Modeling Situations with Differential Equations AB & BC
Modeling means turning a sentence about a rate into an equation with a derivative in it, and nothing here is solved. The phrase that follows "proportional to" is exactly what the constant $k$ multiplies: proportional to $y$ gives $\frac{dy}{dt} = ky$, proportional to the difference between $T$ and $70$ gives $\frac{dT}{dt} = k(T - 70)$ with the parentheses carrying the whole meaning, inversely proportional to $y$ gives $\frac{dy}{dt} = \frac{k}{y}$, and proportional to a product gives $\frac{dy}{dt} = ky(M - y)$.
The constant decides direction: $k > 0$ grows and $k < 0$ decays, so a decay model with a positive $k$ describes the opposite process. $k$ is not the starting amount, not a count per unit time, and not the doubling time; its units are a reciprocal time, so $k = 0.05$ per year means five percent of the current size per year. A finished model can be read without solving: in $\frac{dy}{dt} = 0.3(200 - y)$ the rate is positive below $200$, negative above it, and zero at the equilibrium solution $y = 200$, which every solution approaches and slows down near.
The work
3 ways in · any order
Lesson
Modeling Situations with Differential Equations
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Turns rate language into a derivative and shows exactly what the constant of proportionality multiplies, through direct, inverse, square-root and product forms, then reads the sign and the units of that constant back out and identifies the equilibrium of a finished model without solving it.
Diagnostic
10-item topic check
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Ten items on the two failure modes of modeling: writing an equation about the quantity instead of its rate or attaching k to the wrong expression, and misreading what the constant means once the model is written.
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.