If the gradient is $x(1-x^2)e^{-x^2}$ can we find the stationary points?

If the gradient is $x(1-x^2)e^{-x^2}$ can we find the stationary points?

Question

The curve $C_1$ passes through the origin in the $x\text{-}y$ plane and its gradient is given by $\frac{dy}{dx}=x(1-x^2)e^{-x^2}.$ Show that $C_1$ has a minimum point at the origin and a maximum point at $(1,\frac{1}{2}e^{-1})$. Find the coordinates of the other stationary point. Give a rough sketch of $C_1$.

The curve $C_2$ passes through the origin in the $x\text{-}y$ plane and its gradient is given by $\frac{dy}{dx}=x(1-x^2)e^{-x^3}.$ Show that $C_2$ has a minimum point at the origin and a maximum point at $(1,k)$, where $k>\frac{1}{2}e^{-1}$. (You need not find $k$.)

Suggestion

What condition is required for a stationary point, and how do we classify them?

To find the coordinates of the maxima, we need to know $y$ as a function of $x$. What form do you think the function will take?

Other things to consider when sketching this function:

1. what happens as $x \rightarrow \pm\infty$?

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1. is this function even or odd (or neither)?

Can we find the $x$ coordinates of the stationary points? Can we classify them?

Can we prove the inequality involving $k$ by comparing with the function $C_1$?

Solution

$$\frac{dy}{dx}=x(1-x^2)e^{-x^2}.$$

This is zero when

$x=0,\pm 1$

. To classify these stationary points we need the second derivative:

$$\begin{align*} \frac{d^2y}{dx^2} &= (1-x^2)e^{-x^2}+x(-2x)e^{-x^2}-2x^2(1-x^2)e^{-x^2}\\ &= (1-5x^2+2x^4)e^{-x^2}. \end{align*}$$

Thus

$$\begin{align*} y''(0) &= 1 > 0 \quad \Longrightarrow \quad \text{this is a minimum point,} \\ y''(-1)=y''(1) &= (1-5+2)e^{-1} = -\frac{2}{e} < 0 \quad \Longrightarrow \quad \text{these are maximum points.} \end{align*}$$

Since we know the curve passes through the origin, $(0,0)$ must be a minimum point.

To find the coordinates of the other stationary points we need to know the values of $y(\pm1)$ which requires us to integrate the given expression. There are (at least) two different ways of doing this.

Method 1: integration by parts

Let’s try to integrate

$x(1-x^2)e^{-x^2}$

by parts.

Remember that the formula for integration by parts is $\int u\frac{dv}{dx}\:dx=uv-\int v\frac{du}{dx}\:dx$

Notice that $xe^{-x^2}$ is the derivative of $-\frac{1}{2}e^{-x^2}$. So let’s set $\frac{dv}{dx}=xe^{-x^2}, \implies v=-\frac{1}{2}e^{-x^2},$ and $u=1-x^2, \implies \frac{du}{dx}=-2x.$

So we have that

$$\begin{align*} y=\int x(1-x^2)e^{-x^2}\:dx &= -\frac{1}{2}(1-x^2)e^{-x^2}-\int x e^{-x^2}\:dx\\ &=-\frac{1}{2}(1-x^2)e^{-x^2}+\frac{1}{2}e^{-x^2}=\frac{1}{2}x^2e^{-x^2}. \end{align*}$$

Notice that we don’t need to add a constant because the curve passes though the origin (or we could say the constant of integration is 0).

Method 2: trial solution

We’ll guess at a solution of the form

$y=(ax^2+bx+c)e^{-x^2}$

.

From the work above finding $\frac{d^2y}{dx^2}$ we can see that differentiating an expression of this form tends to increase the degree of the leading polynomial by one. So when integrating, we expect the leading cubic to become a quadratic.

$$\begin{align*} \frac{dy}{dx} &= (2ax+b)e^{-x^2}-2x\left(ax^2+bx+c\right)e^{-x^2} \\ &= \left(-2ax^3-2bx^2+2(a-c)x+b\right)e^{-x^2} \end{align*}$$

Comparing coefficients with $\dfrac{dy}{dx}=x(1-x^2)e^{-x^2}$, we find that

$$\begin{align*} b &= 0,\\ -2a &= -1 ,\quad \Longrightarrow \quad a=\frac{1}{2},\\ 2\left(\frac{1}{2}-c\right) &= 1\quad \Longrightarrow \quad c=0 \end{align*}$$

Thus $y=\frac{1}{2}x^2e^{-x^2},$ which we can see satisfies the boundary condition $x=0,\,y=0.$

Our maxima are therefore found at $\left(\pm 1,\dfrac{1}{2e}\right)$. Furthermore, the function is even, that is $f(-x)=f(x)$ for all $x$; and as $x \rightarrow \pm\infty$, $y \rightarrow 0$.

So now $\frac{dy}{dx}=x(1-x^2)e^{-x^3}.$

Note the change from $-x^2$ to $-x^3$ in the power of $e$ means this becomes a really hard differential equation to solve.

The methods we used above both fail. Fortunately we can complete the question without directly solving this equation.

The stationary points are again at

$x=0,\,\pm 1$

. Classifying these:

$$\begin{align*} \frac{d^2y}{dx^2} &= \left(1-x^2\right)e^{-x^3} -2x^2e^{-x^3} -3x^3\left(1-x^2\right)e^{-x^3} \\ &= \left(1-3x^2-3x^3+3x^5\right)e^{-x^3}. \end{align*}$$

Thus

$$\begin{align*} y''(0) &= 1 > 0 \quad \Longrightarrow \quad \text{this is a minimum point,} \\ y''(\pm1) &= (1-3\mp3\pm3)e^{-1} = -\frac{2}{e} < 0 \quad \Longrightarrow \quad \text{these are maximum points.} \end{align*}$$

Now we need to show the maximum point at $(1,k)$ satisfies $k>\frac{1}{2}e^{-1}$. We cannot evaluate the function directly, but we know that the curve passes through the origin, like $C_1$, and that $C_1$ has a maximum at $(1,\frac{1}{2}e^{-1})$. So we know $C_1$ reaches this point, and we want to show $C_2$ goes above it. For $0<x<1$, we have:

$$\begin{align*} x^2 &> x^3 \\ \Longrightarrow \quad -x^2 &< -x^3 \\ \Longrightarrow x(1-x^2)e^{-x^2} &< x(1-x^2)e^{-x^3}. \end{align*}$$

So the derivative of $C_2$ is greater than that of $C_1$ in this range. That means that $C_2$’s $y$ value is “growing” faster than $C_1$’s between the origin and the maximum point at $x=1$, so its maximum point will have greater $y$-coordinate, that is to say, the maximum occurs at $(1,k)$, where

$$k>\frac{1}{2}e^{-1}$$