
#55
May1111, 08:04 AM

P: 1

Do you think it has (a lot) to do with age?
I'm 14 and I started integration a couple days ago and its fine when I do it; on the spot, but I have trouble remembering it and writing down formal definitions (etc...) later, like the next day. For example, I think the hardest bit of integration I've done yet is finding the area bounded by two curves, one negative and one positive, and while it was fine on the go and I didn't have too much trouble, if I was to look at it now without looking at my notes, I am sure it would take me much longer, or that I would not even find the answer. Thank you 



#56
May1111, 02:28 PM

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For example: Maths that is strongly assosiated with visualization is generally easier to get a hold on than very formal proof structures, for example. And, the capacity for abstract logical thought is still developing during your teens, until the age 1820 or so. (And then, everything goes downhill again..) But, I think you are well underway in developing that capacity, just being 14 and having a good grasp of integration already. 



#57
Apr1012, 05:01 AM

P: 1

My son scores good marks in all subjects except maths. He is afraid of maths. He is not good of analyzing problems. Last night he showed me a problem and told mom i am scared of this bigg problem, such a big problem "In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig). A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?" Can anyone tell me how simple can we explain solution to this problem




#58
Apr1012, 06:03 AM

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Well, you could set it up like this, in order to preserve the "visual element" in the calculation:
Total distance: 2*5+2*(5+3)+2*(5+3+3)+2*(5+3+3+3)+2*(5+3+3+3+3)..and so on Or, that is: 10+2*8+2*11+2*14+2*17 and so on. Here, each term represents the total distance traversed in a particular potatofetching run. Another way of representing this requires a bit of thinking: The "first five metres" are run by all 2*10 runs, so you get 10*2*5 The "next three metres" are run by 2*9 runs, so you get: 9*2*3 The next three metres: 8*2*3 The next three metres: 7*2*3 and so on.. Thus, when adding it all up, you get 100+6*(9+8+7+6+5+4+3+2+1)=100+6*45=380 metres in total 


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