Bioreactor Substrate Yield Log (Chemical & Biomedical Engineering) Solution

Problem Overview

We perform tabular differentiation on discrete bioreactor sensor logs:

Ratio: $Y(t) = \frac{X(t)}{S(t)}$
Index: $A(t) = t \cdot X(t)$

Step-by-Step Mathematical Solution

Step 1: Structural Formula for $\frac{dY}{dt}$ via Quotient Rule

\frac{dY}{dt} = \frac{d}{dt}\left[\frac{X(t)}{S(t)}\right] = \frac{X'(t)S(t) - X(t)S'(t)}{[S(t)]^2}

Step 2: Evaluate $\left.\frac{dY}{dt}\right|_{t=4}$ from Sensor Row $t = 4$

Extract data for $t = 4\text{ hr}$ :

$X(4) = 2.8\text{ g/L}$
$S(4) = 12.5\text{ g/L}$
$X'(4) = 0.9\text{ g/(L}\cdot\text{hr)}$
$S'(4) = -3.5\text{ g/(L}\cdot\text{hr)}$

Substitute into the Quotient Rule:

\left.\frac{dY}{dt}\right|_{t=4} = \frac{(0.9)(12.5) - (2.8)(-3.5)}{(12.5)^2} = \frac{11.25 - (-9.8)}{156.25} = \frac{11.25 + 9.8}{156.25} = \frac{21.05}{156.25} \approx 0.13472\text{ hr}^{-1}

Step 3: Evaluate $A'(6)$ via Product Rule

For $A(t) = t \cdot X(t)$ :

A'(t) = \frac{d}{dt}[t]X(t) + t X'(t) = (1)X(t) + t X'(t)

Extract data for $t = 6\text{ hr}$ :

$X(6) = 5.2\text{ g/L}$
$X'(6) = 1.2\text{ g/(L}\cdot\text{hr)}$

A'(6) = 5.2 + (6)(1.2) = 5.2 + 7.2 = 12.4\text{ g}\cdot\text{hr/(L}\cdot\text{hr)} = 12.4\text{ g/L}

Engineering Conclusion

By applying differentiation rules structurally to discrete telemetry data, bioprocess engineers track dynamic yield conversion efficiencies in real time without needing continuous closed-form equations. The positive yield rate $\frac{dY}{dt} \approx +0.135\text{ hr}^{-1}$ at $t = 4\text{ hr}$ confirms strong metabolic efficiency prior to glucose nutrient depletion.