Topic 7.1 Modeling Situations with Differential Equations: write a differential equation from a verbal description of a rate of change.

What this topic is asking

The College Board (Topic 7.1) asks you to write a differential equation from a verbal description of how a quantity changes. The phrase "rate of change of $y$" is $\frac{dy}{dt}$ (or $\frac{dy}{dx}$), and the description on the right tells you what it equals.

Translating the language

A worked translation

Combined-rate (in/out) models

A frequent AB model is a tank or population with simultaneous inflow and outflow. The net rate of change is the inflow rate minus the outflow rate, each of which may be constant or depend on the current amount. So a tank filling at a fixed $5$ liters per minute while leaking at a rate proportional to the amount gives $\frac{dQ}{dt} = 5 - kQ$. The value where the net rate is zero, here $Q = \frac{5}{k}$, is the equilibrium amount: if the tank starts there it stays there. Identifying inflow and outflow separately, then subtracting, builds these models reliably.

Why modelling comes before solving

This topic deliberately separates writing the differential equation from solving it. Many AB free-response questions award points for the correct equation even when the later solving steps go wrong, so translating the words accurately is worth getting right on its own. The most common modelling error is confusing "proportional to $y$" (which gives $ky$) with a constant rate (which gives just $k$): the first describes a rate that scales with the amount, the second a fixed rate. Read carefully whether the rate depends on the current amount, on a difference, or is constant, and the equation follows.

Equilibrium solutions from the model

Once the model is written, you can read off its equilibrium values without solving: they are the values of the quantity that make $\frac{dy}{dt} = 0$, so the quantity stays constant. For $\frac{dQ}{dt} = 5 - kQ$, the equilibrium is $Q = \frac{5}{k}$; for logistic-style differences like $\frac{dy}{dt} = k(M - y)$, the equilibrium is $y = M$. These constant solutions describe the long-run behavior the system settles toward, and the sign of $\frac{dy}{dt}$ on either side tells you whether the equilibrium is approached or fled. Identifying equilibria straight from the differential equation is a quick, high-value analysis the exam often asks for before any solving.

Reading the units of the model

A correctly built model is dimensionally consistent: the left side $\frac{dy}{dt}$ has units of (amount per time), and every term on the right must match. So in $\frac{dQ}{dt} = 5 - kQ$ with $Q$ in liters and $t$ in minutes, the $5$ is liters per minute and $kQ$ must also be liters per minute, which forces $k$ to have units of per minute. Checking that the units balance is a fast way to catch a mis-built model, for example one where a constant rate was multiplied by the amount when it should not have been. Units also clarify the meaning of the proportionality constant, which is otherwise just an abstract $k$.