Line 398: | Line 398: | ||
text-indent:2em !important; | text-indent:2em !important; | ||
+ | } | ||
+ | |||
+ | .bigphoto{ | ||
+ | margin: auto; | ||
+ | display: block; | ||
} | } | ||
Line 490: | Line 495: | ||
<div class="modelingPartContent" id="partA"> | <div class="modelingPartContent" id="partA"> | ||
− | <p class="content">At first, we assume that E. coli proliferate and die at the same ratio over time, and the value difference is the birth rate ( | + | <p class="content">At first, we assume that E. coli proliferate and die at the same ratio over time, and the value difference is the birth rate (μ<sub>g</sub>). So, we do derivative with this assumption.</p> |
− | <p class="content">Substituting the boundary condition, t = 0, N = N<sub>0</sub>, we then have ∴ e<sup> | + | <img src="" class="bigphoto" width="70%"> |
+ | |||
+ | <p class="content">Substituting the boundary condition, <span style="font-style:italic;">t = 0, N = N<sub>0</sub></span>, we then have ∴ e<sup>C<sub>2</sub>-C<sub>1</sub></sup>=N<sub>0</sub> | ||
Thus, the equation that expresses the relation between bacteria and time is: | Thus, the equation that expresses the relation between bacteria and time is: | ||
</p> | </p> | ||
− | <p class="content">N = N<sub>0</sub>∙e<sup> | + | <p class="content"><span style="font-style:italic;">N = N<sub>0</sub>∙e<sup>μ<sub>g</sub>t</sup></span> |
+ | |||
+ | <p class="content">What’s more, it is useless to say that <span style="font-style:italic;">E. coli</span> consumes their “food”, LB, all the time. Thus, if <span style="font-style:italic;">E. coli</span> consumes their food steadily, the LB consuming rate will be proportional to <span style="font-style:italic;">N</span>, then we can write down the equation:</p> | ||
+ | |||
+ | <img src="" class="bigphoto" width="70%"> | ||
+ | |||
+ | <p class="content">By substituting the boundary condition, we then have 𝐶= − 𝑛<sub>𝐿𝐵<sub>0</sub></sub>/𝑘<sub>𝑐𝑜𝑛</sub>−𝑁<sub>0</sub>/𝜇<sub>𝑔</sub> | ||
</div> | </div> |
Revision as of 16:04, 30 October 2017