When logic and calculus collide

[Lil' Lavery]

A few days ago we were given this problem as a warm-up in calculus,
"Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same velocity."
Well, if you consider the position equation f(t)=g(t)-h(t) (g being the position of one runner and h the other). Then the velocity equation would be f'(t)=g'(t)-h'(t), so if the velocities are the same, f'(t)=0. When t=0 and when the race end (another value t), and possibly other ponts, f(t)=0. And because the derivative of a constant is 0, then f'(t) would also be 0. Another way to look at it would be that both runners ran the same distance in the same amount of time, and therefore had the same average velocity, therefore g'(t)=h'(t), and once again f'(t)=0.
But, logically, does it make sense? Other than at the start (which we assume would have a velocity of 0), would the two runners be running at the same speed, at the same time? They would both have to run at the average velocity at one point or another, but would that (or any other speed) be at the exact same moment?

[Tim Delles]

It is completely logical.

Think about it this way.

One runner starts out fast and the other runner starts out slow.

However as the 2nd runner speeds up, eventually his velocity will have to equal the first runners velocity, and then become greater than his velocity so that they could finish at the same time.

It also can be proven that the first runner could slow down enough that the second runner could run at a constant speed, but still at atleast 1 other point besides the start line they are running at the same velocity

***Not sure if it makes sense to you but it make sense to me***

[Bill_Hancoc]

Originally Posted by Tim Delles

It is completely logical.

Think about it this way.

One runner starts out fast and the other runner starts out slow.

However as the 2nd runner speeds up, eventually his velocity will have to equal the first runners velocity, and then become greater than his velocity so that they could finish at the same time.

It also can be proven that the first runner could slow down enough that the second runner could run at a constant speed, but still at atleast 1 other point besides the start line they are running at the same velocity

***Not sure if it makes sense to you but it make sense to me***

I agree. It seem from a logical standpoint that they would have the same velocity at least once in the race for some point in time. It may be for only 1 millionth of a second but its still the same velocity at the same time

[GRaduns340]

It makes sense.

I'm not sure your assumption of the starting velocities being the same is completely true though. Technically, there is no derivative of the position graph at zero, because the derivative doesn't exist at the endpoints of a graph. Logically, yes, their velocities are both 0, but in order to use a derivative to prove it you would have to assume that the runners were both at rest before the start of the race, but with a lack of that knowledge there is no way to find f'(0).

You'd actually have to prove that it was an intermediate time at which their velocities were the same, like Tim suggested. I don't know if you've done it yet, but the mean value theorem would be useful there. (I think I pulled out the right name).

[Lil' Lavery]

Originally Posted by GRaduns340

It makes sense.

I'm not sure your assumption of the starting velocities being the same is completely true though. Technically, there is no derivative of the position graph at zero, because the derivative doesn't exist at the endpoints of a graph. Logically, yes, their velocities are both 0, but in order to use a derivative to prove it you would have to assume that the runners were both at rest before the start of the race, but with a lack of that knowledge there is no way to find f'(0).

You'd actually have to prove that it was an intermediate time at which their velocities were the same, like Tim suggested. I don't know if you've done it yet, but the mean value theorem would be useful there. (I think I pulled out the right name).

Yeah, we used the Mean Value Theorem to prove it via calculus. And according to calculus, their starting velocities don't have to be equal to 0, but I know calculus can prove it, I was wondering if it worked logically.
I suppose that the two speeds would have to be equal at some point, but I can't help this feeling that there's some way that it could be done with never having the same speed.

[Greg Marra]

Originally Posted by Lil' Lavery

I suppose that the two speeds would have to be equal at some point, but I can't help this feeling that there's some way that it could be done with never having the same speed.

Picture a plot that contains both runners velocities plotted against time.

Velocity is (m/s), right? So when we integrate that against time (the x-axis) we get just (m). This means that the area under the velocity curves for the two runners is the distance they have run. They run an equal distance, so the area under the two curves has to be the same.

Draw two curves such that the area under the two are the same, that one racer finishes before the other, and that they never cross. If the curves cross, it means they have the same velocity.

[KenWittlief]

starting velocity is easy to straighten out. They dont both have to start at the same time. One person might be late for the race, and run up to the starting line and run across it, so his starting velocity was not zero.

But that means he could run faster for the whole race and just catch up at the finish line. They never ran at the same speed at any point.

thinking they both started at the same time is a mistaken assumption. In fact, according to the laws of physics it would be impossible for them both to start at exactly the same instant (point in time). (action and reaction - time for sound to reach the ear of both runners - human response time for sound to cause legs to move... the two runners could never start perfectly synchronised, one will always start before the other).

This kills the proof already.

The second assumption is that both runners ran exactly the same distance. For this to be true one would have to run exactly behind the other, in exactlyt the same path, or they would have to run in a straight line. If one runner wanders off the path of the course the slightest bit, he has to run faster for the race to be a tie, so again, he could run faster for the whole race, and still end in a tie

a third assumption: that to speed up and slow down you must cross through the intermediate speeds - that you ramp up, and ramp down. True when speeding up, but if you run into a lamp post, brick wall, or parked car, your velocity instantly goes to zero. The graph of your velocity would have a discontinuity at that point, and would therefore be undefined (accelerated acceleration, or jerk). At that instant your velocity instantaniously goes from, lets say 12 mph, to 0.

There is no logic to the proof. In theory: yes. In reality: no.

[Richard Wallace]

Originally Posted by Lil' Lavery

"Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same velocity."
Well, if you consider the position equation f(t)=g(t)-h(t) (g being the position of one runner and h the other). Then the velocity equation would be f'(t)=g'(t)-h'(t), so if the velocities are the same, f'(t)=0. When t=0 and when the race end (another value t), and possibly other ponts, f(t)=0.

Everything about your argument is fine up to this point. f'(t) must be zero for some t between the start and finish of the race. This can be proved using the Mean Value Theorem; the result is called Rolle's Theorem.

And because the derivative of a constant is 0, then f'(t) would also be 0.

This does not follow from the problem statement. In practice a runner's velocity at the finish line is rarely zero, and in some kinds of racing (relays, sailing, etc.) the starting velocity is not zero, either. But on a purely mathematical level, the statement "the derivative of a constant is zero" is neither precise nor pertinent here.

Precisely, the derivative of the function f(t) = C is zero for all t. However, in this example you have assumed that the function f(t) = 0 at the beginning and the end of the race, but not for times before the beginning or after the end. What the two runners did before t=0 and after crossing the finish line is not defined, nor is that information needed to prove the result.

[Kevin Sevcik]

Originally Posted by KenWittlief

starting velocity is easy to straighten out. They dont both have to start at the same time. One person might be late for the race, and run up to the starting line and run across it, so his starting velocity was not zero.

But that means he could run faster for the whole race and just catch up at the finish line. They never ran at the same speed at any point.

thinking they both started at the same time is a mistaken assumption. In fact, according to the laws of physics it would be impossible for them both to start at exactly the same instant (point in time). (action and reaction - time for sound to reach the ear of both runners - human response time for sound to cause legs to move... the two runners could never start perfectly synchronised, one will always start before the other).

Ken, you've only shown that this theorem is only true for continuously differentiable functions of position. Your first argument is true if and only if the runner can go from 0 velocity to high velocity instantly, that is, without passing through any velocities in between, with infinite acceleration. That's physically impossible.

Starting time doesn't make a lick of difference. If one starts a second after the gun and runs at speed A, and the other starts 10 seconds late, he has to run at speed A+foo to get to the finish line at the same point. And you can't physically get from 0 to A+foo without going through A at some point.

Arguments about them not starting at the same time or place are beside the point. If you want to argue that the world's not perfect well fine. Toss all your fancy mathematics and physics in the trash because you can come up with some real world situations that are completely different from the stated problem.

[KenWittlief]

well, the problem was stated in terms of humans running a footrace, not in terms of imaginary dots moving along perfectly straight, perfectly parallel lines.

As Rich pointed out, the runners do not have to start from a stand still. In sailing races, and in Nascar the racers are moving when the race begins. As long as you dont cross the starting line before the opening shot your speed is up to you. If you are moving towards the starting line before the race starts, your speed is not zero, and it may not be zero at any point during the race (from 'Go' to the finish line).

Also, runners cant possibly stay exactly on the perfect centerline of a race course, so one person will end up running a longer distance than the other, therefore he could run faster for the entire race, and still finish with a tie.

This problem being worded this way brings up an interesting point: the math we learn in college is based on linear systems, but in the real world almost nothing is linear. So while we try to make the math work out to the nth degree of accuracy and precision (the right answer), in the real world the best we can do with math is approximate the non-linear physical world.

[Lil' Lavery]

Originally Posted by Richard

This does not follow from the problem statement. In practice a runner's velocity at the finish line is rarely zero, and in some kinds of racing (relays, sailing, etc.) the starting velocity is not zero, either. But on a purely mathematical level, the statement "the derivative of a constant is zero" is neither precise nor pertinent here.

f'(t) was the difference of the two derivatives, not the derivative of either of the runners positons, and so could (and would have to be) zero. g'(t) and h'(t) were the velocities of the runners, and niether could have been zero (unless it was a race with absolutely no distance or where niether runner moved). The problem did state that both runners started at the same time, but you're right in that it may not mean they both had a velocity of 0 at the beginning.
I get the Calculus behind it (including the mean value theorem, which was the lesson where we were given this problem), I was trying to justify it logically, and I guess that it must be logically true. I've thought about it a little more and at one point they must both have been travelling the same speed.

[Richard Wallace]

Originally Posted by Lil' Lavery

f'(t) was the difference of the two derivatives, not the derivative of either of the runners positons, and so could (and would have to be) zero. g'(t) and h'(t) were the velocities of the runners, and niether could have been zero (unless it was a race with absolutely no distance or where niether runner moved). The problem did state that both runners started at the same time, but you're right in that it may not mean they both had a velocity of 0 at the beginning.
I get the Calculus behind it (including the mean value theorem, which was the lesson where we were given this problem), I was trying to justify it logically, and I guess that it must be logically true. I've thought about it a little more and at one point they must both have been travelling the same speed.

Sean, I was pretty sure that you understood the calculus assignment, and the MVT. And I'm pretty sure you will be going much further into engineering math during the next few years. That's why I thought it worthwhile to quibble.

My quibble was with the assumption of any knowledge concerning f'(t) or g'(t) or h'(t) before the start and after the finish. Part of the quibble is practical: nothing about these derivatives is given in the problem statement. The other part is mathematical: Rolle's Theorem does not depend on any particular constraints on f'(t) at the endpoints of the interval.

I only jumped in here because I've seen this kind of error in setting up engineering math problems a few times before. The assumption that I'm quibbling over makes no difference to the proof sought in this problem, but there are many engineering problems for which incorrect assumptions about starting and/or ending conditions will lead to the wrong conclusion. And you are going to see them.

When I am a doddering old retiree and you are overseeing the design of infrastructure for the Martian colony, I want your mathematical reasoning to be perfect.

[DonRotolo]

Originally Posted by Richard

When I am a doddering old retiree and you are overseeing the design of infrastructure for the Martian colony, I want your mathematical reasoning to be perfect.

Yes, especially if that's where I'll be living...

Don

[Kaizer007]

I'm sorry this is four days late, but I wanted to ask how the last part of Ken's argument way up there makes sense:

"a third assumption: that to speed up and slow down you must cross through the intermediate speeds - that you ramp up, and ramp down. True when speeding up, but if you run into a lamp post, brick wall, or parked car, your velocity instantly goes to zero. The graph of your velocity would have a discontinuity at that point, and would therefore be undefined (accelerated acceleration, or jerk). At that instant your velocity instantaniously goes from, lets say 12 mph, to 0."

I can't believe, (with only Physics and AP Physics class behind me *both High School Level*) that an instant stop is possible for running into a lamp post. I am to believe that even the slightest part of your body would start slowing down at a tremendous negative acceleration but not one that could be considered undefined. I had thought (from other math classes) that undefined acceleration could not be reached in this physical world.

[KenWittlief]

Originally Posted by Kaizer007

I'm sorry this is four days late, but I wanted to ask how the last part of Ken's argument way up there makes sense:

which part of the runners body would you measure to use as their 'speed' when they hit a brick wall? What system would you use to measure the speed from the instant that, lets say, the skin on their forehead first contacts the wall, and the point when their head comes to a compete stop?

Radar? GPS? a high speed camera? If the impact has a 1µS duration I dont think it can be measured by any means associated with running a foot race. For all practical purposes the acceleration is off the scale (infinite), compared to what a runner normally experiences.

It seems to be a mute point anyway. As others have pointed out, the two runners started 'at the same time', even though the laws of physics say this is impossible

and everyone appears to be assuming the two runners run exactly the same distance, even though this is also impossible.

The 'math class' answer is that they must run at the same speed at some point, including the starting velocity of zero. (because they are really two points moving along straight lines).

The physics (and logical) answer is that they dont have to run the same speed at any instant to end up in a tie.

In fact, as far as I know, there is no rule against a runner approaching the finish line, looping around and going back to the starting line, and looping back to the finish line, to run 3 times the distance as the other runner. This means he could run 18 mph for the entire duration of the race, while the other runner runs at 6mph for the whole race, and they both cross the finish line at the same instant.

This is why I am an engineer and not a mathematician. When you think outside the box on a math test you get an F !