數學請幫忙 Exponential function

2010-06-17 6:00 pm
數學請幫忙

Some people say that all students have a learning curve. For example you retain more information at 8AM than at 2PM. In fact, this learning curve could have an exponential function as its mathematical model. Suppose you retain 45% of the information you encounter at 8AM, you retain 32% at 11AM and you never retain less than 15%

a. Find exponential function that would approximate this data.

b. According to this model, at what time will you retain 25% of teh information you encounter.
更新1:

感謝 rex 大大 相救 但是題目說不能低於 15% 那如何是好 ? 感覺上這種曲線應該比較像 sine wave 24 小時循環 而不是 exponential function. 是嗎 ? 有請 天助 大大 提供高見

更新2:

回 天助 大大 本題是 美國高中 Algebra 2 的老師給的 Final review. 他們只教到 exp. function 所以我也很困惑 也許是老師想耍學生好玩 感謝指導 !

回答 (2)

2010-06-17 7:16 pm
✔ 最佳答案
Some people say that all students have a learning curve. For example you retain more information at 8AM than at 2PM. In fact, this learning curve could have an exponential function as its mathematical model. Suppose you retain 45% of the information you encounter at 8AM, you retain 32% at 11AM and you never retain less than 15%

有人說所有的學生都有條"學習曲線" 例如從早上八點到下午兩點
事實上這學習曲線可以以指數函數當做數學模式
如果你在早上八點是45%而十一點是32%而且從不會在15%以下
~~不懂這學習曲線的定義 大概是指學習的專注程度吧

a. Find exponential function that would approximate this data.

A(t)=A(0)*e^(kt), t:從八點開始算, A(0)=45%
A(3)=45%*e^(3k)=32%, --> k=-0.1136
A(t)=45%*e^(-0.1136t)

b. According to this model, at what time will you retain 25% of teh information you encounter.

25%=45%*e^(-0.1136t), t=5.174
大概在下午一點十分左右

2010-06-17 13:59:51 補充:
我只是猜測可能是這樣解
但是天助大大說似乎不對
我也不是很知道該怎麼做
在這模式下直到下午兩點都不會小於15%

2010-06-17 16:48:20 補充:
這個題目是書本就這樣寫嗎?
如果所謂的exponential function是它指定的~~~
的確學習曲線我見過logistic decay形式
一定要說只能認為題目有點一廂情願
1.它的學習曲線只限於8:00AM~2:00PM
2.它的那句"不會低於15%"有點不知所云

2010-06-17 16:59:22 補充:
修正如下~如果硬要考慮那個15%
1.
A(t)=A(0)e^(kt)+15%, A(0)=30%
A(t)=30%*e^(kt)+15%
A(3)=30%*e^(3k)+15%=32%
e^(3k)=17/30, k=(1/3)ln(17/30)=-0.1893
A(t)=30%*e^(-0.1893t)+15%
2.
25%=30%*e^(-0.1893t)+15%
(1/3)=e^(-0.1893t)
t=1.0986/0.1893=5.8
也就是過了5.8小時後
約在下午一點48分

2010-06-17 17:00:28 補充:
感謝 這樣討論很開心~~~~~
2010-06-17 7:34 pm
本題是logistic decay, 不是 exponential decay

2010-06-17 16:13:50 補充:
提問人匿名,不知程度為何? 要解微分方程式或者直接代公式?

2010-06-17 16:41:23 補充:
本題週期性不是重點(題意未說明8 am之前的學習效率),較合理的學習曲線是
logistic decay,數學模式如下:(解微分方程而得) L(t)=A/[1+Bexp(-kt)]
L(t)>=15, then lim(t->inf)L(t)=15=A
8:00 am時t=0, 45=A/(1+B), then B=- 2/3
11:00 am時t=3, 32=A/[1-(2/3)exp(-3k)], then exp(-3k)=51/64
so, L(t)=15/[1-(2/3)(51/64)^(t/3)]

2010-06-17 16:41:30 補充:
(b)L(t)=25=15/[1-(2/3)(51/64)^(t/3)], t=3[ln(5/3)]/[ln(64/51)]約6.75(hr)=14:45 pm

2010-06-17 16:47:14 補充:
題目指定exponential fn.數學模式: L(t)=A+Bexp(-kt) (解微分方程而得)
L(t) >= 15, then lim(t->inf)L(t)=15=A
8:00 am時, t=0, 45=A+B, then B=30
11:00am時, t=3, 32=A+Bexp(-3k), then exp(-3k)=17/30,
so, L(t)=15+30(17/30)^(t/3)
(b) L(t)=25=15+30(17/30)^(t/3), then t=3ln3/[ln(30/17)]約5.8(hr)=13:48 pm

2010-06-17 16:57:37 補充:
rex先破題,也寫得很好,最佳解就留給rex吧!


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