Q21P Prove product rules (i), (iv), a... [FREE SOLUTION]

Chapter 1: Q21P (page 22)

Prove product rules (i), (iv), and (v)

Short Answer

Expert verified

The product rules (i), (iv), and (v) are proved

Step by step solution

Find the curl of vector v

To prove any rule, simplify its left and right side, and comare them with each other.

Let the vector v be defined as $v_{x} i + v_{y} j + v_{z} k$ and thelocalid="1657359311673" $鈭�$ operator is defined as $鈭� = \frac{鈭�}{鈭� x} i + \frac{鈭�}{鈭� y} j + \frac{鈭�}{鈭� z} k$ . The gradient of vector v is obtaind as

localid="1657346308106" $鈭� . v = (\frac{鈭�}{鈭� x} i + \frac{鈭�}{鈭� y} j + \frac{鈭�}{鈭� z} k) . (v_{x} i + v_{y} j + v_{z} k) = \frac{鈭� v_{x}}{鈭� x} i + \frac{鈭� v_{y}}{鈭� y} j + \frac{鈭� v_{z}}{鈭� z} The curl of vector v is obtaind as 鈭� . v = (\frac{鈭�}{鈭� x} i + \frac{鈭�}{鈭� y} j + \frac{鈭�}{鈭� z} k) . (v_{x} i + v_{y} j + v_{z} k) = (\frac{鈭� v_{x}}{鈭� y} - \frac{鈭� v_{x}}{鈭� z} k) i + (\frac{鈭� v_{y}}{鈭� z} - \frac{鈭� v_{y}}{鈭� x} k) j + (\frac{鈭� v_{z}}{鈭� x} - \frac{鈭� v_{z}}{鈭� y}) k$

Prove∇(fg)=∇g+g∇f

In the expression $鈭� (f g) = f 鈭� g + g 鈭� f$ , where f,g are two dimensional vectors..

Find the gradient of vector g.

$鈭� (g) = (\frac{鈭�}{鈭� x} i + \frac{鈭�}{鈭� x} j) (g) = \frac{鈭�}{鈭� x} i + \frac{鈭�}{鈭� x}$

Find the gradient of vector f.

$鈭� (f) = (\frac{鈭�}{鈭� x} i + \frac{鈭�}{鈭� x} j) (f) = \frac{鈭� f}{鈭� x} i + \frac{鈭� f}{鈭� y} j Now applying the 鈭� perator on the product of f and g . 鈭� (fg) = \frac{鈭� (fg)}{鈭� x} i + \frac{鈭� (fg)}{鈭� y} j = f \frac{鈭� g}{鈭� x} i + \frac{鈭� g}{鈭� y} j + \frac{鈭� f}{鈭� x} i + \frac{鈭� f}{鈭� y} j = f (鈭� g) + g (鈭� f) Thus, it is proved that 鈭� (fg) = f (鈭� g) + g (鈭� f)$

Find∇×(A×B) .

$I n t h e e x p r e s s i o n 鈭� 脳 (A 脳 B) = B (鈭� 脳 A) - A (鈭� 脳 B), w h e r e A, B a r e d e f i n e d a s A = A_{X} I + A_{y} j + A_{z} k a n d B = B_{X} I + B_{y} j + B_{z} k . F i n d t h e c r o s s p r o d u c t o f t h e v e c t o r s A a n d B . A 脳 B = |\begin{matrix} i & j & k \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{matrix}| = i (A_{y} B_{z} - A_{z} B_{y}) - j (A_{x} B_{z} - A_{z} B_{x}) + k (A_{x} B_{y} - A_{y} B_{x}) = i (A_{y} B_{z} - A_{z} B_{y}) - j (A_{z} B_{x} - A_{z} B_{x}) + k (A_{x} B_{y} - A_{y} B_{x}) O b t a i n t h e d i v e r g e n c e o f v e c t o r A 脳 B 鈭� . (A 脳 B) = \frac{鈭�}{鈭� x} (A_{y} B_{z} - A_{z} B_{y}) + \frac{鈭�}{鈭� y} (A_{z} B_{x} - A_{x} B_{z}) + \frac{鈭�}{鈭� x} (A_{x} B_{y} - A_{y} B_{x}) . . . . . (1)$

Find B(∇×A)-A(∇×B)

Find curl of vector A.

$鈭� 脳 A = (\frac{鈭� A_{z}}{鈭� y} - \frac{鈭� A_{y}}{鈭� z}) i - (\frac{鈭� A_{z}}{鈭� x} - \frac{鈭� A_{x}}{鈭� z}) j + (\frac{鈭� A_{y}}{鈭� x} - \frac{鈭� A_{x}}{鈭� y}) k = (\frac{鈭� A_{z}}{鈭� y} - \frac{鈭� A_{y}}{鈭� z}) i + (\frac{鈭� A_{x}}{鈭� z} - \frac{鈭� A_{z}}{鈭� x}) j + (\frac{鈭� A_{y}}{鈭� x} - \frac{鈭� A_{x}}{鈭� y}) k N o w e v a l u a t e B (鈭� 脳 A) B (鈭� 脳 A) = (B_{X} i + B_{y} j + B_{z} k) [(\frac{鈭� A_{z}}{鈭� y} - \frac{鈭� A_{y}}{鈭� z}) i - (\frac{鈭� A_{x}}{鈭� z} - \frac{鈭� A_{z}}{鈭� x}) j + (\frac{鈭� A_{y}}{鈭� x} - \frac{鈭� A_{x}}{鈭� y}) k] = B_{x} (\frac{鈭� A_{z}}{鈭� y} - \frac{鈭� A_{y}}{鈭� z}) i + (\frac{鈭� A_{x}}{鈭� z} - \frac{鈭� A_{z}}{鈭� x}) j + (\frac{鈭� A_{y}}{鈭� x} - \frac{鈭� A_{x}}{鈭� y})$

$F i n d c u r l o f v e c t o r B . 鈭� 脳 B = (\frac{鈭� B_{z}}{鈭� y} - \frac{鈭� B_{y}}{鈭� z}) i - (\frac{鈭� B_{x}}{鈭� z} - \frac{鈭� B_{z}}{鈭� x}) j + (\frac{鈭� B_{y}}{鈭� x} - \frac{鈭� B_{x}}{鈭� y}) k = (\frac{鈭� B_{z}}{鈭� y} - \frac{鈭� B_{y}}{鈭� z}) i + (\frac{鈭� B_{x}}{鈭� z} - \frac{鈭� B_{z}}{鈭� x}) j + (\frac{鈭� B_{y}}{鈭� x} - \frac{鈭� B_{x}}{鈭� y}) k Now evaluate A (鈭� 脳 B) A (鈭� 脳 B) (A_{X} i + A_{y} i + A_{z} i) [= (\frac{鈭� B_{z}}{鈭� y} - \frac{鈭� B_{y}}{鈭� z}) i - (\frac{鈭� B_{x}}{鈭� z} - \frac{鈭� B_{z}}{鈭� x}) j + (\frac{鈭� B_{y}}{鈭� x} - \frac{鈭� B_{x}}{鈭� y})]$

$A_{x} (\frac{鈭� B_{z}}{鈭� y} - \frac{鈭� B_{y}}{鈭� z}) + A_{y} (\frac{鈭� B_{x}}{鈭� z} - \frac{鈭� B_{z}}{鈭� x}) + A_{z} (\frac{鈭� B_{y}}{鈭� x} - \frac{鈭� B_{x}}{鈭� y}) F i n d B (鈭� 脳 A) - A (鈭� 脳 B) B B (鈭� 脳 A) - A (鈭� 脳 B) = B_{x} (\frac{鈭� B_{z}}{鈭� y} - \frac{鈭� B_{y}}{鈭� z}) + B_{y} (\frac{鈭� B_{x}}{鈭� z} - \frac{鈭� B_{z}}{鈭� x}) + B_{z} (\frac{鈭� B_{y}}{鈭� x} - \frac{鈭� B_{x}}{鈭� y}) - A_{x} (\frac{鈭� B_{z}}{鈭� y} - \frac{鈭� B_{y}}{鈭� z}) + A_{y} (\frac{鈭� B_{x}}{鈭� z} - \frac{鈭� B_{z}}{鈭� x}) + A_{z} (\frac{鈭� B_{y}}{鈭� x} - \frac{鈭� B_{x}}{鈭� y}) = \frac{鈭�}{鈭� x} (A_{y} B_{z} - A_{y} B_{z}) + \frac{鈭�}{鈭� y} (A_{z} B_{x} - A_{x} B_{z}) + \frac{鈭�}{鈭� z} (A_{x} B_{y} - A_{y} B_{x})$

鈥�.(2)

From equations (1) and (2) , it can be concluded that

$鈭� 脳 (A 脳 B) = B (鈭� 脳 A) - A (鈭� 脳 B)$

Find

Find the curl of fA,

$鈭� 脳 (f A) = |\begin{matrix} i & j & k \\ \frac{鈭�}{鈭� x} & \frac{鈭�}{鈭� y} & \frac{鈭�}{鈭� z} \\ f A_{x} & f A_{y} & f A_{z} \end{matrix}| = (\frac{鈭�}{鈭� y} (f A_{z}) - \frac{鈭�}{鈭� z} (f A_{y})) i - (\frac{鈭�}{鈭� x} (f A_{z}) - \frac{鈭�}{鈭� z} (f A_{x})) + k (\frac{鈭�}{鈭� x} (f A_{y}) - \frac{鈭�}{鈭� y} (f A_{x})) = [\begin{matrix} (f \frac{鈭�}{鈭� y} (A_{z}) + A_{z} \frac{鈭�}{鈭� z} (f) - f \frac{鈭�}{鈭� y} (A_{y}) - A_{y} \frac{鈭�}{鈭� z} (f)) i - (\begin{matrix} f \frac{鈭�}{鈭� x} (A_{z}) + A_{z} \frac{鈭�}{鈭� x} (f) \\ - f \frac{鈭�}{鈭� z} (A_{z}) + A_{z} \frac{鈭�}{鈭� z} (f) \end{matrix}) j \\ + (f \frac{鈭�}{鈭� y} (A_{y}) + A_{y} \frac{鈭�}{鈭� z} (f) - f \frac{鈭�}{鈭� y} (A_{x}) - A_{x} \frac{鈭�}{鈭� z} (f)) \end{matrix}]$

$= [\begin{matrix} f ((\frac{鈭� A_{z}}{鈭� y} - \frac{鈭� A_{z}}{鈭� y}) i + (\frac{鈭� A_{x}}{鈭� z} - \frac{鈭� A_{z}}{鈭� x}) j + (\frac{鈭� A_{y}}{鈭� x} - \frac{鈭� A_{x}}{鈭� y}) k) \\ - ((A_{y} \frac{鈭� f}{鈭� z} - A_{z} \frac{鈭� f}{鈭� y}) i + (A_{z} \frac{鈭� f}{鈭� x} - A_{z} \frac{鈭� f}{鈭� z}) i (A_{x} \frac{鈭� f}{鈭� y} - A_{y} \frac{鈭� f}{鈭� x}) k) \end{matrix}] Thus using above calculations we can write 鈭� 脳 (fA) = f (鈭� 脳 A) - A 脳 (鈭� f), where f is a constant vector .$

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Prove product rules (i), (iv), and (v)

Short Answer

Step by step solution

Find the curl of vector v

Prove∇(fg)=∇g+g∇f

Find∇×(A×B) .

Find B(∇×A)-A(∇×B)

Find

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