Physics / Calculus Question

I have a few loosely related questions about position functions and their integrals (and derivatives).

  1. For a function position = p(t), is there any significance to the length of the curve of p(t)?

  2. How about the integral of p(t)?

  3. Let’s say p(t) is the position of the mars rover, in meters from the landing zone. Is there any way to determine the total distance the rover travelled in one step (i.e. without splitting p(t) into intervals between direction changes)?

  4. Is there any significance to the length of velocity or acceleration curves (being first and second derivatives of position, respectively)?

My math teacher and physics teacher both don’t know.

If the function is a vector function, then the length of the curve between times t[sub]0[/sub] and t[sub]1[/sub] will be the distance travelled in that time interval. (However, you imply below that you’re not actually considering the full position vector, but rather just distance from a fixed point.)

In the absence of any other information, it doesn’t mean much. However, see Green’s Theorem and the Curl Theorem for ways in which a path integral can be useful. These come up in electrostatics, for example. (Probably not relevant to the Martian rovers though.)

Since the rover is moving in two dimensions — approximately anyway; the local Martian surface is pretty flat — the function you describe can’t give you the information you desire. A mere “distance from the landing site” function throws away too much information. You need both the x and y coordinates as a function of time.

I personally can’t think of any use for those functions, but maybe someone else can.

Oh, sorry - the Mars rover was a bad example. Let’s say it can only move in one dimension.

The basic problem I’m looking to solve is that, as I learned it, you have to split the position function into separate intervals (one interval for each period when the rover was moving in a certain direction) in order to find the total distance something traveled. I’d think there’d be any easier way to do it.

Oh, sorry - the Mars rover was a bad example. Let’s say it can only move in one dimension.

The basic problem I’m looking to solve is that, as I learned it, you have to split the position function into separate intervals (one interval for each period when the rover was moving in a certain direction) in order to find the total distance something traveled. I’d think there’d be an easier way to do it.

Easiest answers first: I don’t know of any significance to the integral of p(t) or the length of the velocity or acceleration curves — or at least I haven’t seen one in the nine years since I started working my way through an education in physics.

As for question (3), I don’t think there is. You essentially want to calculate the total displacement, which is the integral of v(t), but you want to count displacements to the left and to the right as both positive, which means you really want to integrate |v(t)| instead. I don’t think there’s a way to evaluate the integral of the absolute value of a function without splitting it up into regions where the function is positive and regions where it’s negative.

As for question (1): Not in classical mechanics, no. Think of this in terms of units: you can’t add together two quantities that measure different kinds of things. For example, trying to add together a length and a mass just doesn’t make sense. But the curve-length integral has something like

sqrt( 1 + v[sup]2/sup )

in it, which means you need to add together a velocity squared (with units of m[sup]2[/sup]/s[sup]2[/sup]) to 1 (which doesn’t have any units at all.) This doesn’t make physical sense.

Now if you get into Special Relativity, that’s another story. In that case, there’s a natural conversion factor between lengths and times, namely the speed of light, and the “length” of the position-time curve turns out to be something called the proper time. This is pretty far afield from your question, though; if you wouldn’t mind the hijack post again and I’ll try to explain it.

Thanks for all the info, MikeS and Bytegeist. My curiosity has been satisfied.

Sure, if you don’t mind. Sounds interesting.

I’m afraid there isn’t. If you insisted on using an integral to derive the arc-length (distance travelled) function, you’d end up integrating the absolute value of the speed function: Int |dx/dt| dt. But the absolute value function is “unpleasant” to deal with, in that it requires piece-wise integration — that is, integrating, separately, the ranges for which the inner function is positive and negative.

At this point, you might as well bite the bullet and just use the position function you already have, in the manner you’ve already described. If you’d like, you could use the position function’s first derivative (the speed function): the zero-crossings of the speed function will correspond to the local minima and maxima of the position function. It is between these points, of course, that you’d compute each sub-total of the object’s movement.

It is, I’m afraid, a pretty manual process no matter what.

Sorry, Mister V, but I am slightly confused by what you are asking, exactly. To answer what I think you’re asking, I should mention that it is easy to confuse total displacement vs. total distance. If a train travels from New York to Washington and back, the total displacement is zero. You would have to break up the problem into two intervals to find the total distance by summing the each displacement for each of two one-way trips…to find total distance of the round-trip.

As for length of a velocity curve (v vs. t), each point on that curve represents the instantaneous velocity (speed, because we’re graphing the magnitude only) for any one given point in time, t. Put all the points together, and the length of the curve itself doesn’t do much for me. I mean, take the simplest case. Assume velocity is constant for 10 hours. It’s a boring, flat (horizontal) “curve”, isn’t it? The length of that curve would be measured in hours, not inches, wouldn’t it? Assuming it is plotted to scale (and not a freehand sketch), you’d say “Yup, it maintained some constant speed for 10 hrs!” Basically, the same would be true for an acceleration curve. If the “curve” is flat, you’d say “Yup, it maintained the same acceleration for 10 hrs” …even if that line should coincide with the x-axis.

If I am understanding you, I think you’ve simply confused your teachers with this question… - Jinx :confused:

Alrighty, a quick course in curve lengths in Special Relativity coming right up. Hopefully I won’t make a complete mess out of it.

Think about geometry in the plane first. If we have two points on the plane, the distance between them is given by sqrt(([symbol]D[/symbol]x)[sup]2[/sup] + ([symbol]D[/symbol]y)[sup]2[/sup]). This is called an invariant, because it doesn’t depend on the coordinates we choose. As a concrete example of this, suppose we have two surveyors measuring distances on a farm. One of them, though, has a broken compass, so instead of measuring north-south lines he’s actually measuring lines that run from northwest to southeast. Despite this discrepancy, they will still both agree that the distance between the pond and the old oak tree is one-quarter of a mile, since the absolute distance between two points doesn’t depend on the coordinates you use.

Here’s where it gets interesting. In special relativity, we have to introduce the concept of an event; instead of just saying, “I was at position x”, we have to say, “I was at position x at time t.” We can then define something called the “space-time interval” between two events:

([symbol]D[/symbol]s)[sup]2[/sup] = c[sup]2[/sup] ([symbol]D[/symbol]t)[sup]2[/sup] - ([symbol]D[/symbol]x)[sup]2[/sup]

where c is the speed of light. What this says is that if you take the time separation between two events and the distance separation between two events and combine them in this odd way, you get the space-time interval between these two events.

What special relativity says is that ([symbol]D[/symbol]s)[sup]2[/sup] is also an invariant. If someone is whizzing past me in a rocket ship, her coordinates are just as good as mine, and for all she knows, it’s me who’s whizzing past him in the other direction. She’ll measure a different distance-separation between the two events than I will (since she’ll move between the times of the two events), but according to special relativity she’ll measure the same ([symbol]D[/symbol]s)[sup]2[/sup] that I will.

What does this have to do with curve length? Well, suppose my friend blasts off in her rocket ship and follows (according to me) some arbitary curve x(t) through space. According to her, though, she’s staying in one place while the world is moving around her. In particular, if we take a tiny chunk of her curve, the amount of time she sees elapse is just equal to the space-time interval, since according to her the event at the start of this chunk and at the end of this chunk take place exactly where she is at those times. So the amount of time that elapses according to her clock is given by

[symbol]Dt[/symbol] = sqrt(c[sup]2[/sup] ([symbol]D[/symbol]t)[sup]2[/sup] - ([symbol]D[/symbol]x)[sup]2[/sup]) = sqrt(c[sup]2[/sup] - ([symbol]D[/symbol]x/[symbol]D[/symbol]t)[sup]2[/sup]) [symbol]D[/symbol]t

and if we add up all these little chunks to find the total time elapsed on my friend’s clock, we get

[symbol]t[/symbol] = Int sqrt(c[sup]2[/sup] - (dx/dt)[sup]2[/sup]) dt

This is kind of like the curve length, except for the factor of c and the minus sign. It’s the closest thing I could think of in physics to a curve length that has physical importance. There’s a lot of special relativity I didn’t get to here, though. If you’re interested in learning more about it, try reading A Traveler’s Guide to Spacetime by Thomas A. Moore; it’s a good introduction to the subject for someone at your level.