Forming an equation from a rate statement | CAIE 9709 Pure Mathematics 3 Notes

Chapters

Pure Mathematics 3

Mathematics9709

Define symbols, then translate the words

Words decide the sign and factor

increases ∝ P→dP/dt = kP

decreases ∝ V→dV/dt = −kV

increases ∝ 1/N²→dN/dt = k/N²

moves toward S→dT/dt = −k(T−S)

Translate the direction, the proportionality and the changing quantity separately.

If the question does not define its symbols, state them yourself. For example, “let $P$ be the population after $t$ years.” This makes $dP/dt$ and its units unambiguous.

“Rate of change of P” becomes $dP/dt$ when time is the independent variable.
“Proportional to” introduces a constant $k$ .
“Decrease” needs a negative derivative when $k>0$ .
“Inversely proportional to the square” means divide by the square, not by the quantity itself.
State $k>0$ when k represents the size of the proportionality constant; use the sign in the equation for increase or decrease.

Key idea

Keep

k>0

and let the sign describe the direction. This makes growth and decay models easier to compare.

A rate datum can determine k before solving

Worked example

Inverse-square population model

A population density P increases at a rate inversely proportional to $P^2$ . At $t=0$ , $P=2$ and $dP/dt=0.5$ .

\frac{dP}{dt}=\frac{k}{P^2}

0.5=\frac{k}{2^2}\quad\Rightarrow\quad k=2

P^2\,dP=2\,dt

\frac{P^3}{3}=2t+C

The initial condition gives $C=8/3$ .

Answer

P=\sqrt[3]{6t+8}

(Pure Mathematics 2 & 3 Coursebook, Worked Example 10.5)

Recognise direct, inverse and difference models

A phrase such as “moves toward the surrounding temperature $S$ ” depends on the signed gap. One clear form is:

\frac{dT}{dt}=-k(T-S),\qquad k>0

If $T>S$ , the derivative is negative and T falls. If $T<S$ , the derivative is positive and T rises. At $T=S$ , the rate is zero.

Your turnIndependent translation check [6 marks]

Form a differential equation for each statement. Do not solve.

The rate of decrease of h is proportional to h squared.
The rate of increase of N is inversely proportional to N.
A temperature T moves toward a constant surrounding temperature S at a rate proportional to the difference.

Show worked answer

Answer

\dfrac{dh}{dt}=-kh^2

\dfrac{dN}{dt}=\dfrac{k}{N}

\dfrac{dT}{dt}=-k(T-S)

In both equations take $k>0$ .

Common mistake

A negative value of

k

can hide a missing minus sign, but it makes the model harder to read. State

k>0

and put the direction in the equation explicitly.

Define symbols, then translate the words

Words decide the sign and factor

increases ∝ P→dP/dt = kP

decreases ∝ V→dV/dt = −kV

increases ∝ 1/N²→dN/dt = k/N²

moves toward S→dT/dt = −k(T−S)

Translate the direction, the proportionality and the changing quantity separately.

If the question does not define its symbols, state them yourself. For example, “let $P$ be the population after $t$ years.” This makes $dP/dt$ and its units unambiguous.

“Rate of change of P” becomes $dP/dt$ when time is the independent variable.
“Proportional to” introduces a constant $k$ .
“Decrease” needs a negative derivative when $k>0$ .
“Inversely proportional to the square” means divide by the square, not by the quantity itself.
State $k>0$ when k represents the size of the proportionality constant; use the sign in the equation for increase or decrease.

Key idea

Keep

k>0

and let the sign describe the direction. This makes growth and decay models easier to compare.

A rate datum can determine k before solving

Worked example

Inverse-square population model

A population density P increases at a rate inversely proportional to $P^2$ . At $t=0$ , $P=2$ and $dP/dt=0.5$ .

\frac{dP}{dt}=\frac{k}{P^2}

0.5=\frac{k}{2^2}\quad\Rightarrow\quad k=2

P^2\,dP=2\,dt

\frac{P^3}{3}=2t+C

The initial condition gives $C=8/3$ .

Answer

P=\sqrt[3]{6t+8}

(Pure Mathematics 2 & 3 Coursebook, Worked Example 10.5)

Recognise direct, inverse and difference models

A phrase such as “moves toward the surrounding temperature $S$ ” depends on the signed gap. One clear form is:

\frac{dT}{dt}=-k(T-S),\qquad k>0

If $T>S$ , the derivative is negative and T falls. If $T<S$ , the derivative is positive and T rises. At $T=S$ , the rate is zero.

Your turnIndependent translation check [6 marks]

Form a differential equation for each statement. Do not solve.

The rate of decrease of h is proportional to h squared.
The rate of increase of N is inversely proportional to N.
A temperature T moves toward a constant surrounding temperature S at a rate proportional to the difference.

Show worked answer

Answer

\dfrac{dh}{dt}=-kh^2

\dfrac{dN}{dt}=\dfrac{k}{N}

\dfrac{dT}{dt}=-k(T-S)

In both equations take $k>0$ .

Common mistake

A negative value of

k

can hide a missing minus sign, but it makes the model harder to read. State

k>0

and put the direction in the equation explicitly.