## Modeling with First Order ODEs |

**Radioactive Decay:** Let \(N(t)\) be the mass of a radioactive element at time \(t\). The rate of change (decay)
of \(N\) is proportional to its current value.
\[\frac{dN}{dt}=-kN,\]
where \(k > 0\) depends on the element. Solving this ODE we get \(N=ce^{-kt}\). Suppose the initial amount is \(N_0=N(0)\). Then we get \(c=N_0\). Thus the solution is
\[N(t)=N_0e^{-kt}.\]
*Carbon Dating:* There are 2 types of carbon atoms: \(^{12}C\) (the stable nuclide) and radioactive \(^{14}C\) with a halflife of about 5,730 years. The ratio \(^{14}C: ^{12}C\) is approximately constant in nature. A living creature taking carbon from nature is made up with this ratio. After its death \(^{14}C\) starts to decay.

**Example.**
Suppose \(25\%\) of the original amount of \(^{14}C\) remained in a fossil. Find the age of the fossil.

*Solution.* First we find \(k\) for \(^{14}C\) in \(N(t)=N_0e^{-kt}\). We know that \(\frac{N_0}{2}=N_0e^{-5730k}\). Solving we get \(k=\ln 2/5730\) and hence
\[N(t)=N_0e^{-t\ln 2/5730}.\]
For the fossil we want to find \(t\) for which \(N(t)=25N_0/100\). Plugging this into the preceding equation we get
\[25N_0/100=N_0e^{-t\ln 2/5730} \implies t=11,460 \text{ years}\]
The answer is not surprising because of the halflife of \(^{14}C\) (about 5,730 years).

**Newton's Law of Cooling:** Let \(T(t)\) be the temperature of an object at time \(t\). Then
\[\frac{dT}{dt}=-k(T-T_a),\]
where \(k > 0\) is a constant and \(T_a\) is the constant ambient temperature. Solving this ODE we get \(T=T_a+ce^{-kt}\). Suppose the initial temperature is \(T_0=T(0)\). Then we get \(c=T_0-T_a\). Thus the solution is
\[T(t)=T_a+(T_0-T_a)e^{-kt}.\]
Note that \(T(t)=T_a+(T_0-T_a)e^{-kt}\to T_a\) as \(t\to \infty\).

**Kirchhoff's Circuit Law:** Consider an electric circuit with a capacitor, resistor, and battery. Let \(Q(t)\) be the charge of the capacitor at time \(t\). Then
\[R\frac{dQ}{dt}+\frac{Q}{C}=V,\]
where \(R\) is the constant resistant, \(C\) is the constant capacitance, and \(V\) is the constant voltage supplied. Solving this ODE we get \(Q=CV+ke^{-t/RC}\). Suppose the initial charge is \(Q(0)=0\). Then we get \(k=-CV\). Thus the solution is
\[Q(t)=CV(1-e^{-t/RC}).\]
Note that \(Q(t)=CV(1-e^{-t/RC})\to CV\) as \(t\to \infty\).

**Mixing Rate Problems:** The key idea for mixing rate problems is that the effective rate is the rate in minus the rate out.

**Example.**
Suppose a tank contains 200 gal of salt water containing 100 lb salt. Then salt water with 3 lb salt/gal is pumped into the tank at a rate \(2\) gal/min and the "well-stirred" salt water is pumped out of the tank at a rate \(2\) gal/min. Suppose \(Q(t)\) denotes the amount of salt in lb at time \(t\) min. Then
\[\frac{dQ}{dt}=2\cdot 3-2\cdot \frac{Q}{200},\, Q(0)=100\]
Solving this ODE we get \(Q=600+ce^{-t/100}\). Using the initial condition \(Q(0)=100\) we get \(c=-500\). Thus the solution is
\[Q(t)=600-500e^{-t/100}.\]
Note that \(Q(t)=600-500e^{-t/100}\to 600\) as \(t\to \infty\).

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