2013 醫學系期中小考

這是 2013 年臺北醫學大學醫學系的微積分第一次平時考解答。

第 1 題

Suppose a population of bacteria doubles every hour, but that 1.0 × 106 individuals are removed before reproduction to be converted into valuable biological by-products. Suppose the population begins with b0 = 3.0 × 106 bacteria.

  1. Find the population after 1, 2, and 3 hours.
  2. Write the discrete-time dynamical system.

由題意可知

bt+1 = 2.0 (bt − 1.0 × 106)

bt+1 = 2.0 bt − 2.0 × 106

所以

b1 = 4.0 × 106

b2 = 6.0 × 106

b3 = 1.0 × 107

第 2 題

In one simple scenario, mutations occur in only one direction (wild type tum into mutants but not vice versa), but wild type and mutants have different levels of per capita production. Suppose that a fraction 0.1 of wild type mutate each generation, but that each wild type individual produces 2.0 offspring while each mutant produces only 1.5 offspring. In each case, find the following.

  1. The number of wild-type bacteria that mutate.
  2. The number of wild-type bacteria and the number of mutants after mutation.
  3. The number of wild-type bacteria and the number of mutants after reproduction.
  4. The total number of bacteria after mutation and reproduction.
  5. The fraction of mutants after mutation and reproduction.

(Begin with 1.0 × 106 wild type and 1.0 × 105 mutants.)

本題有一些暗示不太容易看出來。從題目的順序暗示在每回合中,細菌先突變再繁殖。此外,細菌一般行分裂生殖,繁殖後原個體消失。

此時由野生種突變的個體有

0.1 (1.0 × 106) = 1.0 × 105

突變後野生種剩下

1.0 × 106 − 1.0 × 105 = 9.0 × 105

而突變種有

1.0 × 105 + 1.0 × 105 = 2.0 × 105

接下來細菌進行繁殖。野生種有

2.0 (9.0 × 105) = 1.8 × 106

突變種有

1.5 (2.0 × 105) = 3.0 × 105

總和為

1.8 × 106 + 3.0 × 105 = 2.1 × 106

回合前突變種的佔比為

如今則是

第 3 題

Consider again a lung that has a volume of 6.0 L and that replaces 0.6 L each breath with ambient air. Suppose that we are tracking oxygen, with an ambient concentration of 21%. Assume that the actual oxygen concentration in exhaled air is approximately 15%. What fraction of oxygen is in fact absorbed?

(6.0 − 0.6) 15% α = 0.6 (21% − 15%)

81α = 3.6

α ≈ 0.044

第 4 題

The model describing the dynamics of the concentration of medication in the bloodstream,

Mt+1 = 0.5Mt + 1.0,

becomes nonlinear if the fraction of medication used is a function of the concentration. In the basic model, half is used no matter row how much [the concentration] is there. More generally,

new concentration = old concentration − (fraction used) × (old concentration) + supplement.

Suppose the fraction used = 0.5 / (1 + 0.1Mt). Write the discrete-time dynamical system and solve for the equilibrium.

由題意可知此系統的迭代式為

當平衡濃度到達時,濃度不再變化。因此設平衡濃度為 M,必滿足

我們求此方程式的解。

0.5M = 1 + 0.1M

M = 2.5

第 5 題

In an excitable heart model. Let Vc be the threshold potential. The discrete-time dynamical system can be describe as

With the parameter values u = Vc = 1, compute the conditions on c for the existence of an equilibrium.

代入 u = Vc = 1。若平衡電位 V 存在,它必滿足

cV > 1,則 c = 1V > 1。但是 c 是衰減係數,應小於 1,故不合。

cV ≤ 1,則

V = cV + 1

因為 cV ≤ 1,所以

c ≤ 0.5