Constant input shifts a decay equilibrium

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

(i) Write down a differential equation relating G,λ,tG,\lambda,t. Solve this differential equation to find an expression of GG in terms of λ,t\lambda,t. marks

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=0t=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 λ=0.005\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. marks

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 μ\mu mg/dL per minute.

(iii) Write down a differential equation relating G,μ,tG,\mu,t to model Neo’s situation. Solve this differential equation to find an expression of GG in terms of μ,t\mu,t. marks

(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. marks

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

[2022/RI/Prelim/I/10]

Total marks: 13

Solution

(i)

Exponential decay

Write down and solve the differential equation for glucose GG with proportional loss λ\lambda.

Hint

  1. dGdt=−λG\frac{dG}{dt}=-\lambda G.

Answer · (i)

G′=−λGG\prime=-\lambda G; G=Ae−λtG=Ae^{-\lambda t}.

(ii)

Time to leave the range

Neo has 80 mg/dL at t=0t=0 and λ=0.005\lambda=0.005. How long until he falls below 70 mg/dL?

Explore this idea

Hint

  1. G=80e−0.005tG=80e^{-0.005t}.

  2. Solve G=70G=70.

Answer · (ii)

Crosses 70 at t≈26.706t\approx26.706 min; below thereafter.

Explore the idea

Glucose without infusion

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of G at t = 080
Read the solution at t26.7
Try an example
G at t = 26.7
70.072
Rate dG/dt there
-0.35
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

Without infusion GG keeps falling towards 0 and leaves the healthy band.

Ideas to try
Read GG at t=26.7t=26.7: it should be 70.
(iii)

Constant infusion

Glucose is injected at constant rate μ\mu. Write down and solve the differential equation.

Hint

  1. dGdt=μ−0.005G=−0.005(G−200μ)\frac{dG}{dt}=\mu-0.005G=-0.005(G-200\mu).

Answer · (iii)

G′=μ−0.005GG\prime=\mu-0.005G; G=200μ+(80−200μ)e−0.005tG=200\mu+(80-200\mu)e^{-0.005t}.

(iv)

Checking a recommendation

Is it recommended to keep injecting at μ=0.7\mu=0.7?

Explore this idea

Hint

  1. With μ=0.7\mu=0.7: G=140−60e−0.005tG=140-60e^{-0.005t}.

Answer · (iv)

No: at μ=0.7\mu=0.7, GG exceeds 100 after≈81.093 min.

Explore the idea

Glucose with μ = 0.7

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of G at t = 080
Read the solution at t81.1
Try an example
G at t = 81.1
99.948
Rate dG/dt there
0.2
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

The equilibrium G=140G=140 is too high: the level leaves the healthy range.

Ideas to try
Read where the curve crosses 100 mg/dL (about 81 minutes).
(v)

Range of μ\mu

Find the maximum range of μ\mu so that GG always stays in the healthy range.

Explore this idea

Hint

  1. The equilibrium 200μ200\mu must lie in [70,100][70,100] (and the start is 80).

Answer · (v)

0.35≤μ≤0.50.35\le\mu\le0.5.

Source prints strict bounds. Inclusive healthy limits admit both endpoints; exclude them only if the limiting equilibrium must also be strictly interior.

Explore the idea

Glucose with μ = 0.4

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of G at t = 080
Read the solution at t100
Try an example
G at t = 100
80
Rate dG/dt there
0
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

With μ=0.4\mu=0.4 the equilibrium is exactly 80 mg/dL: the level stays constant.

Ideas to try
Check that GG stays between 70 and 100 throughout.