An interesting riddle

[iago]

On day 1, a man goes hiking.  The path he takes is ascending, so it's slow going.  He leaves at 8am, and arrives at his destination at 5pm.  

The next day, he goes home.  Since it's downhill, the going is much faster.  He leaves at 8am and arrives home by 3pm.  

The question is: is there some time during both of the days that he was in the same position?  ie, the same position at the same time on both days.


Although I know the answer, I'm going to leave it open for now.  I'm sure it's pretty obvious to certain people what the answer is

[Yoni]

I know the answer as well. To extend the riddle:

What theorem in calculus confirms the answer?

[CrAzY]

Alg 1 student here... Graph it and find intersection points.

[iago]

Alg 1 student here... Graph it and find intersection points.

That's pretty much how I pictured it.  Basically, 2 points on a graph (on a straight line is easiest) moving in opposite directions with different velocity -- do they pass each other at any point?  

But I have no idea how you would confirm it in calculus.

[dxoigmn]

Alg 1 student here... Graph it and find intersection points.

That's pretty much how I pictured it.  Basically, 2 points on a graph (on a straight line is easiest) moving in opposite directions with different velocity -- do they pass each other at any point?  

But I have no idea how you would confirm it in calculus.

The squeeze theorem comes to mind but I'm not entirely sure.

Edit: Added link to mathworld.

[Yoni]

Nope.

[R.a.B.B.i.T]

Send a second man up the second day the exact same way the first man went up on the first day.  The point and time at which they meet is the point and time the first man was/is present on the path both days at the same time.

[edit]
Go abstract math solutions ^^

[edit2]
Am I supposed to use Black color?  Eh...

[iago]

[edit2]
Am I supposed to use Black color?  Eh...

Eh, doesn't hurt

[K]

How about this:

let the distance of the hike = d;
let dx/dt be the rate (distance/time) at which he travels on both days.

given:
x₁(0) = 0; (hasnt gone up yet)
x₂(0) = d; (hasn't gone down yet)

overall times = 9hours, 7hours.
dx/dt₁ = (1/9 * d)
dx/dt₂ = -(1/7 * d) (travelling in the opposite direction)

integrate:
x₁(t) = (1/9 * d * t) + C
x₂(t) = -(1/7 * d * t) + C

solve for C using the givens above:
x₁(t) = (1/9 * d * t)
x₂(t) = - 1/7 * d * t + d = d - (1/7 * d * t)

solve for intersection:

(1/9 * d * t) = d - (1/7 * d * t)
1/9 * t = 1 - 1/7 * t
(1/9 + 1/7) t = 1
16/63 t = 1
t = 63 / 16 = 3.9 hours.

I guess you could call that the fundamental theorem of calculus, ie F(t) = integral(F'(t)dt) + C.  But the integration is hardly necessary if you know the equation of a line with slope m and intercept b

[Yoni]

dx/dt₁ = (1/9 * d)
dx/dt₂ = -(1/7 * d) (travelling in the opposite direction)

Just to nullify K's calculations and previous responses to this thread:

Assume the hiker didn't go up or down at a constant rate.
Some places were harder to hike in, and then he went slower, and some places were easier, and then he went faster.

Also, assume that on the way up, it's possible that he may have gone straight, or even down a few times (for example, when passing a small hill on the way).
Similarly, assume that on the way down, he may have gone straight or up a few times.

Also, assume that it's possible that he didn't take the exact same route going down as he did going up. Someone at the top told him of a faster way, and he chose that instead. (He also went faster because descending is easier.)

Under all these restrictions, he still must have been at the exact same height in both days at least during one time of the day.

What theorem in calculus (which, I'm afraid, hasn't been mentioned in this thread yet) confirms this?

[K]

The intermediate value theorem.

[Yoni]

The intermediate value theorem.

+1i.

[K]

You mentioning that the rate didn't have to be constant and could have increased or decreased jogged my memory.  

Definition:

Given a function f(x) that is continuous over the interval (a, b) with a < b, and a constant k such that k falls between f(a) and f(b)  [ie, f(a) > k > f(b), or f(a) < k < f(b) ], some constant c such that a < c < b is guaranteed to exist such that f(c) = k.

[Adron]

However, since things are quantized and not continous, there are no guarantees he's been at the same height at the same time both days

[Yoni]

However, since things are quantized and not continous, there are no guarantees he's been at the same height at the same time both days

Yes there are. Define "same height" as "within a single quanta(?) of the same height" and then, since the movement is a continuous function multiplied by a quantizing constant (as I would like to believe), you get the result easily using calculus as described.

If not, you'd have to look at the time, stretching and compressing with the changes of the hiker's velocity. This gets annoyingly complicated, stick to Newton mechanics for this please.