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 天生我才 |
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2010/09/29 02:05 |
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無言不必自責過甚。我對數字的沮喪是與生俱來，跟任何的數學老師都無關。我只愛看青蛙蹦蹦跳。
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【無★言】時代悲劇 （二)(junk200) 於 2010-09-29 19:08 回覆： |
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我覺得有些數學蠻有趣的．
那隻青蛙可是我上網找了許久才找到的，應還算可愛吧？
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| 實驗 |
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2010/09/29 00:06 |
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我想.... 這是數學實驗題吧! 我抓隻青蛙實驗後再告知答案....
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【無★言】時代悲劇 （二)(junk200) 於 2010-09-29 19:06 回覆： |
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哈！哈！哈！數學實驗？很有趣的想法．
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| 數學啊？ |
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2010/09/28 01:58 |
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此文令我沮喪，不予推薦。
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【無★言】時代悲劇 （二)(junk200) 於 2010-09-28 04:45 回覆： |
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有人對數學没興趣，我可以理解．人很多種，不必人人志趣相同．然而，依我猜想，對數學沒興趣者，見到數學問題，頂多抛到一旁，不予理會，如此而已，應該不會到達沮喪這麼嚴重的地步．
柏楊說他唸小學（或初中？）時，常挨數學老師的板子．以致於數學興趣全消．敢問妳是否也有類似經驗？
以後若見到我貼出數學問題，還請跳過勿進，根本不理不睬，以免沮喪．見妳沮喪，我覺得很罪過．
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| Problem solving approach |
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2010/09/27 16:10 |
Basically, what you need to do is to calculate the probability of P( | v1 + v2 + v3 | <= r ) v1, v2, v3 are the vectors represent the three jumps. r is the distance for each of the jump. The probability of course is the chance for meeting the conditions over the full extend of v1+v2+v3.
v1, v2, and v3 can be translated into their corresponding polar coordinates, where r and theta1, theta2, theta3, then for calculating the scalar value of the sume vector. The denominator for the calculation of the probability will be the multiple integrals of the scalar value of the sum vector over the 0 to 2 pi for each of the theta1, theta2, and theta3.
To calculate the nominator, you have to use the equation for | v1 + v2 + v3 | = r, and express the theta2 and theta3 in theta1, meaning find the relationship among these polar angles to keep the sum vector to be shorter than r. Then, the calculation of the nominator is the multiple integral of theta1 from 0 to 2 pi, and theta2 and theta 3 in the range conditioned by the equation above. You need to use conditional probabilities to solve theta2 and theta3 in stages, meaning using the relationship obtained from 2 jumps to get to the relationship among theta1, theta2, and theta3, in 3 jumps.
The expansion of both integrals are pretty messy, but I think there are some short-cuts to simplify the calculations. This is where some good trigonometric manipulations and algebraic skills would be useful, which I have not looked into seriously because my calculus is somewhat rusty now. :(
Hope this helps. What I described is the brute force approach for getting the answer. This approach is definitely too messy for a high school math question. There must be some short-cut method which I don't have answers yet.
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| Haha, thought you would ask about the solution. |
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2010/09/27 09:18 |
Could you tell me what available Math skills the students have?
Have they learned polar coordinates, and vectors? How about Calculus?
Do they allow to use any specific calculation assistance? Or just the typical scientific calculators?
Sorry, my kids have long passed the high school math and I don't really have any clues what their math class covers now.
Not sure I can find a suitable solution, but I'll give it a try, if you could let me know where the student stands.
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【無★言】時代悲劇 （二)(junk200) 於 2010-09-27 10:23 回覆： |
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十二年級應該學過極座標、向量、微積分．計算機應該沒用什麼特別的，就是 TI 製的．
我並不敢奢望你花太多時間在這上頭，能指點一個大方向即可．不勝感激之至！
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| 這是AP Math 的問題嗎？ |
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2010/09/27 01:18 |
我想這跳三次應該是 AP Math 的問題對嗎？
基本上如果知道向量的觀念和微積分的方法﹐這題應該有很直接的解法。
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【無★言】時代悲劇 （二)(junk200) 於 2010-09-27 05:50 回覆： |
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這是普通十二年級的數學題，不是AP．
跳三次你會算嗎？我完全沒概念，不知從何著手．
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| 解釋 |
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2010/09/27 01:04 |
第一個答案是對的﹐可是計算方法是捷徑。
第二個答案解決問題的想法比較完整﹐可是以面積代表青蛙第二次落點的機率的假定是錯誤的。
其實青蛙落在白圈和灰環地帶每個點上的機率並不相同﹐就白圈圓周上的任何一個特定位置而言﹐第二跳落於白圈之內的機率還是三分之一﹐第二跳落於白圈之外的機率還是三分之二。如果將這機率分配按照白圈圓周積分一整圈﹐所得的結果仍是白圈內的機率為三分之一﹐灰環內的機率為三分之二。
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【無★言】時代悲劇 （二)(junk200) 於 2010-09-27 05:49 回覆： |
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多謝解答．
有一法則，已忘其名，大意是說簡單直捷的答案通常是對的，繁複冗長的答案通常是錯的．用在此處，正好適用．
當初作此題，跳第二次後，畫出第二個圈，一時算不出二弧的比例，於是想：那就每一點都畫一個圈，算面積比好了．一個圈二弧的比例與面積總和的比例應該相同不變．於是將第二個圈依第一個圈的圓心轉三百六十度．，一下子就算出面積比是四分之一．後來算出二弧比例，居然是三分之一．不同算法所得答案不同，故而發問．
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| 應是第一答案!!! |
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2010/09/22 22:59 |
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當青蛙往前躍時，背後超過一公尺的部份；應是不成立！ 是以該區，應不為機率概算的範圍。
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【無★言】時代悲劇 （二)(junk200) 於 2010-09-26 17:02 回覆： |
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多謝解答．
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| 第一種的方法對 |
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2010/09/20 13:46 |
第一種的方法對
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【無★言】時代悲劇 （二)(junk200) 於 2010-09-21 06:23 回覆： |
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多謝告知．能否說明第二個答案錯於何處？謝謝．
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