Q21P

Question

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

Step-by-Step Solution

Verified

Answer

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

1Step 1: 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 the $\nabla$ operator is defined as $\nabla = \frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k$ . The gradient of vector v is obtaind as

$\nabla . v = (\frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k) . (v_{x} i + v_{y} j + v_{z} k) = \frac{\partial v_{x}}{\partial x} i + \frac{\partial v_{y}}{\partial y} j + \frac{\partial v_{z}}{\partial z} The curl of vector v is obtaind as \nabla . v = (\frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k) . (v_{x} i + v_{y} j + v_{z} k) = (\frac{\partial v_{x}}{\partial y} - \frac{\partial v_{x}}{\partial z} k) i + (\frac{\partial v_{y}}{\partial z} - \frac{\partial v_{y}}{\partial x} k) j + (\frac{\partial v_{z}}{\partial x} - \frac{\partial v_{z}}{\partial y}) k$

2Step 2: Prove ∇ ( fg ) = ∇ g + g ∇ f

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

Find the gradient of vector g.

$\nabla (g) = (\frac{\partial}{\partial x} i + \frac{\partial}{\partial x} j) (g) = \frac{\partial}{\partial x} i + \frac{\partial}{\partial x}$

Find the gradient of vector f.

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

3Step 3: Find ∇ × ( A × B ) .

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

4Step 4: Find B ( ∇ × A ) - A ( ∇ × B )

Find curl of vector A.

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

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

$A_{x} (\frac{\partial B_{z}}{\partial y} - \frac{\partial B_{y}}{\partial z}) + A_{y} (\frac{\partial B_{x}}{\partial z} - \frac{\partial B_{z}}{\partial x}) + A_{z} (\frac{\partial B_{y}}{\partial x} - \frac{\partial B_{x}}{\partial y}) F i n d B (\nabla \times A) - A (\nabla \times B) B B (\nabla \times A) - A (\nabla \times B) = B_{x} (\frac{\partial B_{z}}{\partial y} - \frac{\partial B_{y}}{\partial z}) + B_{y} (\frac{\partial B_{x}}{\partial z} - \frac{\partial B_{z}}{\partial x}) + B_{z} (\frac{\partial B_{y}}{\partial x} - \frac{\partial B_{x}}{\partial y}) - A_{x} (\frac{\partial B_{z}}{\partial y} - \frac{\partial B_{y}}{\partial z}) + A_{y} (\frac{\partial B_{x}}{\partial z} - \frac{\partial B_{z}}{\partial x}) + A_{z} (\frac{\partial B_{y}}{\partial x} - \frac{\partial B_{x}}{\partial y}) = \frac{\partial}{\partial x} (A_{y} B_{z} - A_{y} B_{z}) + \frac{\partial}{\partial y} (A_{z} B_{x} - A_{x} B_{z}) + \frac{\partial}{\partial z} (A_{x} B_{y} - A_{y} B_{x})$

….(2)

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

$\nabla \times (A \times B) = B (\nabla \times A) - A (\nabla \times B)$

5Step 5: Find

Find the curl of fA,

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

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

Q1.22 P

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