Mechanics I: Distance Traveled along a 1-D path

I ran into a rather abstruse question in today's first mechanics recitation.  The question gave the one dimensional position of a particle with respect to time as

$x = 10 - 4t + 2t^3$

It then asked for the distance traveled by the particle between t = 0 and t = 2.  The suggested answer, (from the prof in charge of TAs), was to plot the trajectory of the particle, thereby demonstrating the distance and displacement were different.  Here's the plot:

The idea is that you can see that the particle travelled form 10 to 8 and then back to 18, so the total distance is more than the displacement from 10 to 18 i.e. 8.

The question came up as to how to do this to get the exact answer.  Here goes

What we want to do is add up all the small, (read infinitesimal), distances travelled by the particle between time 0 and time 2.  The phrases, 'adding up', and infinitesimal provide the tip off that we'll do an integral, so:

Getting to the Integral the better way with the chain rule

Since we want to add up all the little pieces of distance that the particle traveled, we want something like:

$distance = \int_{t = 0}^{t = 2} dx$

which is all fine and jolly except we have $dx$ and integration limits in time.  I'd like my integration variable to be the same as the variable that describes the limits of the integral.  To get an expression for dx in terms of dt, we can use the chain rule, taking the total derivatives of both sides of

$x = 10 - 4t + 2t^3$

We arrive at:

$1dx = \left(-4+6t^2\right)dt$

Notice that I took the derivative of each side using the chain rule.  The derivative of $x$ is the derivative of the outside with respect to x, $1$, times the derivative of the variable inside, $dx$. The same things works out for each expression that contains $t$.

The not nearly as good way of looking at getting to dx in terms of dt

Let's do a substitution of variables.  From the original position expression we can get, (by taking a derivative with respet to time),

$$ \dfrac{dx}{dt} = -4 + 6t^2$$

Being the lazy physicists that we are, we multiply both sides by $dt$ to get:

$$ dx = \left(-4 + 6t^2\right) dt$$

Now, On with the derivative

The next big thing to keep in mind is that we need to add up all distances the same regardless of their directions, (the negative vs. the positive x direction).  To achieve this, we can either take the absolute value of dx, or we can use a trick I like better which is to square dx and then take its square root.  I prefer this because when I start dealing with two or more dimensions, I'll already have the Pythagorean theorem.  In other words when I say, 

$$\sqrt{dx^2}$$

I'm thinking

$$\sqrt{dx^2 + dy^2}$$

but in this problem, since we're in one dimension, $$dy = 0$$

To illustrate why we have to use either the absolute value, or the square root of the square, here's our expression for $dx$ vs time without the absolute value/square root:

With the absolute value or the square root/square combination we get the following:

Notice in the first plot, dx goes negative for a while.  This will be counted as negative distance traveled when we integrate.  Since we want to know the total distance traveled, no matter which direction, we need to use the expression shown in the second plot where the 'negative direction' distance has been flipped positive by the absolute value.

Here's a diagram that might help make more sense of the how we're going to sum distance. Notice that the red arrow moves in the backwards direction, but we still count it as posititive distance covered.

In any event, we wind up with the integral

$$distance = \int_{t = 0}^{t = 2}  \sqrt{\left(4 + 6t^2\right)^2} dt$$

Performing this integral in Sage gives:

$12.35$ which is larger than the displacement of $8$ and slightly larger than our initial estimate from the plot of about $12.$

Coming soon, the Sage code here on the page...

The Promised Sage Code

Plot the position vs. time

Show the graph of dx/dt vs. time with all distance positive

More to come...