Exponential Models with Differential Equations AB & BC
Every proportional-rate model solves to $y = y_{0}e^{kt}$, and the constant carries the behaviour: positive $k$ grows, negative $k$ decays, and the units of $k$ are one over time, so it is neither an amount per unit time nor a duration. Doubling time and half-life come from setting the final value to a multiple of the initial one, which cancels $y_{0}$ and leaves $t = \frac{\ln 2}{k}$ in size; a half-life of $5730$ years gives $k \approx -1.21 \times 10^{-4}$ per year, and after $n$ half-lives the fraction left is $\left(\frac{1}{2}\right)^{n}$.
Newton's law of cooling, $\frac{dT}{dt} = k(T - T_{a})$, produces a SHIFTED exponential: the difference from the surroundings decays, not the temperature itself, so $T = T_{a} + Ce^{kt}$ with $C = T_{0} - T_{a}$. Dividing by $T - T_{a}$ to separate discards the constant solution $T = T_{a}$, which this family happens to recover at $C = 0$. The same shape covers any quantity approaching a limit $L$: $y = L + Ce^{kt}$, with $y = L$ as its equilibrium.
The work
3 ways in · any order
Lesson
Exponential Models with Differential Equations
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Solves the proportional-rate equation once and reads everything off the constant: sign for growth or decay, units that rule out the two common misreadings, and a logarithm for half-life and doubling time. Then builds the shifted exponential of Newton's law of cooling and locates the ambient value as the equilibrium the separation removes.
Diagnostic
10-item topic check
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Ten items on exponential models: the sign and meaning of the growth constant, half-life and doubling time computed with a logarithm rather than a division, the shifted exponential for a quantity approaching a limit, and the constant solution that dividing to separate discards.
Targeted Practice
Drill a single misconception
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Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.