|
|
Cepheus
Honorable
Legendary Hero
Far-flung Keeper
|
posted April 14, 2009 12:57 AM |
|
|
|
It seems far too simple an answer and I almost know it's going to be wrong, but I'll be the morbidly embarrased idiot tonight and guess 90 mph.
|
|
Corribus
Hero of Order
The Abyss Staring Back at You
|
|
posted April 14, 2009 12:59 AM |
|
Edited by Corribus at 01:01, 14 Apr 2009.
|
Quote:
So he would have to make 4 miles in 4 minutes...? (generally speaking, x miles in x minutes) That means infinite speed?
I know I'm missing something. Can anyone else help? 
You're not missing anything now. It's sort of a trick question. The answer is that it's impossible for the driver to average 60 mph. Even if he drove at light speed, his average speed wouldn't reach 60 mph.
You can actually plot a function to see this.
RA = 240*R4/(60+4*R4)
Where RA is his average rate of speed and R4 is his rate around leg number 4. You can try to solve for R4 after setting RA = 60, which is impossible, but a better way to do it is limits.
The limit of RA as R4 --> infinity is 60 and the limit of RA as R4 --> 0 is zero. This is easy to interpret. If he goes really slow, he'll never finish, and his average rate around the track is 0. If he goes infinitely fast, it's like he didn't take any time to go around leg 4 at all, meaning he traveled 4 miles in 4 minutes, or 60 mph.
Most people, by the way, would be inclined to say the answer is 90 mph, figuring that 60 + 60 + 30 + 90 averaged is 60. But, you can't average rates like this. Since the total distance traveled is 4 miles, then to average 60 mph, you have to take 4 minutes. But the driver already took 4 minutes to go the first 3 legs, so it's impossible for him to average 60 mph around the 4th leg unless he takes no time to do it (infinite speed).
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
|
|
Cepheus
Honorable
Legendary Hero
Far-flung Keeper
|
|
posted April 14, 2009 01:01 AM |
|
Edited by Cepheus at 01:04, 14 Apr 2009.
|
Knew it.  How stupid of me to take for granted that the problem had an actual answer.
|
|
Corribus
Hero of Order
The Abyss Staring Back at You
|
|
posted April 14, 2009 01:17 AM |
|
|
Well you should have known that 90 mph would have been too obvious.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
|
|
mvassilev
Responsible
Undefeatable Hero
|
|
posted April 14, 2009 02:21 AM |
|
|
Ah, that's clever. I was thinking that since he'd be arriving at the same place he left, his velocity would be 0.
____________
Eccentric Opinion
|
|
Binabik
Responsible
Legendary Hero
|
|
posted April 14, 2009 02:55 AM |
|
|
There is a valid argument for a different answer.
Speed is a unit of distance traveled per unit of time.
So what is "average speed"? You could interpret this as the total distance divided by the total time to travel that distance. (which is what Corribus did) You could also interpret average speed as the average of several speeds. In this case the average speed per leg.
With the second interpretation, (60+30+30+120)/4 = 60
What happens it we change the unit distance?
If the unit were 1/2 leg (and assuming a constant speed throughout the leg) then you'd have (60+60+30+30+30+30+120+120)/8 = 60
Cut the unit of distance in half again and you get the same results. As the unit of distance approaches zero the speed during the last leg still equals 120 does it not? Could this be considered the average instantaneous speed? And if we have an average instantaneous speed of 60 "along the entire route", can we assume that it can actually be done?
Just asking. A little thought experiment for those of use who have forgotten all of our math. It's fun to think about anyway.
|
|
Corribus
Hero of Order
The Abyss Staring Back at You
|
|
posted April 14, 2009 04:05 AM |
|
|
@mvass
Well, yes that'd be true if I asked for velocity. But this isn't the "Let's talk about Physics!!!" thread.
@Binabik
I'll have to think about it. I'll admit, the correct answer of "impossible" seems a bit paradoxical. I had to spend a lot of time convincing myself this was the right answer.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
|
|
Binabik
Responsible
Legendary Hero
|
|
posted April 14, 2009 06:55 AM |
|
|
Yea, this is kinda hard to grasp. Both solutions are obvious. And they are both correct depending on how you interpret the question. I know it's semantics, but I'm having a hard time resolving why the two are not one and the same.
When you said "along the entire route" I assumed you specifically worded it that way to mean the total distance divided by the total time. And I would have solved it the same way you did.
But the other way also seems correct. So why different answers? Average usually implies the arithmetic mean of a set of values. Maybe it can't really be used in the way you did.
Somehow the following doesn't seem right and this is where my forgetfulness of math comes into play. The thread is "let's talk about math", so that's what I'm doing even if it's nonsense.
The instantaneous speed I mentioned is simply the first derivative, right? The curve will have a definite starting and ending point.
So is your solution the same as taking the slope of a straight line connecting the starting and end point of the curve?
And is my solution the same as taking the average of the slopes of a finite number of points along the curve?
Are the two different, and why?
I'm thinking along the lines of:
(sin 30 + sin 60)/2 =/= sin 45
And for the same reason, averaging the slopes of all points along the curve does not equal the slope of start point to end point.
/babbling
|
|
dimis
Responsible
Supreme Hero
Digitally signed by FoG
|
|
posted April 14, 2009 07:01 AM |
|
Edited by dimis at 09:28, 14 Apr 2009.
|
Well, Binabik, there is a problem in your definition which can go away just like that because we tend to omit the "units" of the things that we describe. What do I mean? What does your 60 represent? It is not miles per hour ...  It is "miles per (hour * {unit leg})". What's the meaning of the product "hours * (unit leg)" ? So, you are not computing an average speed really. Besides that's apparent by the whole discussion of having units of leg which you subdivide and so on ...
Hopefully I'll give you another look at the problem that justifies this "impossibility".
A----x--->B
/ |
| |
| |
| |
y=c*x x
| |
| |
| |
\ V
D<---x----C
Say that you don't know how big the last route is from D to A; i.e. y. Clearly, there is a constant "c" such that y = c*x. That's ok, right? I am not cheating I mean ...
How much time do you need for the route A-->B-->C-->D?
Well, x/60 + x/60 + x/30 = 4x/60 = x/15 (hours and "x" is measured in miles)
How much time do you need to reach the finish line from D?
Well, y/v = c*x / v, where v is your speed in miles per hour.
What's your average speed now for the entire trip?
Sum of distances / Sum of times for each distance = (3*x + c*x) / (x/15 + c*x/v).
If you simplify you get:
(3 + c)*15*v / (v + 15*c)
Now you want this quantity to be equal to 60 miles per hour. So by equating these two (note that you have "miles per hour" in both sides of the equality) you get:
(c - 1) * v = 60 * c
Now, clearly you are looking for "v", so you have to divide by (c-1), which means that first of all you have to check if (c-1) = 0 is a solution. Well it's not, since you get 0 = 60.
Note that c = 1 implies y = 1*x = x, which is our problem.
But I don't want to stop here.
For any c > 1 (forget about the barrier of "speed of light" that you have on your "v") you can divide and both the numerator and the end result is positive (has meaning for our constraints). This gives you
v = [ c / (c-1) ] * 60.
In other words, for any y > x there is a solution to the problem. And in fact in the limit, as y --> oo you have to drive with 60 miles per hour till the end of the route. So it doesn't really matter if the guy slowed down for some short distance (short compared to the overall (or what is remaining) - note that c > 1 ==> y > x ).
Finally, I don't know if this will help your intuition in some way, but your idea is somehow close to the following problem.
Problem: Say you have an average of some numbers, a new number appears in front of you and you want to compute their new average based on the average that they had so far.
Solution:
Initially you have the average
average = (s1 + s2 + ... + sN) / N
Now, number s{N+1} comes. The new average is:
new_average = (s1 + s2 + ... + sN + s{N+1}) / (N + 1)
Break the sum in two parts:
new_average = (s1 + s2 + ... + sN) / (N + 1) + s{N+1} / (N + 1)
which is equal to
new_average = ( N / (N+1) ) * (s1 + s2 + ... + sN) / N + s{N+1} / (N+1)
meaning:
new_average = ( N / (N+1) ) * average + s{N+1} / (N+1)
Now, you have expressed your new average based on the previous average, and the (N+1)th number that just arrived.
So, all that you really do in your suggestion, is that you play with this thing and treat everything as pure numbers like I did here.
Note that when you reach D through the route A-->B-->C-->D you have an average of 45 (mph but you forget to state that). And ideally you want the new average to have value 60 again (again in mph which we omit to refer to).
Plugging back in you get (since it's the fourth part of the route):
60 = (3/4) * (150/3) + s4 / 4 ==(multiply by 4 and 3's cancel)==> s4 = 90.
which is what you give. But already there is another error there. What is 150/3?
It's certainly not your average speed in the route A-->B-->C-->D; as we just said above, that's 45 mph.
So, I think the main reason (apart from the fact that "hour * (unit leg)" is not defined), is that you are trying to brake the sum
(s1 + s2 + ... sN) / (t1 + t2 + ... + tN) and you want to make it equal to
s1/t1 + s2/t2 + ... + sN/tN
but as you know, this has no reason what so ever to actually work and make the equality valid (simply because that's not the way we add up fractions).
I don't know if I offered insight or I caused more confusion.
____________
The empty set
|
|
dimis
Responsible
Supreme Hero
Digitally signed by FoG
|
|
posted April 14, 2009 07:27 AM |
|
Edited by dimis at 09:08, 14 Apr 2009.
|
Yeah, our posts kind of coincided, so I'll try to answer some parts of the above post as well (those that I didn't mention initially).
Quote:
The instantaneous speed I mentioned is simply the first derivative, right?
The first derivative of ... distance w.r.t. time. Sorry I am so scholastic, but you know how easily it is for us to mix up apples with oranges.
Quote:
The curve will have a definite starting and ending point.
Yup! Start for t=0 (hours) and end for some t=t1 (again in hours).
Quote:
So is your solution the same as taking the slope of a straight line connecting the starting and end point of the curve?
Nope*. He wants to compute a "box" that has the same "area" as the plot you had above of distance w.r.t. time as a plot of velocity with respect to time (in other words you want to integrate those plots and the end result {which is distance} should be equal). In other words, the slope you refer to is not something unknown. It is precisely zero (in a plot of velocity w.r.t. time - note that it is another plot compared to distance w.r.t. time). What is unknown is how far away from (0,0) should that horizontal line lie (which expresses your average speed and hence we refer to a plot of velocity w.r.t. time). You also know t=0, and t=t1. These form the boundary of your box.
*: Actually yes, what you say is correct there. I am kind of an idiot, because I didn't realize that we had a different idea on plots. But for good intuition read the post below!
Quote:
And is my solution the same as taking the average of the slopes of a finite number of points along the curve?
Are the two different, and why?
I did my best to explain this part in the post above. The best way to grasp a problem is to try to "attack" the arguments made. If you can not find a flaw, then it should be a correct approach. If you do find a flaw though (like I did above), then ....
____________
The empty set
|
|
dimis
Responsible
Supreme Hero
Digitally signed by FoG
|
posted April 14, 2009 07:49 AM |
|
Edited by dimis at 09:47, 14 Apr 2009.
|
Sticking to your plots idea since I liked it!
Initially you have a plot like this (velocity w.r.t. time):
velocity
^
|
|
|xxxxxx
|xxxxxx
|xxxxxx
|xxxxxxxxxxxx
|xxxxxxxxxxxx
|xxxxxxxxxxxx
|------------------------> time
where the boundary gives you your speed during the first 3 intervals of the route (blue first part, green 2nd, and red 3rd).
So you would like to add another box after the red one, so that in the end the total area under your curve is equal to that of another diagram like this one (i.e. start t=0, finish t = t3 + Dt, where t3 is when you finish the 3rd part and Dt expresses the additional time you need to finish the route), which has a straight line all the way through.
In other words, you want to discover a box of some specific height after the red box in the above diagram so that you can fill-in the gap that has been created; because clearly that other box you are about to insert has to be bigger than the red one (and actually bigger than the blue and green ones too), so that in that "new" other diagram you can have a straight line precisely at the height of blues and greens (you want 60 mph, that's why), which finishes in the same time as the above one (after you insert what is missing), and in total have the same area underneath. So, it's fairly intuitive again. It can not happen!
(Intuition: Because the gap you have by the red brick already has the area of the brick you are about to insert next to it, since that area expresses the distance to your destination which is again the gap that you are missing and you have a square==> all bricks equal area and you want a slice after the red one with nonzero width ...)
For other values though, you get the height of the box by that formula
v = c/(c-1) * 60.
That expresses the height of the new "brick" you have to add and do the trick.
edit: Thanks for participating Binabik. You bring up very good points that sometimes are not cleared-up. Unfortunately, others don't follow your path.
@Death: As I said last time, try to "attack" the arguments made on gates and come up with something definitive. Either you accept the truth tables there, or not (regardless of the specific problem; just abstractly in A and B form). If not, which part seems problematic? Simple question. Of course this does not imply that it's going to be easy to come up with the answer.
@Zamfir: I solved the equation, but I am waiting for others to post solution.
____________
The empty set
|
|
Ecoris
Promising
Supreme Hero
|
|
posted April 14, 2009 11:11 AM |
|
Edited by Ecoris at 11:13, 14 Apr 2009.
|
I think I owe dimis an answer to the problem below:
Quote:
New problem: Which part of dimis' solution above incomplete/slightly incorrect?
In your post you show that if
loga(b) = x = natural number
logb(c) = y = natural number
logc(a) = z = natural number
then
a = b = c
A complete solution should mention that the implication the other way is also valid (even though it is obvious). However, that's not what caused me to pose the problem (that would just be nitpicking).
Consider:
Quote:
Substituting back we get
a^1 = b ==> a = b
b^1 = c ==> b = c
c^1 = a ==> c = a
which implies that any triple a = b = c is a solution to the problem.
[emphasis mine] It does not imply that a = b = c is a solution. Rather, it implies that any triple that is a solution must have this form.
____________
|
|
rti
Adventuring Hero
Now known as Rarensu
|
|
posted April 14, 2009 11:17 AM |
|
|
I would like to point out, that while this thread is much mathier than the average conversation, it doesn't contain any real mathematics. It's mostly just Algebra with a splotching of Numerology. Real mathematicians don't use Algebra anymore. These days, its all <set theory this> and [n-space that]. Even someone like me who's gotten an *A* in all the math classes up past Calculus III doesn't have a clue what real mathematics is about.
“The essence of mathematics is not to make simple things complicated, but to make complicated things simple.”
- S. Gudder
Historically, math and physics have been the same subject, because whenever anyone wanted to get a new interesting physics problem done, they would invent a new branch of mathematics to do it with. To measure large objects, Pythagoras invented Geometry. To plot planetary motion, Kepler adapted Descartes's coordinate system. To solve gravity, Newton invented Calculus. Mathematics had no other purpose than to be a tool for physicists.
"A mathematician is a machine for converting coffee into theorems."
- Paul Erdös
However, in the 20th century, all that changed. There was now enough money floating around that millions of educated people could afford to sit around all day doing nothing in particular. So they started thinking "What if BLAH"? Then they would spend years creating beautiful mathematical systems for describing a world with BLAH. They made crazy things, like Hilbert Space, and Boolean Logic, and String Theory. Then all the other mathematicians would say "Ooo... That's so beautiful..." At the end of the day, no real scientist was quite certain what they had done, but it didn't matter anyway, because mathematics was no longer about helping out the Physicists. It had become an art form.
“As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.”
- Albert Einstein
Modern Mathematics is the study of bizarre logical systems. It has very little to do with arithmetic. If you want to get a taste of this without fear of being overwhelmed, read "Gödel, Escher, Bach: an Eternal Golden Braid", by Douglas Hofstadter. It's an introduction to logic and mathematics and natural beauty, but its written to be entertaining and much easier to read than a straight textbook.
____________
Sincerely,
A Proponent of Spelling, Grammar, Punctuation, and Courtesy.
|
|
Ecoris
Promising
Supreme Hero
|
|
posted April 14, 2009 02:41 PM |
|
Edited by Ecoris at 14:44, 14 Apr 2009.
|
Quote:
Real mathematicians don't use Algebra anymore.
Yes they do. What a ridiculous claim.
You say what real mathematics deals with but also mention that you have no clue what it is about. I feel that you are contradicting yourself. And where did Numerology get into this thread? Surely, you mean something else.
There hasn't been much talk about math in the thread so far, it has mostly dealt with posing and solving simple problems. But I still can't see what the point of your post is.
____________
|
|
Corribus
Hero of Order
The Abyss Staring Back at You
|
|
posted April 14, 2009 03:44 PM |
|
Edited by Corribus at 17:58, 14 Apr 2009.
|
@RTI
Quote:
I would like to point out, that while this thread is much mathier than the average conversation, it doesn't contain any real mathematics. It's mostly just Algebra with a splotching of Numerology. Real mathematicians don't use Algebra anymore.
That's ridiculous. It's like saying that just because most physicists don't study Newtonian Mechanics any more, then Newtonian Mechanics isn't "real" physics.
@Binabik and dimis
A lot to read there. I'll try to make time to do so.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
|
|
dimis
Responsible
Supreme Hero
Digitally signed by FoG
|
|
posted April 14, 2009 05:55 PM |
|
Edited by dimis at 18:40, 14 Apr 2009.
|
Quote:
[emphasis mine] It does not imply that a = b = c is a solution. Rather, it implies that any triple that is a solution must have this form.
I agree Ecoris. But I thought the "if" part was given; i.e. it was used as an equivalent expression of "assume this thing holds". For this part (quote), you are right, I didn't use the right expression. Thanks
@RTI: I won't comment the "Algebra is no longer used" part. As of the crazy things that you describe, e.g. Hilbert Spaces, you never know how these things will be applied. An example I am familiar with is Support Vector Machines (this is a really good source). But of course a real mathematician does not really care about the applications. He just does and enjoys to do math. And I hope and believe this is what we are doing here. Just having fun with math.
____________
The empty set
|
|
TheDeath
Responsible
Undefeatable Hero
with serious business
|
|
posted April 15, 2009 02:02 AM |
|
Edited by TheDeath at 02:27, 15 Apr 2009.
|
Going back to the "average speed" issue, I don't think it's paradoxial at all with infinity. This is because the track length is limited. Speed isn't just a measure of length or of time, it's a measure of length/time (so to speak).
BUT that is only if you take the "average speed" as representing "an equal car which travels at X speed constantly in the specified track length". That is, in that example, a 100% 60mph average car would travel the track in 4 minutes (4 miles). So if you take it that the average means THIS then yes, it is infinite. (and impossible)
Quote:
If not, which part seems problematic? Simple question. Of course this does not imply that it's going to be easy to come up with the answer.
To be honest with you, the way I learned logic gates back when I learned assembly, I don't think gates say anything about the statement's validity. The statement's validity is the 'problem' here, not the condition, neither the "then" validity.
In my viewpoint:
Condition False? You don't have to evaluate the "then" part, BUT you also don't know if the statement is true or not.
Condition True? You HAVE to evaluate the "then" part and see if the statement is true or false. (depending on the "then" part).
Summary:
True, False --> False
True, True --> True
False, True --> Unknown
False, False--> Unknown
NOTE that this only talks about the STATEMENT's validity, which is also useless in let's say, CPUs (well the gates I was talking about anyway), as only the conditions are needed, and the "then" parts (only if the condition is true though).
But please do explain if I make a grave error Sorry if I make you repeat yourself.
____________
The above post is subject to SIRIOUSness.
No jokes were harmed during the making of this signature.
|
|
dimis
Responsible
Supreme Hero
Digitally signed by FoG
|
|
posted April 15, 2009 03:50 AM |
|
Edited by dimis at 03:50, 15 Apr 2009.
|
Quote:
To be honest with you, the way I learned logic gates back when I learned assembly, I don't think gates say anything about the statement's validity. The statement's validity is the 'problem' here, not the condition, neither the "then" validity.
All I am asking is if you agree with the truth tables. Did I say anything about validity of statements? NO. That's all. It doesn't matter under which setup you learnt those connectives. Either you agree with the truth tables, or you don't. Pick a side.
In other words, do you accept that the gates will produce as "output" what you see on the last column in each case?
You can substitute F (false) with 0 (zero) and T (true) with 1 in all of the entries.
As I said, simple question. It does not mean the answer comes up simply.
____________
The empty set
|
|
rubycus
Known Hero
-student of the mind-
|
|
posted April 20, 2009 11:51 PM |
|
|
I have a tricky one, at least for the unskilled Did anyone post it yet?:
Give a proof for this equalation:
a^0 = 1
____________
A prudent question is one-half of wisdom.
|
|
Corribus
Hero of Order
The Abyss Staring Back at You
|
|
posted April 21, 2009 01:28 AM |
|
|
Because a^n = a*a^(n-1)
Therefore a^1 = a * a^(1-1) = a * a^0
Since a^1 = a, then a^0 = 1
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
|
| |
|
|
This thread is pages long: 
