Constant input shifts a decay equilibrium · source reference

Printed question

Glucose in the blood stream is reduced at a rate proportional to the amount of glucose present in the blood stream. Let $G$ milligrams (mg) be the amount of glucose in 1 decilitre (dL) of blood stream at time $t$ minutes and let $\lambda$ denote the positive constant of proportionality.

(i) Write down a differential equation relating $G,\lambda,t$. Solve this differential equation to find an expression of $G$ in terms of $\lambda,t$. [3]

Through extensive research, doctors recommended that a healthy glucose level in the blood stream of an adult should be between 70 mg/dL to 100 mg/dL. An adult patient, Neo, has 80 mg/dL of glucose in his blood stream at time $t=0$ minutes. Due to a medical condition, he is not able to extract glucose from the food he eats. As he is not able to replenish the glucose in his blood stream, he is thus losing his glucose in the blood stream at a rate (in mg/dL per minute) corresponding to $\lambda=0.005$ per minute.

(ii) Find the approximate time it would take for Neo’s glucose level in the blood stream to fall below the healthy range if there is no intake of glucose. [2]

In order to maintain Neo’s glucose level in the blood stream in the healthy range, glucose is injected into his blood stream intravenously at a constant rate of $\mu$ mg/dL per minute.

(iii) Write down a differential equation relating $G,\mu,t$ to model Neo’s situation. Solve this differential equation to find an expression of $G$ in terms of $\mu,t$. [5]

(iv) Explain whether it is recommended to keep injecting glucose at a rate of 0.7 mg/dL per minute into Neo’s blood stream. [2]

(v) Find the maximum range of values of $\mu$ so that Neo’s glucose level in the blood stream will always be in the healthy range. [1]

[2022/RI/Prelim/I/10]

Printed working

$dG/dt=-\lambda G$, $\lambda>0$. $\int dG/G=\int-\lambda dt$; $\ln G=-\lambda t+C$ ($G>0$), $G=Ae^{-\lambda t}$, $A=e^C$.

At $t=0,G=80$, $A=80$; $G=80e^{-0.005t}$.

For $G\le70$, $e^{-0.005t}\le0.875$; $-0.005t\le\ln0.875$; $t\ge26.706$; approximate time 26.7 min.

Injection: $G'=\mu-0.005G=-0.005(G-200\mu)$.

$\int dG/(G-200\mu)=\int-0.005dt$; $\ln|G-200\mu|=-0.005t+D$; $G-200\mu=Be^{-0.005t}$, $B=\pm e^D$.

At 0, $G=80$: $B=80-200\mu$. Thus $G=200\mu+(80-200\mu)e^{-0.005t}$.

At $\mu=0.7$, $G=140-60e^{-0.005t}\to140$. Not recommended as after a long time $G$ exceeds 100 mg/dL.

Printed alternative: $140-60e^{-0.005t}\ge100\Rightarrow60e^{-0.005t}\le40\Rightarrow t\ge81.093$. Not recommended after approximately 81 min.

Printed original (v): $70<200\mu<100\Rightarrow0.35<\mu<0.5$. The separate review below gives the endpoint qualification.

Printed answer key

(i) $dG/dt=-\lambda G$, $\lambda>0$; $G=Ae^{-\lambda t}$. (ii) 26.7 min. (iii) $dG/dt=-0.005(G-200\mu)$; $G=200\mu+(80-200\mu)e^{-0.005t}$. (iv) Not recommended as after a long period of time, the glucose level in the blood stream will exceed 100 mg/dL. (v) Printed original: $0.35<\mu<0.5$ (endpoint qualification in the review).

Teacher note 1

Given $dG/dt=f(G)$.

Teacher note 2

Neo’s glucose level falls below the healthy range when $G\le70$ (source wording; the reviewed strict crossing is distinguished below).

Teacher note 3

Given $dG/dt=f(G)$, integrate $\int dG/f(G)=\int1dt$.

Teacher note 4

As $t\to\infty$, $G\to200\mu$. $70<200\mu<100$ would put the long-term equilibrium in the interior of the healthy range.

Mathematical review

(i) $G'=-\lambda G$ gives $G=Ae^{-\lambda t}$.

(ii) Here $G=80e^{-0.005t}$. The threshold is reached at $t=\ln(80/70)/0.005\approx26.706$ minutes; G falls below 70 afterwards.

(iii) $G'=\mu-0.005G=-0.005(G-200\mu)$. Thus $G=200\mu+(80-200\mu)e^{-0.005t}$. Differentiation verifies the DE and the initial value.

(iv) At μ=0.7 the equilibrium is 140 and $G=140-60e^{-0.005t}$. It crosses 100 at $t=\ln(3/2)/0.005\approx81.093$ minutes and then exceeds the interval.

(v) Since $0<e^{-0.005t}\le1$, G is between 80 and its limiting equilibrium. For the inclusive interval $70\le G\le100$, the maximal range is $70\le200\mu\le100$, or $0.35\le\mu\le0.5$. Even if finite-time G must be strictly between 70 and 100, these endpoint input values only approach the endpoints asymptotically and remain strictly inside for all finite times.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 3. Working page(s): 11, 12. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.