Modeling Situations with Differential Equations

AP Calculus · Differential Equations

Modeling Situations with Differential Equations

A differential equation is a sentence about a rate. Translate the English — do not solve it yet.

key idea$\text{“rate of change of } y\text{” }=\ \frac{dy}{dt}\ \ \text{— the sentence \emph{is} the equation}$

Setting up the model

A differential equation is a sentence about a rate. Modelling means nothing more than translating the English into symbols: identify what is changing, call it $y$, and write down what the problem says its rate of change equals. You are not solving anything yet.

Translating a verbal statement

Name the quantity that changes and the variable it changes with respect to — usually time. Write $y$ and $t$ explicitly, with units.
Find the phrase ‘rate of change’, ‘grows’, ‘decays’, ‘increases at’. That phrase is $\dfrac{dy}{dt}$ and nothing else.
Translate proportional to as ‘$=k$ times’. The constant $k$ is unknown and stays as a letter until data determines it.
Translate inversely proportional to as $=\dfrac{k}{\;\cdot\;}$, and jointly proportional as a product of both quantities.
Record any initial condition separately, as $y(0)=\;$…. It is not part of the differential equation — it is the extra fact that later pins down one particular solution.

Differential equationAn equation containing a derivative. A solution is a whole function, not a number.

Proportional toEquals a constant multiple: ‘$P$ grows proportionally to $P$’ means $\dfrac{dP}{dt}=kP$.

Initial conditionA single known value such as $y(0)=50$. It selects one curve from the infinite family of solutions.

General vs particular solutionThe general solution carries an arbitrary constant; a particular solution has had that constant fixed by an initial condition.

Autonomous equationOne whose right-hand side involves only $y$, never $t$ — such as $\dfrac{dy}{dt}=ky$. Its behaviour depends on where you are, not when.

Worked examples

EX 1

Direct proportionality

population growth

A bacterial population grows at a rate proportional to the population present. Write a differential equation.

Name the quantity

Let $P(t)$ be the population at time $t$ hours.

Locate the rate phrase

‘grows at a rate’ $\Rightarrow\dfrac{dP}{dt}$.

Translate the proportionality

‘proportional to the population present’ $\Rightarrow k\cdot P$.

Assemble

$\dfrac{dP}{dt}=kP$, with $k\gt 0$ for growth.

Notice what is missing

$t$ does not appear on the right. The growth rate depends on how many bacteria there are, not on the clock.

EX 2

Newton's law of cooling

proportional to a difference

A cup of coffee at $180^{\circ}$F sits in a $70^{\circ}$F room. Its temperature falls at a rate proportional to the difference between its temperature and the room's. Model it.

Name the quantity

Let $T(t)$ be the coffee's temperature in $^{\circ}$F after $t$ minutes.

Identify the difference

‘the difference between its temperature and the room's’ is $T-70$, in that order.

Write the equation

$\dfrac{dT}{dt}=k(T-70)$.

Sign the constant

The coffee is cooling, so $\dfrac{dT}{dt}\lt 0$ while $T\gt 70$ makes $T-70\gt 0$. Therefore $k\lt 0$.

Add the initial condition

$T(0)=180$, recorded separately from the equation.

EX 3

Inflow minus outflow

a tank problem

Brine flows into a tank at $3$ litres per minute and drains out at $2$ litres per minute. Model the volume $V(t)$.

Name the quantity

$V(t)$ is the volume in litres after $t$ minutes.

Write rate in and rate out

In: $3$ L/min. Out: $2$ L/min. Both are already rates.

Combine as a net rate

$\dfrac{dV}{dt}=\text{in}-\text{out}=3-2=1$.

Read the result

A constant rate of $1$ L/min — a differential equation with no $V$ on the right at all.

Contrast with a leak

Had the outflow been ‘proportional to the volume present’, the model would be $\dfrac{dV}{dt}=3-cV$, and the behaviour would change completely.

EX 4

Reading the sentence exactly

two similar phrasings

Distinguish the models for ‘$y$ increases at a rate inversely proportional to $y$’ and ‘$y$ increases at a rate inversely proportional to $t$’.

Translate the first

Inversely proportional to $y$: $\dfrac{dy}{dt}=\dfrac{k}{y}$.

Translate the second

Inversely proportional to $t$: $\dfrac{dy}{dt}=\dfrac{k}{t}$.

Compare their nature

The first is autonomous — the rate depends only on the current value. The second depends explicitly on elapsed time.

Compare their behaviour

In the first, growth slows as $y$ gets large. In the second, growth slows simply because time passes, whatever $y$ is.

The point

One preposition changed the entire model. Read the sentence word by word before writing anything.

Units & interpretation

$\dfrac{dy}{dt}$units of $y$ per unit of $t$

$k$ in $y'=ky$units of 1/time — a per-unit-time growth rate

A solutiona function, never a number

Verbal phrase	Symbolic form	What it means
“rate of change of $y$”	$\dfrac{dy}{dt}$	the derivative itself
“proportional to $y$”	$ky$	constant multiple of the amount present
“inversely proportional to $y$”	$\dfrac{k}{y}$	large $y$ means slow change
“proportional to the difference from $A$”	$k(y-A)$	change stops when $y=A$
“jointly proportional to $y$ and $M-y$”	$ky(M-y)$	growth limited by a ceiling $M$
“net rate”	in $-$ out	the two rates subtract

If $k$ carries units of $\text{time}^{-1}$, every term on the right must too — a units check catches a mistranslated model instantly, before any solving happens.

Watch out

‘Proportional to’ needs a constant, not a number

Write $kP$, not $2P$. The constant is unknown until data supplies it, and inventing a value invalidates the model.

Difference means subtraction, in the stated order

‘the difference between $T$ and the room’ is $T-70$. Reversing it flips the sign of $k$ and reverses the physics.

Do not fold the initial condition into the equation

$y(0)=50$ is a separate fact. The differential equation describes every member of the solution family; the condition picks one.

The answer to a modelling question is an equation

If the problem says ‘write a differential equation’, stop once you have written it. Solving it earns no extra credit and costs time.

Course map. Topic 7.1 · LO FUN-7.A — interpret verbal statements of problems as differential equations involving a derivative expression (EK FUN-7.A.1).