7.B Population Models

Important Models

Here are the three important differential equations that are also population models.

Exponential Growth/Decay

The differential equation \[\displaystyle{\frac{dP}{dt} = r P}\] has solution \[\displaystyle{P(t) = C e^{r \, t}}.\]

When \(r>0\), we have the exponential growth population model.

When \(r < 0\), we have exponential decay. This includes radioactive decay, and the metabolization of a drug in the bloodstream.

Exponential Growth/Decay with Constant Input/Output

The differential equation \[\frac{dP}{dt} = r (P-A)\] is exponential growth/decay with constant input/output. Its solution is \[P(t) = A + C e^{r \, t}.\] When \(k >0\) and \(H =rA\), we have the population model \[ \frac{dP}{dt} = rP-H \] for exponential growth at rate \(r\) with constant harvesting at rate \(H\).

Constrained Growth

The differential equation \[\frac{dP}{dt} = r \, P \left(1 - \frac{P}{K}\right)\] has the logistic function as its solution \[P(t) = \frac{K}{1+ C e^{-r \, t}}.\] When \(r>0\) and \(K>0\), we have constrained growth at rate \(r\) with carrying capacity \(K\).

Newton’s Law of Cooling

Newton’s Law of Cooling states that the rate of change of temperature of an object \(\frac{dT}{dt}\) is related to the ambient temperature \(A\) by the equation: \[ \frac{dT}{dt}=r(T-A).\] Note that this is the same differential equation as exponential growth with constant harvesting.

Activities

Drug Metabolization

After a drug is administered in a single injection, the rate at which the drug leaves the body is proportional to the quantity \(Q\) present.
1. What differential equation models this process?
2. Suppose that 10 mg of a drug is injected into the bloodstream. 3 hours later, only 2 mg remains in the bloodstream. Find the formula for the amount of the drug \(Q(t)\), in mg, after \(t\) hours.
Suppose that a drug is being administered at a constant rate of 60 mg/hour and its decreasing at a rate of 20% of the amount present.
1. What differential equation models this process?
2. Assuming that \(Q(t)=0\), find the formula for the amount of drug \(Q(t)\), in mg, after \(t\) hours.
3. What value does \(Q(t)\) converge to for large values of \(t\)?

Elementary, My Dear Watson

In a mystery novel, the body of a murder victim is found at noon in a room with a constant temperature of \(20^\circ C\). At noon, the body is \(35^\circ C\) and at 2 pm the body is \(33^\circ C\). Normal body temperature is \(37^\circ C\). What time did the murder occur?

Optimal Coffee Temperature

In my \(22^\circ C\) office, my coffee starts at \(80^\circ C\), but cools to \(75^\circ C\) after 10 minutes. If I like my coffee between \(63^\circ C\) and \(70^\circ C\), how long is my window to drink it?

Constrained Rabbit Population

The yearly population \(P(t)\) of rabbits is modeled by \[ \frac{dP}{dt} = 0.45 P \left(1 - \frac{P}{90} \right). \]

If the current population is \(P=120\), will the population increase or decrease? How do you know?
For what values of \(P\) is \(\frac{dP}{dt}=0\)?
What value does \(P(t)\) converge to for large values of \(t\)?

Constrained Deer Population

The annual population \(P(t)\) of deer is modeled by the constrained growth equation \[ \frac{dP}{dt} = 0.22 P - 0.0275 P^2. \]

What is the growth rate \(r\)?
What is the carrying capacity \(K\)?
How large is the deer population when that population is growing at its fastest rate? How fast is the population growing at that time?

Solutions

Drug Metabolization

This is an exponential decay function.
1. \(\displaystyle={\frac{dQ}{dt} = rQ}\)
2. The solution is expoential decay \(Q(t) = Ce^{rt}\). We have \(Q(0) = 10 = C e^{r\cdot 0} = C\). So \(Q(t) = 10e^{rt}\), Next we have \(Q(3) = 2 = 10 e^{3r}\) so \(r = \frac{1}{3} \ln (1/5) = -0.537\). Our final formula is \(Q(t) = 10 e^{-0537t}\).
Suppose that a drug is being administered at a constant rate of 60 mg/hour and its decreasing at a rate of 20% of the amount present.
1. \(\displaystyle={\frac{dQ}{dt} = -0.2 Q + 60}\)
2. We need to rewrite the differential equation as \(\displaystyle={\frac{dQ}{dt} = -0.2 (Q - 300)}\). The solution is \(Q(t) = 300 - 300 e^{-0.2t}\).
3. \(Q(t)\) converges to \(300\)

Elementary, My Dear Watson

We use Newton’s Law of Cooling \(\frac{dT}{dt} = r(T-A)\) where \(A=20\). Let’s set \(t=0\) to be noon. We have \(T(0)=35\) and \(T(2)=33\). The initial condition tell us that the solution is \(T(t) = 20 + 15e^{rt}\). Next, we have \(T(2) = 33 = 20 + 15e^{2r}\) and so \(r = \frac{1}{2} \ln(13/15) = -0.072.\) Our temperature function is \(T(t) = 20+15e^{-0.072t}.\)

Finally, we solve \(37 = 20+15e^{-0.072t}\) which is equivalent to \(17/15 = e^{-0.072t}\). Our time of death is \(t= -1/(0.072) \ln(17/15) = -1.74\). So the time of death was 10:15 am.

Optimal Coffee Temperature

We have \(T(t) = 22 + 58 e^{rt}\). Solving \(75 = 22 + 58e^{10r}\) gives \(r= \frac{1}{10} \ln (53/58) = -0.009\). The temperature formula is \(22 + 68 e^{-0.009 t}\).

Next, we solve \(63 = 22 + 58 e^{-0.009 t}\) to find that \(t= -1/0.009 \ln(41/58)=38.54\). Finally we solve \(70 = 22 + 58 e^{-0.009 t}\) to find that \(t= -1/0.009 \ln(48/58)=21.03\). So the coffee is drinkable from minute 21 to minute \(38.5\).

Constrained Rabbit Population

The yearly population \(P(t)\) of rabbits is modeled by \[ \frac{dP}{dt} = 0.45 P \left(1 - \frac{P}{90} \right). \]

We have \(\frac{dP}{dt} = 0.45 \cdot 120 \left(1 - \frac{120}{90} \right) < 0\). The population will decrease because the rate of change is negative.
\(P'(t)=0\) when \(P=0\) and when \(P=90\).
The carrying capacity is 90.

Constrained Deer Population

The annual population \(P(t)\) of deer is modeled by the constrained growth equation \[ \frac{dP}{dt} = 0.22 P - 0.0275 P^2. \]

The growth rate is \(r=0.23\).
We rewrite the derivative as \[ \frac{dP}{dt} = 0.22 P\left( 1 - \frac{0.0275}{0.22} P \right) = 0.22 P\left( 1 - \frac{P}{8} \right) \] So the carrying capacity is \(K=8\).
The rate of change is largest when \(P=K/2=4\). The rate of change is \(0.22 \cdot 4 \cdot \frac{1}{2} = 0.44\).