An Application of the Chain Rule and the Importance of Graphs - TSV 4.35

Here we are given $a$ as a function of $x$:

$$a(x)=24-6x^2$$

Applying the chain rule we have

$$\begin{eqnarray*} \int a(x)dx &=& \int vdv\\ \int (24-6x^2)dx &=& \int vdv\\ 24x-2x^3 &=& \frac{v^2}{2}+C \end{eqnarray*}$$

If we apply our initial condition we find that $C=24x_0-2x_0^3-v_0^2/2=0$ and so we have

$$v=\pm\sqrt{48x-4x^3}$$

Thus we have multiple solutions for each of the remaining questions:

$$v(x=2.5)=\pm7.58\textrm{m}/\textrm{s}^2$$

$$v=0 \quad \textrm{when} \quad x=0,\pm2\sqrt{3}\textrm{in}$$

$$v \quad \textrm{maximum when} \quad x=\pm2 \textrm{in}$$

Where in the last one we find the maximum by $a(x)=0$ and solving for x.

However, we shall see that not all the solutions are applicable by looking at some graphs.

Figure 1: Acceleration and Velocity as functions of $x$

$\begin{array}{c@{\hspace{1in}}c} \includegraphics[scale=.5]{aofx.eps} & \includ... ...cs[scale=.5]{vofx.eps} \\ a(x)=24-6x^2 & v(x) = \pm\sqrt{48x-4x^3} \end{array}$

Here are $a$ and $v$ as functions of $x$ for the initial condition specified ($x=0$ when $v=0$). What we must notice is that once we start at the origin, we will always remain on this oval part on the right - we cannot ever end up on that part on the left - they are disconnected. We say that this motion is periodic - after a definate amount of time we have returned to where we started. Thus we are stuck to the x values $0\le x \le 2\sqrt{3}$. Thus we must reject the solution $x=-2\sqrt{3}\textrm{in}$ for when $v=0$, as this position is not accessible. Likewise we reject $x=-2$ for a position when $v$ is extremal.

Now, does this mean we can never have $x<0$? Of course we can, but that requires a different initial condition! This next graph shows several velocity plots for different values of the initial value parameter $C$, as we will see on the next page.

Figure 2: Velocity family, C = -50, -32, -20, 0, +20

$$v(x)=\pm\sqrt{48x-4x^3+2C}$$

Here, a positive $C$ value means $v=0$ at some $x>0$ (see the equation for $C$ above to convince yourself of this). Thus the allowed region shrinks away from the origin (we cannot now have $x=0$ because then $v^2<0$ which is impossible).

Conversely, a negative $C$ means we have some positive $v$ at $x=0$. We can increase this value of $v_0$, and move through a region of periodic motion where we can have $x<0$, until we see the ``$\times$'' region in the graph at $x=-2$ (this is for $C=-32$). The motion on this curve is no longer periodic since we have both $v=0$ and $a=0$ at the point $x=-2$. What happens is that as we approach this point we continually slow down, but never reach it as t $\to \infty$. The motion is asymptotic.

Finally, if we decrease $C$ beyond $-32$ our two regions connect and the behavior is quite different. Since we always have $a<0$ when $x<-2$, then if we're on the top velocity branch, our velocity decreases and we eventually get pushed toward the right half. However, once we're on the bottom half the velocity becomes more negative and we get pushed away, infinitely far to the left. Thus our motion is now unbounded.

One last note - this is a problem where we cannot find $x(t)$ analytically. We have

$$\frac{dx}{dt}=v(x)=\pm\sqrt{48x-4x^3+2C}$$

and upon separating our variables we obtain

$$\int dt=\pm \int \frac{dx}{\sqrt{48x-4x^3+2C}}$$

and at this point we are stuck - there is no expression for this integral in terms of elementary functions (nor even non-elementary ones, so far as I and Maple are aware), except for possibly some particular $C$ values. So we cannot even write down $t(x)$, let alone invert it for $x(t)$.