Prove that a radial force field is conservative

Question

A radial or "central" force field $\mathbf{F}$ in the plane can be written in the form $\mathbf {F}(x,y)=f(r)\mathbf{r} $, where $\mathbf {r}=x\mathbf{i}+y\mathbf {j}$ and $r=\Vert \mathbf {r}\Vert$. Show that such a force field is conservative.

This problem is an exercise of Apostol Vol. II and came immediately after the Second Fundamental Theorem of Calculus for Line Integrals (SFTCLI) (the first fundamental theorem is ennounced after the SFTCLI and also after this problem ).
It is said that a vector field $\mathbf {f}$ is conservative if it is a gradient of some scalar field $\varphi$, which is called the potential function of $\mathbf {f}$. It is all I know to prove this, by now I don't know how to prove that $\mathbf {F}$ is a gradient. I verified that $\varphi_{xy}=\varphi_{yx}$ (where $\varphi$ is supposed to be the potential function of $\mathbf{F}$) but I know that is not a sufficient argument to say that $\mathbf{F}$ is a gradient. $$\mathbf {F}(x,y)= xf(r)\mathbf {i} +yf(r)\mathbf {j}=\nabla_{\varphi}$$ $$\varphi_{x}=xf(r), \varphi_{y}=yf(r) $$ $$\varphi_{xy}=xf'(r)\frac{\partial r}{\partial {y}}=\frac{xyf'(r)}{\sqrt{x^2+y^2}}$$ $$\varphi_{yx}=yf'(r)\frac{\partial r}{\partial {x}}=\frac{xyf'(r)}{\sqrt{x^2+y^2}}$$

What can I do?

Answer 1

Observe an anti-derivative of $rf(r)$ is given by \begin{align} F(r) = \int^r_0 \tau f(\tau)\ d\tau. \end{align} In particular, if $r = \|\mathbf{r}\|$, then it follows from FTC and chain rule that \begin{align} \nabla F(r) = rf(r)\nabla\mathbf{r}. \end{align} Moreover, observe that \begin{align} \nabla \mathbf{r} = \frac{x}{r}\mathbf{i} +\frac{y}{r}\mathbf{j}, \end{align} then it follows \begin{align} \nabla F(r) = f(r)(x\mathbf{i}+y\mathbf{j}). \end{align}

Answer 2

The easiest way to verify this is to use the following theorem:

Let $\mathbf{F}(x,y)=P(x,y)\mathbf{i} + Q(x,y)\mathbf{j}$. If $P,Q$ are both continuously differentiable on a simply connected open region (the entire plane is such a region), then $\mathbf{F}(x,y)$ is a gradient if and only if

$$P_y = Q_x \quad \text{ for all }(x,y)$$

EDIT- As pointed out by Ted in the comments below this theorem will actually NOT work for this particular instance. I will keep the answer here because frequently this WILL be the theorem you use to show that a given vector field is a gradient.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \nabla \times \bracks{\mrm{f}\pars{r}\vec{r}} & = \bracks{\nabla\mrm{f}\pars{r}}\times\vec{r}\ +\ \mrm{f}\pars{r}\nabla\times\vec{r} = \mrm{f}'\pars{r}\,\ \overbrace{{\vec{r} \over r} \times \vec{r}}^{\ds{\vec{0}}}\ +\ \mrm{f}\pars{r}\ \overbrace{\nabla\times\nabla\pars{{1 \over 2}\,r^{2}}}^{\ds{\vec{0}}} \\[5mm] & = \bbx{\vec{0}} \end{align}