Pim, if i say the width is 1.231m approx., you should not be surprised.Please Register or Log in to view the hidden image!
everneo: that is indeed the correct answer. Would you care to explain how you obtained it? Please Register or Log in to view the hidden image!
the link to the answer is also available in that page. sorry i could not overcome the temptation..Please Register or Log in to view the hidden image! *ducks and runs away*
Well, as everneo pointed out the answer is on the same page as the picture..(thnx for that). There is also an explanation of how it is derived. It is in Dutch though, good luck Please Register or Log in to view the hidden image!
Well, i hope people will try to solve it without looking there Please Register or Log in to view the hidden image!
The answer is 1.2311857 meters. How did I do it? First, I set up the distance forumla for both lines. Then, I made a linear equation for both. I knew f(x) = g(x) = 1, so I solved for 'x'. I set both x's equal to eachother, and solved for 'w,' the width. As far as anyone wanting to see the work, there is not much to it. Like I said in a previous thread, I visualize most stuff... With all the work, you land to this: 1/sqrt(9-w^2) + 1/sqrt(4-w^2) - 1 = 0 James Sibley
Back to the desert! The Car Problem(tm): Geez, this has taken me forever. Using calculus and separating into variables, dg/dx (change in gas per unit distance) = - (amount of gas needing to be transfered) / 20, the transfer size, * the MPG. The rate is defined to be negative. I am transfering the whole load as I go, shuttling it back and forth a meter at a time, so the number of trips I take is almost entirely dependent upon how much gas I have, not how far I am traveling. Then 200/G dg = - dx 200 ln g = - x + C, a constant using the point (g, x) = (20, 300), having ... is that where I screwed up. Well, the latest place. (I was off by a factor of 10, couldn't figure out why, etc. etc., and now I see I did it in another place too.) 20, 300, because once you get there you can just take off and can stop worrying about this back-and-forth crap. Anyway, you get 200 (ln 20) + 300 = C Solving for g from the previous equation, g = e^([C - x] / 200) x = 0 should give the original amount of gas before setting out with my tons of fuel tanks. g = e^( [200 (ln 20) + 300] / 200 - 0 ) = e^( ln 200 + 300/200 ), where the ratio 300/200 is derived from the "final" location of 300 miles and the product of 20 gallons * mpg. g = 20 e^1.5 Which is, presumably, incorrect. (It's about 89 gallons.) Can anyone tell me what I did wrong? If you say that you can only leave 10 gallons at a time, your fuel load is effectively 10 gallons (the unused 9.99 gallons just stay in your tank), and you get g = 20 e^(3/1) = 401. Just to doublecheck -- this was when I was still getting "reasonable" values of about 430 and 460 gallons for the 20/10 tank size calculations, I tried writing a program for it. Please Register or Log in to view the hidden image! I based it off of Persol's sequence, so I'd get 20, 35, 65... gallons needed at 300, 250, 200... miles respectively, working backwards. Here is the relevant code, after literally hours of finding at isolating bugs: Code: Const curMileage As Currency = 10 Const curLoad As Currency = 20 Dim curMyLocation As Currency Dim curStepsize As Currency Dim curMyGas As Currency curStepsize = Val(txtInputStep.Text) curMyGas = 20 curMyLocation = 300 Do Until curMyLocation <= 0 curMyGas = curMyGas + DeltaGas(curMyGas, curStepsize) curMyLocation = curMyLocation - curStepsize Loop Private Function DeltaGas(curMyGas As Currency, curStepsize As Currency) 'calculate gas needed to iterate the step Dim curDelta As Currency 'stepsize / mileage = gas per trip 'tip: next time, dimension #of trips as a separate variable! If Round(curMyGas / (curLoad - curStepsize / curMileage)) < curMyGas / (curLoad - curStepsize / curMileage) Then curDelta = curStepsize / curMileage * (2 * (Round(curMyGas / (curLoad _ - 2 * curStepsize / curMileage) + 1)) - 1) '2 * single-trip = roundtrip for TotalLoad/perload, 'then subtract one because don't go back. ElseIf Round(curMyGas / (curLoad - curStepsize / curMileage)) > curMyGas / (curLoad - curStepsize / curMileage) Then curDelta = curStepsize / curMileage * (2 * (Round(curMyGas / (curLoad _ - 2 * curStepsize / curMileage))) - 1) 'rounding is STUPID in VB! have to do two cases since I don't know the language very well. Else curDelta = curStepsize / curMileage * (2 * curMyGas / (curLoad _ - 2 * curStepsize / curMileage) - 1) End If DeltaGas = curDelta End Function It's in Visual Basic -- sorry, hehe, but it's the only programming language I am learning. btw, I managed to get Persol's 965 value somewhere along the way, but that was without rounding up to the nearest whole number of one-way trips... grrr. Something like that. It's why I had to code the stupid rounding bits in the code. Anyway, some sample values are: Stepsize (distance between stations) | gallons required at start: 0.5 | 422 1 | 425 5 | 462 10 | 508 20 | 735 25 | 566.67 (has to do with the non-linear nature of it. how many refueling trips to make, how many total stations, etc. One more station can make a significant difference.) 30 | 740 33 | 1019 40 | 1532 49 | 2348 50 | 1240 59 | 4174 60 | 2150 75 | 4385 90 | 190910 98 | 123,774,510 99 |1,990,297,029.8 Kind of surprising, since the peak of distance * fuel deposited does peak at 50 miles. I will not guarantee these values are accurate, but until someone finds a hole in my logic or coding ineptitude I will view them as precedent. Too bad my calculus is faulty... gr. peace. :m:
86 drums of fuel It takes 86 drums to cross the desert. 3 drums to place a drum at 100 miles. 15 drums to place a drum at 200 miles. 63 drums to place a drum at 300 miles. 3 + 15 + 63 + 3(drums at each station) + 2 (drums from starting point) = 86 drums of fuel. This follows the method that ryans posted earlier. The problem with the formula to get 42 drums of fuel is the distance unit. The maximum distance you can travel to allow you to shuttle fuel between stations is 100 miles, not the 200 miles that was used. Phoenix.
You are right; the original equation should have been dG = (2 * dx) (G/20) [<=the number of trips taken, no units] * 1 gallon / 10 mi. Unit analysis is miles * gallons / miles. And that is assuming that the amount of gas spent in shuttling fuel around is negligible when compared to how much there is in all, since in reality less trips would have to be made. That still gives an end equation of 20 ln (e^3), or the 401 gallons from the "conservative" estimate. This is still different from the one from the program, lower than the answer on the website, and much lower than anything anyone else was able to come up with! so what am I doing wrong? Please Register or Log in to view the hidden image!
This is another problem i have been thinking about for a while. Let a coordinate system be given. A man is standing in point (0,0). His dog is standing in (0,2). The man starts walking along the positive x-axis with constant speed. The dog starts walking in the direction of the man with the same speed. The dog hence walks along a curved line. The question now is if an explicit expression can be determined for this curved line, that is, a function y in terms of x. What do you think?
No response to my method... lol. Oh well. I am curious how everyone else did it. I would like to see other methods. As for the man/dog problem... I am thinking about that one. I got something close... something like f(x) = sqrt[ (r/r_1)^2 - 1 ]*x + 2 I just have to get rid of r/r_1 and convert it to something involving x... JS
4DHyperCubix: You mean your solution to the ladder problem? I think it's a very elegant way to do it like that Please Register or Log in to view the hidden image! I used geometrical properties of the triangles in the picture to compute the width. Don't feel like posting that here though Please Register or Log in to view the hidden image!
What kind of geometrical properties? Knowing that will be good enough for me Please Register or Log in to view the hidden image! How far along have you got with that man/dog problem?
Well, i don't know the english word for it..i think it's something like congruent triangles? Does that makes sense? ("Gelijkvormig" in Dutch) I haven't been able to come any further with the dog-man problem. I tried to determine the equation for the derivative of the curve the dog travels, but that is not that easy since the dog moves too...
Thick after coffee Gentlemen, I guess I am just thick even after coffee. Can sombody explain to me why we are talking about a curve in the dog/man problem? It seems to me that the line created is a linear or straight line since both have the same speed. It matters not what direction and for that matter that they have the same speed. Plotting coordinates of motion of two uniformly moving bodies seems to result in a straight line.Please Register or Log in to view the hidden image! What am I missing here?
MacM: since the man is moving, the direction of the dog changes every moment, and thus the dog walks along a curve..If it would be a straight line, the dog would walk in the same direction all the time. which does not make sense because the man is not at the same point all the time, he is moving
