Chinahand Posted February 2, 2017 Share Posted February 2, 2017 Isn't it interesting which scenario sounds worse. You have a 190 year lease, you pay £250 at the start and increase it at 6.97% per annum, or you start at £250 per year and keep increasing it in such a manner as to end up paying £1.3 billion over the 190 years. They are identical... now I wonder why the BBC went with the second angle ... ;-) Quote Link to comment Share on other sites More sharing options...
Bobbie Bobster Posted February 2, 2017 Share Posted February 2, 2017 Long term inflation assumption within the range of 3-4% would be defensible, but you are correct it has a big impact on the result. Better to consider the real rate of return on ground rents, rather than the absolute number for inflation, but casual interest (no pun intended) only goes so far and I didn't research it. Quote Link to comment Share on other sites More sharing options...
Chinahand Posted May 18, 2017 Share Posted May 18, 2017 A hint for solving some of them might be Huygens. Quote Link to comment Share on other sites More sharing options...
Albert Tatlock Posted May 18, 2017 Share Posted May 18, 2017 Here's a real toughie... Complete the following sequence of numbers: 1966. Quote Link to comment Share on other sites More sharing options...
Chinahand Posted June 1, 2017 Share Posted June 1, 2017 Erm this one is a toughy. You've the equation 1+cos(x) between 0 and pi. And you rotate it 360 degrees about the y axis. You get sort of inverted dish or the shape of a World War II british soldiers helmet. I've been able to work out the volume of the helmet - pi x (pi^2 -4) But what is the surface area? My understanding is that it is the integral between 0 and pi of 2 * pi * x * sqrt(1+(sin(x))^2) dx But what the heck is that!! Here's the Wolfram Alpha go at it: http://www.wolframalpha.com/input/?i=rotate+1%2Bcos+x+about+y+axis,+0+<%3D+x+<%3D+pi, It tells me the answer is 37.7038 but that isn't particularly helpful. I'm actually wanting to generalize the equation for helmets of different height and diameter: ie the equation H x (1+cos(pi / Radius x X)) And don't know how to manipulate the 37.7038 to do that without the full equation showing how it was derived. Anyone any idea? The shape is the shape of a test sample I'm working on. Quote Link to comment Share on other sites More sharing options...
Tarne Posted June 1, 2017 Share Posted June 1, 2017 You've got the answer right there haven't you? If you click on the formula in Wolfram, you can see more detail and change it around if you need to http://www.wolframalpha.com/input/?i=2+π+abs(x)+sqrt(1+%2B+sin^2(x))&lk=1&rawformassumption="ClashPrefs"+->+{"Math"} Quote Link to comment Share on other sites More sharing options...
wrighty Posted June 1, 2017 Share Posted June 1, 2017 (edited) This is, I think, an elliptic integral, and therefore not integrable using standard functions. I had a go at manipulating your general formula with height H and radius r, and ended up with a factor of (pi/r)^2 inside the square root term which I couldn't shift. By general principles, you'd expect the surface area to increase by a factor of H/2 (given the native curve is height 2), and a factor of (r/pi) since you're simply stretching in two dimensions by those factors. It'd work for a cylinder. Does this tie in with numerical approximations on Wolfram? ETA - No, that's all wrong. Far more complex than that. The integral actually has a factor of (H*pi/2r)^2 inside the square root, which makes it worse. For large H and small r the 1 could be discarded which makes it solvable. Edited June 1, 2017 by wrighty Quote Link to comment Share on other sites More sharing options...
La Colombe Posted June 1, 2017 Share Posted June 1, 2017 1 hour ago, Chinahand said: I'm actually wanting to generalize the equation for helmets of different height and diameter: Snigger.... It's a hard one alright... 1 Quote Link to comment Share on other sites More sharing options...
Chinahand Posted June 2, 2017 Share Posted June 2, 2017 Wrighty - do you think I can solve it via integration by parts? I think it might help, but the choice of which is u and which is dv/dx isn't immediately obvious. Integration by parts is Integration[u * dv/dx dx] = u*v - integration[v * du/dx dx] a) if u = x and dv/dx =[1+sin^2(x)]^1/2 I've still got to integrate the dv/dx to find v which looks intimidating, but maybe doable. With u = x du/dx = 1, so I've then got to integrate whatever v is again - hard, but if I've been able to do it the first time, you'd hope I could do it again! b) or if u = [1+sin^2(x)]^1/2 and dv/dx = x I've got to differentiate [1+sin^2(x)]^1/2 - which doesn't look too bad, but then I've got to multiply it by 1/2 x^2 and integrate it again - which I suspect is a nightmare. So I suspect the way to go is a). I'm totally overloaded with other things, and so will try tonight, but maybe it is possible. What do you think? Is it worth an attempt? Quote Link to comment Share on other sites More sharing options...
Chinahand Posted June 2, 2017 Share Posted June 2, 2017 Looks like it aint! Quote Link to comment Share on other sites More sharing options...
wrighty Posted June 2, 2017 Share Posted June 2, 2017 4 hours ago, Chinahand said: Looks like it aint! I already said that! I remember at school trying to calculate the arc length of an ellipse by integration, using all the tricks I knew, and it just didn't work. I then found out it was impossible, so I didn't feel so bad about failing. It was Liouville, I think, who established which functions can be integrated. Whereas you can differentiate any smooth function, integration is a harder problem and there are many very simple looking functions that can't be integrated to give other elementary functions. Things like e^(x^2), sin (x^2) are examples. 1 Quote Link to comment Share on other sites More sharing options...
The Well-Manicured Man Posted June 3, 2017 Share Posted June 3, 2017 On 4/13/2015 at 11:47 PM, pongo said: Can you solve the maths question for Singapore schoolkids that went viral? I'd rather just employ a Singapore servant to do it for me. Quote Link to comment Share on other sites More sharing options...
Chinahand Posted June 3, 2017 Share Posted June 3, 2017 A fool and his money are easily parted. 1 Quote Link to comment Share on other sites More sharing options...
Chinahand Posted February 3, 2018 Share Posted February 3, 2018 A ship which can sail at 2 x pi miles an hour is steaming around a 1 mile radius circle. The sea in this area has a uniform current flowing from north to south at a speed of 0.5 x pi miles an hour. How long does it take for the ship to sail around the circle? I don't think the effect of the current just cancels out because if the current was 2 pi miles an hour the boat wouldn't be able to do the into the current leg. Vectors ... what fun! I do enjoy that this is a reasonable analogy of a work problem I'm doing at the moment ... the joy of maths! Any ideas ... anyone!? Quote Link to comment Share on other sites More sharing options...
quilp Posted February 3, 2018 Share Posted February 3, 2018 Which part of the ship travels the furthest? Quote Link to comment Share on other sites More sharing options...
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