Exponential Models — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

Bacteria in a dish, money in an account, a medicine leaving the blood, a radioactive sample: each changes at a rate proportional to how much there is. One differential equation,

\frac{dy}{dt} = ky

, covers them all, and its solution

y = y_0e^{kt}

is worth knowing by heart, along with what

k

means and how to find it from data.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The general solution of $\frac{dy}{dt} = ky$ is

$y = Ae^{kt}$
$y = e^{kt} + C$
$y = \frac{k}{2}y^2 + C$
$y = kt + C$

Show the answer

$\ln|y| = kt + C$ , so $y = Ae^{kt}$ .

Correct: $y = Ae^{kt}$ :
$y = e^{kt} + C$ :
$y = \frac{k}{2}y^2 + C$ :
$y = kt + C$ :

2. For $\frac{dy}{dx} = 2y$ with $y(0) = 5$ , the constant $A$ in $y = Ae^{2x}$ is

5
2
10
$\ln 5$

Show the answer

At $x = 0$ , $y = Ae^0 = A$ , so $A = 5$ .

Correct: 5:
2:
10:
$\ln 5$ :

Part 4 · See it

See it first

Left: P = 50e^(0.2t) for t from 0 to 10, starting at 50 and doubling to 100 at t ≈ 3.47 and to 200 at t ≈ 6.93. Right: A = 80e^(−0.3t), starting at 80 and halving to 40 at t ≈ 2.31 and to 20 at t ≈ 4.62. — Left: $\frac{dP}{dt} = 0.2P$ with $P(0) = 50$ grows and doubles every $\frac{\ln 2}{0.2}$ time units. Right: $\frac{dA}{dt} = -0.3A$ with $A(0) = 80$ decays and halves every $\frac{\ln 2}{0.3}$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A quantity changes at a rate proportional to its size $\frac{dy}{dt} = ky$
Separate, integrate and use y(0) $y = y_0e^{kt}$ , the exponential model
$k > 0$ or $k < 0$ growth or decay; $|k|$ is the rate as a fraction of the amount, per unit time
A second data point $y(t_1) = y_1$ is known $e^{kt_1} = \frac{y_1}{y_0}$ , so $k$ $= \frac{1}{t_1}\ln\frac{y_1}{y_0}$

Part 6 · Key ideas

Key ideas

$\frac{dy}{dt} = ky$ and $y(0) = y_0$ give $y = y_0e^{kt}$ . You may quote this result.
$k = \frac{dy/dt}{y}$ : the rate of change as a fraction of the current amount, in "per unit time" (per hour, per year).
Doubling time ( $k > 0$ ) and half-life ( $k < 0$ ) are both $\frac{\ln 2}{|k|}$ ; they do not depend on the starting amount.
To find $k$ from two values, divide first: $\frac{y_1}{y_0} = e^{kt_1}$ , then take $\ln$ .
Cooling toward room temperature $T_r$ is exponential in $T - T_r$ , not in $T$ .

Part 7 · Misconception

A common mistake

The wrong idea: A population goes from 400 to 1200 in 3 hours, so $k = \frac{1200 - 400}{3}$ .

What actually happens: That is an average rate, measured in cells per hour: a linear idea. For $y = 400e^{kt}$ : $1200 = 400e^{3k}$ , so $e^{3k} = 3$ and $k = \frac{\ln 3}{3}$ per hour.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A bacteria culture

A culture grows at a rate proportional to its size; two counts
$t$ (hours)	0	3
$P(t)$ (bacteria)	400	1200

1. $\frac{dP}{dt} = kP$ . Find the growth constant $k$ , exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$1200 = 400e^{3k}$ , so $e^{3k} = 3$ , $3k = \ln 3$ and $k = \frac{\ln 3}{3} \approx 0.366$ per hour.

Answer: $\frac{\ln(3)}{3}$ ≈ 0.366

2. Write $P(t)$ for the culture.

Type an expression for $P(t) =$ .

Show the answer

$P(t) = 400e^{\frac{\ln 3}{3}t}$ , which can also be written $400 \cdot 3^{t/3}$ .

Answer: $P(t) =$ $400e^{\frac{\ln 3}{3}t} = 400 \cdot 3^{t/3}$

3. What does $k = \frac{\ln 3}{3} \approx 0.366$ per hour mean for this culture?

At each moment, the population grows at about 36.6% of its current size per hour.
The population grows by about 0.366 bacteria each hour.
The population triples each hour.
It is the starting population, in hundreds.

Show the answer

$\frac{dP}{dt} = kP$ means the growth rate is $k$ times the current size: about 0.366 of the population per hour.

Correct: At each moment, the population grows at about 36.6% of its current size per hour.: $k = \frac{dP/dt}{P}$ , the rate as a fraction of the amount.
The population grows by about 0.366 bacteria each hour.: $k$ is a fraction of the amount, not a count of bacteria.
The population triples each hour.: It triples every 3 hours.
It is the starting population, in hundreds.: The starting population is $P(0) = 400$ .

4. Which differential equation has the solution $y = 7e^{-2t}$ ?

$\frac{dy}{dt} = 7y$
$\frac{dy}{dt} = -2t$
$\frac{dy}{dt} = -14y$
$\frac{dy}{dt} = -2y$

Show the answer

For $y = y_0e^{kt}$ , $\frac{dy}{dt} = ky$ ; here $k = -2$ .

$\frac{dy}{dt} = 7y$ : 7 is the starting amount, not the constant k.
$\frac{dy}{dt} = -2t$ : The rate is proportional to y, not to t.
$\frac{dy}{dt} = -14y$ : $-14y = -98e^{-2t}$ , seven times too big.
Correct: $\frac{dy}{dt} = -2y$ : $\frac{dy}{dt} = -14e^{-2t} = -2y$ .

5. A town's population $P$ satisfies $\frac{dP}{dt} = kP$ , with $t$ in years. What are the units of $k$ ?

people per year
per year
years
people

Show the answer

$k = \frac{dP/dt}{P}$ : (people per year) ÷ people = per year.

people per year: That is the unit of $\frac{dP}{dt}$ ; dividing by $P$ removes "people".
Correct: per year: $\frac{dP}{dt}$ is people per year and $P$ is people, so $k$ is $\frac{1}{\text{year}}$ .
years: $k \cdot t$ must have no units, so $k$ is per year.
people: k multiplies P to give a rate, so it carries "per year".

6. A pie at 90 °F cools in a 20 °F freezer: $T - 20 = 70e^{kt}$ , $t$ in minutes. After 10 minutes it is at 55 °F. Find $k$ exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$35 = 70e^{10k}$ , so $e^{10k} = \frac{1}{2}$ and $k$ $= \frac{\ln(1/2)}{10}$ $= -\frac{\ln 2}{10}$ per minute.

Answer: $\frac{-\ln(2)}{10}$ ≈ -0.069

Part 9 · Summary

Summary

When the rate of change is proportional to the amount,

\frac{dy}{dt} = ky

and

y = y_0e^{kt}

.

k > 0

is growth,

k < 0

decay;

k

is the rate as a fraction of the amount per unit time. Find

k

by dividing two values and taking

\ln

; doubling time and half-life are

\frac{\ln 2}{|k|}

.

Part 10 · Up next

What comes next

Part 11 · Connections