Pharmacokinetic Drug Elimination Profile (Biomedical Engineering) Solution

1. Initial Baseline Concentration ($y$-Intercept)

Evaluate $C(t)$ at $t = 0$:

2. Horizontal Asymptote & Metabolic Clearance

1. Name the Governing Formula: The Exponential Clearance Decay Theorem:
   $$\lim_{t \to \infty} e^{-kt} = 0 \quad (\text{for } k > 0)$$
   (Formalized in Horizontal Asymptotes.)
2. Evaluate Limit as $t \to \infty$:
   $$\lim_{t \to \infty} C(t) = \lim_{t \to \infty} 20(e^{-0.5t} - e^{-2t}) = 20(0 - 0) = 0\ \mu\text{g/mL}$$
   The line $C = 0$ ($t$-axis) is a horizontal asymptote. All drug active ingredients are completely cleared from the bloodstream over time.

3. Peak Plasma Concentration via First Derivative

1. Name the Governing Formula: The Exponential Derivative Rule and the First Derivative Test for Local Extrema:
   $$\frac{d}{dt}[e^{-kt}] = -k e^{-kt}$$
   (Formalized in First Derivative Test.)
2. Compute $C'(t)$:
   $$C'(t) = 20\left(-0.5e^{-0.5t} - (-2)e^{-2t}\right) = 20\left(-0.5e^{-0.5t} + 2e^{-2t}\right) = -10e^{-0.5t} + 40e^{-2t}$$
3. Locate Critical Number: Set $C'(t) = 0$:
   $$40e^{-2t} = 10e^{-0.5t} \implies \frac{e^{-0.5t}}{e^{-2t}} = \frac{40}{10} = 4$$
   $$e^{1.5t} = 4 \implies 1.5t = \ln(4) \implies t_{\max} = \frac{\ln(4)}{1.5} = \frac{2\ln(2)}{1.5} \approx \frac{1.3863}{1.5} \approx 0.924\text{ hours}$$
4. Classification & Peak Concentration:
   • For $t < 0.924$: $40e^{-2t} > 10e^{-0.5t} \implies C'(t) > 0$ (Increasing plasma absorption).
   • For $t > 0.924$: $40e^{-2t} < 10e^{-0.5t} \implies C'(t) < 0$ (Decreasing clearance). By the First Derivative Test, $t_{\max} \approx 0.924\text{ h}$ is a local and global maximum.
     $$C_{\max} = 20\left(e^{-0.5(0.924)} - e^{-2(0.924)}\right) = 20\left(e^{-\ln(4)/3} - e^{-4\ln(4)/3}\right)$$
     Using exact powers: $e^{-\ln(4)/3} = 4^{-1/3} \approx 0.62996$, and $e^{-4\ln(4)/3} = 4^{-4/3} = \frac{1}{4 \cdot 4^{1/3}} \approx 0.15749$.
     $$C_{\max} = 20(0.62996 - 0.15749) = 20(0.47247) \approx 9.45\ \mu\text{g/mL}$$
     The peak therapeutic concentration is $9.45\ \mu\text{g/mL}$ achieved at $t \approx 55.4\text{ minutes}$.

4. Inflection Point via Second Derivative

1. Name the Governing Formula: The Second Derivative Inflection Criterion:
   $$C''(t) = 0 \quad \text{with concavity sign transition}$$
   (Formalized in Concavity and the Second Derivative Test.)
2. Compute $C''(t)$:
   $$C''(t) = \frac{d}{dt}[-10e^{-0.5t} + 40e^{-2t}] = -10(-0.5)e^{-0.5t} + 40(-2)e^{-2t} = 5e^{-0.5t} - 80e^{-2t}$$
3. Solve for Zero Concavity:
   $$5e^{-0.5t} - 80e^{-2t} = 0 \implies 5e^{-0.5t} = 80e^{-2t} \implies \frac{e^{-0.5t}}{e^{-2t}} = \frac{80}{5} = 16$$
   $$e^{1.5t} = 16 \implies 1.5t = \ln(16) \implies t_{\text{inf}} = \frac{\ln(16)}{1.5} = \frac{4\ln(2)}{1.5} \approx \frac{2.7726}{1.5} \approx 1.848\text{ hours}$$
4. Evaluate Concentration at Inflection:
   $$C(1.848) = 20\left(16^{-1/3} - 16^{-4/3}\right) = 20\left(\frac{1}{16^{1/3}} - \frac{1}{16 \cdot 16^{1/3}}\right) = 20 \times \frac{15}{16 \cdot 2.5198} \approx 7.44\ \mu\text{g/mL}$$
   The inflection point is $(1.848, 7.44)$.

5. Information Consolidation Table & Pharmacokinetic Transition

Engineering / Kinetic Interpretation

Between $t = 0$ and $t = 1.848\text{ h}$, the curve is concave down ($\cap$); the clearance rate is decelerating as drug absorption subsides. Beyond $t = 1.848\text{ h}$, the curve becomes concave up ($\cup$); clearance follows pure exponential decay, asymptotically approaching zero. This inflection point marks the onset of the pure elimination phase, instructing the infusion pump when to deliver maintenance micro-doses.