 2.3 Volume integral
- 2.4 Gauss’s Divergence theorem (without proof)
- 2.4.1 Application of Gauss’s Divergence theorem
- 2.5 Green’s theorem (without proof)
- 2.5.1 Application of Green’s theorem
- 2.6 Stoke’s theorem (without proof)
- 2.6.1 Application of Stoke’s theorem
- E-resources
Vector Differential Operator Del ( ∇ ): It is defined as:
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
Gradient of a scalar function: Let ∅(𝑥, 𝑦, 𝑧) be a scalar function, then the vector
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
is called the gradient of a scalar function ∅.
Thus, 𝑔𝑟𝑎𝑑 ∅ = 𝛻∅
1.1.1 Geometrical Interpretation of Gradient: If a surface ∅(𝑥, 𝑦, 𝑧) = 𝑐
passes through a point P. The value of function at each point of the surface is the same as at P.
Then such a surface is called a level surface through P.
Example. If ø(x,y,z) represent potential at the P. Then equipotential surface ø(x,y,z) = c is a level
surface.
Note: Two level surfaces can’t intersect.
Let the level surface passes through P at which the value of function is ø.
Consider another level surface passing through Q , Where the value of function ø + dø.
Let 𝑟⃗ 𝑎𝑛𝑑 𝑟⃗ + 𝛿𝑟⃗ be the position vector of P and Q then 𝑃𝑄
𝜕𝜙
𝜕𝑥
𝜕𝜙
𝜕𝑦
𝜕𝜙
𝜕𝑧
𝜕𝜙
𝜕𝑥
𝜕𝜙
𝜕𝑦
𝜕𝜙
𝜕𝑧
If Q lies on the level surface of P , then 𝑑𝜙 = 0
From equation (1), we get
∇𝜙. 𝑑𝑟⃗ = 0 , then ∇𝜙 ⊥ 𝑡𝑜 𝑑𝑟⃗ (𝑡𝑎𝑛𝑔𝑒𝑛𝑡)
Hence ∇𝜙 is the Normal to the surface𝜙
1 .1.2 Properties of gradient
(1) If 𝜙 is a constant scalar point function, then ∇𝜙 = 0
(2) If 𝜙
1
and 𝜙
2
are two scalar point functions, then
(a) ∇
1
2
1
2
(b) ∇
1
1
2
2
1
1
2
2
, where 𝑐
1
2
are constants.
(c) ∇
1
2
1
2
2
1
(d) ∇ (
𝜙 1
𝜙 2
𝜙 2
∇𝜙 1
−𝜙 1
∇𝜙 2
𝜙 2
2
2
Example 1. Find grad ø, when ø is given by ∅ = 3 𝑥
2
3
2
at the point ( 1 , − 2 , − 1 ).
Solution. Here 𝜙 = 3 𝑥
2
3
2
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
2
3
2
𝜕
𝜕𝑥
2
3
2
𝜕
𝜕𝑦
2
3
2
𝜕
𝜕𝑧
2
3
2
2
2
2
3
]
( 1 ,− 2 ,− 1
)
Example 2. What is the greatest rate of increase of 𝑢 = 𝑥𝑦𝑧
2
at the point ( 1 , 0 , 3 )?
Solution. Here, we have
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
2
2
2
Now, 𝑔𝑟𝑎𝑑 𝑢] ( 1 , 0 , 3 )
The greatest rate of increase of 𝑢 at the point
Example 3. Find a unit vector normal to the surface 𝑦𝑥
2
Solution. Let ∅ = 𝑦𝑥
2
2
2
Now, 𝑔𝑟𝑎𝑑 ∅] ( 2 ,− 2 , 3
)
and |𝑔𝑟𝑎𝑑 ∅| = √
Now, 𝑔𝑟𝑎𝑑 𝑢. (𝑔𝑟𝑎𝑑 𝑣 × 𝑔𝑟𝑎𝑑 𝑤) = |
Hence, 𝑔𝑟𝑎𝑑 𝑢, 𝑔𝑟𝑎𝑑 𝑣 𝑎𝑛𝑑 𝑔𝑟𝑎𝑑 𝑤 𝑎𝑟𝑒 𝑐𝑜𝑝𝑙𝑎𝑛𝑎𝑟 𝑣𝑒𝑐𝑡𝑜𝑟𝑠.
Exercise
1. Show that ∇
= 𝑎⃗ where 𝑟⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘
and 𝑎⃗ is a constant vector.
2. Find a unit vector normal to the surface 𝑥
3
3
3. Find a unit vector normal to the surface 𝑥𝑦
3
2
= 4 at the point (− 1 , − 1 , 2 ).
4. Find a unit vector normal to the surface 𝑧
2
2
2
at the point
5. Calculate the angle between the normal to the surface 𝑥𝑦 = 𝑧
2
at the points
Answers
1
√ 14
1
√ 11
1
√
5
) 5. cos
− 1
1
√ 22
1. 2 Directional derivative
Directional derivative of a scalar field 𝑓 at a point 𝑃
in the direction of unit vector 𝑎̂ is
given by
𝑑𝑓
𝑑𝑠
Example 1: Find the directional derivative of ∅ =
2
2
2
−
1
2
at the point 𝑃
in the
direction of the vector 𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘
Solution. Here, ∅ =
2
2
2
−
1
2
⇒ 𝑔𝑟𝑎𝑑 ∅ = −
𝑥𝑖̂ +𝑦𝑗̂ +𝑧𝑘
̂
( 𝑥
2
+𝑦
2
+𝑧
2
)
3
2
Let 𝑎̂ be the unit in the given direction, then 𝑎̂ =
𝑦𝑧𝑖̂ +𝑧𝑥𝑗̂ +𝑥𝑦𝑘
̂
√𝑥
2
+𝑦
2
+𝑧
2
𝑔𝑟𝑎𝑑 ∅]
( 3 , 1 , 2
)
3 𝑖̂ +𝑗̂ + 2 𝑘
̂
14 √ 14
and 𝑎̂ ]
( 3 , 1 , 2
)
2 𝑖̂ + 6 𝑗̂+ 3 𝑘
̂
7
Directional derivative =
𝑑∅
𝑑𝑠
6 + 6 + 6
98 √
14
𝟗
𝟒𝟗 √
𝟏𝟒
Example 2: Find the directional derivative of ∅ = 5 𝑥
2
2
5
2
2
𝑥 at the point 𝑃
in the direction of the line
𝑥− 1
2
𝑦− 3
− 2
𝑧
1
Solution. Here, ∅ = 5 𝑥
2
2
5
2
2
5
2
2
2
2
]
( 1 , 1 , 1
)
25
2
Let 𝑎⃗ be a vector in the direction of the line
𝑥− 1
2
𝑦− 3
− 2
𝑧
1
, then
2 𝑖̂ − 2 𝑗̂ +𝑘
̂
3
and 𝑎̂
]
( 1 , 1 , 1 )
2 𝑖̂ − 2 𝑗̂ +𝑘
̂
3
Directional derivative =
𝑑∅
𝑑𝑠
25
3
10
3
35
6
Example 3: Find the directional derivative of ∅ = 𝑥
2
2
2
at the point 𝑃
in the
direction of the line 𝑃𝑄 where 𝑄 is the point
In what direction it will be maximum? Find also the magnitude of this maximum.
Solution. Here, ∅ = 𝑥
2
2
2
𝛻∅]
( 1 , 2 , 3
)
and 𝑃𝑄
Let 𝑎̂ be a unit vector in the direction of 𝑃𝑄
, then
4 𝑖̂ − 2 𝑗̂ +𝑘
̂
√
21
and 𝑎̂
]
( 1 , 2 , 3 )
4 𝑖̂ − 2 𝑗̂+𝑘
̂
√
21
Directional derivative =
𝑑∅
𝑑𝑠
8 + 8 + 12
√ 21
28
√ 21
Directional derivative will be maximum in the direction of normal to the given surface i.e., in the
direction of 𝛻∅.
The maximum value of this directional derivative = |𝛻∅| = 2 √
Example 4 : Find the directional derivative of ∅ = 𝑥𝑦
2
3
at the point 𝑃
in the
direction of the normal to the surface 𝑥 log 𝑧 − 𝑦
2
Solution. Here, ∅ = 𝑥𝑦
2
3
2
3
2
Let ∅
1
≡ 𝑥 log 𝑧 − 𝑦
2
1
= log 𝑧 𝑖̂ − 2 𝑦𝑗̂ +
𝑥
𝑧
𝛻∅]
( 2 ,− 1 , 1 )
and 𝛻∅
1
]
( 2 ,− 1 , 1 )
Let 𝑎̂ be a unit vector in the direction of the normal to the surface ∅
1
at the point
, then
𝛻∅
1
]
( 2 ,− 1 , 1 )
|𝛻∅
1
]
( 2 ,− 1 , 1 )
|
2 𝑗̂ + 2 𝑘
̂
2 √
2
Required Directional derivative = (𝛻∅). 𝑎̂
2 𝑗̂+ 2 𝑘
̂
2 √
2
− 6 − 6
2 √ 2
6
√ 2
Exercise
1. If the directional derivative of ∅ = 𝑎𝑥
2
2
2
𝑥 at the point
has maximum
magnitude 15 in the direction parallel to the line
𝑥− 1
2
𝑦− 3
− 2
𝑧
1
find the values of 𝑎, 𝑏 𝑎𝑛𝑑 𝑐.
2. For the function ∅ =
𝑦
𝑥
2
+𝑦
2
, find the magnitude of the directional derivative making an angle
30° with the positive 𝑥 − 𝑎𝑥𝑖𝑠 at the point
3. Find the values of constants 𝑎, 𝑏, 𝑐 so that the maximum value of the directional derivative of
2
2
3
at
has a magnitude 64 in the direction parallel to 𝑧 −
4. Find the directional derivative of 𝑓
2
2
2
at the point 𝑃
in the
direction of the vector 𝑎⃗ = 𝑖̂ − 2 𝑘
5. Find the directional derivative of Ψ
𝑥+ 5 𝑦− 13 𝑧
at the point
in the direction
towards the point
Answers
20
9
55
9
50
9
1
2
4
√ 5
− 28
Gradient in Polar Form
Sometimes the surface is given in the form of ∅ = 𝑓(𝑟, 𝜃). We can find 𝑔𝑟𝑎𝑑 ∅ directly without
changing into cartesian form.
Let 𝑢 𝑟
𝜃
be the unit vectors along and perpendicular to 𝑟⃗.
Directional derivative along
𝑟
𝑟
Thus
𝜕∅
𝜕𝑠
𝜕∅
𝜕𝑟
𝑟
----------------- (1) [𝑑𝑠 = 𝑑𝑟 in the direction of 𝑟⃗ ]
Directional derivative along (𝑢
𝜃
𝜃
Thus
𝜕∅
𝜕𝑠
𝜕∅
𝑟𝑑𝜃
𝜃
- ---------------- (2) [𝑑𝑠 = 𝑟𝑑𝜃 in the direction of 𝑟⃗ ]
Now, 𝛻∅ =
𝑟
𝑟
𝜃
𝜃
Or 𝛻∅ =
𝜕∅
𝜕𝑟
𝑟
𝜕∅
𝑟𝑑𝜃
𝜃
[ from equation (1) and (2)]
Example 1. Find (i) 𝛻 (
𝑒
𝜇𝑟
𝑟
) (ii) 𝛻
2
(iii) 𝛻 log 𝑟
𝑛
(iv) 𝑔𝑟𝑎𝑑𝑓
′
× 𝑟⃗
Solution. (i) 𝛻 (
𝑒
𝜇𝑟
𝑟
𝜕
𝜕𝑟
𝑒
𝜇𝑟
𝑟
) 𝑟̂ = [
𝑟𝜇𝑒
𝜇𝑟
−𝑒
𝜇𝑟
𝑟
2
]
𝑟⃗
𝑟
𝑒
𝜇𝑟
( 𝜇𝑟− 1
) 𝑟⃗
𝑟
3
(ii) 𝛻
2
2
𝜕
𝜕𝑟
2
𝑟⃗
𝑟
(iii) 𝛻 log 𝑟
𝑛
𝜕
𝜕𝑟
log 𝑟
𝑛
1
𝑟
𝑛
𝑛− 1
𝑟⃗
𝑟
𝑛𝑟⃗
𝑟
2
(iv) 𝑔𝑟𝑎𝑑𝑓
′
× 𝑟⃗ =
[
𝜕
𝜕𝑟
′
]
× 𝑟⃗ = 𝑓
′′
𝑟⃗
𝑟
× 𝑟⃗ = 0
Example 2. If 𝑉
1
2
log
2
2
prove that 𝑔𝑟𝑎𝑑𝑉 =
𝑟⃗ −𝑘
⃗⃗ ( 𝑘
⃗⃗
.𝑟⃗
)
{𝑟⃗ −𝑘
⃗⃗ (𝑘
⃗⃗ .𝑟⃗ )}.{𝑟⃗ −𝑘
⃗⃗ (𝑘
⃗⃗ .𝑟⃗ )}
Solution. Here, 𝑉 =
1
2
log
2
2
1
2
log 𝑟
2
= log 𝑟
L.H.S. = 𝑔𝑟𝑎𝑑 𝑉 = 𝛻𝑉 =
𝜕
𝜕𝑟
log 𝑟
1
𝑟
𝑟⃗
𝑟
𝑟⃗
𝑟
2
R.H.S. =
𝑟⃗ −𝑘
⃗⃗ (𝑘
⃗⃗ .𝑟⃗ )
{𝑟⃗ −𝑘
⃗⃗ (𝑘
⃗⃗ .𝑟⃗ )}.{𝑟⃗ −𝑘
⃗⃗ (𝑘
⃗⃗ .𝑟⃗ )}
𝑟⃗ − 0
{𝑟⃗ − 0 }.{𝑟⃗ − 0 }
[ as 𝑘
= 0 ]
𝑟⃗
𝑟⃗ .𝑟⃗
𝑟⃗
𝑟
2
1. 3 Divergence of a vector point function
Introduction: The divergence of a vector field is the extent to which the vector field flux
behaves like a source/sink at a given point.
Definition: The divergence of a differentiable vector point function 𝑉
is denoted by 𝑑𝑖𝑣𝑉
and
defined as: 𝑑𝑖𝑣𝑉
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝜕𝑉
⃗⃗⃗
𝜕𝑥
𝜕𝑉
⃗⃗⃗
𝜕𝑦
𝜕𝑉
⃗⃗⃗
𝜕𝑧
Hence, 𝑑𝑖𝑣 𝑉
gives the rate of outflow per unit volume at a point of the fluid.
Note: If 𝑑𝑖𝑣 𝑉
= 0 , then (i) 𝑉
is called solenoidal vector (ii) fluid is called compressible
1. 4 Curl of a vector point function
Introduction: The curl is a vector operator that describes the infinitesimal rotation of a vector
field in three-dimensional space.
Definition: The curl of a differentiable vector point function 𝐹
is denoted by curl 𝐹
and defined
as:
= 𝛻 × 𝐹
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
) × 𝐹
1 .4.1 Physical Interpretation of Curl
Let us consider a rigid body rotating about a fixed axis through 𝑂 with uniform angular velocity
1
2
3
. Then we have,
𝑣⃗ = 𝜔⃗⃗⃗ × 𝑟⃗ , where 𝑣⃗ is the linear velocity at any point and 𝑟⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘
is a position vector
for that point.
∴ 𝑣⃗ = 𝜔⃗⃗⃗ × 𝑟⃗ = |
1
2
3
2
3
3
1
1
2
No𝑐𝑢𝑟𝑙 𝑣⃗ = |
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
2
3
3
1
1
2
1
2
3
1
2
Thus, the angular velocity at any point is equal to the half of the 𝑐𝑢𝑟𝑙 of the linear velocity at that
point of the body.
Note: If 𝑣⃗ = 0
, then 𝑣⃗ is said to be an irrotational vector.
Example 1. Find the divergence and curl of 𝑟⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘
Solution. 𝑑𝑖𝑣 𝑟⃗ = 𝛻. 𝑟⃗
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑐𝑢𝑟𝑙 𝑟⃗ = 𝛻 × 𝑟⃗ = (𝑖̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
) × (𝑥𝑖̂ + 𝑦𝑗 ̂+ 𝑧𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
Example 2. Find the divergence and curl of the vector.
2
2
2
Solution. 𝑑𝑖𝑣 𝑅
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
). [(𝑥
2
2
2
]
= 𝛻 × 𝑅
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
) × [
2
2
2
]
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
2
2
2
Example 3. Prove that
2
2
is both solenoidal and irrotational.
Solution. Let 𝑉
2
2
is solenoidal
= 𝛻 × 𝑉
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
2
2
is irrotational.
Example 4. Show that the vector field 𝐴
, where 𝐴
2
2
irrotational. And find a scalar ∅ such that 𝐴
Proof. 𝑐𝑢𝑟𝑙(∅𝑎⃗ ) = ∑ 𝑖̂ ×
𝜕
𝜕𝑥
(∅𝑎⃗ ) = ∑ 𝑖̂ × (∅
𝜕𝑎⃗⃗
𝜕𝑥
𝜕∅
𝜕𝑥
𝑖̂ ×
𝜕𝑎⃗⃗
𝜕𝑥
𝜕∅
𝜕𝑥
) × 𝑎⃗ = ∅𝑐𝑢𝑟𝑙𝑎⃗ +
× 𝑎⃗
(3) 𝑑𝑖𝑣(𝑎⃗ × 𝑏
Proof. 𝑑𝑖𝑣(𝑎⃗ × 𝑏
𝜕
𝜕𝑥
(𝑎⃗ × 𝑏
𝑖̂. (𝑎⃗ ×
𝜕𝑏
⃗⃗
𝜕𝑥
𝜕𝑎⃗⃗
𝜕𝑥
× 𝑏
𝜕𝑏
⃗⃗
𝜕𝑥
× 𝑎⃗ ) + ∑ 𝑖̂. (
𝜕𝑎⃗⃗
𝜕𝑥
× 𝑏
𝜕𝑎⃗⃗
𝜕𝑥
× 𝑏
𝜕𝑏
⃗⃗
𝜕𝑥
× 𝑎⃗ )
𝑖̂ ×
𝜕𝑎⃗⃗
𝜕𝑥
(𝑖̂ ×
𝜕𝑏
⃗⃗
𝜕𝑥
(4) 𝑐𝑢𝑟𝑙(𝑎⃗ × 𝑏
Proof. 𝑐𝑢𝑟𝑙
𝑎⃗ × 𝑏
𝑖̂ ×
𝜕
𝜕𝑥
𝑎⃗ × 𝑏
𝑖̂ × (𝑎⃗ ×
𝜕𝑏
⃗⃗
𝜕𝑥
𝜕𝑎⃗⃗
𝜕𝑥
× 𝑏
= ∑ 𝑖̂ × (𝑎⃗ ×
𝜕𝑏
⃗⃗
𝜕𝑥
) + ∑ 𝑖̂ × (
𝜕𝑎⃗⃗
𝜕𝑥
× 𝑏
[(𝑖̂.
𝜕𝑏
⃗⃗
𝜕𝑥
𝜕𝑏
⃗⃗
𝜕𝑥
] +
[(𝑖̂. 𝑏
𝜕𝑎⃗⃗
𝜕𝑥
𝜕𝑎⃗⃗
𝜕𝑥
]
𝜕𝑏
⃗⃗
𝜕𝑥
𝜕𝑏
⃗⃗
𝜕𝑥
𝜕𝑎⃗⃗
𝜕𝑥
𝜕𝑎⃗⃗
𝜕𝑥
𝜕
𝜕𝑥
𝜕
𝜕𝑥
2
Proof. 𝑑𝑖𝑣
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
𝜕
𝜕𝑥
𝜕∅
𝜕𝑥
𝜕
𝜕𝑦
𝜕∅
𝜕𝑦
𝜕
𝜕𝑧
𝜕∅
𝜕𝑧
𝜕
2
∅
𝜕𝑥
2
𝜕
2
∅
𝜕𝑦
2
𝜕
2
∅
𝜕𝑧
2
𝜕
2
𝜕𝑥
2
𝜕
2
𝜕𝑦
2
𝜕
2
𝜕𝑧
2
2
2
is known as Laplacian operator
(6) 𝑐𝑢𝑟𝑙(𝑔𝑟𝑎𝑑∅) = 𝛻 × (𝛻∅) = 0
Proof. curl
= 𝛻 ×
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
×
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
𝜕
2
∅
𝜕𝑦𝜕𝑧
𝜕
2
∅
𝜕𝑧𝜕𝑦
Proof. Let 𝑉
1
2
3
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
1
2
3
𝜕𝑉
3
𝜕𝑦
𝜕𝑉
2
𝜕𝑧
𝜕𝑉
1
𝜕𝑧
𝜕𝑉
3
𝜕𝑥
𝜕𝑉
2
𝜕𝑥
𝜕𝑉
1
𝜕𝑦
Now, 𝑑𝑖𝑣
𝜕
𝜕𝑥
𝜕𝑉
3
𝜕𝑦
𝜕𝑉
2
𝜕𝑧
𝜕
𝜕𝑦
𝜕𝑉
1
𝜕𝑧
𝜕𝑉
3
𝜕𝑥
𝜕
𝜕𝑧
𝜕𝑉
2
𝜕𝑥
𝜕𝑉
1
𝜕𝑦
𝜕
2
𝑉
3
𝜕𝑥𝜕𝑦
𝜕
2
𝑉
2
𝜕𝑥𝜕𝑧
𝜕
2
𝑉
1
𝜕𝑦𝜕𝑧
𝜕
2
𝑉
3
𝜕𝑦𝜕𝑥
𝜕
2
𝑉
2
𝜕𝑧𝜕𝑥
𝜕
2
𝑉
1
𝜕𝑧𝜕𝑦
Example 1. If 𝑢 = 𝑥
2
2
2
and 𝑉
, show that 𝑑𝑖𝑣
Solution. 𝑑𝑖𝑣(𝑢𝑉
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
𝜕𝑢
𝜕𝑧
2
2
2
2
2
2
Example 2. If 𝑟⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘
, prove that
(i) 𝑐𝑢𝑟𝑙
𝑛
(ii) 𝛻
2
𝑛
𝑛− 2
Solution. (i) 𝑐𝑢𝑟𝑙
𝑛
𝑛
𝑛
× 𝑟⃗
𝑛− 1
× 𝑟⃗ = 𝑛𝑟
𝑛− 1
𝑟⃗
𝑟
× 𝑟⃗ = 0
(ii) 𝛻
2
𝑛
𝑟⃗ ) = 𝛻[𝛻. (𝑟
𝑛
𝑟⃗ )] = 𝑔𝑟𝑎𝑑[𝑑𝑖𝑣(𝑟
𝑛
𝑟⃗ )]
= 𝑔𝑟𝑎𝑑[𝑟
𝑛
𝑛
. 𝑟⃗ ]
= 𝑔𝑟𝑎𝑑[ 3 𝑟
𝑛
𝑛− 1
. 𝑟⃗ ] = 𝑔𝑟𝑎𝑑 [ 3 𝑟
𝑛
𝑛− 1
𝑟⃗
𝑟
. 𝑟⃗ ]
= 𝑔𝑟𝑎𝑑 [ 3 𝑟
𝑛
𝑛− 1
𝑟
2
𝑟
] = 𝑔𝑟𝑎𝑑
[(
𝑛
]
𝑛− 1
𝑟⃗
𝑟
𝑛− 2
1. 5 Tangent and Normal Planes
Let ∅
= 𝑐 be the equation of a level surface. Let 𝑟 = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘
be the position vector
of any point 𝑃(𝑥, 𝑦, 𝑧) on this surface.
Then
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
is a vector along the normal to the surface ∅ at 𝑃.
i.e. 𝛻∅ is perpendicular to the tangent plane at 𝑃.
Tangent Plane at 𝑷 :
Let 𝑅 = 𝑋𝑖̂ + 𝑌𝑗̂ + 𝑍𝑘
be the position vector of any moving point 𝑄(𝑋, 𝑌, 𝑍) on the tangent
plane.
and 𝑃𝑄
lies in the tangent plane at 𝑃. Therefore, it is perpendicular to the vector 𝛻∅.
[(
]
[
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
]
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
This is the equation of tangent plane.
Normal at 𝑷 :
Let ∅(𝑥, 𝑦, 𝑧) = 𝑐 be the equation of a level surface. Let 𝑟 = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘
be the position vector
of any point 𝑃(𝑥, 𝑦, 𝑧) on this surface. Then
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
is a vector along the normal to the surface ∅ at 𝑃.
i.e. 𝛻∅ is perpendicular to the tangent plane at 𝑃.
Let 𝑅 = 𝑋𝑖̂ + 𝑌𝑗̂ + 𝑍𝑘
be the position vector of any moving point 𝑄
on the normal at 𝑃.
and 𝑃𝑄
lies along the normal of 𝑃 to the surface ∅. Therefore, it is parallel to the vector 𝛻∅.
× 𝛻∅ = 0
× 𝛻∅ = 0
[vector form of normal]
Cartesian form of Normal:
The vectors (𝑋 − 𝑥)𝑖̂ + (𝑌 − 𝑦)𝑗̂ + (𝑍 − 𝑧)𝑘
and 𝛻∅ =
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
will be parallel, if
= 𝜆 [
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
]
Equating the coefficient of 𝑖̂ , 𝑗̂ and 𝑘
, we get
𝜕∅
𝜕𝑥
𝜕∅
𝜕𝑦
𝜕∅
𝜕𝑧
OR
𝑋−𝑥
𝜕∅
𝜕𝑥
𝑌−𝑦
𝜕∅
𝜕𝑦
𝑍−𝑧
𝜕∅
𝜕𝑧
, which are equations of Normal.
Example 1. Find the equations of the tangent plane and normal to the surface.
2
− 3 𝑥𝑦 − 4 𝑥 = 7 at the point
Solution. Let ∅ ≡ 2 𝑥𝑧
2
2
⇒ 𝛻∅]
1 ,− 1 , 2
Equation of tangent at the point ( 1 , − 1 , 2 ) is given by
[
]. ( 7 𝑖̂ − 3 𝑗̂ + 8 𝑘
Or 7 𝑋 − 3 𝑌 + 8 𝑍 = 26
Equations of the normal to the given surface at the point
are:
𝑋−𝑥
𝜕∅
𝜕𝑥
𝑌−𝑦
𝜕∅
𝜕𝑦
𝑍−𝑧
𝜕∅
𝜕𝑧
i.e.
𝑋− 1
7
𝑌+ 1
− 3
𝑍+ 2
8
Example 2. Find the equations of the tangent plane and normal to the surface 𝑥𝑦𝑧 =4 at the
point
Solution. Let ∅ ≡ 𝑥𝑦𝑧 − 4 = 0
⇒ 𝛻∅]
1 , 2 , 2
Equation of tangent at the point
is given by
