📚 Case Study Practical Drill: Pharmacokinetics Modelling | 案例分析实战演练：药代动力学建模
In this case study, we apply first-order differential equations from Cambridge Year 13 Further Mathematics to model how the concentration of an intravenously infused drug changes over time. You will set up the model, solve it analytically, use experimental data to estimate patient-specific parameters and validate predictions. This practical drill emphasises the translation of a real clinical problem into mathematical language and the critical evaluation of model assumptions.
在本案例分析中，我们运用剑桥13年级进阶数学中的一阶微分方程知识，模拟静脉输注药物在体内浓度随时间的变化。你将建立模型、解析求解，并利用实验数据估算患者个体化参数，验证预测。本实战演练强调将真实临床问题转化为数学语言，并对模型假设进行批判性评估。
1. Understanding the Clinical Scenario | 理解临床情境
A patient is given a constant-rate intravenous infusion of an antibiotic. The drug enters the bloodstream at a fixed rate, distributes rapidly throughout the body, and is eliminated by the kidneys at a rate proportional to its current plasma concentration. Clinicians need to predict how the concentration builds up to a steady state to ensure the level stays within the therapeutic window.
患者接受恒定速率的静脉输注抗生素。药物以固定速率进入血液，迅速分布到全身，同时以与当前血浆浓度成正比的速度被肾脏清除。临床医生需要预测浓度如何逐步上升至稳态，以确保药物水平维持在治疗窗内。
This is modelled as a one-compartment system with a single well-mixed volume. The key quantities are the infusion rate R (mg h⁻¹), the volume of distribution V (L) and the elimination rate constant kₑ (h⁻¹). The concentration C(t) is the state variable.
这被建模为一个单室系统，容积充分混合。关键量包括输注速率 R（mg·h⁻¹）、表观分布容积 V（L）和消除速率常数 kₑ（h⁻¹）。浓度 C(t) 是状态变量。
2. Mathematical Model: One-Compartment IV Infusion | 数学模型：单室静脉输注
Let C(t) be the drug concentration (mg L⁻¹) at time t (h). The net rate of change of C is the difference between the constant input and the first-order elimination:
令 C(t) 为时间 t（h）时的药物浓度（mg·L⁻¹）。C 的净变化率为恒定输入与一级消除之差：
dC/dt = k₀ – kₑ C
nt. The initial condition is C(0)=0 because no drug is present before the infusion starts.
其中 k₀ = R/V 为零级输入速率常数（mg·L⁻¹·h⁻¹），kₑ 为一级消除速率常数。初始条件为 C(0)=0，因为输注开始前体内无药物。
3. Solving the Differential Equation | 求解微分方程
The equation is separable. Rearrange and integrate:
该方程可分离变量。整理并积分：
∫ dC/(k₀ – kₑ C) = ∫ dt
Using the substitution u = k₀ – kₑ C, we obtain –(1/kₑ) ln|k₀ – kₑ C| = t + constant. Applying C(0)=0 gives the particular solution:
使用换元 u = k₀ – kₑ C，得到 –(1/kₑ) ln|k₀ – kₑ C| = t + 常数。代入 C(0)=0 求得特解：
C(t) = (k₀/kₑ)(1 – exp(–kₑ t))
This expression shows that the concentration starts at zero and approaches an upper limit exponentially.
该表达式表明，浓度从零开始，按指数规律趋近于一个上限。
4. Interpreting the Solution and Steady State | 解释解与稳态
As t → ∞, exp(–kₑ t) → 0, so C(t) → k₀/kₑ. The steady-state concentration is denoted by Cₛ:
当 t → ∞ 时，exp(–kₑ t) → 0，因此 C(t) → k₀/kₑ。稳态浓度记为 Cₛ：
Cₛ = k₀/kₑ
depends solely on kₑ. For instance, the half-life t₁/₂ = ln 2 / kₑ is the time to reach 50% of the steady state after a single bolus, but in constant infusion the time to reach 90% of Cₛ is approximately 3.3 half-lives.
达到 Cₛ 某一比例所需的时间仅取决于 kₑ。例如，单次推注后的半衰期 t₁/₂ = ln 2 / kₑ，而在恒速输注中达到 90% Cₛ 约需 3.3 个半衰期。
Physiologically, kₑ reflects renal and hepatic clearance. A smaller kₑ means slower elimination and a higher Cₛ for the same infusion rate, potentially leading to toxicity.
从生理学看，kₑ 反映肾和肝的清除能力。kₑ 越小，消除越慢，相同输注速率下 Cₛ 越高，可能引起中毒。
5. Case Data and Parameter Estimation | 案例数据与参数估计
istribution V = 10 L. Hence k₀ = 3.2 mg L⁻¹ h⁻¹. Blood samples were drawn to measure C(t):
考虑一位患者接受 IV 输注，R = 32 mg·h⁻¹，估计分布容积 V = 10 L，因此 k₀ = 3.2 mg·L⁻¹·h⁻¹。采血测定 C(t)：
| t (h) | C (mg L⁻¹) |
|---|---|
| 0 | 0 |
| 1 | 2.64 |
| 2 | 4.41 |
| 4 | 6.38 |
| 6 | 7.27 |
| 8 | 7.67 |
| 10 | 7.85 |
| 12 | 7.93 |
The data suggest Cₛ ≈ 8.0 mg L⁻¹. We can estimate kₑ directly from the steady state:
数据提示 Cₛ ≈ 8.0 mg·L⁻¹。可由稳态直接估算 kₑ：
kₑ = k₀ / Cₛ = 3.2 / 8.0 = 0.40 h⁻¹
However, using only the plateau gives no information about the model’s dynamic fit. A more robust method is to linearise the model.
然而，仅使用平台期无法获知模型的动态拟合优度。更稳健的方法是将模型线性化。
6. Linear Regression Approach | 线性回归方法
Rearrange the solution to isolate the exponential term:
重排解的形式以分离指数项：
1 – C/Cₛ = exp(–kₑ t)
Take natural logarithms:
取自然对数：
ln(1 – C/Cₛ) = –kₑ t
Plotting ln(1 – C/Cₛ) against t should yield a straight line through the origin with slope –kₑ. Using the estimated Cₛ = 8.0, compute the transformed values:
绘制 ln(1 – C/Cₛ) 对 t 的图应得到一条过原点的直线，斜率为 –kₑ。利用估计的 Cₛ = 8.0，计算转换值：
| t (h) | C | 1 – C/Cₛ | ln(1 – C/Cₛ) |
|---|---|---|---|
| 1 | 2.64 | 0.670 | –0.400 |
| 2 | 4.41 | 0.449 | –0.800 |
| 4 | 6.38 | 0.202 | –1.599 |
| 6 | 7.27 | 0.091 | –2.396 |
| 8 | 7.67 | 0.041 | –3.194 |
Fitting a straight line through the origin gives a slope of approximately – = 0.40 h⁻¹. The high linear correlation (R² ≈ 1) validates the one-compartment assumption.
拟合一条过原点的直线得到斜率约为 –0.40，证实 kₑ = 0.40 h⁻¹。高度的线性相关性（R² ≈ 1）验证了单室假设。
7. Model Validation and Prediction | 模型验证与预测
With k₀ = 3.2 and kₑ = 0.40, the predicted concentration is C(t) = 8 (1 – exp(–0.40 t)). Compare the model with the data:
采用 k₀ = 3.2 及 kₑ = 0.40，预测浓度为 C(t) = 8 (1 – exp(–0.40 t))。将模型与数据对比：
- At t=10 h, predicted C = 7.85 vs observed 7.85 – excellent agreement.
- At t=2 h, predicted 4.41 vs observed 4.41 – no discrepancy.
- t=10 h 时，预测值 7.85 对比观测值 7.85 —— 高度吻合。
- t=2 h 时，预测值 4.41 对比观测值 4.41 —— 无偏差。
The model can now forecast the time to reach 95% of Cₛ: solve 1 – exp(–0.40 t) = 0.95 → exp(–0.40 t) = 0.05 → t = ln(0.05)/(–.40) ≈ 7.5 h. After about 7.5 hours, the concentration is therapeutically safe and stable.
模型现可预测达到 95% Cₛ 的时间：解 1 – exp(–0.40 t) = 0.95 → exp(–0.40 t) = 0.05 → t = ln(0.05)/(–0.40) ≈ 7.5 h。约 7.5 小时后，浓度达到治疗安全稳定水平。
8. Clinical Implications and Dosage Adjustment | 临床意义与剂量调整
× kₑ = 10 × 0.40 = 4.0 mg L⁻¹ h⁻¹. Since V = 10 L, the new infusion rate is R = 40 mg h⁻¹.
若目标稳态浓度为 10 mg·L⁻¹，保持 kₑ = 0.40 不变，所需 k₀ 为：k₀ = Cₛ × kₑ = 10 × 0.40 = 4.0 mg·L⁻¹·h⁻¹。因 V = 10 L，新输注速率为 R = 40 mg·h⁻¹。
The bolus dose D = Cₛ × V = 10 × 10 = 100 mg. The model then becomes C(t) = (k₀/kₑ) + (D/V – k₀/kₑ) exp(–kₑ t), illustrating superposition.
可静脉推注负荷剂量以立即达到目标。负荷剂量 D = Cₛ × V = 100 mg。此时模型变为 C(t) = (k₀/kₑ) + (
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