Introduction: The Power of Proportional Growth

In our study of differential equations so far, we have looked at how to find solutions using separation of variables (Topic 7.6 & 7.7). In this chapter, we focus on a very specific and incredibly common type of relationship: Exponential Models.

Whether it is a colony of bacteria doubling every hour or a radioactive isotope slowly disappearing, these processes follow a simple rule: the faster the thing grows (or shrinks), the more of it there is. This "proportionality" is the heart of exponential modeling. Don't worry if this seems a bit abstract at first; once you see the pattern, these problems become some of the most predictable on the AP exam!

The Differential Equation: \(\frac{dy}{dt} = ky\)

The "Law of Exponential Change" is defined by a simple differential equation. When a problem says that "the rate of change of a quantity is proportional to the quantity itself," it is giving you this equation:

\(\frac{dy}{dt} = ky\)

Let's break down what these symbols mean:

  • \(y\): The amount of the quantity present at time \(t\).
  • \(\frac{dy}{dt}\): The rate at which the quantity is changing with respect to time.
  • \(k\): The constant of proportionality (also called the growth or decay constant).

The Two Flavors of \(k\)

1. Exponential Growth: If \(k > 0\), the rate of change is positive, and the quantity increases over time (like a population).
2. Exponential Decay: If \(k < 0\), the rate of change is negative, and the quantity decreases over time (like radioactive decay).

The Solution: The Exponential Function

Using the technique of separation of variables, we can solve \(\frac{dy}{dt} = ky\) to find the general solution. While you should know how to separate the variables (\(\int \frac{1}{y} dy = \int k dt\)), for this specific model, you are allowed to recognize the solution immediately:

\(y(t) = y_0 e^{kt}\)

Where:

  • \(y(t)\) is the value at time \(t\).
  • \(y_0\) is the initial value (the amount at \(t = 0\)). This is our "constant of integration" \(C\) that we find using initial conditions.
  • \(e\) is Euler's number (approx. \(2.718\)).
  • \(k\) is the growth/decay constant.

Quick Tip: If a problem gives you the initial value \(y(0) = 5\), you can immediately plug that in as \(y_0\). No complicated solving for \(C\) required!

Step-by-Step: Solving an Exponential Model Problem

Most AP questions involving Topic 7.8 follow a standard three-step process. Let's walk through an example.

Example: A population of fruit flies increases at a rate proportional to the current population. Initially, there are \(20\) flies. After \(3\) days, there are \(100\) flies. Find the population after \(5\) days.

Step 1: Write the Equation

Because the rate is proportional to the population, we use \(\frac{dy}{dt} = ky\), which we know leads to \(y = y_0 e^{kt}\).

Step 2: Use Initial Conditions to Find \(y_0\) and \(k\)

We know \(y_0 = 20\) (the initial amount). So, our equation is \(y = 20e^{kt}\).
Now, use the second piece of information: at \(t = 3\), \(y = 100\).
\(100 = 20e^{k(3)}\)
\(5 = e^{3k}\)
\(\ln(5) = 3k\)
\(k = \frac{\ln(5)}{3}\)

Step 3: Solve the Final Question

The question asks for the population at \(t = 5\).
\(y(5) = 20e^{(\frac{\ln(5)}{3})(5)}\)
Using a calculator (required for this type of calculation on Part B of the exam):
\(y(5) \approx 292.401\)

Key Takeaway: Always find your \(k\) value first before trying to answer the specific question asked!

Important Concepts for AP Success

1. "Half-Life" and "Doubling Time"

In many problems, you won't be given a specific value for \(k\). Instead, you might be given a half-life (the time it takes for half of a substance to decay) or doubling time (the time it takes for a population to double).

  • To find \(k\) for doubling: \(2 = e^{kt} \implies k = \frac{\ln(2)}{\text{time to double}}\)
  • To find \(k\) for half-life: \(0.5 = e^{kt} \implies k = \frac{\ln(0.5)}{\text{half-life}}\)

2. Calculator Usage and Rounding

On the AP exam, if you are using your graphing calculator for these problems, remember Skill 4.E: "Apply appropriate rounding procedures." Always round your final answer to three decimal places unless the problem specifies otherwise. Avoid rounding your \(k\) value midway through the problem—keep the exact value in your calculator to ensure your final answer is accurate!

3. Common Mistake: Mixing up the Rate and the Amount

A common trap is confusing the rate of change (\(\frac{dy}{dt}\)) with the amount (\(y\)). If the question asks "How fast is the population growing at \(t=5\)?", it is asking for the derivative (\(\frac{dy}{dt}\)). You can find this by calculating \(k \cdot y(5)\).

Summary Checklist

  • Does the problem say "rate is proportional to the amount"? If so, use \(\frac{dy}{dt} = ky\).
  • The general solution is always \(y = y_0 e^{kt}\).
  • \(y_0\) is the amount at \(t = 0\).
  • If \(k > 0\), it's growth. If \(k < 0\), it's decay.
  • Use the natural log (\(\ln\)) to solve for \(k\) or \(t\) when they are in the exponent.
  • Always provide answers to three decimal places on the AP exam!

Note: Advanced "Logistic Models" (where growth levels off) are for BC students only (Topic 7.9). For AP Calculus AB, you only need to master this "unrestricted" exponential growth and decay!