Unit 7 · Topic 7.1 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Calculus AB 7.1: Modeling Situations with Differential Equations

Q: What is the difference between a differential equation and an ordinary equation?

An ordinary equation like x2 = 9 has numbers as its solutions. A differential equation contains a derivative of an unknown function, and its solutions are functions.

Q: How do I know which variable is independent?

Read the derivative. In dydt, y depends on t, so t is independent. In words, it is the variable after "with respect to", usually time in an applied problem.

Q: Does k have to be positive?

Not unless the problem says so. If it does say k > 0 and the quantity is falling, write the minus sign into the equation, as in dhdt = -kh. If it does not, k takes whatever sign the data give it.

Q: Do I need to know the time to use a differential equation?

Only if the time appears on the right side. In dydt = 0.4y, the rate depends only on y, so knowing y is enough. In dydt = t + y, you need both.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 7: 5 to 10% of the multiple-choice section

What you'll be able to do

Interpret a verbal statement about how a quantity changes as a differential equation, an equation that involves a derivative of the unknown function.

Key vocabulary

Term	What it means	The exam trap
Differential equation	An equation that relates a function of an independent variable to one or more of that function's derivatives.	Look for a derivative in the equation; an equation relating two variables without one is a different thing.
Independent variable	The input the function depends on, the variable written after "with respect to" and on the bottom of a derivative such as $\frac{d y}{d t}$ .	Do not assume it is $x$ ; read the derivative to see which variable it is.
Rate of change of $y$ with respect to $t$	How fast $y$ is changing as $t$ changes, written $\frac{d y}{d t}$ .	A sentence about a rate puts the derivative on the left side, not $y$ itself.

The depth h of water in a draining barrel, in inches, plotted against time t in minutes. The curve falls from 36 inches at t = 0 to empty at t = 40, getting less steep as it goes. The depth obeys dh/dt = -0.3 times the square root of h, and short tangent segments mark the rate at three depths: -1.8 inches per minute at 36 inches, -1.2 at 16 inches and -0.6 at 4 inches.

Part 1 Think first How fast will the frogs multiply?

Make a guess before the lesson starts. Committing to an answer, even a wrong one, helps the right one stick.

Predict first

A pond has 200 frogs, and right now the population is growing at 30 frogs per year. A biologist says the population grows at a rate proportional to the number of frogs in the pond. Some years later there are 400 frogs. If she is right, how fast is the population growing at that moment?

In short. A differential equation is an equation that relates a function of an independent variable to one or more of that function's derivatives, such as $\frac{d y}{d t} = k \sqrt{y}$ for "the rate of change of $y$ with respect to $t$ is proportional to the square root of $y$ ."

"The rate of change of $y$ with respect to $t$ " is $\frac{d y}{d t}$ , and the variable after "with respect to" is the independent variable.
"Proportional to $Q$ " means a constant $k$ times $Q$ , and "inversely proportional to $Q$ " means $\frac{k}{Q}$ . The right side uses whatever the words say the rate depends on.
Substituting a state into the equation gives the rate at that state, and one known rate is enough to find $k$ . A falling quantity has a negative rate.

Part 2 Learn it Three short ideas, each checked as you go

7a. What a differential equation says

Most equations you have solved ask for a number. A differential equation asks for a function. It is an equation that relates a function, such as $y$ as a function of an independent variable $t$ , to one or more of that function's derivatives.

Take $\frac{d P}{d t} = 0.03 P$ , where $P$ is a town's population $t$ years from now. The unknown is the function $P (t)$ . The independent variable is $t$ , the variable on the bottom of $\frac{d P}{d t}$ . The equation does not give you $P$ directly. It tells you how fast $P$ is changing once you know $P$ : with 10,000 people, the town grows at $0.03 (10,000) = 300$ people per year.

The right side can hold the independent variable, the function, or both: $\frac{d y}{d x} = 2 x$ , $\frac{d y}{d x} = y - 4$ and $\frac{d y}{d x} = x + y$ are all differential equations. One can also hold a second derivative, as $\frac{d^{2} s}{d t^{2}} = - 9.8$ does for an object falling near the ground.

The derivative is what makes it a differential equation. $y = 3 x^{2} + 1$ is a formula for a function, and $x^{2} + y^{2} = 25$ relates $x$ and $y$ , but neither has a derivative in it. When a problem writes $y^{'}$ , the independent variable is whatever it says $y$ is a function of.

Quick check · Sort it

No calculator. Tap an equation, then tap whether it is a differential equation.

A differential equation

Not a differential equation

7b. Turning words into an equation

Applied problems describe a rate in words, and your job is to write the equation. These phrases do most of the work:

"The rate of change of $y$ with respect to $t$ " is $\frac{d y}{d t}$ . The variable after "with respect to" is the independent variable.
" $A$ is proportional to $Q$ " means $A = k Q$ for some constant $k$ , the constant of proportionality.
" $A$ is inversely proportional to $Q$ " means $A = \frac{k}{Q}$ .

Put them together. "The rate at which $y$ changes with respect to $t$ is proportional to the square root of $y$ " becomes

\frac{d y}{d t} = k \sqrt{y} .

The left side is the derivative, because the sentence is about a rate. The right side is built from whatever the sentence says the rate depends on: here $y$ , the amount, not $t$ , the time.

A rate can also be proportional to a difference. "A bowl of soup cools at a rate proportional to the difference between its temperature $S$ and the room's 21 degrees Celsius" gives $\frac{d S}{d t} = k (S - 21)$ .

Watch the direction of change. If a quantity is decreasing and the problem says $k$ is a positive constant, put the minus sign in yourself: a draining tank with depth $h$ might be $\frac{d h}{d t} = - k \sqrt{h}$ . If the problem does not say, $k$ carries the sign, and it comes out negative.

Quick check · Match

No calculator. In each statement $y$ is a function of $t$ and $k$ is a constant. Tap a statement, then the equation it describes.

7c. Reading a rate from the equation

A differential equation is a rule: give it the state, and it gives the rate. To find the rate at a particular moment, substitute what you know into the right side.

The barrel in the picture at the top drains so that its depth $h$ , in inches, satisfies $\frac{d h}{d t} = - 0.3 \sqrt{h}$ , with $t$ in minutes. When the water is 36 inches deep,

\frac{d h}{d t} = - 0.3 \sqrt{36} = - 1.8 inches per minute .

The negative sign says the depth is falling. The units of $\frac{d h}{d t}$ are the units of $h$ over the units of $t$ . Notice that you never needed the time. This rate depends only on the depth, so 36 inches of water drains at the same rate at 9 a.m. or at noon. When the right side holds $t$ as well, you need the time too.

When the problem gives one rate and leaves $k$ unknown, that one fact pins $k$ down. Say $\frac{d y}{d t} = k \sqrt{y}$ , and $y$ is falling at 3 units per minute when $y = 9$ . Falling means the rate is $- 3$ , so $- 3 = k \sqrt{9}$ and $k = - 1$ . Now the equation gives the rate at every other state: at $y = 4$ , $\frac{d y}{d t} = - 1 \cdot \sqrt{4} = - 2$ .

Quick check · Calculate

No calculator. The number of bacteria in a dish, $B$ , in thousands, satisfies $\frac{d B}{d t} = 0.5 \sqrt{B}$ , with $t$ in hours. What is $\frac{d B}{d t}$ , in thousands of bacteria per hour, when $B = 16$ ?

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Building the right side from the wrong variable, such as $k \sqrt{t}$ for a rate proportional to the square root of the amount $y$ .	The amount is the function $y$ and the time is $t$ . The sentence names what the rate depends on, and the two give different equations with different rates.	Find the words right after "proportional to" and write exactly that quantity on the right.
Writing $y = k \sqrt{y}$ , with no derivative, for a statement about a rate.	"The rate of change of $y$ " is $\frac{d y}{d t}$ , not $y$ . Without the derivative the equation is not about change at all.	Write $\frac{d y}{d t}$ on the left whenever the sentence says rate, changes, grows, cools or drains.
Using a positive rate for a quantity that is falling.	A decreasing quantity has a negative derivative. Cooling at 3.5 degrees per minute means $\frac{d H}{d t} = - 3.5$ , and a positive value gives $k$ the wrong sign and every later rate the wrong direction.	Decide the sign from the words (cools, drains, decays, leaks) before you substitute.

Part 3 Watch it done From a sentence to a rate

Watch each step, then finish the last one yourself.

A cup of tea at 90 degrees Celsius is set down in a room kept at 20 degrees Celsius. The temperature of the tea, $H (t)$ degrees Celsius $t$ minutes later, changes at a rate proportional to the difference between the tea's temperature and the room's temperature. At the moment the tea is 90 degrees, it is cooling at 3.5 degrees per minute. How fast is its temperature changing when the tea is 60 degrees?

Step 1 · Name the variables
The function is $H$ , the tea's temperature, and the independent variable is $t$ , the time in minutes. The rate of change of $H$ with respect to $t$ is $\frac{d H}{d t}$ .
Step 2 · Translate the sentence
The rate is proportional to the difference between $H$ and 20, so $\frac{d H}{d t} = k (H - 20)$ for some constant $k$ .
Step 3 · Use the known rate to find k
When $H = 90$ , the tea is cooling at 3.5 degrees per minute, so $\frac{d H}{d t} = - 3.5$ . Then $- 3.5 = k (90 - 20) = 70 k$ , and $k = - 0.05$ . The equation is $\frac{d H}{d t} = - 0.05 (H - 20)$ .
Why this step?
Why does $k$ come out negative here?
The tea is hotter than the room, so $H - 20$ is positive, and the tea is cooling, so $\frac{d H}{d t}$ is negative. A positive number times $k$ can only be negative if $k$ is negative.
Your step · Calculate

No calculator. Finish it. The tea's temperature satisfies $\frac{d H}{d t} = - 0.05 (H - 20)$ . What is $\frac{d H}{d t}$ , in degrees Celsius per minute, when $H = 60$ ?

Your answer

Part 4 Try it A decaying sample and a spreading rumor

Try it · Calculate

No calculator. A sample of a radioactive substance decays at a rate proportional to the amount present. When 80 grams remain, the amount is decreasing at 4 grams per day. What is $\frac{d A}{d t}$ , in grams per day, when 30 grams remain? Here $A (t)$ is the amount in grams and $t$ is in days.

Your answer

Try it · Multiple choice

No calculator. A town has 2000 residents. $N (t)$ is the number who have heard a rumor $t$ days after it starts. The rate at which $N$ changes with respect to $t$ is proportional to the number of residents who have not yet heard the rumor. Which differential equation models this, where $k$ is a positive constant?

Part 5 Explain it Write it as a free-response answer

Free response on the exam is written by hand, so do this one on paper. Nothing you type here leaves the page. When you are done, open the model answer and score yourself against each point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. A patient receives a drug through an IV drip at a constant rate of 30 milligrams per hour. At the same time, the patient's body removes the drug from the bloodstream at a rate proportional to the amount of the drug in the bloodstream. Let $A (t)$ be the amount of the drug in the bloodstream, in milligrams, $t$ hours after the drip starts.

(a) Write a differential equation for $\frac{d A}{d t}$ in terms of $A$ and a positive constant $k$ .

(b) At the moment when $A = 100$ , the amount of the drug in the bloodstream is increasing at 18 milligrams per hour. Find the value of $k$ .

(c) Find $\frac{d A}{d t}$ at the moment when $A = 200$ . Using correct units, interpret this value in the context of the problem.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) The drug comes in at 30 milligrams per hour and leaves at $k A$ milligrams per hour, so $\frac{d A}{d t} = 30 - k A$ .

(b) When $A = 100$ , $\frac{d A}{d t} = 18$ , so $18 = 30 - 100 k$ , which gives $100 k = 12$ and $k = 0.12$ .

(c) $\frac{d A}{d t} = 30 - 0.12 (200) = 30 - 24 = 6$ . At the moment the bloodstream holds 200 milligrams of the drug, the amount of the drug is increasing at 6 milligrams per hour.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

In (a), writes $\frac{d A}{d t}$ on the left with 30 as the rate in.
In (a), subtracts $k A$ for the rate out, giving $\frac{d A}{d t} = 30 - k A$ .
In (b), substitutes $A = 100$ and $\frac{d A}{d t} = 18$ into the equation and gets $k = 0.12$ .
In (c), gets $\frac{d A}{d t} = 6$ and says the amount of the drug is increasing at 6 milligrams per hour when $A = 200$ .

Part 6 Check yourself A colony's growth rate in a table

All three case questions use this table. Read every value you need from it.

A biologist grows a colony of bacteria. $P (t)$ is the number of bacteria $t$ hours after the start. She measures the growth rate $\frac{d P}{d t}$ , in bacteria per hour, at three moments, and records the size of the colony at each one.

Size of the colony and its growth rate
$P$ (bacteria)	$\frac{d P}{d t}$ (bacteria per hour)
100	2
400	4
900	6

Case question 1 · Decide and justify

No calculator. A classmate says the colony's growth rate is proportional to $P$ , because the rate goes up whenever $P$ goes up. Is the classmate right?

Which reason supports your answer?

Case question 2 · Multiple choice

No calculator. Which differential equation agrees with every row of the table?

Case question 3 · Calculate

No calculator. The colony's growth rate is proportional to a power of $P$ . Using the table, what is $\frac{d P}{d t}$ , in bacteria per hour, when the colony has 1600 bacteria?

Your answer

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A can of soda at 6 degrees Celsius is left in a car at 30 degrees Celsius. Its temperature rises at a rate proportional to the difference between the car's temperature and the soda's. At the moment the soda is 6 degrees, it is warming at 1.2 degrees per minute. A student finds how fast it is warming when it reaches 18 degrees. Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. A balloon leaks so that the volume of air in it, $V$ liters, $t$ seconds after the leak starts, satisfies $\frac{d V}{d t} = - 0.2 V$ . Tap a blank, then tap an answer.

In this differential equation, the independent variable is , the unknown function is , and the equation relates that function to .

Part 7 Keep it Bring back earlier ideas

A differential equation is a statement about a derivative in context, and the area under a rate is the change it builds up. Both come back in this unit.

Bring it back · Calculate

No calculator. A cyclist's velocity is $v (t) = 2 + t$ meters per second for $0 \leq t \leq 6$ , with $t$ in seconds. Using the area between the graph of $v$ and the $t$ -axis, how many meters does the cyclist travel from $t = 0$ to $t = 6$ ?

Your answer

Bring it back · Multiple choice

No calculator. $W (t)$ is the amount of water in a reservoir, in millions of gallons, $t$ days after June 1. Which statement is the best interpretation of $W^{'} (5) = - 3$ ?

In three lines

A differential equation relates a function of an independent variable to one or more of its derivatives.
"The rate of change of y is proportional to Q" means dy/dt = kQ; read which variable Q is, and give a falling quantity a negative rate.
Substitute a state into the equation to read the rate there, and use one known rate to find k.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a free-response question asks for a differential equation, write the derivative on the left with the right letters, such as $\frac{d A}{d t}$ , and keep the constant $k$ in it until the problem gives you a rate to find it. When you interpret a value of a derivative, give the moment, the units and whether the quantity is increasing or decreasing.

Questions students ask

What is the difference between a differential equation and an ordinary equation?

An ordinary equation like $x^{2} = 9$ has numbers as its solutions. A differential equation contains a derivative of an unknown function, and its solutions are functions.

How do I know which variable is independent?

Read the derivative. In $\frac{d y}{d t}$ , $y$ depends on $t$ , so $t$ is independent. In words, it is the variable after "with respect to", usually time in an applied problem.

Does k have to be positive?

Not unless the problem says so. If it does say $k > 0$ and the quantity is falling, write the minus sign into the equation, as in $\frac{d h}{d t} = - k \sqrt{h}$ . If it does not, $k$ takes whatever sign the data give it.

Do I need to know the time to use a differential equation?

Only if the time appears on the right side. In $\frac{d y}{d t} = 0.4 y$ , the rate depends only on $y$ , so knowing $y$ is enough. In $\frac{d y}{d t} = t + y$ , you need both.

Will I have to solve these equations?

Not in this topic. Here you write them and read rates from them. Later topics check whether a function satisfies one, sketch its slope field and then solve the ones that separate.

Course alignment, for teachers

AP Calculus AB topic 7.1, Unit 7: Differential Equations.

Skill 2.C: Identify a re-expression of mathematical information presented in a given representation.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.