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Estimate K based on the data Differential equation: dL/dt = K(1-L) using separation of variables We let L (t) be the fraction of the list

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Estimate K based on the data

Differential equation: dL/dt = K(1-L) using separation of variables

We let L(t) be the fraction of the list already committed to memory at time t. So L=0 corresponds to knowing none of the list, and L=1 corresponds to knowing the entire list. The differential equation is dL/dt=k(1-L). Compile your data in a graph with t , the amount of time spent studying, on the horizontal axis, and L, the fraction of the list learned, on the vertical axis.

2. Use this data to approximate your personal k-value and compare your data with the predictions of the model. You may use numeric or analytic methods, but be sure to carefully explain your work. Estimate how long it would take you to learn a list of 50 and 100 three-digit numbers.

3. Repeat the process in Part 1 on two of the other lists and compute your k-value on these lists. Is your personal k-value really constant, or does it improve with practice? If k does improve with practice, how would you modify the model to include this? list 2 and 3 repeat to estimate K

List 2 List 3 t | 1 | 3 It 1 | 2 | 3 4 5 6 17 List 1 L K 0.2 0.2231 0.35 0.2154 0.5 0.2310 0.45 0.1495 0.65 0.2100 0.9 0.3838 0.95 0.4280 1 none - |- |- 0.2630 |L 0.2 0.3 0.4 0.6 0.6 | K 0.2331 0.1782 0.1703 0.2290 0.1833 ILK 0.2 0.2231 0.25 01438 0.4 0.1703 0.45 0.1495 0.5 0.1386 0.5 0.1156 0.65 0.8 0.9 1 0.1750 0.2300 0.2878 0.4 . 0.0730 | 9 | 10 8 | 9 none 8 9 Avg. | 10 - 0.5 0.0866 0.55 0.0887 0.45 0.0598 0.1250 Avg. 0.2096 | 10 Avg. List 2 List 3 t | 1 | 3 It 1 | 2 | 3 4 5 6 17 List 1 L K 0.2 0.2231 0.35 0.2154 0.5 0.2310 0.45 0.1495 0.65 0.2100 0.9 0.3838 0.95 0.4280 1 none - |- |- 0.2630 |L 0.2 0.3 0.4 0.6 0.6 | K 0.2331 0.1782 0.1703 0.2290 0.1833 ILK 0.2 0.2231 0.25 01438 0.4 0.1703 0.45 0.1495 0.5 0.1386 0.5 0.1156 0.65 0.8 0.9 1 0.1750 0.2300 0.2878 0.4 . 0.0730 | 9 | 10 8 | 9 none 8 9 Avg. | 10 - 0.5 0.0866 0.55 0.0887 0.45 0.0598 0.1250 Avg. 0.2096 | 10 Avg

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