Math Detours · Grade 12 · Sharpen · 11

The equation that says "rate ∝ amount"

dy/dt = ky solved by separating variables. Radioactive decay, a cooling coffee, and why half-lives never reach zero.

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Tonight's idea

Many quantities change at a rate proportional to how much there is. Radioactive atoms decay independently, so a fixed fraction of them decays each second. Bacteria divide. Money earns interest. In each case

\(\dfrac{dy}{dt} = ky\): the rate of change of \(y\) is \(k\) times \(y\). With \(k > 0\) it grows; with \(k \lt 0\) it decays.

This is a differential equation: it gives the slope, not the function. To solve it, treat \(dy/dt\) as a fraction and put all the \(y\) on one side, all the \(t\) on the other:

\[\frac{dy}{y} = k\,dt \quad\Longrightarrow\quad \int\frac{dy}{y} = \int k\,dt \quad\Longrightarrow\quad \ln|y| = kt + C.\]

Exponentiate: \(y = e^Ce^{kt} = Ae^{kt}\), and \(t = 0\) shows \(A = y(0)\). This is separation of variables. Check: the derivative of \(Ae^{kt}\) is \(kAe^{kt} = ky\).

For decay write \(y = y_0e^{-kt}\) with \(k > 0\). The half-life \(T\) is the time to halve: \(y_0e^{-kT} = \tfrac12y_0\), so \(e^{-kT} = \tfrac12\) and

\(T = \dfrac{\ln 2}{k}\).

Halving takes the same time from any level, so after \(T, 2T, 3T\) you have \(\tfrac12, \tfrac14, \tfrac18\) of the start. Half of something is never nothing.

y₀y₀/2y₀/4y₀/8 T2T3Tt
\(y = y_0e^{-kt}\). Each half-life \(T = \ln 2/k\) halves whatever is left, from any starting level, so the curve never reaches zero.

Watch one: coffee in a cool room

Newton's law of cooling: a drink loses heat at a rate proportional to the gap between its temperature and the room's. With \(u = \text{temperature} - \text{room}\), that is \(du/dt = -ku\), the same equation. Coffee at 90 °C in a 20 °C room is 60 °C after ten minutes. When does it reach 40 °C?

\(u_0 = 70\) and \(u(10) = 40\), so \(70e^{-10k} = 40\) and \(k = \tfrac{1}{10}\ln\tfrac74 \approx 0.0560\) per minute. For 40 °C we need \(u = 20\):

\[e^{-kt} = \tfrac{20}{70} = \tfrac27, \qquad t = \frac{\ln 3.5}{k} \approx 22.4 \text{ minutes}.\]

The first 30 degrees took ten minutes; the next 20 take more than twelve. Cooling slows as the gap closes.

Your turn

  1. Solve \(dy/dt = -0.2y\) with \(y(0) = 50\). Find \(y(10)\) and the half-life.
  2. A medical isotope has a half-life of 8 days. (a) Find \(k\) in \(y = y_0e^{-kt}\). (b) What fraction remains after 24 days? After 30 days?
  3. A population satisfies \(dP/dt = kP\) with \(k > 0\) and doubles every 20 years. (a) Show that the doubling time is \(\ln 2/k\). (b) How long does tripling take? Do you need to know \(k\)?
  4. A body is found at midnight at 30 °C in a room held at 20 °C; an hour later it is 28 °C. A living body is at 37 °C. Use Newton's law of cooling to estimate the time of death. (Find \(k\) from the two readings, then run the clock backwards.)

Stretch: an equation that does reach zero

Water drains from a tank at a rate proportional to the square root of its depth: \(dh/dt = -k\sqrt h\). Separate variables and solve. A tank 1 m deep is half empty after 10 minutes. When is it completely empty, and why does this one reach zero when \(e^{-kt}\) never does?

Answers

  1. \(y = 50e^{-0.2t}\). \(y(10) = 50e^{-2} \approx 6.77\). \(T = \ln 2/0.2 \approx 3.47\).
  2. (a) \(k = \ln 2/8 \approx 0.0866\) per day. (b) 24 days is three half-lives: \(\tfrac18 = 12.5\%\). 30 days: \(2^{-30/8} = e^{-30k} \approx 0.074\), about 7.4%.
  3. (a) \(P_0e^{kT} = 2P_0\) gives \(kT = \ln 2\). (b) \(e^{kt} = 3\) gives \(t = \ln 3/k = \dfrac{\ln 3}{\ln 2}\times 20 \approx 31.7\) years. You never need \(k\): tripling always takes \(\ln 3/\ln 2 \approx 1.585\) doubling times.
  4. Write \(\theta\) for the body's temperature (we used \(T\) for the half-life) and let \(u = \theta - 20\). At midnight \(u = 10\), an hour later \(u = 8\), so \(e^{-k} = 0.8\) and \(k = \ln 1.25 \approx 0.223\) per hour. At death \(u = 17\). The time \(t\) from death to midnight satisfies \(17e^{-kt} = 10\), so \(t = \ln 1.7/k \approx 2.38\) hours, about 2 h 23 min: death at roughly 9:37 pm. The estimate is only as good as its assumptions (a constant room temperature, 37 °C at death, and the simple law itself).

Stretch. \(\dfrac{dh}{\sqrt h} = -k\,dt\) gives \(2\sqrt h = -kt + C\); with \(h(0) = 1\), \(C = 2\), so \(\sqrt h = 1 - kt/2\). Half empty at \(t = 10\): \(\sqrt{1/2} = 1 - 5k\), \(k = (2 - \sqrt2)/10 \approx 0.0586\). Empty when \(\sqrt h = 0\): \(t = 2/k = 20/(2 - \sqrt2) \approx 34.1\) minutes. Near zero, \(\sqrt h\) is much larger than \(h\), so the outflow does not shrink fast enough to stall. With rate proportional to \(h\) itself, the closer you get to zero the slower you go, and you never arrive.

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