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Vector Algebra Part 3

The document discusses curl and its properties including the definition and calculation of curl for vector point functions. Examples are provided to calculate the curl of specific vector functions and to show that a given field is conservative by showing its curl is equal to zero.

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Akanksha Thakur
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37 views10 pages

Vector Algebra Part 3

The document discusses curl and its properties including the definition and calculation of curl for vector point functions. Examples are provided to calculate the curl of specific vector functions and to show that a given field is conservative by showing its curl is equal to zero.

Uploaded by

Akanksha Thakur
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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AIAS

B.Tech., 1st Semester

Applied Mathematics-I [MATH144]

Module IV: Vector Calculus

Lecture 3: Curl and its properties

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Curl of a Vector Point Function: The curl of a Vector Point Function 𝐹 = F1 i + F2 j + F3 k

is written as curl 𝐹 and defined as

𝑐𝑢𝑟𝑙 𝐹 = ∇ × 𝐹
∂ф ∂ф ∂ф
𝑐𝑢𝑟𝑙 𝐹 = i +j +k × F1 i + F2 j + F3 k
∂x ∂y ∂z

𝑖 𝑗 𝑘
∂ ∂ ∂
=
∂x ∂y ∂z
F1 F2 F3
∂ ∂ ∂ ∂ ∂ ∂
= ∂y
F3 − ∂z F2 𝑖+ ∂z
F1 − ∂x F3 𝑗 + [∂x F2 − ∂y F1 ] 𝑘

There 𝑐𝑢𝑟𝑙 𝐹 is a vector quantity.

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Further, if 𝑐𝑢𝑟𝑙 𝐹 = ∇ × 𝐹 = 0

The field 𝐹 is said to be irrotational i.e. there is no rotational motion in the field.

Note: A field which is not irrotational is called rotational or vortex field.

Example: Calculate curl of a Vector Point function 𝐹 = xyzi + 3x 2 yj + (xz 2 − y 2 z)k.

Solution: here 𝐹 = xyzi + 3x 2 yj + (xz 2 − y 2 z)k

And we know that 𝑐𝑢𝑟𝑙 𝐹 = ∇ × 𝐹


∂ф ∂ф ∂ф
𝑐𝑢𝑟𝑙 𝐹 = i ∂x
+j ∂y
+k ∂z
× xyzi + 3x 2 yj + (xz 2 − y 2 z)k

𝑖 𝑗 𝑘
∂ ∂ ∂
= ∂x ∂y ∂z
2
xyz 3x y (xz 2 − y 2 z)
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= −2𝑦𝑧 𝑖 − 𝑧 2 − 𝑥𝑦 𝑗 + [6𝑥𝑦 − 𝑥𝑧] 𝑘

= −2𝑦𝑧 𝑖 + 𝑥𝑦 − 𝑧 2 𝑗 + [6𝑥𝑦 − 𝑥𝑧] 𝑘 Answer

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Example: Show that the field given by force:

𝐹 = (𝑦 2 cos x + z 3 ) i + (2y sin x − 4) j + (3xz 2 + 2)k is conservative and find its scalar
potential.

Solution: we know that 𝑐𝑢𝑟𝑙 𝐹 = ∇ × 𝐹


∂ ∂ ∂
𝑐𝑢𝑟𝑙 𝐹 = i +j +k × F1 i + F2 j + F3 k
∂x ∂y ∂z

∂ ∂ ∂
= i ∂x
+j ∂y
+k ∂z
× [(𝑦 2 cos x + z 3 ) i + (2y sin x − 4) j + (3xz 2 + 2)k]

𝑖 𝑗 𝑘
∂ ∂ ∂
= ∂x ∂y ∂z
(𝑦 2 cos x + z 3 ) (2y sin x − 4) 2
(3xz + 2)

= 𝑖 0 − 0 + 𝑗 0 + 3𝑧 2 − 3𝑧 2 + 𝑘(2𝑦 cos 𝑥 − 2𝑦 cos 𝑥 − 0)

=0 5
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Therefore, 𝐹 represents a conservative (or irrotational) field. The corresponding scalar


potential ф is given by ∇ф = 𝐹 = (𝑦 2 cos x + z 3 ) i + (2y sin x − 4) j + (3xz 2 + 2)k

Also we have total differential is given by


∂ф ∂ф ∂ф
dф = dx + dy + dz = ∇ф. 𝑑𝑟
∂x ∂y ∂z

Therefore, dф = (𝑦 2 cos x + z 3 ) dx + (2y sin x − 4) dy + (3xz 2 + 2)dz

= d(𝑦 2 sin x + xz 3 − 4y + 2z)

Hence ф = (𝑦 2 sin x + xz 3 − 4y + 2z + c) Answer

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Practice Questions: AIAS

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Thank you!

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