📚 Edexcel Maths 3.2 Transport in Animals: Modelling Blood Flow and Exchange | Edexcel 数学 3.2 动物运输：血流与交换建模
In Edexcel Maths papers, the topic 'Transport in Animals' is often used as a real-world modelling context rather than a pure biology recall exercise. Students are expected to extract quantitative relationships from descriptions of circulation, blood flow, diffusion and gas exchange, then apply algebraic, graphical or calculus techniques to solve problems. This article bridges the biological content and the mathematical methods you are likely to meet under this heading.
在 Edexcel 数学试卷中，“动物运输”这一主题通常作为现实世界建模背景出现，而不是单纯的生物学记忆题。学生需要从循环、血流、扩散和气体交换的描述中提取数量关系，然后运用代数、图形或微积分技巧来解决问题。本文连接了生物学内容与你在该标题下可能遇到的数学方法。
1. From Biological Transport to Mathematical Models | 从生物运输到数学模型
Biological transport systems move substances such as oxygen, carbon dioxide, nutrients and waste around an animal body. To make these processes examinable in mathematics, we convert them into variables, equations and rates of change. Typical quantities include flow rate Q, velocity v, cross-sectional area A, pressure P, concentration C and time t.
生物运输系统在动物体内运输氧气、二氧化碳、营养物质和废物等物质。为了使这些过程在数学中可考，我们将其转化为变量、方程和变化率。典型量包括流量 Q、速度 v、截面积 A、压力 P、浓度 C 和时间 t。
| Symbol | Meaning | Common units |
|---|---|---|
| Q | volumetric flow rate | cm³ s⁻¹ or L min⁻¹ |
| v | flow velocity | cm s⁻¹ or m s⁻¹ |
| A | cross-sectional area | cm² or m² |
| P | pressure | Pa or mmHg |
| R | vascular resistance | mmHg min L⁻¹ |
| r | vessel radius | m or cm |
| η | viscosity | Pa s |
| L | vessel length | m |
| C | concentration | mol m⁻³ |
| D | diffusion coefficient | m² s⁻¹ |
| J | diffusive flux | mol m⁻² s⁻¹ |
Once these variables are defined, the task is to identify the governing equation. Most problems reduce to direct proportion, inverse proportion, exponential decay, or simple differential equations.
一旦定义了这些变量，任务就是识别控制方程。大多数问题可归结为正比例、反比例、指数衰减或简单微分方程。
2. Cardiac Output: A Core Rate Equation | 心输出量：核心速率方程
Cardiac output is the volume of blood pumped by one ventricle per minute. It is calculated by multiplying stroke volume, the volume ejected per beat, by heart rate. This gives a simple linear relationship that appears frequently in modelling questions.
心输出量是一个心室每分钟泵出的血液体积。它由每搏输出量乘以心率计算得出。这个简单的线性关系在建模题中频繁出现。
CO = SV × HR
Here CO is cardiac output in L min⁻¹, SV is stroke volume in L or cm³, and HR is heart rate in beats per minute. Careful unit conversion is often the first mathematical skill tested.
这里 CO 是心输出量，单位为 L min⁻¹；SV 是每搏输出量，单位为 L 或 cm³；HR 是心率，单位为每分钟心跳次数。单位换算通常是第一个测试的数学技能。
If a heart has a stroke volume of 70 cm³ and beats 72 times per minute, the cardiac output is:
如果心脏每搏输出量为 70 cm³，心率为每分钟 72 次，则心输出量为：
CO = 70 cm³ × 72 min⁻¹ = 5040 cm³ min⁻¹ = 5.04 L min⁻¹
equired heart rate is HR = CO ÷ SV = 6300 ÷ 90 = 70 beats per minute after unit agreement.
该方程也可以变形。若运动时目标心输出量为 6.3 L min⁻¹，每搏输出量为 90 cm³，则所需心率为 HR = CO ÷ SV = 6300 ÷ 90 = 每分钟 70 次，需先统一单位。
In exam questions, you may be given two of the three values and asked to find the third. Always express SV and CO in the same volume unit before dividing.
在考试题中，可能给出三个值中的两个并要求求第三个。计算前务必把 SV 和 CO 表示为相同的体积单位。
3. Continuity and Flow Velocity | 连续性与流速
For an incompressible fluid like blood, the volumetric flow rate Q is conserved along a tube when no branching occurs. This is the continuity equation. It links flow rate, average velocity and cross-sectional area.
对于血液这样的不可压缩流体，当管道无分支时，体积流量 Q 沿管道守恒。这就是连续性方程。它把流量、平均流速和截面积联系起来。
Q = v × A
Because the total cross-sectional area of the capillary bed is much larger than that of the aorta, the same cardiac output produces a much lower velocity in capillaries. This is mathematically identical to the relationship between speed and area in fluid dynamics problems.
由于毛细血管床的总截面积远大于主动脉，相同的心输出量在毛细血管中产生的流速要低得多。这与流体动力学问题中速度与面积的关系在数学上完全相同。
t 8.1 L min⁻¹ if all the cardiac output passed through the aorta at that moment. In reality, flow is pulsatile, so averages are used.
假设主动脉截面积为 4.5 cm²，血流速度为 30 cm s⁻¹。根据 CO = 5.04 L min⁻¹，流量是一致的，因为 Q = 30 × 4.5 = 135 cm³ s⁻¹，约等于 8.1 L min⁻¹，前提是当时所有心输出量都通过主动脉。实际上血流是脉动的，所以使用平均值。
es in daughter branches equals the parent flow rate. This is written as Q₀ = Q₁ + Q₂ + Q₃.
对于分支网络，守恒意味着子分支流量之和等于母分支流量。写作 Q₀ = Q₁ + Q₂ + Q₃。
Q₀ = Q₁ + Q₂ + Q₃
You may be asked to calculate an unknown branch velocity after finding its flow rate from area data.
你可能会被要求利用面积数据先求流量，再计算未知分支的流速。
4. Pressure, Resistance and Poiseuille's Law | 压力、阻力与泊肃叶定律
Blood flows from high pressure to low pressure. The pressure difference ΔP is the driving force, and vascular resistance R opposes flow. Their relationship is analogous to Ohm's law in electricity.
血液从高压流向低压。压力差 ΔP 是驱动力，血管阻力 R 阻碍流动。它们的关系类似于电学中的欧姆定律。
ΔP = Q × R
Poiseuille's law gives resistance for laminar flow in a cylindrical vessel. The most important mathematical feature is that resistance depends on the fourth power of the radius.
泊肃叶定律给出了圆柱形血管中层流流动的阻力。最重要的数学特征是阻力与半径的四次方有关。
Q = π ΔP r⁴ / (8 η L)
Here r is the vessel radius, η is blood viscosity and L is vessel length. Since r is raised to the fourth power, a small change in radius produces a large change in flow.
这里 r 是血管半径，η 是血液粘度，L 是血管长度。由于 r 被提高到四次方，半径的微小变化会引起流量的巨大变化。
If radius decreases by 10%, the new flow as a fraction of the original is 0.9⁴ = 0.6561, meaning a 34.4% ins why arterioles control local blood flow effectively through small changes in diameter.
如果半径减少 10%，新流量是原来的 0.9⁴ = 0.6561，即流量减少了 34.4%。这解释了为什么小动脉通过直径的微小变化就能有效地控制局部血流。
Qnew / Qold = (rnew / rold)⁴
When using Poiseuille's law, convert radius to metres if η is in Pa s and ΔP in Pa, or keep units consistent within a ratio. Exam questions often avoid full substitution and instead ask for proportional changes, reducing arithmetic errors.
使用泊肃叶定律时，如果 η 用 Pa s、ΔP 用 Pa，则半径要换算为米；或者在比值中保持单位一致。考试题通常避免完整代入，而是要求比例变化，从而减少算术错误。
5. Surface Area to Volume Ratio | 表面积与体积比
Exchange surfaces in animals rely on a large surface area relative to volume. Mathematically, for a sphere of radius r, the ratio of surface area to volume decreases as r increases.
动物的交换表面依赖于相对于体积较大的表面积。数学上，对于半径为 r 的球体，表面积与体积之比随 r 增大而减小。
SA:V = 4πr² / (4/3 πr³) = 3/r
For a cylinder, which is a better model of many capillaries and transport vessels, the ratio is approximately 2/r when length is much greater than radius.
对于圆柱体，它是许多毛细血管和运输血管的更好模型，当长度远大于半径时，比值约为 2/r。
SA:V ≈ 2/r
hile a radius of 5 cm gives only 0.4 cm⁻¹. The mathematical comparison is direct inverse proportion.
这意味着较小的血管单位体积具有大得多的表面积，有利于交换。半径为 0.5 cm 的圆柱体 SA:V = 4 cm⁻¹，而半径为 5 cm 时只有 0.4 cm⁻¹。数学上这是直接的反比例关系。
You may be asked to calculate the total surface area of a capillary network given the number of capillaries and their average radius and length. Use the curved surface area of a cylinder, 2πrL, and multiply by the total number.
你可能会被要求根据毛细血管数量、平均半径和长度计算毛细血管网络的总表面积。使用圆柱的侧面积 2πrL，再乘以总数。
Total SA = n × 2πrL
6. Fick's Law and Diffusion Across Membranes | 菲克定律与跨膜扩散
Diffusion is the passive movement of particles down a concentration gradient. Fick's first law describes the rate of diffusion through a membrane or tissue layer.
扩散是粒子沿浓度梯度被动的运动。菲克第一定律描述了通过膜或组织层的扩散速率。
J = -D × ΔC / Δx
In this equation, J is the flux, D is the diffusion coefficient, ΔC is the concentration difference, and Δx is the diffusion distance. The negative sign indicates movement from high to low concentration, but for magnitude calculations it is often omitted.
该方程中，J 是通量，D 是扩散系数，ΔC 是浓度差，Δx 是扩散距离。负号表示从高浓度向低浓度运动，但在大小计算中通常省略负号。
For a given membrane, diffusion rate increases if surface area increases, concentration gradient increases, or diffusion distance decreases. This is often summarised as Fick's law in a simplified proportional form.
对于给定膜，如果表面积增大、浓度梯度增大或扩散距离减小，扩散速率就会增大。这通常被总结为菲克定律的简化比例形式。
Rate of diffusion ∝ (Surface area × Concentration gradient) / Thickness
Suppose oxygen diffuses across a membrane of thickness 0.5 µ ⁻⁹ m² s⁻¹. The flux magnitude is J = (1.8 × 10⁻⁹ × 40) / (0.5 × 10⁻⁶) = 0.144 mol m⁻² s⁻¹.
假设氧气穿过厚度为 0.5 µm 的膜，ΔC = 40 mol m⁻³，D = 1.8 × 10⁻⁹ m² s⁻¹。通量大小为 J = (1.8 × 10⁻⁹ × 40) / (0.5 × 10⁻⁶) = 0.144 mol m⁻² s⁻¹。
This kind of substitution tests careful handling of powers of ten and unit prefixes such as µm. Always convert micrometres to metres before substitution when SI units are used.
这类代入计算考察对十的幂次和微米等单位前缀的仔细处理。使用 SI 单位时，代入前务必将微米转换为米。
7. Oxygen Delivery and Consumption | 氧气输送与消耗
Oxygen delivery to tissues depends on both flow rate and the concentration of oxygen in the blood. The delivery rate is the product of cardiac output and arterial oxygen content.
组织氧气输送取决于血流速率和血液中的氧浓度。输送速率是心输出量与动脉血氧含量的乘积。
DO₂ = CO × CaO₂
Arterial oxygen content CaO₂ is partly bound to haemoglobin and partly dissolved in plasma. A simplified model is:
动脉血氧含量 CaO₂ 一部分与血红蛋白结合，一部分溶解在血浆中。简化模型为：
CaO₂ = (Hb × 1.34 × SaO₂) + (0.003 × PaO₂)
Here Hb is haemoglobin concentration in g dL⁻¹, 1.34 is the oxygen-binding capacity in mL g⁻¹, SaO₂ is the fractional saturation, and PaO₂ is the partial pressure of oxygen. The coefficient 0.003 represents dissolved oxygen in mL dL⁻¹ per mmHg.
这里 Hb 是血红蛋白浓度，单位 g dL⁻¹；1.34 是氧结合能力，单位 mL g⁻¹；SaO₂ 是血氧饱和度；PaO₂ 是氧分压。系数 0.003 表示每 mmHg 溶解氧的 mL dL⁻¹ 量。
8 ≈ 20 mL dL⁻¹. With CO = 5 L min⁻¹ = 50 dL min⁻¹, oxygen delivery is about 1000 mL min⁻¹.
如果 Hb = 15 g dL⁻¹，SaO₂ = 0.98，PaO₂ = 100 mmHg，则 CaO₂ = 15 × 1.34 × 0.98 + 0.003 × 100 = 19.698 + 0.3 = 19.998 ≈ 20 mL dL⁻¹。当 CO = 5 L min⁻¹ = 50 dL min⁻¹ 时，氧输送约为 1000 mL min⁻¹。
DO₂ = 50 dL min⁻¹ × 20 mL dL⁻¹ = 1000 mL min⁻¹
Questions may ask how a change in heart rate or haemoglobin concentration affects oxygen delivery. Since the relationship is multiplicative, a 10% increase in CO gives a 10% increase in DO₂ if CaO₂ is unchanged.
题目可能问心率或血红蛋白浓度的变化如何影响氧输送。由于关系是乘积形式的，如果 CaO₂ 不变，CO 增加 10%，DO₂ 就增加 10%。
8. Modelling Heart Rate Response | 心率响应建模
Heart rate rises during exercise and falls during recovery. A simple recovery model assumes the heart rate decays exponentially toward its resting value.
心率在运动时升高，恢复期下降。一个简单的恢复模型假设心率指数式衰减到静息值。
HR(t) = HR_rest + (HR_peak - HR_rest) × e⁻ᵏᵗ
since exercise stopped. The expression HR_peak - HR_rest is the initial excess heart rate.
这里 HR_peak 是运动结束时的心率，HR_rest 是静息心率，k 是恢复常数，t 是运动停止后的时间。HR_peak - HR_rest 是最初的超额心率。
est = 60 bpm, HR_peak = 150 bpm and k = 0.02 s⁻¹, then after 30 seconds the heart rate is:
如果 HR_rest = 60 bpm，HR_peak = 150 bpm，k = 0.02 s⁻¹，则 30 秒后的心率为：
HR(30) = 60 + 90 × e⁻⁰·⁶ = 60 + 90 × 0.5488 ≈ 109 bpm
= ln(0.4444) ÷ (-0.02) ≈ 40.5 s. This tests natural logarithms and exponential equations.
要求心率降到某一值所需的时间，需要将式子改写为对数形式。若目标心率为 100 bpm，则 100 = 60 + 90e⁻⁰·⁰²ᵗ，所以 40/90 = e⁻⁰·⁰²ᵗ，得 t = ln(0.4444) ÷ (-0.02) ≈ 40.5 s。这考察自然对数和指数方程。
The same structure appears in cooling curves, capacitor discharge and radioactive decay, so recognising the shared form is valuable.
同样的结构出现在冷却曲线、电容器放电和放射性衰变中，因此识别共同形式很有价值。
9. Differential Equations for Clearance | 清除过程的微分方程
Many transport-related processes are described by first-order differential equations. If a substance is cleared from the blood at a rate proportional to its concentration, the concentration C follows exponential decay.
许多与运输相关的过程用一阶微分方程描述。如果一种物质从血液中清除的速率与其浓度成正比，则浓度 C 遵循指数衰减。
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