Okay I have this problem I have to do, but the teacher never explained how to do it...apparently we're supposed to teach ourselves in this class. :shrug:

Anyway... A plane goes from Chicago to St. Louis in five hours, then goes back in the same amount of time. I have to show that there is some point the plane will pass at the same time both ways.

LUIS9

Sep 27th, 2005, 03:02 AM

Okay I have this problem I have to do, but the teacher never explained how to do it...apparently we're supposed to teach ourselves in this class. :shrug:

Anyway... A plane goes from Chicago to St. Louis in five hours, then goes back in the same amount of time. I have to show that there is some point the plane will pass at the same time both ways.

This sounds more like a physics problem to me, draw the vectors out and maybe apply the law of cosines, actually forget it. Nonetheless, it does sound like good old kinematics. since you arent given any directions I guess the vector free diagrams wont be of help. Sorry I wish I could help, throw me in some derivatives or integrals I'd probably still be able to tackle those.

Rtael

Sep 27th, 2005, 03:03 AM

I've already done the derivative problems, and we haven't learned integrals yet. :p

Rtael

Sep 27th, 2005, 03:36 AM

*bump*

416_Man

Sep 27th, 2005, 03:39 AM

That is why I took data management and probability. :p ;)

Rtael

Sep 27th, 2005, 03:41 AM

Loser. :p

LUIS9

Sep 27th, 2005, 03:49 AM

Loser. :p

I wouldnt call someone a loser just because they opted on taking some other mathematical course to replace calculus. Its really only useful to a handful of people, scientists, engineers(civil, structural, mechanical, actually probably all engineers, not so much computer engineers, there more logic oriented). However, the challenge of taking a calculus course is very rewarding especially if your mathematical skills arent very exhaustive.

Rtael

Sep 27th, 2005, 03:56 AM

I was just joking, dear.

LUIS9

Sep 27th, 2005, 04:06 AM

I was just joking, dear.

I am sure you were!;)

backhanddtl4

Sep 27th, 2005, 05:52 AM

AHHH I'm doing the same stuff. But this problem is...rather out of the ordinary. We are doing derivatives and soon, integrals...PUKE

drake3781

Sep 27th, 2005, 06:22 AM

Okay I have this problem I have to do, but the teacher never explained how to do it...apparently we're supposed to teach ourselves in this class. :shrug:

Anyway... A plane goes from Chicago to St. Louis in five hours, then goes back in the same amount of time. I have to show that there is some point the plane will pass at the same time both ways.

Hmmm... I want to help but I don't understand your explanation. Is this the correct problem? Was it written and can you add any information?

LUIS9

Sep 27th, 2005, 06:25 AM

AHHH I'm doing the same stuff. But this problem is...rather out of the ordinary. We are doing derivatives and soon, integrals...PUKE

Integrals are better than derivatives trust me. No more implicit differentiation, thats the worst of the derivatives, well the applied forms can get funky like the related rate problems and max min applications. Trut me you'l love integrals until you get to the applications, volumes, area problems beyond those you will get on the typical standarized calculus tests Ap and IB.

LUIS9

Sep 27th, 2005, 06:26 AM

Hmmm... I want to help but I don't understand your explanation. Is this the correct problem? Was it written and can you add any information?

Right, it does seem like info is missing certainly some direction at least or Initial rates or velocities.

Prizeidiot

Sep 27th, 2005, 07:51 AM

I'm assuming its a physical application question.

Yeah... there's definitely something missing. If you can find that missing bit, maybe we can help.

Rtael

Sep 27th, 2005, 11:10 AM

Well I guess I forgot to add that the plane takes the same route both ways, but I thought that would be assumed. Everything else is there. I'm supposed to show that there is a point on the route through which the plane passes at the same time on both trips.

BlazeII

Sep 27th, 2005, 11:32 AM

Well I guess I forgot to add that the plane takes the same route both ways, but I thought that would be assumed. Everything else is there. I'm supposed to show that there is a point on the route through which the plane passes at the same time on both trips.

I we also assuming that the velocity remained constant through out?

Rtael

Sep 27th, 2005, 11:58 AM

It doesn't say, so I would assume not?

Rtael

Sep 27th, 2005, 12:09 PM

Hmm...I think it has something to do with position functions and the intermediate value theroem.

hingis-seles

Sep 27th, 2005, 01:11 PM

I'd love to help, but I suck at Calculus. I took a course this summer and only got an 88% :(

middy

Sep 27th, 2005, 01:57 PM

I have an Engineering degree and I still think that there is something missing or I'm just not seeing the whole picture :scratch: let me think about it some more

moby

Sep 27th, 2005, 02:29 PM

Let f(t) be the displacement along the route measured from Chicago after t hours, when the plane travels from Chicago to St. Louis.

Let g(t) be the displacement along the route measured from Chicago after t hours, when the plane travels from St. Louis to Chicago.

Each of f(t) and g(t) lie between 0 and M inclusive, where M is the total length of route.
Also, both functions are only defined for t between 0 and 5 inclusive.

f(0) = 0, f(5) = M => f is strictly increasing
g(0) = M, g(5) = 0 => g is strictly decreasing

Therefore f(t) and g(t) must intersect at some point on the graph of displacement along the route measured from Chicago against t, for some value of t.

I'm too lazy to draw a graph, and I don't seem to have used Calculus. :p
I'm also sure there are more elegant solutions around.

moby

Sep 27th, 2005, 02:47 PM

Hmmm... I thought of another proof, simpler and better.

Define P to be a point along the route from Chicago (C) to St. Louis (S).
Let the distance CP measured along the route be x.
As before, M is the total length of route.

Let f(x) be the time taken in hours to travel PS + SP.
f(0) = 10, and f(M) = 0
Clearly f(x) is strictly decreasing, so there exists some value of x between 0 and M that gives f(x) = 5.

Since the time taken to travel CP = 5 - time taken to travel PS,
when f(x) = 5, time taken to travel CP = time takes to travel SP.

geewhiz

Sep 27th, 2005, 04:07 PM

I'd say it's probably the first proof that they are doing, since it gives us both position functions and IVT.

Chicago to St. Louis = c, St Louis to Chicago = s. t = time. c(t) and s(t) are the position functions either way. Distance between Chicago and St Louis (M) is fixed.

c(0) = 0 (as no distance has been travelled yet)
c(5) = M (as after 5 hours the plane has travelled the whole distance)

If we arbitrarily say that the distance between Chicago and St Louis (M) is 100 miles, then you get:
f(0) = c(0)-s(0) => 0-100 = -100
f(5) = 100-0 = 100

c and s must be continuous functions, simply because it's a plane. That means f is also a continuous function (the result of subtracting two continuous functions), therefore from f(0) to f(5) it covers all points from -100 to +100.

For the IVT proof, if you made a graph with the arbitrary distance (M) on the y axis, running from -100 to +100, and time (t) on the x axis, running 0 to 5 hours, then there must be a point where your line had to cross the x axis. Therefore there must be a time when f(t)=0 which means c(t) and s(t) were the same, in other words, the plane was in the same place, whether it was travelling to St Louis or to Chicago. :)

TF Chipmunk

Sep 28th, 2005, 01:39 AM

Well the only place where it will pass at the same point going both ways would be when velocity = 0 (when it's at it's farthest point, the place where it turns back to Chicago).

So all you have to do is get the derivative of s(t), and set that equal to 0, and solve for t, and you'll have it :)

Rtael

Sep 28th, 2005, 01:46 AM

Well all the answers came too late, but thanks anyway guys. ;)

The teacher ended up writing the answer on the board... :p

TF Chipmunk

Sep 28th, 2005, 04:20 AM

Well all the answers came too late, but thanks anyway guys. ;)

The teacher ended up writing the answer on the board... :p
What was the right way to do it? Was it my way? I hope it was, otherwise that means I don't understand this stuff :o

Rtael

Sep 28th, 2005, 04:22 AM

It was something like the first thing moby posted.

beauty_is_pink

Sep 28th, 2005, 04:24 AM

ughhhh im so glad im done with calc... i hate it sooooooooo much.. i rather lick a toliet seat then go back to a calc class!

ys

Sep 28th, 2005, 05:37 AM

I still did not understand "the same time" bit. If can "one" plane pass the same point at the same time..

The problem would make sense if are talking about TWO planes starting simultaneously from both ends... With one plane, I still do not understand what it meant.. :confused: Or if it would say "the same time after take off", it would make some sense too..

Wigglytuff

Sep 28th, 2005, 05:42 AM

Okay I have this problem I have to do, but the teacher never explained how to do it...apparently we're supposed to teach ourselves in this class. :shrug:

Anyway... A plane goes from Chicago to St. Louis in five hours, then goes back in the same amount of time. I have to show that there is some point the plane will pass at the same time both ways.

i wish i could help. but the only thing i know about calculus is that A squared plus B squared equals C squared.

maybe that will help try that: A squared plus B squared equals C squared.

The Crow

Sep 28th, 2005, 09:49 AM

The problem would make sense if are talking about TWO planes starting simultaneously from both ends...

Basically that's what it is if I understood correctly.

controlfreak

Sep 28th, 2005, 11:09 AM

You can show it with a very simple graph of location vs time.

x-axis is time, from 0 (start of flight) to 1 (end of flight)
y-axis is location: mark one point on the axis for Chicago and one for St Louis (their positions are arbitrary)

Now draw a line, any line, it can be straigt or curvy, from Chicago @ t=0 to St Louis @ t=1.
Draw another line, straight or curvy, from St Louis @ t=0 to Chicago @ t=1.

The lines must cross. (Your graph should look something like a squiggly X shape)

For a full proof you can use one of moby's methods; you are right in thinking it is related to intermediate value theorem.