Ivan Shishkin, Rye (1878)

Problems/Differential equationUnreviewed

Coffee now, regret later

by Nugget·
24
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A tired engineer is working late on a project and drinks a coffee containing 200200 mg of caffeine at noon. They would like to be able to fall asleep by 23:0023{:}00.

We make the following simplifying assumptions:

  • the amount of caffeine in the body decreases at a rate proportional to the amount currently present;
  • the proportionality constant remains constant over time;
  • no other source of caffeine is consumed;
  • caffeine is absorbed instantaneously.
  • sleep is assumed not to be significantly affected once the amount of caffeine in the body falls below 5050 mg;
  • the half-life of caffeine is 55 hours.

Let C(t)C(t) denote the amount of caffeine, in milligrams, present in the engineer’s body tt hours after drinking the coffee.

  1. Explain why CC can be modeled by a differential equation of the form
    C′(t)=−kC(t),C'(t)=-kC(t),where k>0k>0.

  2. Solve this differential equation and show that
    C(t)=C0e−kt.C(t)=C_0e^{-kt}.

  3. Use the initial condition to determine C0C_0.

  4. Using the half-life of caffeine, determine the constant kk.

  5. Determine how long it takes for the amount of caffeine in the body to fall below 5050 mg.

  6. At what time does this threshold occur?

  7. The next day, the engineer plans to go to bed at 23:0023{:}00. What is the latest time they could drink the same coffee so that the model predicts less than 5050 mg of caffeine remaining at bedtime?

  8. The engineer is still tired and considers having a second identical coffee at 16:0016{:}00. Assuming the effects of the two coffees add together, determine whether the 5050 mg threshold will be reached before 23:0023{:}00.

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Solution by Nugget

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Let C(t)C(t) denote the amount of caffeine, in milligrams, present in the engineer’s body tt hours after drinking the coffee.

  1. Since the rate at which caffeine is eliminated is assumed to be proportional to the amount currently present in the body, we have
    C′(t)=−kC(t),C'(t)=-kC(t),where k>0k>0. The negative sign indicates that the amount of caffeine decreases over time.

  2. Separating variables gives
    dCC=−k dt.\frac{dC}{C}=-k\,dt.Integrating,
    ln⁡C=−kt+c,\ln C=-kt+c,

C(t)=C0e−kt,C(t)=C_0e^{-kt},where C0>0C_0>0 is a constant.

  1. Immediately after drinking the coffee, the engineer has 200200 mg of caffeine in their body. Hence
    C(0)=200.C(0)=200.Since
    C(0)=C0,C(0)=C_0,we obtain
    C0=200.C_0=200.Thus
    C(t)=200e−kt.C(t)=200e^{-kt}.

  2. The half-life is 55 hours, so after 55 hours the amount of caffeine has been divided by two:
    C(5)=100.C(5)=100.Therefore
    100=200e−5k,100=200e^{-5k},so
    e−5k=12.e^{-5k}=\frac12.Taking logarithms,
    −5k=ln⁡(12)=−ln⁡2,-5k=\ln\left(\frac12\right)=-\ln 2,and hence
    k=ln⁡25.\boxed{k=\frac{\ln 2}{5}}.

The model is therefore
C(t)=200e−ln⁡25t.\boxed{C(t)=200e^{-\frac{\ln 2}{5}t}}.

  1. We want to determine when the amount of caffeine reaches 5050 mg:
    200e−ln⁡25t=50.200e^{-\frac{\ln 2}{5}t}=50.Thus
    e−ln⁡25t=14.e^{-\frac{\ln 2}{5}t}=\frac14.Since
    14=2−2,\frac14=2^{-2},we obtain
    −ln⁡25t=−2ln⁡2,-\frac{\ln 2}{5}t=-2\ln 2,and therefore
    t=10 hours.\boxed{t=10\text{ hours}}.

  2. The coffee was consumed at noon. Ten hours later, the caffeine level reaches 5050 mg at
    22:00.\boxed{22{:}00}.

  3. To have at most 5050 mg of caffeine remaining by 23:0023{:}00, the engineer must allow at least 1010 hours between drinking the coffee and going to bed.

Therefore, the latest possible time to drink the coffee is
23:00−10 hours=13:00.23{:}00-10\text{ hours}=13{:}00.

Hence,
13:00.\boxed{13{:}00}.

  1. Suppose the engineer drinks one coffee at 12:0012{:}00 and another identical coffee at 16:0016{:}00.

At 23:0023{:}00, the caffeine remaining from the first coffee is
C1=200e−ln⁡25⋅11=200⋅2−11/5.C_1=200e^{-\frac{\ln 2}{5}\cdot 11} =200\cdot 2^{-11/5}.

The second coffee has been in the body for 77 hours, so its contribution is
C2=200e−ln⁡25⋅7=200⋅2−7/5.C_2=200e^{-\frac{\ln 2}{5}\cdot 7} =200\cdot 2^{-7/5}.

Thus the total amount of caffeine at 23:0023{:}00 is
Ctotal=200(2−11/5+2−7/5).C_{\mathrm{total}} = 200\left(2^{-11/5}+2^{-7/5}\right).

Numerically,
Ctotal≈43.5+75.8≈119.3 mg.C_{\mathrm{total}}\approx 43.5+75.8 \approx119.3\text{ mg}.

Therefore,
Ctotal>50 mg.\boxed{C_{\mathrm{total}}>50\text{ mg}}.

According to our model, the engineer will not reach the 5050 mg threshold before 23:0023{:}00. So perhaps that second coffee was not such a good idea.

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