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B.Tech., 1st Semester
Applied Mathematics-I [MATH144]
Module IV: Vector Calculus
Lecture 2: Directional derivative, Divergence and its properties
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Directional derivative of a scalar Function AIAS
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or in other words
Directional derivative: The component of ∇ф in the direction of vector 𝑑 is equal to ∇ф. 𝑑
and is called the directional derivative of ф in the direction of vector𝑑 .
Divergence of Vector Function:
Divergence of a vector point function: If 𝐹 = F1 i + F2 j + F3 k denote a vector point
function then ∇. 𝐹 denotes the divergence of 𝐹 and it is written as
∂ф ∂ф ∂ф
𝑑𝑖𝑣 𝐹 = i ∂x
+j ∂y
+k ∂z
. F1 i + F2 j + F3 k
∂F1 ∂F2 ∂F3
= + +
∂x ∂y ∂z
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This shows that 𝑑𝑖𝑣 𝐹 is a scalar point function.
If 𝑑𝑖𝑣 𝐹 = 0 , Vector function F is called solenoidal vector function.
If 𝑉 = V1 i + V2 j + V3 k denote the velocity of a fluid at Point P(x,y,z) and fluid is
incompressible,there can be no gain or loss in volume element and we have
𝒅𝒊𝒗 𝑽 = 𝟎
It is also known as equation of continuity or conservation of mass.
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Example: Find the directional derivative of the scalar function 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 2 + 𝑥𝑦 + 𝑧 2 at
the point A (1, -1, -1) in the direction of line AB where B has coordinates (3, 2, 1).
Solution: we know that 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 2 + 𝑥𝑦 + 𝑧 2
∂ ∂ ∂
Therefore, directional derivative is ∇f = i ∂x + j ∂y + k ∂z f
∂ ∂ ∂
= i ∂x + j ∂y + k ∂z (𝑥 2 + 𝑥𝑦 + 𝑧 2 )
= 2x + y i + xj + zk
directional derivative at the point A (1, -1, -1) is given by
= 𝑖 + 𝑗 − 2𝑘
Now 𝐴𝐵 = 𝐵 − 𝐴 = 3𝑖 + 2𝑗 + 𝑘 − 𝑖 − 𝑗 − 𝑘 = 2𝑖 + 3𝑗 + 2𝑘
Therefore, directional derivative at the point A (1, -1, -1) in the direction of line 𝐴𝐵 is
(2𝑖+3𝑗 +2𝑘 ) 1 1
= 𝑖 + 𝑗 − 2𝑘 . 4+9+4
= 17
2+3−4 = 17
Answer
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Example: Find the directional derivative of the divergence of the vector function
𝑓 𝑥, 𝑦, 𝑧 = 𝑥𝑦𝑖 + 𝑥𝑦 2 𝑗 + 𝑧 2 𝑘 at the point (2, 1, 2) in the direction of outer normal to the
sphere, 𝑥 2 + 𝑦 2 + 𝑧 2 = 9.
Solution: we know that 𝑓 𝑥, 𝑦, 𝑧 = 𝑥𝑦𝑖 + 𝑥𝑦 2 𝑗 + 𝑧 2 𝑘
Therefore, divergence of the vector function f is
∂ ∂ ∂
∇. f = i +j +k .f
∂x ∂y ∂z
∂ ∂ ∂
= i ∂x + j ∂y + k ∂z . (𝑥𝑦𝑖 + 𝑥𝑦 2 𝑗 + 𝑧 2 k)
= y + 2xy + 2z
directional derivative of divergence of the vector function f is
∂ ∂ ∂
= i ∂x + j ∂y + k ∂z (y + 2xy + 2z)
= 2𝑦𝑖 + (1 + 2𝑥)𝑗 + 2k
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Now, we know that ф 𝑥, 𝑦, 𝑧 = 𝑥 2 + 𝑦 2 + 𝑧 2 = 9
And ∇ф is a vector normal to the surface of sphere ф 𝑥, 𝑦, 𝑧 = 𝑐
∂ ∂ ∂
Therefore, normal vector is ∇ф = i ∂x + j ∂y + k ∂z ф
∂ ∂ ∂
= i ∂x + j ∂y + k ∂z (𝑥 2 + 𝑦 2 + 𝑧 2 − 9)
= i 2x + j 2𝑦 + k(2𝑧)
Normal to sphere at the point (2, 1, 2) = 4𝑖 + 2𝑗 + 4𝑘
Directional derivative along normal at the point (2, 1, 2)
4𝑖 +2𝑗 +4𝑘 1
= 2𝑖 + 5𝑗 + 2𝑘 . 16+4+16
= 6
8 + 10 + 8
13
= 3
Answer
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Example: Find the value of a for which the vector function 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 + 3𝑦 𝑖 +
𝑦 − 2𝑧 𝑗 + (𝑥 + 𝑎𝑧)𝑘 is solenoidal vector.
Solution: we know that 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 + 3𝑦 𝑖 + 𝑦 − 2𝑧 𝑗 + (𝑥 + 𝑎𝑧)𝑘
And If 𝑑𝑖𝑣 𝑓 = 0 , Vector function f is called solenoidal vector function
∂ ∂ ∂
∇. f = i ∂x + j ∂y + k ∂z . f = 0
∂ ∂ ∂
i ∂x + j ∂y + k ∂z . 𝑥 + 3𝑦 𝑖 + 𝑦 − 2𝑧 𝑗 + 𝑥 + 𝑎𝑧 𝑘 = 0
∂ ∂ ∂
𝑥 + 3𝑦 + ∂y 𝑦 − 2𝑧 + ∂z 𝑥 + 𝑎𝑧 = 0
∂x
1+1+a= 0
Hence a = - 2 for f to be solenoidal. Answer
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Practice Questions: AIAS
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Thank you!
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