Exponential models with differential equations

Q: What are the components of the differential equation \(\frac{dy}{dt} = ky\)?

\(\frac{dy}{dt}\) is the rate of change, \(k\) is the constant of proportionality, and \(y\) is the amount present.

Q: In the solution \(y = Ce^{kt}\), what does the constant \(C\) represent?

\(C\) is the initial value of the quantity (the amount at \(t=0\)).

Q: How do you determine the value of the growth constant \(k\) in a contextual problem?

Use a secondary data point \((t, y)\) and solve for \(k\) using logarithms.

Q: What determines whether an exponential model represents growth or decay?

The sign of the constant \(k\).

Q: Why is it important to avoid early rounding of the constant \(k\) in calculator-active problems?

Early rounding can lead to inaccuracies in the final answer, which must be correct to three decimal places.

Q: If \(\frac{dP}{dt} = 0.03P\), what is the specific growth constant and general solution?

The growth constant is \(k = 0.03\) and the solution is \(P(t) = P_0 e^{0.03t}\).

AP Calculus BC · AP (Advanced Placement) · 4 min read · Updated: 7 Sept 2026

Welcome to Exponential Models!

In the world of calculus, some of the most famous patterns involve things that grow or shrink based on how much of them is already there. Think of a population of rabbits: the more rabbits you have, the faster they reproduce. Or think of a cooling cup of coffee: the hotter it is compared to the room, the faster it loses heat. In this chapter, we explore how to translate these "proportional" relationships into differential equations and find their solutions.

Don't worry if differential equations sound intimidating—for this specific chapter, we are looking at one very specific and very powerful pattern that you might already recognize from your earlier algebra days!

The "Core Secret" of Exponential Growth

The heart of this chapter is a single sentence: "The rate of change of a quantity is proportional to the quantity itself."

If you see that phrase in a word problem, your brain should immediately jump to this differential equation:

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

Let's break down what this math means in plain English:

\(\frac{dy}{dt}\): The rate of change of the quantity \(y\) over time \(t\).
\(k\): The constant of proportionality (essentially, the "speed" of the growth or decay).
\(y\): The amount of the stuff currently present.

Analogy: Imagine your bank account. If the bank pays you interest, the amount of money you gain (\(\frac{dy}{dt}\)) depends on how much money is already in the account (\(y\)). The interest rate is your \(k\).

Key Takeaway:

Whenever a rate is proportional to the amount present, you are dealing with an exponential model.

From Equation to Solution

While we learned how to solve differential equations using Separation of Variables in Topic 7.6, the AP exam expects you to recognize the solution to \(\frac{dy}{dt} = ky\) immediately. It always looks like this:

\(y = Ce^{kt}\)

In this solution:

\(C\) is the initial value (the amount at time \(t=0\)). You will often see it written as \(y_0\).
\(e\) is Euler's number (approximately \(2.718\)).
\(k\) is the growth constant (if \(k > 0\), it's growing; if \(k < 0\), it's decaying).
\(t\) is time.

Did you know? This is exactly the same as the "PERT" formula (\(A = Pe^{rt}\)) you likely used in Algebra 2 for continuous interest!

Step-by-Step: Solving a Contextual Problem

When you encounter a word problem on the AP exam, follow these steps to stay organized:

Step 1: Identify the variables. Look for the initial amount (where \(t=0\)) and any other data point (like "after 5 hours, the population is 200").

Step 2: Set up the general solution. Write down \(y = Ce^{kt}\).

Step 3: Solve for \(C\). Plug in the initial condition. If you are told the population starts at 50, then \(C = 50\). Now your equation is \(y = 50e^{kt}\).

Step 4: Solve for \(k\). Use the second data point. For example, if at \(t=5\), \(y=200\), you would write:
\(200 = 50e^{k(5)}\)
\(4 = e^{5k}\)
\(\ln(4) = 5k\)
\(k = \frac{\ln(4)}{5}\)

Step 5: Answer the specific question. Usually, the exam will ask for the amount at a specific time or when the amount will reach a certain level. Plug your \(C\) and \(k\) back into the formula to solve.

Growth vs. Decay

The behavior of the model depends entirely on the sign of \(k\):

Exponential Growth (\(k > 0\)): The quantity increases faster and faster over time. Think of bacteria in a petri dish or an unregulated population.
Exponential Decay (\(k < 0\)): The quantity decreases, slowing down as it approaches zero. Think of radioactive isotopes or a medication being cleared from your bloodstream.

Quick Review Box:

Differential Equation: \(\frac{dy}{dt} = ky\)
General Solution: \(y = Ce^{kt}\)
At \(t=0\): \(y = C\)

Common Pitfalls to Avoid

Even top students can trip up on these small details:

Units of Measure: The AP exam loves to ask you to "interpret the meaning" of your answer. Always include units (e.g., "The population is increasing at a rate of 500 bacteria per hour").
Mixing up \(k\) and \(\frac{dy}{dt}\): Remember that \(k\) is a constant (a fixed number), while \(\frac{dy}{dt}\) is a rate that changes as \(y\) changes.
Calculator Rounding: If you are in the calculator-active section, do not round your value for \(k\) early in the problem. Keep the full decimal in your calculator until the very final answer to ensure accuracy to three decimal places.

Check for Understanding

If you see the equation \(\frac{dP}{dt} = 0.03P\), what can you conclude?

Answer: This is an exponential growth model where the population \(P\) is growing at a rate proportional to its size, with a growth constant of \(0.03\) (or \(3\%\)). The solution would be \(P(t) = P_0 e^{0.03t}\).

Note: In the next chapter, we will look at Logistic Models, which are similar but include a "carrying capacity" that prevents the population from growing forever! For now, master the pure exponential form.

Summary Table

Concept: Proportional Rate of Change
Equation: \(\frac{dy}{dt} = ky\)
Solution: \(y = Ce^{kt}\)
Key Skill: Using initial conditions to find \(C\) and \(k\).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the differential equation for a quantity that grows at a rate proportional to its size?

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

What are the components of the differential equation \(\frac{dy}{dt} = ky\)?

\(\frac{dy}{dt}\) is the rate of change, \(k\) is the constant of proportionality, and \(y\) is the amount present.

What is the general solution to the differential equation \(\frac{dy}{dt} = ky\)?

\(y = Ce^{kt}\)

In the solution \(y = Ce^{kt}\), what does the constant \(C\) represent?

\(C\) is the initial value of the quantity (the amount at \(t=0\)).

How do you determine the value of the growth constant \(k\) in a contextual problem?

Use a secondary data point \((t, y)\) and solve for \(k\) using logarithms.

What determines whether an exponential model represents growth or decay?

The sign of the constant \(k\).

Why is it important to avoid early rounding of the constant \(k\) in calculator-active problems?

Early rounding can lead to inaccuracies in the final answer, which must be correct to three decimal places.

If \(\frac{dP}{dt} = 0.03P\), what is the specific growth constant and general solution?

The growth constant is \(k = 0.03\) and the solution is \(P(t) = P_0 e^{0.03t}\).

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