Problem 50:  More curve sketching ($✓$ $✓$ $✓$) 2001 Paper II

(i)
The curve ${C}_{1}$ passes through the origin in the $x$$y$ plane and its gradient is given by $\frac{\mathrm{d}y}{\mathrm{d}x}=x\left(1-{x}^{2}\right){e}^{-{x}^{2}}.$

Show that ${C}_{1}$ has a minimum point at the origin and a maximum point at $\left(1,\frac{1}{2}\phantom{\rule{0.3em}{0ex}}{e}^{-1}\right)\phantom{\rule{0.3em}{0ex}}$. Find the coordinates of the other stationary point. Give a rough sketch of ${C}_{1}\phantom{\rule{0.3em}{0ex}}$.

(ii)
The curve ${C}_{2}$ passes through the origin and its gradient is given by $\frac{\mathrm{d}y}{\mathrm{d}x}=x\left(1-{x}^{2}\right){e}^{-{x}^{3}}.$

Show that ${C}_{2}$ has a minimum point at the origin and a maximum point at $\left(1,k\right)\phantom{\rule{0.3em}{0ex}}$, where $k>\frac{1}{2}\phantom{\rule{0.3em}{0ex}}{e}^{-1}.$ (You need not ﬁnd $k\phantom{\rule{0.3em}{0ex}}$.)

No work is required to ﬁnd the $x$ coordinate of the stationary points, but you have to integrate the differential equation to ﬁnd the $y$ coordinate. For the second part, you cannot integrate the equation — other than numerically, or in terms of rather obscure special functions that you almost certainly haven’t come across, such as the incomplete gamma function deﬁned by

$\Gamma \left(x,a\right)={\int }_{0}^{a}{t}^{x-1}{\mathrm{e}}^{-t}\mathrm{d}t\phantom{\rule{0.3em}{0ex}}.$

However, you can obtain an estimate, which is all that is required, by comparing the gradients of ${C}_{1}$ with ${C}_{2}\phantom{\rule{0.3em}{0ex}}$ and thinking of the graphs for $-1\le x\le 1\phantom{\rule{0.3em}{0ex}}$. This is perhaps a bit tricky; an idea that you may well not alight on under examination conditions.

Solution to problem 50

${C}_{1}$ has stationary points when $\frac{\mathrm{d}y}{\mathrm{d}x}=0$, i.e. when $x=0$, $x=+1$ or $x=-1$. To ﬁnd the $y$ coordinates of the stationary points, we integrate the differential equation, using integration by parts:

$y=\int \left(1-{x}^{2}\right)x{\mathrm{e}}^{-{x}^{2}}\mathrm{d}x=\left(1-{x}^{2}\right)\left(-\frac{1}{2}{\mathrm{e}}^{-{x}^{2}}\right)-\int x{\mathrm{e}}^{-{x}^{2}}\mathrm{d}x=-\frac{1}{2}\left(1-{x}^{2}\right){\mathrm{e}}^{-{x}^{2}}+\frac{1}{2}{\mathrm{e}}^{-{x}^{2}}+\text{const}=\frac{1}{2}{x}^{2}{\mathrm{e}}^{-{x}^{2}}\phantom{\rule{0.3em}{0ex}},$

where we have used the fact that ${C}_{1}$ passes through the origin to evaluate the constant of integration. The coordinates of the stationary points are therefore $\left(1,\frac{1}{2}\phantom{\rule{0.3em}{0ex}}{\mathrm{e}}^{-1}\right)$, $\left(0,0\right)$ and $\left(-1,\frac{1}{2}\phantom{\rule{0.3em}{0ex}}{\mathrm{e}}^{-1}\right)$.

One way of classifying the stationary points is to look at the second derivative:

$\frac{\mathrm{d}y}{\mathrm{d}x}=\left(x-{x}^{3}\right){e}^{-{x}^{2}}⇒\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^{2}}=\left(1-3{x}^{2}\right){\mathrm{e}}^{-{x}^{2}}-2{x}^{2}\left(1-{x}^{2}\right){\mathrm{e}}^{-{x}^{2}}=\left(1-5{x}^{2}+2{x}^{4}\right){\mathrm{e}}^{-{x}^{2}}\phantom{\rule{0.3em}{0ex}},$

which is positive when $x=0$ (indicating a minimum) and negative when $x=1$ and $x=-1$ (indicating maxima).

Since $y\to 0$ as $x\to ±\infty$, the sketch is:

For ${C}_{2}$, the stationary points are again at $x=0$ and $x=±1$. To classify the stationary points, we calculate the second derivative:

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^{2}}=\left(1-3{x}^{2}-3{x}^{3}+3{x}^{5}\right){\mathrm{e}}^{-{x}^{3}}\phantom{\rule{0.3em}{0ex}},$

which is positive at $x=0$ (a minimum) and negative at $x=1$ (a maximum).

Since we cannot integrate this differential equation explicitly, we must compare ${C}_{2}$ with ${C}_{1}$ to see whether the value of $y$ at the maxima is indeed greater for ${C}_{2}$ than for ${C}_{1}\phantom{\rule{0.3em}{0ex}}$.

For $0, ${x}^{3}<{x}^{2}$, so ${\mathrm{e}}^{{x}^{3}}<{\mathrm{e}}^{{x}^{2}}$ and ${\mathrm{e}}^{-{x}^{3}}>{\mathrm{e}}^{-{x}^{2}}$. Thus the gradient of ${C}_{2}$ is greater than the gradient of ${C}_{1}$. Since both curves pass through the origin, we deduce that ${C}_{2}$ lies above ${C}_{1}$ for $0 and therefore the maximum point $\left(1,k\right)$ on ${C}_{2}$ has $k>\frac{1}{2}{\mathrm{e}}^{-1}\phantom{\rule{0.3em}{0ex}}$.

Post-mortem

You were probably surprised that the examiners didn’t ask you to sketch ${C}_{2}$; they usually do. It looks as if it could be done, because you know that ${C}_{2}$ passes through the origin and you can relate its slope, roughly at least, to that of ${C}_{1}$, which you did sketch. The difficulty lies in determining its behaviour as $x\to \infty$; does $y\to 0$ as for ${C}_{1}$? In fact, it doesn’t; no reason why it should. Instead, $y\to 0.153...$ . We shouldn’t have expected ${C}_{1}$ to asymptote to the $x$-axis either; the fact that it does so is due to a rather delicate balancing of the two terms that determine its gradient.