Elizabeth Polidan Week4: Difference between revisions
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(Added figures to table) |
(Fixed problem with table.) |
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u2 = 60 | u2 = 60 | ||
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| [[Image: | | [[Image:Week4_Fig5.jpg]] | ||
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| '''Figure 5''' | | '''Figure 5''' | ||
Revision as of 23:44, 7 February 2013
Elizabeth Polidan
BIOL 398.03 / MATH 388
- Loyola Marymount University
- Los Angeles, CA, USA
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Figure 1
Original script t0 = 0; % initial time t1 = 100; % final time c0 = 0; % initial nutrient conc N0 = 30; c20 = 0; S0 = [c0;N0;c20]; q = 0.15 u = 120 r = 1.0 K = 5 V = 0.5 u2 = 60 |
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Figure 2
t0 = 0; % initial time t1 = 100; % final time c0 = 0; % initial nutrient conc N0 = 30; c20 = 0; S0 = [c0;N0;c20]; q = 0.15 u = 30 r = 1.0 K = 5 V = 0.5 u2 = 60 |
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Figure 3
t0 = 0; % initial time t1 = 100; % final time c0 = 0; % initial nutrient conc N0 = 30; c20 = 0; S0 = [c0;N0;c20]; q = 0.15 u = 120 r = 1.0 K = 5 V = 0.25 u2 = 60 |
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Figure 4
t0 = 0; % initial time t1 = 100; % final time c0 = 0; % initial nutrient conc N0 = 30; c20 = 0; S0 = [c0;N0;c20]; q = 0.30 u = 120 r = 1.0 K = 5 V = 0.5 u2 = 60 |
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Figure 5
t0 = 0; % initial time t1 = 100; % final time c0 = 0; % initial nutrient conc N0 = 30; c20 = 0; S0 = [c0;N0;c20]; q = 0.45 u = 120 r = 1.0 K = 5 V = 0.5 u2 = 60 |
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Figure 6
t0 = 0; % initial time t1 = 100; % final time c0 = 0; % initial nutrient conc N0 = 30; c20 = 0; S0 = [c0;N0;c20]; q = 0.15 u = 120 r = 1.0 K = 5 V = 0.5 u2 = 30 |
