The Scenario
A biologist places 100 bacteria in a petri dish at 9 AM. She returns an hour later and counts 200. By 11 AM there are 400. The population has doubled in the first hour, then doubled again in the second hour.
The biologist observes: “The bacteria multiply faster when there are more of them — the growth rate depends on the current population.”
At 9 AM the population grew by 100 in one hour. At 10 AM it grew by 200 in one hour. More bacteria → faster growth. How do we capture this observation as a precise mathematical equation?
The Question
The biologist says: “The rate of increase of bacteria is proportional to the number of bacteria present.”
How do you write this observation as a mathematical equation?
Think about: What is changing? What is the rate? What is it proportional to?
💡 The Bridge (click to reveal)
Let $N$ = number of bacteria at time $t$.
“Rate of increase of bacteria” = $\dfrac{\text{d}N}{\text{d}t}$ (the derivative of $N$ with respect to $t$).
“Proportional to the number present” = $k \cdot N$ (where $k$ is a positive constant).
Combining: $\dfrac{\text{d}N}{\text{d}t} = kN$
This is a differential equation — an equation that relates a function to its derivative. Today we learn how to translate ANY such word description into a differential equation.
A differential equation is an equation that relates a quantity to its rate of change. Today we focus on formulating them — translating a word description of a rate of change into a mathematical equation. We do NOT solve them yet (that is the next lesson).
A word problem says: “The rate of change of $y$ is proportional to [something].” We translate this to:
$$\boxed{\;\dfrac{\text{d}y}{\text{d}t} = k \cdot f(y)\;}$$where $k$ is the constant of proportionality and $f(y)$ is the function describing what $y$ is proportional to.
Follow these five steps to translate any word description into a differential equation of the form $\dfrac{\text{d}y}{\text{d}t} = k \cdot f(y)$:
1. Name the quantity. Identify what is changing and give it a symbol (e.g. $y$, $N$, $P$, $m$, $h$, $e$).
2. Write the rate. Express the rate of change as a derivative with respect to the correct independent variable: $\dfrac{\text{d}y}{\text{d}t}$ (time) or $\dfrac{\text{d}y}{\text{d}x}$ / $\dfrac{\text{d}e}{\text{d}\theta}$ (another variable).
3. Translate the proportion. “proportional to $f$” $\rightarrow k \cdot f$; “inversely proportional to $f$” $\rightarrow \dfrac{k}{f}$.
4. Identify $f(y)$. Pick out the function the rate is proportional to: $y$, $y^2$, $\sqrt{y}$, $\dfrac{1}{y}$, $\ln(\cdots)$, etc.
5. Fix the sign. Increasing quantity $\rightarrow +k \cdot f(y)$; decreasing quantity $\rightarrow -k \cdot f(y)$. Always take $k > 0$.
The phrase “proportional to” can describe different functions of $y$. Here are the five common types in this chapter:
| Type | Phrase | Differential equation | Behaviour |
|---|---|---|---|
| Direct | “proportional to $y$” | $\dfrac{\text{d}y}{\text{d}t} = ky$ | rate grows linearly with $y$ |
| Inverse | “inversely proportional to $y$” | $\dfrac{\text{d}y}{\text{d}t} = \dfrac{k}{y}$ | as $y$ grows, rate shrinks |
| Square | “proportional to $y^2$” | $\dfrac{\text{d}y}{\text{d}t} = ky^2$ | rate grows much faster than $y$ |
| Square root | “proportional to $\sqrt{y}$” | $\dfrac{\text{d}y}{\text{d}t} = k\sqrt{y}$ | rate grows, slower than $y$ |
| Logarithm | “proportional to $\ln(\cdots)$” | $\dfrac{\text{d}y}{\text{d}t} = k\ln(H-y)$ | rate slows as $y$ nears $H$ |
Variants like “inversely proportional to the square of $y$” give $\dfrac{\text{d}y}{\text{d}t} = \dfrac{k}{y^2}$ — the same inverse pattern, with $y^2$ in the denominator.
The constant $k$ is always taken as positive; the direction of change is shown by a separate sign:
| Situation | Sign of $\dfrac{\text{d}y}{\text{d}t}$ | Form |
|---|---|---|
| Increasing | positive | $\dfrac{\text{d}y}{\text{d}t} = +k \cdot f(y)$ |
| Decreasing | negative | $\dfrac{\text{d}y}{\text{d}t} = -k \cdot f(y)$ |
Key words to watch for: “increases / grows” $\rightarrow +$; “decreases / decays / shrinks” $\rightarrow -$.
In pairs, link every scenario to the differential equation that models it. Click a scenario card, then click the equation you think it pairs with — a numbered line joins them. Link all six pairs, then press Check to score. Press Reset to shuffle and play again. Talk each link through before you click: which words give you the derivative, and which words give you $k \cdot f(y)$?
Reading tip: “rate of increase / decrease” → $\dfrac{\text{d}y}{\text{d}t}$; “proportional to $f$” → $k \cdot f$; “inversely proportional to $f$” → $\dfrac{k}{f}$; “decreases” → a minus sign.
Work on your own. Each mission describes a rate of change and shows a dotted target graph of $\dfrac{\text{d}y}{\text{d}t}$ against $y$. Choose the proportionality type, the direction and the value of $k$ so that your solid curve matches the target, then press Check model. Predict first, then adjust, then explain what each control does to the curve.
Dotted amber curve = the target given by the mission. Solid emerald curve = the model your controls are building right now.
The differential equation $\dfrac{\mathrm{d}v}{\mathrm{d}t} = 5v^2$ models the motion of a particle, where $v$ is the velocity of the particle in $\mathrm{m\,s^{-1}}$ and $t$ is the time in seconds. Explain the meaning of $\dfrac{\mathrm{d}v}{\mathrm{d}t}$ and what the differential equation tells you about the motion of the particle.
👁 Reveal Solution
$\dfrac{\text{d}v}{\text{d}t}$ is the rate of change of velocity with respect to time — this is the acceleration of the particle.
Its units are $\dfrac{\text{m/s}}{\text{s}} = \text{m/s}^2$.
The equation is $\dfrac{\text{d}v}{\text{d}t} = 5v^2$.
Since $v^2 \geq 0$ for all real $v$, and $5 > 0$, we have $\dfrac{\text{d}v}{\text{d}t} \geq 0$ always.
This means the velocity is always increasing (or zero) — the particle never decelerates.
The acceleration $= 5v^2$ is proportional to the square of the velocity.
When $v = 0$: acceleration $= 0$ (no change in velocity).
When $v = 1$: acceleration $= 5\,\text{m/s}^2$.
When $v = 2$: acceleration $= 20\,\text{m/s}^2$.
As the particle moves faster, its acceleration grows rapidly — the velocity increases without bound (in this model).
$\dfrac{\text{d}v}{\text{d}t}$ is the acceleration of the particle. The equation tells us that the acceleration equals $5v^2$, which is always non-negative and proportional to the square of the velocity. The particle's velocity always increases, and the rate of increase grows as the velocity grows.
In a chemical reaction in which substance A is converted into substance B, the rate of increase of the mass of substance B is inversely proportional to the mass of substance B present. Find a differential equation to model this situation.
👁 Reveal Solution
Let $m$ = mass of substance B present at time $t$ (in grams or kg).
“Rate of increase of the mass of B” = $\dfrac{\text{d}m}{\text{d}t}$.
The word “increase” tells us this rate is positive.
“Inversely proportional to the mass $m$” means $\dfrac{k}{m}$, where $k > 0$ is the constant of proportionality.
Note: “inversely proportional” ≠ “proportional”. Inverse means the variable goes in the denominator.
$$\dfrac{\text{d}m}{\text{d}t} = \dfrac{k}{m}$$
where $k$ is a positive constant.
The mass $m$ is increasing (the reaction produces B), so $\dfrac{\text{d}m}{\text{d}t} > 0$. Since $k > 0$ and $m > 0$, we have $\dfrac{k}{m} > 0$. ✓
As $m$ grows, $\dfrac{k}{m}$ shrinks, so the rate of increase slows down — this makes physical sense (less reactant A left to convert).
Attempt each question on paper first, then click Reveal Solution to check your formulation. Questions are ordered from easiest to hardest.
A spark from a firework is moving in a straight line at a speed which is inversely proportional to the square of the distance which the spark has travelled from the firework. Find an expression for the speed (i.e. the rate of change of distance travelled) of the spark.
💡 Hint
Let $x$ = distance travelled from the firework.
Speed = rate of change of distance = $\dfrac{\text{d}x}{\text{d}t}$.
“Inversely proportional to the square of the distance” = $\dfrac{k}{x^2}$.
👁 Reveal Solution
Let $x$ = distance travelled by the spark from the firework (in metres), at time $t$.
Speed = rate of change of distance = $\dfrac{\text{d}x}{\text{d}t}$.
The question tells us this directly: “the rate of change of distance travelled”.
“Inversely proportional to the square of the distance” = $\dfrac{k}{x^2}$.
“Inverse” → denominator. “Square of the distance” → $x^2$.
$$\boxed{\dfrac{\text{d}x}{\text{d}t} = \dfrac{k}{x^2}}$$
where $k$ is a positive constant.
The rate at which a sunflower increases in height is proportional to the natural logarithm of the difference between its final height $H$ and its height $h$ at a particular time. Find a differential equation to model this situation.
💡 Hint
The variable is $h$ (current height), and the final height $H$ is a constant.
“Rate of increase in height” = $\dfrac{\text{d}h}{\text{d}t}$.
“Natural logarithm of the difference” = $\ln(H - h)$.
Is the height increasing or decreasing? What sign should the rate have?
👁 Reveal Solution
Let $h$ = height of the sunflower at time $t$.
$H$ = final height (a constant).
“Rate at which the sunflower increases in height” = $\dfrac{\text{d}h}{\text{d}t}$.
The word “increases” means this rate is positive.
“Proportional to the natural logarithm of the difference between $H$ and $h$”
Difference between $H$ and $h$: $H - h$ (positive, since $h < H$ while growing).
Natural logarithm: $\ln(H - h)$.
Proportional to: $k \cdot \ln(H - h)$.
$$\boxed{\dfrac{\text{d}h}{\text{d}t} = k\ln(H - h)}$$
where $k$ is a positive constant and $H$ is the final height.
After a major advertising campaign, anngineering company finds that its profits are increasing at a rate proportional to the square root of the profits at any given time. Find an expression to model this situation.
💡 Hint
Let $P$ = profits at time $t$.
“Profits are increasing at a rate” → $\dfrac{\text{d}P}{\text{d}t}$ (positive).
“Proportional to the square root of the profits” → $k\sqrt{P}$.
👁 Reveal Solution
Let $P$ = profits at time $t$.
“Profits are increasing at a rate” = $\dfrac{\text{d}P}{\text{d}t}$.
“Increasing” means the rate is positive.
“Proportional to the square root of the profits”
Square root of profits: $\sqrt{P}$.
Proportional to: $k\sqrt{P}$.
$$\boxed{\dfrac{\text{d}P}{\text{d}t} = k\sqrt{P}}$$
where $k$ is a positive constant.
The coefficient of restitution $e$ of a squash ball increases with respect to the ball's temperature $\theta$ at a rate proportional to the temperature, for typical playing temperatures. (The coefficient of restitution is a measure of how elastic, or bouncy, the ball is. Its value lies between zero and one, zero meaning that the ball is not at all elastic and one meaning that it is perfectly elastic.) Find a differential equation to model this situation.
💡 Hint
Here the variables are $e$ (coefficient of restitution) and $\theta$ (temperature).
The rate is with respect to $\theta$, not time. So the derivative is $\dfrac{\text{d}e}{\text{d}\theta}$.
“Increases” means positive rate. “Proportional to the temperature” = $k\theta$.
👁 Reveal Solution
$e$ = coefficient of restitution (dimensionless, between 0 and 1).
$\theta$ = temperature of the ball.
Note: the independent variable here is $\theta$ (temperature), not time $t$.
“Increases with respect to the ball's temperature $\theta$”
The rate of change of $e$ with respect to $\theta$ is $\dfrac{\text{d}e}{\text{d}\theta}$.
“Increases” means this rate is positive.
“Proportional to the temperature” = $k\theta$.
Since the rate is positive (increasing) and $k > 0$, we need $\theta > 0$ for typical playing temperatures (which is physically reasonable).
$$\dfrac{\text{d}e}{\text{d}\theta} = k\theta$$
where $k$ is a positive constant.
For typical playing temperatures $\theta > 0$, so $k\theta > 0$ and $\dfrac{\text{d}e}{\text{d}\theta} > 0$ — the coefficient of restitution increases with temperature. ✓
This matches the physical observation: a warm squash ball is bouncier than a cold one.
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.