Mathematics 9709 · Pure Mathematics 3

Forming a DE from a rate statement, and interpreting the solution

Before you can solve, you sometimes have to build the DE from words.
After solving, you sometimes have to read the solution back.
These two sit either side of the solving. They usually show up as the first and last parts of a multi-part question.

Turn the words into symbols: “rate of change of $X$ ” $=\frac{dX}{dt}$ ; “proportional to” $=k\times$ ; “inversely proportional to” $=k\div$ ; a falling quantity gets a minus sign. Then a second bit of data fixes $k$ .

Forming: turn the sentence into a rate equation

“The rate of increase of $x$ is $k$ times $x$ ” becomes $\frac{dx}{dt}=kx$ straight away.
The “with respect to time” is almost always left out, so you add it yourself.

The words say…	Which means…
rate of change of $P$	$\dfrac{dP}{dt}$
proportional to $x$	$=kx$
inversely proportional to the square of $P$	$=\dfrac{k}{P^2}$
rate of decrease of $V$ , proportional to $V$	$\dfrac{dV}{dt}=-kV$

If the question does not name a variable, pick your own (the textbook tip).
Say what each letter stands for. Examiners need it.

Worked example

Textbook 10.6 (motorbike value): rate of decrease $\propto V$ ; show $V=10000\,e^{-kt}$

Form the DE. Value falls, so a minus: $\dfrac{dV}{dt}=-kV$ ( $k>0$ ).
Separate and integrate: $\dfrac1V\,dV=-k\,dt\;\Rightarrow\;\ln V=-kt+C$ .
At $t=0$ the value is $10000$ , so $C=\ln 10000$ ; un-log to $V=10000\,e^{-kt}$ .
Use the second datum to pin $k$ : the value halves after 3 years, so $5000=10000\,e^{-3k}\Rightarrow e^{-3k}=\tfrac12\Rightarrow k=\tfrac13\ln 2$ .

Answer

V=10000\,e^{-kt},\quad k=\tfrac13\ln 2

Common mistake

On a “show that” forming part the marks are easy to win, but easy to drop. Watch for three slips: missing the minus sign for a falling quantity, not saying what your variables and $k$ mean, and misreading the proportionality (inverse, or the square of it).

The classic shape: show the DE, then solve it

Worked example

9709/32 Oct–Nov 2024 Q10 (balloon): show $\dfrac{dr}{dt}=\dfrac{50-r}{5r^2}$ , then solve for $t$ [parts]

A round balloon: air pumped in at the steady rate $40\pi$ and leaking out at rate $0.8\pi r$ . With $V=\tfrac43\pi r^3$ , the chain rule gives $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ . The net inflow is $40\pi-0.8\pi r$ , which rearranges to part (a)'s $\dfrac{dr}{dt}=\dfrac{50-r}{5r^2}$ .

Separate: $\dfrac{5r^2}{50-r}\,dr=dt$ .
The pre-step: $\frac{5r^2}{50-r}$ is improper, so divide first (polynomial division): $\dfrac{5r^2}{50-r}=-5r-250+\dfrac{12500}{50-r}$ .
Integrate: $t=-\tfrac52 r^2-250r-12500\ln(50-r)+C$ .
At $t=0,\;r=0$ : $C=12500\ln 50$ , so $t=12500\ln\dfrac{50}{50-r}-\tfrac52 r^2-250r$ .

Answer

t=12500\ln\!\dfrac{50}{50-r}-\tfrac52 r^2-250r

Checked. At $r=12$ this gives $t\approx 70.5$ . The top-heavy-fraction divide is the step students skip. But $\frac{5r^2}{50-r}$ will not integrate until you split off that polynomial.

Interpreting: reading the solved function back

The last part often asks what the solution means, not for more algebra, in three common ways:

Long-term behaviour: “what happens to $x$ as $t$ becomes large?” Look at the limit. For a decay $V=10000\,e^{-kt}$ , as $t\to\infty$ the exponential $\to 0$ , so $V\to 0$ (see the curve below).
Limiting / maximum value: read the value the curve levels off at (e.g. a maximum temperature).
Model fit: compare predicted values against a data table and comment, does the model track the data, or drift in the long term? (Coursebook P2&3 Worked Example 10.5)

Exponential decay V = 10000 e^(−kt): as t grows the curve flattens toward 0. That long-term behaviour is what an interpret part asks about.

Newton's law of cooling says “rate of change of temperature is proportional to the gap from the surroundings”: $\frac{dT}{dt}=-k(T-s)$ (with $k>0$ ). This is just one more “proportional to a difference” case. Separate and solve the same way.
It is not a special method with its own name.

Your turn— tap to reveal the worked answer (Translate-only drills (textbook Ex 10B Q1, do not solve))

Turn each sentence straight into a DE:

“rate of decrease of height $h$ is proportional to the square of the height” $\Rightarrow\;\frac{dh}{dt}=-kh^2$ .
“rate of increase of $N$ is inversely proportional to $N$ ” $\Rightarrow\;\frac{dN}{dt}=\frac{k}{N}$ .

Answer

\tfrac{dh}{dt}=-kh^2,\qquad \tfrac{dN}{dt}=\tfrac{k}{N}

Common mistake

The interpret part is the most-skipped mark in the chapter. If asked “what happens as

t

becomes large” or “comment on the model”, write a sentence. An empty box scores zero even when your algebra was perfect.

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Forming a DE from a rate statement, and interpreting the solution

Before you can solve, you sometimes have to build the DE from words.
After solving, you sometimes have to read the solution back.
These two sit either side of the solving. They usually show up as the first and last parts of a multi-part question.

Forming: turn the sentence into a rate equation

“The rate of increase of $x$ is $k$ times $x$ ” becomes $\frac{dx}{dt}=kx$ straight away.
The “with respect to time” is almost always left out, so you add it yourself.

The words say…	Which means…
rate of change of $P$	$\dfrac{dP}{dt}$
proportional to $x$	$=kx$
inversely proportional to the square of $P$	$=\dfrac{k}{P^2}$
rate of decrease of $V$ , proportional to $V$	$\dfrac{dV}{dt}=-kV$

If the question does not name a variable, pick your own (the textbook tip).
Say what each letter stands for. Examiners need it.

Worked example

Textbook 10.6 (motorbike value): rate of decrease $\propto V$ ; show $V=10000\,e^{-kt}$

Form the DE. Value falls, so a minus: $\dfrac{dV}{dt}=-kV$ ( $k>0$ ).
Separate and integrate: $\dfrac1V\,dV=-k\,dt\;\Rightarrow\;\ln V=-kt+C$ .
At $t=0$ the value is $10000$ , so $C=\ln 10000$ ; un-log to $V=10000\,e^{-kt}$ .
Use the second datum to pin $k$ : the value halves after 3 years, so $5000=10000\,e^{-3k}\Rightarrow e^{-3k}=\tfrac12\Rightarrow k=\tfrac13\ln 2$ .

Answer

V=10000\,e^{-kt},\quad k=\tfrac13\ln 2

Common mistake

The classic shape: show the DE, then solve it

Worked example

9709/32 Oct–Nov 2024 Q10 (balloon): show $\dfrac{dr}{dt}=\dfrac{50-r}{5r^2}$ , then solve for $t$ [parts]

Separate: $\dfrac{5r^2}{50-r}\,dr=dt$ .
The pre-step: $\frac{5r^2}{50-r}$ is improper, so divide first (polynomial division): $\dfrac{5r^2}{50-r}=-5r-250+\dfrac{12500}{50-r}$ .
Integrate: $t=-\tfrac52 r^2-250r-12500\ln(50-r)+C$ .
At $t=0,\;r=0$ : $C=12500\ln 50$ , so $t=12500\ln\dfrac{50}{50-r}-\tfrac52 r^2-250r$ .

Answer

t=12500\ln\!\dfrac{50}{50-r}-\tfrac52 r^2-250r

Checked. At $r=12$ this gives $t\approx 70.5$ . The top-heavy-fraction divide is the step students skip. But $\frac{5r^2}{50-r}$ will not integrate until you split off that polynomial.

Interpreting: reading the solved function back

The last part often asks what the solution means, not for more algebra, in three common ways:

Long-term behaviour: “what happens to $x$ as $t$ becomes large?” Look at the limit. For a decay $V=10000\,e^{-kt}$ , as $t\to\infty$ the exponential $\to 0$ , so $V\to 0$ (see the curve below).
Limiting / maximum value: read the value the curve levels off at (e.g. a maximum temperature).
Model fit: compare predicted values against a data table and comment, does the model track the data, or drift in the long term? (Coursebook P2&3 Worked Example 10.5)

Exponential decay V = 10000 e^(−kt): as t grows the curve flattens toward 0. That long-term behaviour is what an interpret part asks about.

Newton's law of cooling says “rate of change of temperature is proportional to the gap from the surroundings”: $\frac{dT}{dt}=-k(T-s)$ (with $k>0$ ). This is just one more “proportional to a difference” case. Separate and solve the same way.
It is not a special method with its own name.

Your turn— tap to reveal the worked answer (Translate-only drills (textbook Ex 10B Q1, do not solve))

Turn each sentence straight into a DE:

“rate of decrease of height $h$ is proportional to the square of the height” $\Rightarrow\;\frac{dh}{dt}=-kh^2$ .
“rate of increase of $N$ is inversely proportional to $N$ ” $\Rightarrow\;\frac{dN}{dt}=\frac{k}{N}$ .

Answer

\tfrac{dh}{dt}=-kh^2,\qquad \tfrac{dN}{dt}=\tfrac{k}{N}

Common mistake

The interpret part is the most-skipped mark in the chapter. If asked “what happens as

t

becomes large” or “comment on the model”, write a sentence. An empty box scores zero even when your algebra was perfect.