Q24P

Question

Derive the three quotient rules.

Step-by-Step Solution

Verified

Answer

The three quotient rules $\nabla (\frac{f}{g}) = \frac{g (\nabla f) - f (\nabla g)}{g^{2}}, \nabla (\frac{\overset{⇀}{A}}{g}) = \frac{g (\nabla \overset{⇀}{A}) - \overset{⇀}{A} (\nabla g)}{g^{2}} and \nabla \times (\frac{\overset{⇀}{A}}{g}) = \frac{g (\nabla \times \overset{⇀}{A}) - \overset{⇀}{A} \times (\nabla g)}{g^{2}} have been derived .$

1Step 1: Derive first quiotinet rule

To compute an expression substitute the vectors and other required expression and then simplify

The first quiotient rule is $\nabla (\frac{f}{g}) = \frac{g (\nabla f) - f (\nabla g)}{g^{2}},$ The $\nabla$ operator is defined as $\nabla = (\frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k) Compute the value of \nabla (\frac{f}{g}), as : \nabla (\frac{f}{g}) = (\frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k) (\frac{f}{g}) = i \frac{\partial}{\partial x} (\frac{f}{g}) + j \frac{\partial}{\partial y} (\frac{f}{g}) k + \frac{\partial}{\partial z} (\frac{f}{g}) = (\frac{g \frac{\partial f}{\partial x} - f (\frac{\partial g}{\partial x})}{g^{2}}) + j (\frac{g \frac{\partial f}{\partial y} - f (\frac{\partial g}{\partial y})}{g^{2}}) + k (\frac{g \frac{\partial f}{\partial z} - f (\frac{\partial g}{\partial z})}{g^{2}}) = \frac{1}{g} (\frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j + \frac{\partial f}{\partial z} k) - \frac{f}{g^{2}} (\frac{\partial g}{\partial x} i + \frac{\partial g}{\partial y} j + \frac{\partial g}{\partial z} k)$

$Substitiute \nabla for \frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k into equation (1), as :$

$\nabla (\frac{f}{g}) = \frac{1}{g} (\frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j + \frac{\partial f}{\partial z} k) - \frac{f}{g^{2}} (\frac{\partial g}{\partial x} i + \frac{\partial g}{\partial y} j + \frac{\partial g}{\partial z} k) = \frac{1}{g} (\nabla f) - \frac{1}{g^{2}} (\nabla g) = \frac{g (\nabla f) - f (\nabla g)}{g^{2}}$

Thus the first quoitient rule is proven.

2Step 2: Derive second quiotient rule

To compute an expression substitute the vectors and other required expression and then simplify

The second quiotient rule is $\nabla (\frac{\overset{⇀}{A}}{g}) = \frac{g (\nabla \overset{⇀}{A}) - \overset{⇀}{A} (\nabla g)}{g^{2}}$ . Let the vector $\overset{⇀}{A}$ is defined as The operator is defined as $\overset{⇀}{A} = A_{x} i + A_{y} j + A_{z} k The \nabla operator is defined as \nabla = \frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k .$

$Compute the value of \nabla (\frac{\overset{⇀}{A}}{g}), as : \nabla (\frac{\overset{⇀}{A}}{g}) = (\frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k) (\frac{A_{x} i + A_{y} i + A_{z} k}{g}) = i \frac{\partial}{\partial x} (\frac{A_{x}}{g}) + j \frac{\partial}{\partial y} (\frac{A_{y}}{g}) + j \frac{\partial}{\partial z} (\frac{A_{z}}{g}) = \frac{g \frac{\partial A_{x}}{\partial x} - A_{x} (\frac{\partial g}{\partial x})}{g^{2}} + \frac{g \frac{\partial A_{y}}{\partial y} - A_{y} (\frac{\partial g}{\partial y})}{g^{2}} + \frac{g \frac{\partial A_{z}}{\partial z} - A_{z} (\frac{\partial g}{\partial z})}{g^{2}} = \frac{1}{g} (\frac{\partial A_{x}}{\partial x} + \frac{\partial A_{y}}{\partial y} + \frac{\partial A_{z}}{\partial z}) - \frac{f}{g^{2}} (\frac{\partial g}{\partial x} + A_{y} \frac{\partial g}{\partial y} + A_{Z} \frac{\partial g}{\partial z}) . . . . . . (1)$

The dot product of $\nabla$ and vector $\overset{⇀}{A}$ is obtained as ,

$\nabla . \overset{⇀}{A} = (\frac{\partial}{\partial X} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial Z} k) (A_{x} i + A_{y} j + A_{z} k) = \frac{A_{x}}{\partial X} i + \frac{\partial A_{y}}{\partial y} j + \frac{\partial A_{z}}{\partial Z} k The dot product of \nabla g and vector \overset{⇀}{A} is obtained as, \overset{⇀}{A} . (\nabla g) = (A_{x} i + A_{y} j + A_{z} k) (\frac{\partial}{\partial X} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial Z} k) g = (A_{x} i + A_{y} j + A_{z} k) (\frac{\partial g}{\partial x} i + \frac{\partial g}{\partial y} j + \frac{\partial g}{\partial z} k) = A_{x} \frac{\partial g}{\partial x} + A_{y} \frac{\partial g}{\partial y} + A_{x} \frac{\partial g}{\partial z}$

$Substitiute \overset{⇀}{A} - (\nabla g) for A_{x} \frac{\partial g}{\partial x} + A_{y} \frac{\partial g}{\partial y} + A_{y} \frac{\partial g}{\partial y} and \nabla - \overset{⇀}{A} for \frac{A_{x}}{\partial x} + A_{y} \frac{\partial A_{y}}{\partial y} + A_{y} \frac{\partial A_{z}}{\partial z} into equation (1), as : \nabla (\frac{\overset{⇀}{A}}{g}) = \frac{1}{g} (\frac{\partial A_{x}}{\partial x} i + \frac{\partial A_{y}}{\partial y} j + \frac{\partial A_{z}}{\partial z} k) - \frac{f}{g^{2}} (A_{x} \frac{\partial g}{\partial x} + A_{y} \frac{\partial g}{\partial y} + A_{z} \frac{\partial g}{\partial z}) = \frac{1}{g} (\nabla . \overset{⇀}{A}) - \frac{f}{g^{2}} (\overset{⇀}{A} . \nabla g) = \frac{g (\nabla . \overset{⇀}{A}) - \overset{⇀}{A} (\nabla g)}{g^{2}}$

Thus the second quoitient rule is proven.

3Step 3: Derive third quiotient rule

To compute an expression substitute the vectors and other required expression and then simplify

The second quiotient rule is $\nabla \times (\frac{\overset{⇀}{A}}{g}) = \frac{g (\nabla \times \overset{⇀}{A}) - \overset{⇀}{A} (\nabla g)}{g^{2}}$ . Let the vector $\overset{⇀}{A}$ is defined as $\overset{⇀}{A} = A_{x} i + A_{x} j + A_{x} k$ The operator is defined as $\nabla = \frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k$ .

Compute the value of $\nabla \times (\frac{\overset{⇀}{A}}{g})$ , as:

$\nabla (\frac{\overset{⇀}{A}}{g}) = |\begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{A_{x}}{g} & \frac{A_{y}}{g} & \frac{A_{z}}{g} \end{matrix}| = i (\frac{\partial}{\partial y} (\frac{A_{z}}{g}) - \frac{\partial}{\partial z} (\frac{A_{y}}{g})) + j (\frac{\partial}{\partial z} (\frac{A_{x}}{g}) - \frac{\partial}{\partial x} (\frac{A_{z}}{g})) + k (\frac{\partial}{\partial x} (\frac{A_{y}}{g}) - \frac{\partial}{\partial y} (\frac{A_{x}}{g})) = \frac{1}{g} (i (\frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z}) + j (\frac{\partial A_{x}}{\partial z} - \frac{\partial A_{z}}{\partial x}) + k (\frac{\partial A_{y}}{\partial x} - \frac{\partial A_{x}}{\partial y})) = \frac{g (\nabla \times \overset{⇀}{A}) - \overset{⇀}{A} \times (\nabla g)}{g^{2}}$

Thus the third quoitient rule is proven.

Q23P

Q25P

Other exercises in this chapter

Q22P

(a) If A and B are two vector functions, what does the expression A·∇B mean?(That is, what are its x, y, and z components, in terms of the Carte

View solution

Q23P

(For masochists only.) Prove product rules (ii) and (vi). Refer to Prob. 1.22 for the definition of (A . V) B.

View solution

Q25P

(a) Check product rule (iv) (by calculating each term separately) for the functions A=xx^+2yy^+3z z^ &#

View solution

Q26P

Calculate the Laplacian of the following functions:(a) (b)(c) .(d)

View solution