# Scientific curves

## Problem

Curve sketching is an essential art in the application of mathematics to science. A good sketch of a curve does not need to be accurately plotted to scale, but will encode all of the key information about the curve: turning points, maximum or minimum values, asymptotes, roots and a sense of the scale of the function.

Sketch $V(r)$ against $r$ for each of these tricky curves, treating $a, b$ and $c$ as unknown constants in each case. As you make your plots, ask yourself: do different shapes of curve emerge for different ranges of the constants, or will the graphs look similar (i.e. same numbers of turning points, regions etc.) for the various choices?

1. An approximation for the potential energy of a system of two atoms separated by a distance $r$

$$V(r) = a\left[\left(\frac{b}{r}\right)^{12}-\left(\frac{b}{r}\right)^6\right]$$

2. A radial probability density function for an electron orbit

$$V(r) = ar^2e^{-\frac{r}{b}}$$

3. Potential energy for the vibrational modes of ammonium

$$

V(r)=ar^2+be^{-cr^2}

$$

## Student Solutions

1. A radial probability density function for an electron orbit

$$f(r) = Nr^2e^{-\frac{2r}{a}}$$

I feel that the easiest way to think of this curve is to first consider the curves $r^2$ and $e^-r$ separately, then consider the shape of the curve produced when multiplying the two together.

$r^2$ grows with r whilst $e^{-r}$decays to 0 with increasing r

when $r$ is very small $e^{-r}$ $\approx$ 1, therefore the curve is effectively $r^2$.

when $r \to \infty$ then $e^{-r} \to 0$, when we multiply the two numbers together a very small number will therefore result.

At $r = 0$, $f(r) = f(0) = 0$

at $f(r) = 0$

either $r^2= 0$ therefore $r = 0$

or $e^{\frac{-2r}{a}} = 0 $ therefore $r =\ln(0) = \infty$

as $r \to \infty$ we have $f(r) \to 0$

If we now differentiate the function we will be able to find its turning points

$$ \frac{d}{dr} Nr^2e^{-\frac{2r}{a}}= 2Nr(1 - \frac{r}{a})e^{\frac{-2r}{a}}$$

When $ \frac{d}{dr} = 0$ we find $r =0$, $r= a$ and $r = \infty$.

If we now differentiate once more to determine the nature of each of the turning points we find that:

$$ \frac{d^2f}{dr^2}= \frac{d}{dr}2Nr(1 - \frac{r}{a})e^{\frac{-2r}{a}}

=2(1-\frac{r}{a} + )(2r)(\frac{-1}{a}))Ne^{\frac{-2r}{a}} + (\frac{-2}{a}(2r)(1 - \frac{r}{a})Ne^{\frac{-2r}{a}}$$

at $r = 0$: $ \frac{d^2f}{dr^2}=2N$ which is greater than $0$, therefore $r = 0$ is a minimum

at $r = \infty$: $ \frac{d^2f}{dr^2}= 0$

at $r = a$: $ \frac{d^2f}{dr^2}=-2Ne^{-2}$ which is less than $0$, therefore $r = a$ is a maximum.

From the above information we can deduce the general shape of $f(r)$ as shown below.

(Note: The graph below has been plotted assuming all constants equal 1 )

2. Potential energy for the vibrational modes of ammonium

$$ V(x)=\frac{1}{2}kx^2+be^{-ax^2} $$

When $x$ is very small $v(x) \approx be^{-ax^2}$

and when $x$ is large $e^{-ac^2} \to 0$, $v(x) \approx \frac{1}{2}kx^2$

when $x = 0$, $v(0) = b$ by inspection it can be deduced that $v(x)$ can never reach $0$ as $be^{-ax^2}$ is always greater than $0$ and $\frac{1}{2}kx^2$ is greater than $0$ at all $x$ except $x=0$, but at this point $be^{-ax^2}= b$

as $x \to \infty$ we have $v(x) \to \frac{1}{2}kx^2 \to \infty$

If we now differentiate the function we will be able to determine its turning points.

$$ \frac{dv}{dx} = xk - 2abxe^{-ax^2} = x\left(k-2abe^{-ax^2}\right)$$

when $ \frac{dv}{dx} = 0$ we find $x = 0$ and $(k-2abe^{-ax^2}) =0$

therefore $x = \sqrt{\frac{-1}{a}\ln\left(\frac{k}{2ab}\right)}$

If we now differentiate once more to determine the nature of each of the turning points we find that:

$$ \frac{d^2v}{dx^2}= k - 2abe^{-ax^2} + 4a^2bx^2e^{-ax^2}$$

at $x = 0$, $ \frac{d^2v}{dx^2}= k- 2ab$ , the nature of this turning point will therefore depend on the relative magnitudes of the constants $a$, $b$ and $k$. If $k$ is greater than $2ab$, $x=0$ will be a minimum but if $2ab$ is greater than $k$ it will be a maximum.

at $x= \sqrt{\frac{-1}{a}\ln\left(\frac{k}{2ab}\right)}$ we should expect a minimum (or possibly a point of inflexion) as a consequence of the fact that as there are only two turning points and as $x \to \infty$ $v(x) \to \infty$ (it therefore cannot be a maximum).

## Teachers' Resources

This problem is aimed at interested and enthusiastic students: although we don't recommend it for mainstream classroom use it provides enrichment and will develop mathematical thinking and knowledge.

Like all NRICH problems it has been carefully designed to be full of mathematical interest and as such might be used in a variety of ways: you might suggest that keen students consider the problem in their own time, use it for a maths club at school or use it in class as a vehicle to foster discussion and mathematical thinking. See our Teachers Guide to Getting Started with rich tasks for more suggestions and ideas.

This problem will be of particular interest to students considering a university course containing an element of mathematical physics.