Question

Let\(r = xi + yj + zk\)and\(r = \left| r \right|\).

Verify each identity.

\(\left( a \right)\nabla r = \frac{r}{r} \)\(\left( b \right)\nabla \times r = 0 \)\(\left( c \right)\nabla \left( {\frac{1}{r}} \right) = \frac{{ - r}}{{{r^3}}}\) \(\left( d \right)\nabla lnr = \frac{r}{{{r^2}}}\)

Short answer

1. The solution is\(\nabla r = \frac{r}{r}{\rm{ }}\)
2. The solution is\(\nabla \times r = 0 \)
3. The solution is\(\nabla \left( {\frac{1}{r}} \right) = \frac{{ - r}}{{{r^3}}}\)
4. The solution is\(\nabla lnr = \frac{r}{{{r^2}}}\)

Step-by-step solution

Given information

\(\left( a \right)\nabla r = \frac{r}{r} \)\(\left( b \right)\nabla \times r = 0 \)\(\left( c \right)\nabla \left( {\frac{1}{r}} \right) = \frac{{ - r}}{{{r^3}}}\)\(\left( d \right)\nabla lnr = \frac{r}{{{r^2}}}\)

It is given that\(r = |r|\)and also that\(r = xi + yj + zk\)

Therefore,\(r = \sqrt {{x^2} + {y^2} + {z^2}} \)

(a)

Substitute the expression

By definition

\(\begin{array}{l}\nabla r = \frac{{\partial r}}{{\partial x}}i + \frac{{\partial r}}{{\partial y}}j + \frac{{\partial r}}{{\partial z}}k\\\nabla r = \frac{{\partial \left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}}{{\partial x}}i + \frac{{\partial \left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}}{{\partial y}}j + \frac{{\partial \left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}}{{\partial z}}k\end{array}\)

\(\nabla r = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}i + \frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}j + \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{k}}\)

Take\(\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\)common from the three terms

\(\begin{array}{l}\nabla r = \frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}(xi + yj + zk) \\\nabla r = \frac{{(xi + yj + zk)}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\end{array}\)

Substitute\(\sqrt {{x^2} + {y^2} + {z^2}} = r and xi + yj + zk = r\), to get

\(\nabla r = \frac{r}{r}{\rm{ }}\)

Hence verified.

(b)

Verify the expression

\(\begin{array}{c}\nabla \times r = \left| {\begin{array}{*{20}{c}}i&j&k\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\x&y&z\end{array}} \right|\\ = i\left| {\begin{array}{*{20}{c}}{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\y&z\end{array}} \right| - j\left| {\begin{array}{*{20}{c}}{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial z}}}\\x&z\end{array}} \right| + k\left| {\begin{array}{*{20}{c}}{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}\\x&y\end{array}} \right|\\ = \left( {\frac{{\partial z}}{{\partial y}} - \frac{{\partial y}}{{\partial z}}} \right)i - \left( {\frac{{\partial z}}{{\partial x}} - \frac{{\partial x}}{{\partial z}}} \right)j + \left( {\frac{{\partial y}}{{\partial x}} - \frac{{\partial x}}{{\partial y}}} \right)k\\ = (0 - 0)i - (0 - 0)j + (0 - 0)k\end{array}\)\(\)

\(\nabla \times r = 0\)

Hence verified.

(c)

Use the chain rule 

\(\nabla (1/r) = \frac{{\partial (1/r)}}{{\partial x}}i + \frac{{\partial (1/r)}}{{\partial y}}j + + \frac{{\partial (1/r)}}{{\partial z}}k\)

We can use chain rule, to rewrite as follows

\(\nabla (1/r) = \frac{{\partial (1/r)}}{{\partial r}} \times \frac{r}{x}i + \frac{{\partial (1/r)}}{{\partial r}} \times \frac{{\partial r}}{{\partial y}}j + \frac{{\partial (1/r)}}{{\partial r}} \times \frac{{\partial r}}{{\partial z}}{\bf{k}}\)

You can take\(\frac{{\partial (1/r)}}{{\partial r}}\)common,

\(\begin{array}{l}\nabla (1/r) = \frac{{(1/r)}}{r} \times \left( {\frac{{\partial r}}{{\partial x}}i + \frac{{\partial r}}{{\partial y}}j + \frac{{\partial r}}{{\partial z}}k} \right)\\\nabla (1/r) = \left( { - \frac{1}{{{r^2}}}} \right)\nabla r\end{array}\)

From part (a), we know that \(\nabla r = \frac{r}{r}{\rm{ }}\)

Therefore,

\(\begin{array}{l}\nabla (1/r) = \left( { - \frac{1}{{{r^2}}}} \right)\left( {\frac{r}{r}} \right) \\\nabla (1/r) = - \frac{r}{{{r^3}}}\end{array}\)

Hence verified.

(d)

To rewrite as follows, use the chain rule

\(\nabla lnr = \frac{{\partial lnr}}{{\partial x}}i + \frac{{\partial lnr}}{{\partial y}}j + + \frac{{\partial lnr}}{{\partial z}}k\)

We can use chain rule, to rewrite as follows

\(\begin{array}{l}\nabla lnr = \frac{{\partial lnr}}{{\partial r}} \times \frac{{\partial r}}{x}i + \frac{{\partial \ln r}}{{\partial r}} \times \frac{{\partial r}}{{\partial y}}j + \frac{{\partial lnr}}{{\partial r}} \times \frac{{\partial r}}{{\partial z}}{\bf{k}}\\\end{array}\)

You can take\(\frac{{\partial lnr}}{{\partial r}}\)common,

\(\begin{array}{l}\nabla lnr = \frac{{lnr}}{r} \times \left( {\frac{{\partial r}}{{\partial x}}i + \frac{{\partial r}}{{\partial y}}j + \frac{{\partial r}}{{\partial z}}k} \right)\\\nabla lnr = \left( {\frac{1}{r}} \right)\nabla r\end{array}\)

Hence verified.