Question

Let ${}^{\mathbf{r}}$be the separation vector from a fixed point ${}^{\mathbf{(}\mathbf{x}\mathbf{\text{'}}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{,}\mathbf{}\mathbf{z}\mathbf{\text{'}}\mathbf{)}}$to the point localid="1654317524404" $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$, and let r be its length. Show that

(a)localid="1654317730952" ${}^{\mathbf{\nabla }\mathbf{}\mathbf{\left(}{\mathbf{r}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{r}}$

(b) ${}^{\mathbf{\nabla }\mathbf{}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{r}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{/}{\mathbf{r}}^{\mathbf{2}}}$

(c) What is the general formula for localid="1654317981268" ${}^{\mathbf{\nabla }\mathbf{}\mathbf{(}\mathbf{}{\mathbf{r}}^{\mathbf{n}}\mathbf{}\mathbf{)}}$

Short answer

(a) It is proved that ${}^{\nabla ({\mathrm{r}}^2)=2\overline{\mathrm{r}}}$.

(b) It is proved thatlocalid="1654318458278" ${}^{\nabla \left(\frac{1}{\mathrm{r}}\right)=\frac{-1}{\left({\mathrm{r}}^2\right)}\left\{\hat{\mathrm{r}}\right\}}$.

(c) The general expression islocalid="1654318707337" ${}^{\nabla \left({\mathrm{r}}^{\mathrm{n}}\right)={\mathrm{nr}}^{\mathrm{n}-1}\hat{\mathrm{r}}}$

Step-by-step solution

Write the given information.

Write the divergence of the scalar function.

$\mathbf{\nabla }\mathbf{}\mathbf{=}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\widehat{\mathbf{i}\mathbf{}}\mathbf{+}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}\hat{\mathbf{j}}\mathbf{+}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}\hat{\mathbf{k}}$

Consider the vector function as,

$\mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\mathbf{}{\mathbf{F}}_{\mathbf{1}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{+}{\mathbf{F}}_{\mathbf{2}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{+}{\mathbf{F}}_{\mathbf{3}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$

Here, the components of the function are localid="1654319596190" ${}^{F_{1,}{F}_{2,}}$and${}^{F_3}$.

Then, the divergence of the vector function as,

$\mathbf{F}\mathbf{}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{1}}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{2}}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{3}}}{\mathbf{\partial }\mathbf{z}}$

Derive the proof for ∇ (1/r)=-r/r2.

(a)

Write the expression for the separation for the separation vector.

$\stackrel{\rightharpoonup }{\mathrm{r}}=(\mathrm{x}-{\mathrm{x}}^{\text{'}})\hat{\mathrm{x}}+(\mathrm{y}-{\mathrm{y}}^{\text{'}})\hat{\mathrm{y}}+(\mathrm{z}-{\mathrm{z}}^{\text{'}})\hat{\mathrm{z}}$

Since, r is the length solve further as,

Calculate the divergence of the function.

$$\begin{gathered} \nabla \left({\mathrm{r}}^2\right)=\nabla \left[{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\right] \\ =\left\{\begin{array}{c}\frac{\partial }{\partial \mathrm{x}}\left\{{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\right\}\hat{\mathrm{x}}+\frac{\partial }{\partial \mathrm{y}}\left\{{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\right\}\hat{\mathrm{y}}\\ +\frac{\partial }{\partial \mathrm{z}}\left\{{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\right\}\hat{\mathrm{z}}\end{array}\right\} \\ =2{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2\hat{\mathrm{x}}+2{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2\hat{\mathrm{y}}+{2(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\hat{\mathrm{z}} \\ =2\stackrel{\rightharpoonup }{\mathrm{r}} \end{gathered}$$

Therefore,it is proved that $\nabla \left(r^2\right)=2\overline{r}$

Derive the proof for∇ ( r2 )=2r . 

Write the expression for the separation for the separation vector.

$\nabla \left(\frac{1}{\mathrm{r}}\right)=\nabla \left[\frac{1}{\sqrt{{(\mathrm{x}-{\mathrm{x}}^{\text{'}})}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2}}\right]$

Calculate the divergence of the function.

$$\begin{gathered} \nabla \left(\frac{1}{r}\right)=\left\{\begin{array}{c}\frac{\partial }{\partial x}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{x}\\ +\frac{\partial }{\partial y}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{y}\\ +\frac{\partial }{\partial z}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{z}\end{array}\right\} \\ =\frac{1}{2\left\{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2\right\}}\left\{2(x-x^{\text{'}})\hat{x}+(y-y^{\text{'}})\hat{y}+(z-z^{\text{'}})\hat{z}\right\} \\ =\frac{-1}{{\left(r^2\right)}^{{\displaystyle \frac{3}{2}}}}\left\{\stackrel{\rightharpoonup }{r}\right\} \\ =\frac{-1}{\left(r^3\right)}\left\{r\hat{r}\right\} \end{gathered}$$

Solve further as,

$\nabla \left(\frac{1}{\mathrm{r}}\right)=\frac{-1}{\left({\mathrm{r}}^2\right)}\left\{\hat{\mathrm{r}}\right\}$

Therefore, it is proved that ${}^{\nabla \left(\frac{1}{\mathrm{r}}\right)=\frac{-1}{\left({\mathrm{r}}^2\right)}\left\{\hat{\mathrm{r}}\right\}}$.

Derive the general formula for ∇ ( rn ).

(c)

Write the expression for the separation for the separation vector.

$r^n={\left[{\left(x-x^{\text{'}}\right)}^2+{\left(y-y^{\text{'}}\right)}^2+{\left(z-z^{\text{'}}\right)}^2\right]}^{\frac{n}{2}}$

Calculate the divergence of the function.

$$\begin{gathered} \nabla \left(r^n\right)=\left\{\begin{array}{c}\frac{\partial }{\partial x}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{x}\\ +\frac{\partial }{\partial y}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{y}\\ +\frac{\partial }{\partial z}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{z}\end{array}\right\} \\ =\left\{\begin{array}{c}\frac{n}{2}{\left[{(x-x^{\text{'}})}^2\hat{x}+{(y-y^{\text{'}})}^2\hat{y}+{(z-z^{\text{'}})}^2\hat{z}\right]}^{\frac{n}{2}-1}\\ \left\{2{(x-x^{\text{'}})}^2\hat{x}+{(y-y^{\text{'}})}^2\hat{y}+{(z-z^{\text{'}})}^2\hat{z}\right\}\end{array}\right\} \\ =\frac{n}{2}{\left(r^2\right)}^{\frac{n}{2}-1}\left(2\stackrel{\rightharpoonup }{r}\right) \\ =n\left(r^n\right)\left(r^{-2}\right)r\hat{r} \end{gathered}$$

Solve further as,

$$\begin{gathered} \nabla \left(r^n\right)=n\left(r^n\right)\left(r^{-2}\right)r \\ =n{r}^{n-1}\hat{r} \end{gathered}$$

Therefore, the general expression is ${}^{\nabla \left({\mathrm{r}}^{\mathrm{n}}\right)=\mathrm{n}{\mathrm{r}}^{\mathrm{n}-1}\hat{\mathrm{r}}}$.