Models & Qualitative Behavior

Why this form: a population far below capacity grows like $rP$, as if resources were unlimited; the factor $1 - \tfrac PK$ is the fraction of room still available, which throttles growth to $0$ as $P$ reaches $K$ and makes it negative above $K$. The rate $f(P) = rP\left(1 - \tfrac PK\right)$ is a parabola with zeros at $0$ and $K$, so its vertex is midway, at $P = \tfrac K2$, where the rate is $\tfrac{rK}{4}$. With $r = 1$ and $K = 10$ the fastest growth is $2.5$ per unit time, at $P = 5$.

Why the excess: Newton's law says heat flows at a rate proportional to the temperature difference, $T' = -k(T - T_{\text{amb}})$. The difference $u = T - T_{\text{amb}}$ satisfies $u' = -ku$, so it — not $T$ — decays exponentially, and equal times multiply it by equal factors. For a tank, bookkeeping gives the equation: salt enters at (concentration in)$\times$(flow in) and leaves at (current concentration $S/V$)$\times$(flow out). With equal flows, the equilibrium $S' = 0$ is where the tank's concentration matches the incoming one, so $S = (\text{incoming concentration})\times V$.

Why it undershoots here: the solution $4e^t$ bends upward, so its tangent line at $t = 0$ lies below it, and one step along the tangent ends below the curve: $4 + 0.5\cdot4 = 6$ against $4e^{0.5} \approx 6.59$, an error of about $0.59$. Two steps of $h = 0.25$ re-aim halfway: $4 \to 5 \to 6.25$, error about $0.34$. The error has not quite halved; it approaches exact halving as $h$ shrinks, because Euler's error is proportional to $h$ only for small $h$ (see the proof below).

Why units first: every term must have units of amount per time (grams per minute, degrees per minute); a term that does not is a modelling error, caught before any calculus. Why the equilibrium: it is the one number the model announces without solving, and the exponential archetype then says exactly how the gap to it closes — by the factor $e^{-k}$ per unit time, with time constant $\tfrac1k$.

Step 1 — the excess went from $-20$ to $-10$ in $2$ minutes: it halves every $2$ minutes, so $u = -20\cdot2^{-t/2}$ and $T = 25 - 20\cdot2^{-t/2}$. No need to compute $k$ — the halving time carries the same information ($k = \tfrac{\ln 2}{2} \approx 0.347$ per minute).

Step 2 — $24^\circ$ is an excess of $-1$: $20\cdot2^{-t/2} = 1$, so $2^{t/2} = 20$ and $t = 2\log_2 20 \approx 8.64$ minutes.

Step 3 — check by halvings: $20 \to 10 \to 5 \to 2.5 \to 1.25$ takes $8$ minutes, and $1.25 \to 1$ needs a bit under half of another $2$-minute period.

Step 4 — the last degree takes longest: going from $15^\circ$ to $20^\circ$ took $2$ minutes, but $20^\circ$ to $24^\circ$ takes more than $4$ more, because the driving difference keeps shrinking.

Step 1 — the equation is linear: $S' + \dfrac{2}{50 + t}S = 6$. Its integrating factor is $e^{\int 2/(50+t)\,dt} = (50 + t)^2$.

Step 2 — multiply through: $\big((50+t)^2S\big)' = 6(50+t)^2$, so $(50+t)^2S = 2(50+t)^3 + C$.

Step 3 — the start: $2500\cdot10 = 2\cdot125000 + C$ gives $C = -225000$, so $S = 2(50 + t) - \dfrac{225000}{(50+t)^2}$.

Step 4 — the tank holds $100$ L at $t = 50$: $S = 200 - \dfrac{225000}{10000} = 177.5$ g, a concentration of $1.775$ g/L.

Step 5 — read the formula: $2(50 + t) = 2V$ is the salt the tank would hold at the incoming concentration $2$ g/L; the second term is the initial deficit, fading as $\tfrac{1}{V^2}$. Treating $V$ as constant here (using $\tfrac{S}{50}$ for the outflow concentration) would be wrong — see the sidenote.

Step 1 — the excess over equilibrium, $w = v - 49$, satisfies $w' = -0.2w$, the same archetype as cooling. So $w = -49e^{-0.2t}$ and $v = 49\left(1 - e^{-0.2t}\right)$.

Step 2 — $90\%$ of terminal: $1 - e^{-0.2t} = 0.9$, so $e^{-0.2t} = 0.1$ and $t = 5\ln 10 \approx 11.5$ s.

Step 3 — read the time constant $\tfrac{1}{0.2} = 5$ s: each $5$ s closes $63\%$ of the remaining gap, and each $5\ln 10$ s closes $90\%$.

Step 4 — the model says terminal velocity is approached, never reached; a real skydiver is within a fraction of a percent after about $30$ s ($e^{-6} \approx 0.25\%$).

Why $A$ has that value: at $t = 0$ the formula gives $\dfrac{K}{1 + A} = P_0$. Why the midpoint is the inflection: $P'' = f'(P)\,P'$, and $f(P) = rP(1 - P/K)$ has $f'(P) = 0$ exactly at $P = \tfrac K2$, so the curve switches from bending up to bending down there. Why the symmetry: $\tfrac1P - \tfrac1K$ is a pure exponential, so the curve's shape depends only on $t - t^*$.

Step 1 — half capacity: $\dfrac{1000}{1 + 9e^{-0.5t}} = 500$ needs $9e^{-0.5t} = 1$, so $t = 2\ln 9 \approx 4.39$. This is the inflection point, where growth is fastest ($P' = \tfrac{rK}{4} = 125$ per unit time).

Step 2 — $P(10) = \dfrac{1000}{1 + 9e^{-5}} \approx 942.8$.

Step 3 — the S-shape is symmetric about the inflection point: $P$ rises from $100$ to $500$ in $4.39$ time units, and from $500$ to $900$ in another $4.39$ (solve $9e^{-0.5t} = \tfrac19$: $t = 8\ln 3 \approx 8.79 = 2\cdot 4.39$).

Step 4 — a population started above capacity, say $P(0) = 1500$, has $A = -\tfrac13$ and decreases toward $1000$; one started at exactly $1000$ never moves.

Step 1 — the reciprocals are $\tfrac1{10}$, $\tfrac1{25}$, $\tfrac1{50}$. Their successive differences are $\tfrac1{25} - \tfrac1{10} = -\tfrac3{50}$ and $\tfrac1{50} - \tfrac1{25} = -\tfrac1{50}$, and these also shrink by $q$, so $q = \tfrac13$.

Step 2 — find $\tfrac1K$ from $\tfrac1{25} - \tfrac1K = q\left(\tfrac1{10} - \tfrac1K\right)$: multiplying by $3$, $\tfrac3{25} - \tfrac3K = \tfrac1{10} - \tfrac1K$, so $\tfrac2K = \tfrac3{25} - \tfrac1{10} = \tfrac1{50}$ and $K = 100$.

Step 3 — the rate per interval is $r = \ln 3$, and $A = \tfrac{K}{P_0} - 1 = 9$.

Step 4 — predict: $\dfrac1{P_3} = \dfrac1{100} + q^3\cdot\dfrac{9}{100} = \dfrac1{100}\left(1 + \dfrac{9}{27}\right)$, so $P_3 = 75$. The counts $10, 25, 50, 75$ then approach $100$, the next ones being $90$ and $96.4$.

Halving in equal time intervals is exactly the property of exponentials, which is why the "excess halves every 10 minutes" phrasing works. $\blacksquare$

Mark arrows on a number line and stability is visible: arrows converging on an equilibrium make it stable (perturb, return), diverging make it unstable (perturb, flee). Logistic growth $y' = ry(1 - y/K)$: zeros at $0$ and $K$; $f > 0$ between them, so arrows point from $0$ toward $K$ — the origin is unstable, the carrying capacity stable, and every positive solution S-curves to $K$, all without solving the equation. The phase line is the qualitative theory in miniature: solve only when you need numbers; sketch when you need understanding.

Step 1 — model: $y' = ky$ with $y = 10e^{kt}$.

Step 2 — pin $k$ from the half-life: $y(6) = 5$ gives $e^{6k} = \tfrac12$, so $k = -\tfrac{\ln2}{6} \approx -0.1155$.

Step 3 — solve $10e^{kt} = 1$: $e^{kt} = \tfrac1{10}$, so $t = \dfrac{\ln 10}{\ln 2}\cdot6 = 6\log_2 10 \approx 19.9$ hours.

Step 4 — sanity check in halvings: $10 \to 5 \to 2.5 \to 1.25$ after three half-lives ($18$ h), and $1.25 \to 1$ needs about a third more of one half-life — $18 + 2 \approx 20$ h ✓.

Step 5 — the modeling lesson: the answer depended only on the ratio $10:1$, so every decade of decay costs the same $\log_2 10 \approx 3.32$ half-lives, whatever the starting dose.

Step 1 — build the equation from bookkeeping: rate in $= 2 \cdot 5 = 10$ g/min; rate out $=$ (concentration)(flow) $= \tfrac{S}{100}\cdot5 = \tfrac{S}{20}$; so $S' = 10 - \tfrac{S}{20}$.

Step 2 — equilibrium first: $S' = 0$ at $S = 200$ g — which is just (incoming concentration)(volume) $= 2\cdot100$, as it must be.

Step 3 — solve: linear with $\mu = e^{t/20}$: $(e^{t/20}S)' = 10e^{t/20}$, so $e^{t/20}S = 200e^{t/20} + C$, and $S(0) = 0$ gives $C = -200$: $S(t) = 200\left(1 - e^{-t/20}\right)$.

Step 4 — read the time scale: the "time constant" is $20$ min; after $20$ min the tank is at $200(1 - e^{-1}) \approx 126$ g, after an hour $\approx 190$ g — the exponential approach to equilibrium that governs tanks, cooling coffee, and charging capacitors alike.