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 $k<0$, then $\frac{dy}{dt}<0$ whenever $y>0$, so the quantity decays.

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 $k<0$, slopes are negative but their magnitude shrinks as $y$ shrinks, so the graph decreases quickly at first, then levels off toward $0$.

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 $C<0$). Also note that $y=0$ is a solution too; it corresponds to $C=0$.

Why the solution is exponential

The key reason exponential functions appear is that $\frac{d}{dt}(e^{kt}) = ke^{kt}$. Exponentials are (essentially) the only functions that are proportional to their own derivatives, which is exactly what $\frac{dy}{dt} = ky$ demands.

Worked example 1: Solve and use an initial condition

Suppose a population $P(t)$ satisfies

$\frac{dP}{dt} = 0.3P$

and $P(0)=200$.

Step 1: General solution

$P(t) = Ce^{0.3t}$

Step 2: Apply the initial condition

At $t=0$:

$P(0) = Ce^{0} = C = 200$

So

$P(t) = 200e^{0.3t}$

Interpretation: the population grows exponentially with continuous relative growth rate $0.3$ per unit time.

Worked example 2: Decay

A substance decays according to

$\frac{dm}{dt} = -0.08m$

with $m(0)=50$.

By the same steps:

$m(t) = 50e^{-0.08t}$

Notice how the negative $k$ produces exponential decay.

Exam Focus

• Typical question patterns
  • Solve $\frac{dy}{dt} = ky$ and express the solution using an initial condition like $y(0)=y_0$.
  • Given a particular solution, verify it satisfies the differential equation by differentiating and substituting.
  • Use the solution to compute a value at a given time.
• Common mistakes
  • Dropping the absolute value when integrating $\int \frac{1}{y}dy$ and then getting stuck with sign issues. Using $y=Ce^{kt}$ avoids this.
  • Writing $y = e^{kt + C}$ and forgetting to combine constants (you need the multiplicative constant $C$).
  • Mixing up whether $t$ is in years, hours, days, etc., and then misinterpreting $k$.

Finding the Constant $k$ from Information

In many AP problems, the differential equation form is given but $k$ is not. Instead, you’re told something like “the population doubles in 5 years” or “the rate at $t=2$ is 12 when the amount is 200.” The skill is translating that information into an equation for $k$.

Method A: Use a known point on the solution

If $y=Ce^{kt}$ and you know two values like $y(t_1)$ and $y(t_2)$, you can divide to eliminate $C$.

Suppose you know $y(t_1)=y_1$ and $y(t_2)=y_2$. Then:

$y_1 = Ce^{kt_1}$

$y_2 = Ce^{kt_2}$

Divide the second by the first:

$\frac{y_2}{y_1} = e^{k(t_2-t_1)}$

Then take natural logs:

$\ln\left(\frac{y_2}{y_1}\right) = k(t_2-t_1)$

So

$k = \frac{1}{t_2-t_1}\ln\left(\frac{y_2}{y_1}\right)$

This is one of the cleanest ways to find $k$ because the constant $C$ cancels.

Method B: Use the differential equation at a specific time

Sometimes you’re told the instantaneous rate and the amount at the same time. Since

$\frac{dy}{dt} = ky$

at that time,

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

Be careful with units: if $\frac{dy}{dt}$ is “people per day” and $y$ is “people,” then $k$ is “per day.”

Worked example 1: Using a doubling statement

A quantity satisfies $\frac{dy}{dt} = ky$ and triples in 10 hours. Find $k$.

Triples means

$\frac{y(10)}{y(0)} = 3$

But

$\frac{y(10)}{y(0)} = e^{k(10-0)} = e^{10k}$

So

$e^{10k} = 3$

Take ln:

$10k = \ln(3)$

Thus

$k = \frac{\ln(3)}{10}$

Worked example 2: Using rate and amount at a time

At time $t=4$ minutes, a bacteria culture has $y=500$ bacteria and is growing at $\frac{dy}{dt}=75$ bacteria per minute. Assuming exponential growth, find $k$.

Using $\frac{dy}{dt}=ky$ at that instant:

$75 = k(500)$

So

$k = 0.15$

The units are “per minute.”

A common mistake is to think $k$ changes with time because the growth rate changes; in an exponential model, the _absolute_ growth rate changes, but $k$ stays constant.

Exam Focus

• Typical question patterns
  • “Doubles in $T$ units of time” or “half-life is $T$” and you must find $k$.
  • Given $y(t_1)$ and $y(t_2)$, find $k$ (often by taking a ratio).
  • Given $y$ and $\frac{dy}{dt}$ at the same time, compute $k$ quickly.
• Common mistakes
  • Using $\log$ base 10 instead of $\ln$ (natural log) when solving for $k$.
  • Forgetting to subtract times: the exponent involves $t_2-t_1$.
  • Interpreting “triples” as “adds 3” instead of “multiplies by 3.”

Doubling Time, Half-Life, and Time-to-Reach Calculations

Once you have $y(t)=Ce^{kt}$, many questions focus on how long it takes to reach a certain multiple (double, half, 10% remaining, etc.). These are especially common because they test your ability to manipulate exponential equations and interpret them.

Doubling time

The doubling time $T_d$ is the time it takes for the quantity to multiply by 2:

$\frac{y(t+T_d)}{y(t)} = 2$

For $y(t)=Ce^{kt}$:

$\frac{Ce^{k(t+T_d)}}{Ce^{kt}} = e^{kT_d} = 2$

So

$kT_d = \ln(2)$

and

$T_d = \frac{\ln(2)}{k}$

This only makes sense for growth where $k>0$.

Half-life

The half-life $T_{1/2}$ is the time it takes to multiply by $\frac{1}{2}$:

$e^{kT_{1/2}} = \frac{1}{2}$

So

$kT_{1/2} = \ln\left(\frac{1}{2}\right) = -\ln(2)$

Thus

$T_{1/2} = \frac{-\ln(2)}{k}$

For decay, $k<0$, so this produces a positive half-life.

Time to reach a specific value

If you want the time when $y(t)=y^*$, solve

$y^* = Ce^{kt}$

Assuming $C\neq 0$:

$\frac{y^*}{C} = e^{kt}$

Take ln:

$\ln\left(\frac{y^*}{C}\right) = kt$

So

$t = \frac{1}{k}\ln\left(\frac{y^*}{C}\right)$

A frequent error is to try to “cancel the exponent” without using logs. Exponential equations are solved with logarithms.

Worked example: Half-life from a differential equation

A radioactive material satisfies

$\frac{dM}{dt} = -0.04M$

Find its half-life.

Here $k=-0.04$. Use

$T_{1/2} = \frac{-\ln(2)}{k}$

So

$T_{1/2} = \frac{-\ln(2)}{-0.04} = \frac{\ln(2)}{0.04}$

That value is in the same time units as $t$ (for example, years if $t$ is in years).

Exam Focus

• Typical question patterns
  • Compute doubling time or half-life from a given $k$.
  • Find $t$ when the quantity reaches a threshold (often a percent of the original amount).
  • Compare two exponentials by comparing their $k$ values or doubling times.
• Common mistakes
  • Forgetting that half-life uses a factor of $\frac{1}{2}$, not “subtract half.”
  • Getting a negative time because of sign errors with $k$ and logarithms.
  • Mixing up “time to double” with “growth rate”; they’re inversely related.

Exponential Approach to an Equilibrium: $\frac{dy}{dt} = k(y-L)$

Not all exponential behavior is “proportional to the amount itself.” Another extremely important exponential model is when the rate is proportional to how far you are from a long-term equilibrium value.

The idea (what it is)

Suppose a quantity $y(t)$ is being pulled toward a constant level $L$. The further you are from $L$, the faster you change; as you get closer, the change slows down. That idea is captured by:

$\frac{dy}{dt} = k(y-L)$

Here, $L$ is the **equilibrium (limiting) value**. If $k<0$, then:

• when $y>L$, the term $y-L$ is positive, so $\frac{dy}{dt}$ is negative and $y$ decreases toward $L$
• when $y<L$, the term $y-L$ is negative, so $\frac{dy}{dt}$ is positive and $y$ increases toward $L$

So $k<0$ produces stable “approach toward equilibrium.”

This model matters because it shows up in classic AP Calculus AB applications like Newton’s Law of Cooling/Heating, and it reinforces the same separation-of-variables technique while adding interpretation.

How to solve it

Let $u = y-L$. Then $\frac{du}{dt} = \frac{dy}{dt}$, and the differential equation becomes

$\frac{du}{dt} = ku$

So

$u = Ce^{kt}$

Substitute back $u=y-L$:

$y-L = Ce^{kt}$

Therefore,

$y = L + Ce^{kt}$

This is the “shifted exponential” form: the graph levels off at $y=L$ instead of at $0$.

Newton’s Law of Cooling/Heating (a key application)

Newton’s Law of Cooling/Heating says the rate of change of an object’s temperature is proportional to the difference between the object’s temperature and the surrounding (ambient) temperature.

If $T(t)$ is the object temperature and $T_s$ is constant ambient temperature, then

$\frac{dT}{dt} = k(T-T_s)$

Typically, $k<0$ because if the object is hotter than the surroundings, it cools down (decreases), and if it’s colder, it warms up (increases).

The solution is

$T(t) = T_s + Ce^{kt}$

A helpful way to interpret this: the “temperature difference” $T(t)-T_s$ decays exponentially toward $0$.

Worked example: Cooling to room temperature

A mug of coffee is in a room at $20$ degrees (same units throughout). It starts at $90$ degrees. After 10 minutes, it is $60$ degrees. Find $T(t)$.

Step 1: Set up the model and solution form

$\frac{dT}{dt} = k(T-20)$

So

$T(t) = 20 + Ce^{kt}$

Step 2: Use $T(0)=90$ to find $C$

$90 = 20 + Ce^{k\cdot 0} = 20 + C$

So

$C = 70$

Thus

$T(t) = 20 + 70e^{kt}$

Step 3: Use $T(10)=60$ to find $k$

$60 = 20 + 70e^{10k}$

Subtract 20:

$40 = 70e^{10k}$

Divide by 70:

$\frac{4}{7} = e^{10k}$

Take ln:

$\ln\left(\frac{4}{7}\right) = 10k$

So

$k = \frac{1}{10}\ln\left(\frac{4}{7}\right)$

Final model:

$T(t) = 20 + 70e^{\left(\frac{1}{10}\ln\left(\frac{4}{7}\right)\right)t}$

You could also leave it as $T(t)=20+70e^{kt}$ with $k$ as above.

Common interpretation check: $\frac{4}{7}<1$, so $\ln\left(\frac{4}{7}\right)<0$, making $k<0$, which matches cooling.

Exam Focus

• Typical question patterns
  • Solve or use a model of the form $\frac{dy}{dt} = k(y-L)$, especially in temperature contexts.
  • Use two data points to determine $k$ and write the explicit function.
  • Find when the temperature reaches a specified value (solve for $t$ using logs).
• Common mistakes
  • Using $\frac{dy}{dt}=k(L-y)$ but then mixing that up with $k(y-L)$ without adjusting the sign of $k$.
  • Forgetting that the exponential applies to the difference from ambient: $T-T_s$, not $T$ alone.
  • Solving for $k$ but not checking its sign against the physical situation.

Interpreting Parameters, Units, and Graph Behavior

Exponential differential-equation models are not just about “getting an equation.” AP questions often test whether you understand what the constants and variables mean.

What $k$ represents

In $\frac{dy}{dt}=ky$, divide both sides by $y$:

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

So $k$ is the **constant relative growth rate** (per unit time). If $k=0.05$ and $t$ is measured in years, the quantity has an instantaneous growth rate of 5% per year at every moment.

• Units: if $t$ is in days, then $k$ is “per day.”
• Magnitude: larger $|k|$ means faster growth/decay.

In $\frac{dy}{dt}=k(y-L)$, $k$ is the constant relative rate at which the “distance from equilibrium” changes.

Initial conditions determine the vertical scale

For $y=Ce^{kt}$, the constant $C$ is simply $y(0)$ (if the initial time is $0$):

$y(0)=Ce^{0}=C$

If the initial time is some other value $t=t_0$, then you solve for $C$ using $y(t_0)=y_0$:

$y_0 = Ce^{kt_0}$

So

$C = y_0 e^{-kt_0}$

A common mistake is to treat $C$ as if it were always the initial amount; that’s only true when the initial time is $0$.

Concavity and “increasing faster”

For $y=Ce^{kt}$, the second derivative is

$\frac{d^2y}{dt^2} = k^2Ce^{kt} = k^2y$

Since $k^2\ge 0$, if $y>0$ then $\frac{d^2y}{dt^2}>0$, so the graph is concave up. That matches the idea that exponential growth accelerates.

For exponential decay with $y>0$, the function decreases but remains concave up, which can surprise students. The slope is negative, but it becomes less negative over time.

Worked example: Interpreting a model statement

A problem says: “The rate at which a savings account grows is proportional to the amount in the account.”

That statement is precisely

$\frac{dA}{dt} = rA$

• $A(t)$ is the account balance.
• $\frac{dA}{dt}$ is the instantaneous growth in dollars per unit time.
• $r$ is the proportionality constant (per unit time).

If you’re also told “continuous compounding at 6% per year,” then $r=0.06$ when $t$ is measured in years.

Exam Focus

• Typical question patterns
  • Interpret $k$ (or $r$) in context, including sign and units.
  • Use an initial condition at a nonzero time $t_0$.
  • Reason about the shape of the solution graph from the differential equation (growth vs decay, approach to equilibrium).
• Common mistakes
  • Claiming exponential decay must be concave down because it “goes down.” Concavity depends on how the slope changes.
  • Ignoring units and using a per-year rate with $t$ measured in months without converting.
  • Confusing the equilibrium value $L$ with the initial value in $y=L+Ce^{kt}$.

Building an Exponential Model from a Word Problem (Modeling Workflow)

Many AP questions are really testing a repeatable modeling process. The algebra and calculus are important, but the bigger skill is translating between words, differential equations, and functions.

Step 1: Define variables clearly

State what $y(t)$ represents and what units $t$ uses. This prevents later confusion when interpreting $k$.

Step 2: Translate the rate statement into a differential equation

Common translations:

• “Rate is proportional to amount”:

$\frac{dy}{dt} = ky$

• “Rate is proportional to difference from equilibrium $L$”:

$\frac{dy}{dt} = k(y-L)$

Step 3: Solve the differential equation

• For $\frac{dy}{dt}=ky$:

$y=Ce^{kt}$

• For $\frac{dy}{dt}=k(y-L)$:

$y=L+Ce^{kt}$

Step 4: Use given conditions to find constants

Use initial values like $y(0)=y_0$ or two data points to solve for $C$ and $k$.

Step 5: Interpret the result

Check:

• Does the sign of $k$ match growth/decay?
• Does the solution approach the correct limiting value (0 or $L$)?
• Are units consistent?

Worked example: Continuous growth model

A culture starts with 1000 cells. After 3 hours, it has 1800 cells. Assume growth rate is proportional to the number of cells.

1) Model

Let $N(t)$ be cells after $t$ hours.

$\frac{dN}{dt} = kN$

2) Solve

$N(t)=Ce^{kt}$

3) Use $N(0)=1000$

$C=1000$

So

$N(t)=1000e^{kt}$

4) Use $N(3)=1800$ to find $k$

$1800=1000e^{3k}$

$1.8=e^{3k}$

$\ln(1.8)=3k$

$k=\frac{\ln(1.8)}{3}$

5) Final model

$N(t)=1000e^{\left(\frac{\ln(1.8)}{3}\right)t}$

Interpretation: $k>0$, so it grows; $k$ is the continuous relative growth rate per hour.

Exam Focus

• Typical question patterns
  • Turn a verbal description into a differential equation, then into an explicit function.
  • Use two time-value data points to determine parameters.
  • Use the model to predict a future value or solve for a time when a threshold is reached.
• Common mistakes
  • Writing $y=C(1+k)^t$ (a discrete model) when the situation is explicitly continuous or given as a differential equation.
  • Plugging data points into $\frac{dy}{dt}=ky$ instead of into $y=Ce^{kt}$ when the data points are function values (not derivative values).
  • Solving for $k$ but rounding too early, causing noticeable error later when solving for time.