ME11 - Exponential growth and decay models | OCR A-Level Mathematics B (MEI)

Use exponential models to describe proportional growth and decay, estimate rates and solve threshold problems. Interpret half-lives, compounding and long-term values, then compare predictions with evidence and explain sensible refinements.

Constant proportional change

A drug's concentration falls from $48$ to $24$ units in three hours. If the same model continues, does it disappear after another three hours? No: equal-time decay removes the same fraction, leaving $12$ , then $6$ , then $3$ . The amount becomes small without ever becoming exactly zero in this model.

You already know

$\mathrm e^{u+v}=\mathrm e^u\mathrm e^v$ , and $\mathrm e^u>0$ for every finite real $u$ . The inverse relation is $\mathrm e^u=v\Leftrightarrow u=\ln v$ , for $v>0$ .

Here is the surprise: a second dose that doubles the concentration would not change the time needed to halve it, once the same decay process resumes. The multiplier depends on elapsed time, not on the amount you start with. A model of one dose must not silently continue through a new dose without being adjusted.

Exponential model

A positive exponential model has form $Q=Q_0\mathrm e^{kt}$ , with $Q_0>0$ . Equal time intervals give equal multiplying factors. The model grows for $k>0$ , decays for $k<0$ , and is constant for $k=0$ . After a stated initial time, normally use $t\ge0$ .

The exponent has no units. If $t$ is measured in hours, $k$ has units per hour; the numerical value changes if you change the time unit.

Interpreting the rate

For $Q=Q_0\mathrm e^{kt}$ ,

\frac{Q(t+h)}{Q(t)}=\mathrm e^{kh},\qquad \frac{\mathrm dQ}{\mathrm dt}=kQ.

The known exponential gradient links the model to an instantaneous rate proportional to the current amount. Over one time unit the percentage change is $100(\mathrm e^k-1)\%$ . For decay, the positive percentage decrease is $100(1-\mathrm e^k)\%$ .

Not in the formula sheet: learn it.

An instantaneous proportional rate is not the same as a whole-hour fractional change. Use the exponential multiplier to move between them.

Worked example 1: a decaying concentration

Question. A concentration is modelled by $C=48\mathrm e^{-0.16t}$ , with $t$ in hours. Find its one-hour percentage decrease and its value at $t=3$ , giving numerical answers to 3 significant figures.

Plan. Calculate the one-hour multiplier and evaluate the model at the requested elapsed time.

\begin{aligned} q&=\mathrm e^{-0.16}\approx0.852144 &&\text{one-hour multiplier}\\ 100(1-q)&\approx14.8\% &&\text{percentage lost}\\ C(3)&=48\mathrm e^{-0.48}\approx29.7 &&\text{substitute three hours} \end{aligned}

Why this step? The multiplier gives the fraction remaining, so subtract it from $1$ to find the fraction lost.

Answer. About $14.8\%$ per hour; concentration about $29.7$ units after $3$ hours.

Check. $0<C(3)<48$ , and three hourly multiplications give $(\mathrm e^{-0.16})^3=\mathrm e^{-0.48}$ .

Change only the sign of the rate. The same coefficient now describes growth, so subtract $1$ from the multiplier instead.

Worked example 2: the opposite direction of change

Question. An amount is modelled by $C=48\mathrm e^{0.16t}$ . Find its one-hour percentage increase and its value at $t=3$ , to 3 significant figures.

Plan. Use the positive rate to find a multiplier greater than $1$ .

\begin{aligned} q&=\mathrm e^{0.16}\approx1.17351 &&\text{one-hour multiplier}\\ 100(q-1)&\approx17.4\% &&\text{percentage gained}\\ C(3)&=48\mathrm e^{0.48}\approx77.6 &&\text{three hours of growth} \end{aligned}

Why this step? Reversing the continuous rate gives reciprocal multipliers, not equal and opposite whole-hour percentages.

Answer. About $17.4\%$ growth per hour; amount about $77.6$ at $t=3$ .

Check. $\mathrm e^{0.16}\mathrm e^{-0.16}=1$ , so the two unit-time multipliers undo one another.

A population prediction can be non-integer: retain its value in calculations, then round appropriately when reporting a count of individuals. A concentration should instead use precision appropriate to the measurements or the question's instruction.

Watch out

Writing a $16\%$ hourly change for a rate constant $0.16$ skips the exponential. Calculate the actual unit-time multiplier first.

Interpret another decay model without mistaking its rate for its percentage loss.

Fitting rates and finding threshold times

Fit a rate using the ratio of two positive amounts. The time difference matters: the first observation may occur after the clock started.

Recovering a fitted exponential

For $Q=Q_0\mathrm e^{kt}$ , if $Q(t_1)=Q_1>0$ , $Q(t_2)=Q_2>0$ and $t_2\ne t_1$ ,

\begin{aligned} \frac{Q_2}{Q_1}&=\mathrm e^{k(t_2-t_1)}\\ k&=\frac{\ln(Q_2/Q_1)}{t_2-t_1}\\ Q_0&=Q_1\mathrm e^{-kt_1} \end{aligned}

Not in the formula sheet: learn it.

Dividing cancels the coefficient. Two fitted observations can determine the parameters of a chosen model, but do not provide independent evidence that the model is suitable.

Worked example 3: fitting a decay threshold

Question. A concentration is $48$ units at $t=0$ and $24$ at $t=3$ hours. Assuming $C=C_0\mathrm e^{kt}$ , determine its model and when it first falls to $10$ units.

Plan. Fit the rate from the halving ratio, then solve the target equation with the unrounded rate.

\begin{aligned} C_0&=48 &&\text{initial observation}\\ k&=\frac{\ln(24/48)}3=-\frac{\ln2}3 &&\text{use the amount ratio}\\ 10&=48\mathrm e^{kt} &&\text{set the threshold}\\ t&=\frac{\ln(10/48)}k=\frac{3\ln4.8}{\ln2}\approx6.79 &&\text{solve in hours} \end{aligned}

Why this step? Both the target log and the decay rate are negative, so their quotient gives positive elapsed time.

Answer. $C=48\mathrm e^{-(\ln2)t/3}$ , reaching $10$ after about $6.79$ hours.

Check. $C(6)=12>10$ and $C(7)\approx9.52<10$ . If observations occur only at whole hours, the first reading below $10$ is at hour $7$ .

Change only the observation times, moving both two hours later. The rate stays the same; the time-zero amount and threshold clock time change.

Worked example 4: fitting away from time zero

Question. A concentration is $48$ units at $t=2$ and $24$ at $t=5$ hours. Determine its exponential model and when it reaches $10$ .

Plan. Use the three-hour interval for the rate and measure the threshold interval from time $2$ .

\begin{aligned} k&=\frac{\ln(24/48)}{5-2}=-\frac{\ln2}3 &&\text{use elapsed time}\\ C_0&=48\mathrm e^{-2k}\approx76.2 &&\text{recover the time-zero coefficient}\\ 10&=48\mathrm e^{k(t-2)} &&\text{start from the known observation}\\ t&=2+\frac{\ln(10/48)}k\approx8.79 &&\text{restore the given clock} \end{aligned}

Why this step? The first observed value is not automatically the model's initial coefficient. Here the clock already reads $2$ .

Answer. $C=48\mathrm e^{-(\ln2)(t-2)/3}$ ; the threshold occurs at about $t=8.79$ hours.

Check. The model returns $48$ at $t=2$ and $24$ at $t=5$ . Its backward prediction at $t=0$ needs contextual justification if it is to describe a real concentration.

A decreasing model cannot reach a larger amount after the start. A negative computed time describes a backward extension, and is not an acceptable future threshold when $t\ge0$ . Round up only for a stated discrete observation rule, rather than for every time answer.

Exam tip

Keep the exact logarithmic rate, or enough guard digits, through a threshold calculation. A rounded intermediate rate can noticeably shift a later predicted time.

Use a different decay ratio and a target reached after another interval.

Half-lives, doubling times and time constants

Half-life is a time interval, not a date on the clock or a fixed amount lost. Each half-life halves whatever remains.

Fixed multiplying times

For $Q=Q_0\mathrm e^{kt}$ , with $Q_0>0$ ,

H=\frac{\ln2}{|k|}\quad(k<0),\qquad D=\frac{\ln2}{k}\quad(k>0).

After $j$ half-lives the remaining fraction is $2^{-j}$ . With $k=0$ , neither a finite half-life nor a finite doubling time exists for the positive constant amount.

Not in the formula sheet: learn it.

The initial coefficient cancels from the equation for a fixed ratio. For instance, a model $160\mathrm e^{-(\ln2)t/6}$ has half-life $6$ hours; after $18$ hours it gives $160/8=20$ , whatever units the amount uses.

Worked example 5: a half-life from a rate

Question. A mass is modelled by $M=80\mathrm e^{-0.15t}$ , with $t$ in days. Determine its half-life and its one-day percentage decrease, to 3 significant figures.

Plan. Set the amount ratio to $1/2$ , then find the unit-time loss separately.

\begin{aligned} \tfrac12&=\mathrm e^{-0.15H} &&\text{halving ratio}\\ H&=\frac{\ln2}{0.15}\approx4.62 &&\text{solve for positive time}\\ 100(1-\mathrm e^{-0.15})&\approx13.9\% &&\text{one-day decrease} \end{aligned}

Why this step? The half-life is found from a ratio, so it does not depend on the initial mass $80$ .

Answer. Half-life about $4.62$ days; daily decrease about $13.9\%$ .

Check. At the exact half-life, $M=80\mathrm e^{-\ln2}=40$ .

Change only the sign of the rate constant. The relevant fixed-ratio time is now a doubling time.

Worked example 6: a doubling time from the opposite rate

Question. For $M=80\mathrm e^{0.15t}$ , determine its doubling time and its one-day percentage increase, to 3 significant figures.

Plan. Set the ratio to $2$ and calculate the daily multiplier.

\begin{aligned} 2&=\mathrm e^{0.15D} &&\text{doubling ratio}\\ D&=\frac{\ln2}{0.15}\approx4.62 &&\text{take logs}\\ 100(\mathrm e^{0.15}-1)&\approx16.2\% &&\text{daily increase} \end{aligned}

Why this step? Reversing the signed rate preserves the fixed-ratio time but changes the whole-day percentage.

Answer. Doubling time about $4.62$ days; daily increase about $16.2\%$ .

Check. At the exact doubling time, $M=80\mathrm e^{\ln2}=160$ .

In $Q=Q_0\mathrm e^{-t/\tau}$ , the positive parameter $\tau$ is the time constant. At $t=\tau$ , the remaining fraction is $\mathrm e^{-1}\approx0.368$ . Hence $\tau=1/|k|$ and $H=\tau\ln2$ : these derived relations are not in the formula sheet. Do not substitute a time constant into a calculation that asks for a half-life.

Watch out

Halving the initial amount does not halve the half-life. The same multiplier acts on any initial amount, so the initial coefficient cancels.

Recover a rate from a half-life rather than finding the time from the rate.

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Pure Mathematics

Statistics

Mechanics

ME11 - Exponential growth and decay models

Constant proportional change

Fitting rates and finding threshold times

Half-lives, doubling times and time constants

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