a math q

If a car travels 150 km in 2 hours. How can you proove that it has to travel at 75 km/h at least once?

What you must prove is that some amount less than 1/2 plus some other amount less than 1/2 must always be less than 1.

Given:
a = 150 (or whatever)
b = a / 2
0 < x < b
0 < y < b

Proof:
a = (b - x) + (b - y) ' Less than 75 plus less than 75
a = 2b - (x + y)
a = 2(a / 2) - (x + y) ' Replace b with a / 2
a = (2a / 2) - (x + y)
a = a - (x + y) ' Is false...that's the proof

To drive it home:
(x + y) > 0
a <> a - (x + y)

(Use the equals with the line through it; not <>)

The following

Quote:

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a = (b - x) + (b - y)

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is invalid unless (x+y) = zero. For b = a/2, the above is a complicated way of saying "a = a -(x+y)"

The proof needs a bit more work.

Well that was a logical arguent explained in a mathematical way :) If one of the two equalsize parts are greater than 75 then the other must be smaller and vice versa, you may have to generalize it over all intervals.

if you have a basket full of water and put your hand in it, the water will flow over.

you could also show that theres at least one derivative that equals the average dy/dx of any function that startx-endx=dx and starty-endy=dy

Love you alot with many hearts Kedaman.

Wonder Woman, I wish you loved me as much as you love Kedaman.

There are many problems in mathematics which seem intuitively obvious, but are tuff to prove. This is an example of one of them.

To do it with a formal proof requires (I think) Integral Calculus and something called the Mean Value Theorem. I do not think anybody is interested in this approach. Besides, I would have to do some research to get it right.

An informal approach starts by expressing the 150 km as the sum of a lot of little distances using velocity times time for each little distance.

Code:

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150 = V1*T1 + V2*T2 + V3*T3 . . . . . + Vn*Tn

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The V's & T's are velocities and times for various parts of the 150 kilometer trip. All the T's must add up to 2 hours.

If all the velocities are equal to 75 km/hr, then obviously 75 km/hr happened once. If not 75 km/hr all the way, then there are three possibilities: (1) Always less than 75km/hr, (2) always greater, or (3) sometimes less & sometimes greater.

Assume all the velocities are less than 75 km/hr. Then the above is equivalent to

Code:

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150 = (75 - X1)*T1 + (75 - X2)*T2 . . . + (75 - Xn)*Tn

or 150 = 75*(T1 + T2 . . . + Tn) - (X1*T1 + X2*T2 . . . + Xn*Tn)

or 150 = 75*2 -(some value).

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The above contradiction shows that all velocities cannot be less than 75 km/hr. A similar analysis shows that that all values cannot be greater than 75 km/hr.

Now we know that all velocities must be 75 km/hr or some must be greater than 75 km/hr and some must be less.

If some are greater and some are less, you must pass through 75km/hr at least once.

Thanks Wonder Woman, i think she loves you to Guv.

Mean value theorem, thats what it was called(i was thinking of rolles theorem which was a bit similar), i didn't have the book right there so i couldn't check it out but i remember you could prove it, there was a picture of a graph presenting the speed of the car, and it showed there was always at least one tangent that was parallell to the dy/dx line. Well anyway that got my attention :p since i thought that the hand in the basket theorem was more cool :D

I think the basic way to do it is to assume that the speed of the car is continuouse and differentiable and use this sort of logic


1 imagine a second car traveling at a constant speed of 75 km/h from the start to the finish, starting and finishing at the same time as the first car.

The 2 cars must be neck and neck and the start, call the next place they are neck and neck point c. (they are neck and neck at the finish so point c must exist)

There are 2 cases that may have happened between the start and point c.

1 The first car started off faster than the second car and then slowed down to be caught up.


2 The first car started off slower than the second car and then sped up to catch it up.

in case 1 at the start point the first car was going at a greater speed than 75 km/h at the start point and slower than 75km/h at point c. and hence must have been traveling at exactly 75 km/h somewhere in the braking process (because the speed is continuous and differentiable)

in case 2 at the start point the cars speed was less than 75 km/h and at point c it was graeter than 75 m/h so somewhere between point a and point c the car must have been going at 75 km/h.


That's in words, you'll need to translate that into maths, but I think that's essentially the proof. It's quite easy to extend that to the mean value theorum and even Cauchy's mean value theorum.

hope it helps.

Sounds like a pancake problem, to me.

DerFarm