How do I know a problem is exponential and not logistic?

Look for the phrase "proportional to the amount present" with no carrying capacity or upper limit. If the rate depends only on y as ky, it is exponential. If growth slows as y approaches a maximum L (giving dy/dt=ky(1-y/L)), it is logistic, which is covered in U7.9.

Do I need to re-derive y = y_0 e^kt every time?

On a free-response question, if asked to solve the differential equation you should show separation of variables. But once you have justified the general solution, you can quote y = y_0 e^kt for the applications. Knowing the formula cold saves time on multiple-choice questions.

Why is half-life independent of the starting amount?

When you set 1/2y_0 = y_0 e^kt, the y_0 cancels, leaving e^kt=1/2. Since the equation for t never involves y_0, the halving time depends only on k. The same cancellation explains why doubling time is constant.

What is the difference between k in cooling versus plain growth?

In plain growth/decay, k multiplies y directly. In Newton's law of cooling, k multiplies the difference T - T_s. After substituting y = T - T_s, the math is identical, but you must add T_s back at the end to report the real temperature.

AP-CALCBC-7.8

U7.8 Exponential Models

Master AP Calculus BC exponential models: solve dy/dt = ky, and apply the solution to half-life, doubling time, Newton's law of cooling, and continuous interest.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U7.8 Exponential Models, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U7.8 Exponential Models free

What this lesson covers

Whenever a quantity changes at a rate proportional to its current size, you get exponential growth or decay. Population, radioactive decay, cooling coffee, and compound interest all obey the same simple differential equation:

\frac{dy}{dt} = ky

. In this lesson you will learn to recognize this equation instantly, solve it, and pin down the constant

k

from real data. You already know separation of variables from U7.6, so here the goal is to turn that skill into a reusable formula and apply it fluently to the four classic model types the exam loves. Master this and half the differential-equation FRQs become routine.

The Equation and Its General Solution

The defining feature of an exponential model is that the rate of change is proportional to the amount present. In symbols,

\frac{dy}{dt} = ky

, where

k

is a constant. If

k > 0

the quantity grows; if

k < 0

it decays.

You could solve this by separation of variables every time, but it is faster to memorize the result. Separating gives

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

, so

\ln|y| = kt + C

, and exponentiating yields

y = y_0 e^{kt}

Here

y_0

is the value of

y

t = 0

, because

y(0) = y_0 e^{0} = y_0

. This single formula is the backbone of every application in this topic.

A crucial conceptual point tested repeatedly:

\frac{dy}{dt} = ky

says the relative growth rate

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

is constant. That is NOT the same as a constant rate of change. A common misconception is to treat exponential growth as linear. If the problem says "grows proportionally to its size," you have an exponential model, not a linear one.

The exam may hand you the differential equation and ask you to verify the solution, or hand you a word problem and expect you to build the equation. Practice recognizing the trigger phrase "proportional to the amount present."

Finding k From Data

Almost every exponential problem gives you two pieces of information: an initial value and one later data point. Use them in sequence.

Step one: identify

y_0

from the condition at

t = 0

. Step two: plug the second data point into

y = y_0 e^{kt}

and solve for

k

using a logarithm.

For example, if

y_0 = 100

and

y = 300

when

t = 5

, then

300 = 100 e^{5k}

, so

e^{5k} = 3

, giving

k = \frac{\ln 3}{5}

. Leave answers in exact logarithmic form unless a decimal is requested; AP graders accept

\frac{\ln 3}{5}

.

A frequent error is mixing up which value is

y_0

. The initial amount is whatever occurs at

t = 0

, so always reset your clock so that the starting moment is

t = 0

. If a problem describes an event "3 hours later," that means

t = 3

, not a new starting point.

Given	What to do
Value at $t=0$	Set equal to $y_0$
A second point $(t_1, y_1)$	Solve $y_1 = y_0 e^{k t_1}$ for $k$
Need a future value	Plug the desired $t$ into the finished formula
Need a time	Set $y$ equal to target, solve for $t$ with $\ln$

Half-Life and Doubling Time

Half-life and doubling time are just special target values of

y

. Because

k

is fixed, the time to halve or double is the same no matter where you start — a hallmark of exponential behavior.

Doubling time

T_d

satisfies

2y_0 = y_0 e^{kT_d}

, so

e^{kT_d} = 2

and

T_d = \frac{\ln 2}{k}

. Half-life

T_h

(with

k<0

for decay) satisfies

\frac{1}{2}y_0 = y_0 e^{kT_h}

, giving

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

. Since

k

is negative for decay,

T_h

comes out positive.

A useful shortcut: if you are told the half-life, you can find

k

immediately. From

T_h = \frac{-\ln 2}{k}

you get

k = \frac{-\ln 2}{T_h}

. For a carbon-14 half-life of 5730 years,

k = \frac{-\ln 2}{5730}

.

The exam tests whether you understand that these times are independent of the starting amount. If asked "how long until only a quarter remains," recognize that a quarter is two half-lives, so the answer is

2T_h

— or just solve

\frac{1}{4}y_0 = y_0 e^{kt}

directly. Both approaches must agree, which is a nice self-check.

Newton's Law of Cooling and Compound Interest

Two applications add a small twist. Newton's law of cooling says the temperature

T

of an object changes at a rate proportional to the difference between the object and its surroundings:

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

, where

T_s

is the ambient temperature. This is not quite

\frac{dy}{dt}=ky

, but the substitution

y = T - T_s

fixes it: since

T_s

is constant,

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

. So

T - T_s = (T_0 - T_s)e^{kt}, \qquad T = T_s + (T_0 - T_s)e^{kt}.

Here

k < 0

so the object approaches ambient temperature: as

t \to \infty

e^{kt} \to 0

and

T \to T_s

. Recognizing this limiting temperature is a common exam question.

Continuous compound interest is pure exponential growth: money

A

satisfies

\frac{dA}{dt} = rA

, so

A = A_0 e^{rt}

, where

r

is the annual rate as a decimal. This is the continuous version of periodic compounding; the exam often contrasts "compounded continuously" (use

e^{rt}

) with discrete compounding.

The main misconception is forgetting to shift back in cooling problems — students solve for

y

but forget to add

T_s

to report the actual temperature.

Key terms

Exponential model.: A quantity whose rate of change is proportional to its current amount, obeying $\frac{dy}{dt}=ky$ .
Growth/decay constant $k$ .: The proportionality constant; $k>0$ gives growth, $k<0$ gives decay. Equal to the constant relative rate $\frac{1}{y}\frac{dy}{dt}$ .
General solution.: The formula $y = y_0 e^{kt}$ , where $y_0$ is the value at $t=0$ .
Half-life.: The time for a decaying quantity to fall to half its value, $T_h = \frac{-\ln 2}{k}$ ; independent of starting amount.
Doubling time.: The time for a growing quantity to double, $T_d = \frac{\ln 2}{k}$ ; independent of starting amount.
Newton's law of cooling.: $\frac{dT}{dt}=k(T-T_s)$ , with solution $T = T_s + (T_0 - T_s)e^{kt}$ , approaching ambient $T_s$ .
Continuous compound interest.: Money satisfying $\frac{dA}{dt}=rA$ , giving $A=A_0 e^{rt}$ for annual rate $r$ .

Worked example

A cup of coffee at 200°F is placed in a room at 70°F. After 5 minutes its temperature is 150°F. Write the temperature as a function of time, and find when the coffee reaches 100°F.

This is Newton's law of cooling with ambient temperature

T_s = 70

. The model is

T = 70 + (T_0 - 70)e^{kt}

.

The initial temperature is

T_0 = 200

, so

T = 70 + 130 e^{kt}

.

Use the data point

T = 150

t = 5

to find

k

150 = 70 + 130 e^{5k}

, so

130 e^{5k} = 80

, giving

e^{5k} = \frac{80}{130} = \frac{8}{13}

. Then

k = \frac{1}{5}\ln\frac{8}{13}

, which is negative as expected for cooling.

So the temperature function is

T(t) = 70 + 130 e^{\left(\frac{1}{5}\ln\frac{8}{13}\right)t}

.

Now set

T = 100

100 = 70 + 130 e^{kt}

, so

130 e^{kt} = 30

and

e^{kt} = \frac{30}{130} = \frac{3}{13}

. Take logs:

kt = \ln\frac{3}{13}

, so

t = \frac{\ln(3/13)}{k} = \frac{\ln(3/13)}{\frac{1}{5}\ln(8/13)} = \frac{5\ln(3/13)}{\ln(8/13)}

.

Numerically,

\ln(3/13) \approx -1.466

and

\ln(8/13) \approx -0.485

, so

t \approx \frac{5(-1.466)}{-0.485} \approx 15.1

minutes. The coffee reaches 100°F after about 15.1 minutes.

Practice questions

A radioactive sample decays according to

\frac{dy}{dt} = ky

and has a half-life of 8 days. What is the value of

k

$k = \frac{\ln 2}{8}$
$k = -\frac{\ln 2}{8}$
$k = \frac{8}{\ln 2}$
$k = -8\ln 2$

Answer: $k = -\frac{\ln 2}{8}$

Half-life means

\frac{1}{2}y_0 = y_0 e^{8k}

, so

e^{8k} = \frac{1}{2}

and

8k = \ln\frac{1}{2} = -\ln 2

. Therefore

k = -\frac{\ln 2}{8}

. The sign must be negative because the sample is decaying.

A bacterial culture grows so that its rate of increase is proportional to the amount present. It starts with 500 cells and grows to 2000 cells in 3 hours. Find the number of cells after 7 hours, showing your setup.

Answer: $500 \cdot 4^{7/3} \approx 12{,}699$ cells.

Use

y = 500 e^{kt}

. From

2000 = 500 e^{3k}

we get

e^{3k} = 4

, so

k = \frac{\ln 4}{3}

. At

t = 7

y = 500 e^{7k} = 500 e^{7\ln 4 / 3} = 500\cdot 4^{7/3}

. Since

4^{7/3} \approx 25.4

, the population is about

12{,}699

cells. Notice you never need

k

as a decimal — writing

e^{k}=4^{1/3}

keeps the arithmetic clean.

Money is deposited in an account earning interest compounded continuously at annual rate

r

. If the balance doubles in 12 years, what is

r

Answer: $r = \frac{\ln 2}{12} \approx 0.0578$ , or about 5.78%.

Continuous compounding gives

A = A_0 e^{rt}

. Doubling means

2A_0 = A_0 e^{12r}

, so

e^{12r} = 2

and

12r = \ln 2

, giving

r = \frac{\ln 2}{12} \approx 0.0578

. This is the same structure as doubling time:

T_d = \frac{\ln 2}{r}

rearranged.

FAQ

How do I know a problem is exponential and not logistic?: Look for the phrase "proportional to the amount present" with no carrying capacity or upper limit. If the rate depends only on $y$ as $ky$ , it is exponential. If growth slows as $y$ approaches a maximum $L$ (giving $\frac{dy}{dt}=ky(1-\frac{y}{L})$ ), it is logistic, which is covered in U7.9.
Do I need to re-derive $y = y_0 e^{kt}$ every time?: On a free-response question, if asked to solve the differential equation you should show separation of variables. But once you have justified the general solution, you can quote $y = y_0 e^{kt}$ for the applications. Knowing the formula cold saves time on multiple-choice questions.
Why is half-life independent of the starting amount?: When you set $\frac{1}{2}y_0 = y_0 e^{kt}$ , the $y_0$ cancels, leaving $e^{kt}=\frac{1}{2}$ . Since the equation for $t$ never involves $y_0$ , the halving time depends only on $k$ . The same cancellation explains why doubling time is constant.
What is the difference between $k$ in cooling versus plain growth?: In plain growth/decay, $k$ multiplies $y$ directly. In Newton's law of cooling, $k$ multiplies the difference $T - T_s$ . After substituting $y = T - T_s$ , the math is identical, but you must add $T_s$ back at the end to report the real temperature.

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