Growth, Decay, and Continuous Models

The phrase 'rate of change of P is proportional to P' translates directly to

• $k > 0$ gives growth (population, bacteria, invested principal).
• $k < 0$ gives decay (radioactivity, moisture loss, cooling) — write it as $\dfrac{dP}{dt} = -kP$ with $k>0$ to keep signs honest.

This is a separable, first-order, first-degree equation. Separate the variables and integrate: $\displaystyle\int\dfrac{dP}{P} = \int k\,dt$, giving $\log P = kt + c$.

Solving $\dfrac{dP}{dt} = kP$ with $P(0)=P_0$ gives the master formula:

• The initial value fixes $P_0$.
• A second data point $(t_1, P_1)$ fixes $k$: $\dfrac{P_1}{P_0} = e^{kt_1} \Rightarrow k = \dfrac{1}{t_1}\log\dfrac{P_1}{P_0}$.
• Often you never need $k$ alone — dividing two instances of $P_0 e^{kt}$ cancels $P_0$, and the ratio form $\dfrac{P_2}{P_1}=e^{k(t_2-t_1)}$ does all the work.

For growth $P = P_0 e^{kt}$:

• Doubling in period $T$: $e^{kT}=2$. After $n$ such periods $(t=nT)$, $P = P_0\cdot 2^{n}$. No logs needed when $t$ is a whole multiple of $T$.
• Percentage increase: 'increases by $p\%$ in time $T$' means $\dfrac{P(T)}{P_0} = 1 + \dfrac{p}{100}$. A 20% rise is a factor $\tfrac65$; a 10% rise is $\tfrac{11}{10}$. Then $e^{kT} = 1+\tfrac{p}{100}$ fixes $k$.
• Finding the start $P_0$: given two later readings, divide to get $k$, then back-substitute one reading to recover $P_0$.

Decay model $\dfrac{dm}{dt} = -km$ with solution $m = m_0 e^{-kt}$.

• Half-life link: at $t=h$, $m = \tfrac{m_0}{2}$, so $e^{-kh}=\tfrac12 \Rightarrow k = \dfrac{\log 2}{h}$.
• After $n$ half-lives $(t = nh)$: $m = m_0\left(\tfrac12\right)^{n}$. Just count half-lives when $t$ is a whole multiple of $h$.
• Initial decay rate: $\left.\dfrac{dm}{dt}\right|_{t=0} = -km_0 = -\dfrac{m_0\log 2}{h}$ — negative because mass is falling.

Continuous growth of a principal:

• 'Doubles in $T$ years' gives $e^{rT}=2$, so $rT = \log 2$ — used to find either $r$ or a doubling-based amount.
• 'Rate $x\%$, doubles in $T$': $\dfrac{x}{100}\cdot T = \log 2 \Rightarrow x = \dfrac{100\log 2}{T}$.

Pure proportional loss (moisture, cooling of the simplest kind): $\dfrac{dP}{dt} = -kP \Rightarrow \log\dfrac{P}{N} = -kt$. 'Loses half in the first hour' gives $k = \log 2$; then solve for the time to lose any fraction. Mixed model with a constant $\dfrac{dP}{dt} = kP - c$: rewrite as $\dfrac{dP}{kP - c} = dt$ and integrate to $\dfrac1k\log|kP - c| = t + C$. Fix $C$ from $P(0)$, then substitute the target $P$. (For $\dfrac{dP}{dt}=0.5P-450$, $kP-c = 0.5(P-900)$, so $2\log|P-900| = t + C$.)

Square-root rate (assets shrinking, tank draining): $\dfrac{dx}{dt} = -k\sqrt{x}$. Separate $\dfrac{dx}{\sqrt x} = -k\,dt$ and integrate: