Unit 7 Notes: Exponential Models via Differential Equations (AP Calculus AB)

Proportional Growth and the Differential Equation

An exponential model shows up whenever the rate of change of a quantity is proportional to the quantity itself. In plain language: the more you have, the faster it grows (or the faster it decays). This “self-reinforcing” idea is what makes exponential behavior so common in population growth, radioactive decay, and continuously compounded interest.

What it means (and why it matters)

If y(t) is the amount of something at time t, the statement “the rate of change of y is proportional to y” translates directly into a differential equation:

\frac{dy}{dt} = ky

Here, proportional means there is some constant multiplier k such that the derivative is always that constant times the current amount.

• If k>0, then \frac{dy}{dt}>0 whenever y>0, so the quantity grows.
• If k

This matters in AP Calculus because it’s one of the most important “modeling bridges” between derivatives and functions: you start with a rate statement (a derivative), build a differential equation, and then solve it to get an explicit function you can interpret.

How the model behaves (before solving anything)

Even before solving, you can predict key behavior:

• If y>0 and k>0, slopes get steeper as y gets bigger, so the graph increases faster and faster.
• If y>0 and k

A common misconception is to think “constant percent change” is different from “proportional rate.” They are actually the same idea: proportional rate means

\frac{1}{y}\frac{dy}{dt} = k

The left side is the instantaneous relative growth rate (instantaneous percent rate, as a decimal). So exponential models are exactly the models where the relative growth rate is constant.

Notation you’ll see (same structure, different letters)

The letter changes, but the structure “derivative equals constant times the function” is the key.

Exam Focus

• Typical question patterns
  • You’re told “rate is proportional to amount” (or “grows/decays at a rate proportional to current amount”) and asked to write the differential equation and/or solve it.
  • You’re given a model \frac{dy}{dt} = ky and an initial value, then asked to find y(t) or evaluate y at some time.
  • You’re asked to interpret k in context (including its units and what its sign implies).
• Common mistakes
  • Treating k as a percentage rather than a decimal per unit time (for example, using 5 instead of 0.05).
  • Forgetting that “proportional to” implies multiplication by the current amount y, not by time t.
  • Assuming growth is linear because the rate is “constant.” In exponential models, the relative rate is constant, not the absolute rate.

Solving \frac{dy}{dt} = ky (Separation of Variables)

What solving means here

To solve a differential equation is to find a function y(t) whose derivative satisfies the given relationship. For exponential models, the solving technique is usually **separation of variables**, because you can rearrange the equation to group all y-terms on one side and all t-terms on the other.

How separation works, step by step

Start with:

\frac{dy}{dt} = ky

1. Divide both sides by y (assuming y\neq 0; we’ll address y=0 as a special solution):

\frac{1}{y}\frac{dy}{dt} = k

2. Multiply both sides by dt to separate variables:

\frac{1}{y}dy = k\,dt

3. Integrate both sides:

\int \frac{1}{y}dy = \int k\,dt

That gives:

\ln|y| = kt + C

4. Solve for y by exponentiating:

|y| = e^{kt+C} = e^C e^{kt}

Let A = e^C (a positive constant), then

|y| = A e^{kt}

This is usually rewritten as a single constant that can be any nonzero real number:

y = Ce^{kt}

This form automatically includes negative solutions (if C