Meaning of $\vec{r}^n$ and proof for $\nabla(\vec{r}^n)=nr^{n-2}\vec r$

Question

I have to prove $\nabla(\vec{r}^n)=nr^{n-2}\vec r$, where $\vec{r}=x\hat i+y\hat j+z\hat k$.

But how can I define the powers of vector? $r^2$ is $r.r$. Is $r^3=(r.r)r$? But then how will I define the gradient of $r^3$ since its a vector?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \partiald{r^{n}}{x} & = nr^{n - 1}\,\partiald{r}{x} = {1 \over 2}\,nr^{n - 2}\,\partiald{r^{2}}{x} = {1 \over 2}\,nr^{n - 2}\,\partiald{\pars{x^{2} + y^{2} + z^{2}}}{x} = {1 \over 2}\,nr^{n - 2}\,\pars{2x} = nr^{n - 2}\,x \end{align} Similarly for $\ds{y\ \mbox{and}\ z}$ 'components'. So, $\bbox[#ffe,10px,border:1px dashed navy]{\,% \ds{\nabla r^{n} = nr^{n - 2}\ \vec{r}}}\,$.

Answer 2

If $f$ is an scalar then $$\nabla f = \dfrac{\partial f}{\partial r} \hat{r} + \dfrac{1}{r} \ \dfrac{\partial f}{\partial \theta} \hat{\theta} + \dfrac{1}{r \sin \theta} \ \dfrac{\partial f}{\partial \phi} \hat{\phi} $$ so for the case of $f=f(r)$ we have $\nabla f(r) = \dfrac{\partial f}{\partial r} \dfrac{\vec r}{r} = \dfrac{\partial r^n}{\partial r} \dfrac{\vec r}{r} = nr^{n-2}\vec r$.

PS For $\vec f$ being a vector we may define the "Tensor derivative". See e.g. Wiki.

Answer 3

The detailed proof is as follows : $$\vec{r}=x\hat i+ y\hat j + z\hat k$$ $$|\vec r|^n=r^n=\sqrt[n]{x^2+y^2+z^2}$$ $$\nabla =\frac{\partial}{\partial x}\hat i+\frac{\partial}{\partial y}\hat j +\frac{\partial}{\partial z}\hat k $$

$$\nabla r^n =\frac{\partial r^n}{\partial x}\hat i+\frac{\partial r^n}{\partial y}\hat j +\frac{\partial r^n}{\partial z}\hat k $$

$$\frac{\partial r^n}{\partial x}=\frac{\partial (x^2+y^2+z^2)^{n/2}}{\partial x}= \frac{n}{2}(x^2+y^2+z^2)^{\frac{n}{2}-1}\frac{\partial}{\partial x} (x^2+y^2+z^2)$$

$$= \frac{n}{2}(x^2+y^2+z^2)^{\frac{n}{2}-1}2x$$

$$= n(x^2+y^2+z^2)^{\frac{n}{2}-1}x$$ $||y$

$$\frac{\partial r^n}{\partial y}=\frac{\partial (x^2+y^2+z^2)^{n/2}}{\partial y}= n(x^2+y^2+z^2)^{\frac{n}{2}-1}y$$

and $$\frac{\partial r^n}{\partial z}=\frac{\partial (x^2+y^2+z^2)^{n/2}}{\partial z}= n(x^2+y^2+z^2)^{\frac{n}{2}-1}z$$ Finally, $$\nabla r^n=n(x^2+y^2+z^2)^{n/2-1}(x\hat i+ y\hat j + z\hat k)$$

$$\nabla r^n=n(x^2+y^2+z^2)^{n/2-1}\vec r$$

$$\nabla r^n=n(x^2+y^2+z^2)^{\frac{n-2}{2}}\vec r$$

$$\nabla r^n=nr^{n-2}\vec r$$