Topic 7.8 Exponential Models with Differential Equations: derive and apply the exponential growth and decay model from a proportional-rate differential equation.

What this topic is asking

The College Board (Topic 7.8) is the headline application of Unit 7: the exponential model. Whenever the rate of change of a quantity is proportional to the quantity itself, $\frac{dy}{dt} = ky$, the solution is exponential, $y = y_0 e^{kt}$. You derive it by separation and apply it to growth, decay, and half-life problems.

Deriving the model

A worked growth problem

Half-life and doubling time

For decay ($k < 0$), the half-life $t_{1/2}$ is the time for the amount to fall to half. From $\frac{1}{2}y_0 = y_0 e^{k\,t_{1/2}}$ you get $e^{k\,t_{1/2}} = \frac{1}{2}$, so $t_{1/2} = \frac{\ln(1/2)}{k} = \frac{-\ln 2}{k}$ (positive since $k < 0$). For growth ($k > 0$), the doubling time comes from $2 = e^{k\,t_d}$, giving $t_d = \frac{\ln 2}{k}$. A useful shortcut is to work in terms of the ratio $e^{kt}$ directly: if the amount multiplies by $r$ over a known time, then over $n$ such periods it multiplies by $r^n$, avoiding repeated exponential arithmetic, as in the decay worked example where $A(10) = 200(0.75)^2$.

Why proportional rate forces exponential behavior

The defining feature is that the rate scales with the amount: more material means faster change. This self-reinforcing structure is exactly what produces exponential growth or decay, and it is why so many natural processes (population, radioactivity, continuously compounded interest, cooling toward zero) follow the same model. Recognizing the phrase "rate proportional to the amount" as the signal for $y = y_0 e^{kt}$ lets you write the answer immediately, then fit $y_0$ and $k$ to the data. The most common error is sign confusion on $k$: decay needs $k < 0$, and a positive $k$ in a decay problem gives growth, so check that your model decreases when the quantity should decrease.

What the exponential model cannot do

A pure exponential model grows without bound, which makes it realistic only over a limited range. Real populations and many physical quantities eventually level off as resources run out or a ceiling is reached, behavior the exponential model does not capture (the bounded and logistic models that do are BC topics). On the AB exam this matters when a question asks you to comment on the model's limitations: an answer noting that $y = y_0 e^{kt}$ predicts unlimited growth, which is unrealistic over long times, shows the understanding the practices reward. Knowing the model's domain of validity, accurate for early growth or steady decay but not for the long run, is part of using it well.

The differential equation versus the solution

It helps to keep clear which object you are working with. The differential equation $\frac{dy}{dt} = ky$ describes the rule the quantity obeys; the solution $y = y_0 e^{kt}$ is the explicit function that obeys it. The exam may ask for either: "write a differential equation modelling the situation" wants the rule, while "find a formula for the amount at time $t$" wants the solution. You move from the rule to the solution by separation of variables, and you can always check a solution by differentiating it and confirming it satisfies the rule. Distinguishing the two prevents the error of giving the explicit formula when only the differential equation was requested, or vice versa.