1.06i - Exponential growth and decay models | OCR A-Level Mathematics A

Build growth and decay models from equal proportional changes, fit their rates and find meaningful threshold times. Use half-lives, continuous compounding and nonzero baselines, then check predictions against observations and refine the assumptions when needed.

Constant proportional change

If a culture grows by the same percentage each month, how long before its size doubles? Start with $500$ organisms and a monthly multiplier $\mathrm e^{0.12}$ . The first month's increase is about $64$ , but the second month's increase is larger: the same percentage acts on a larger population. Equal additions would describe a different model.

You already know

$\mathrm e^{u+v}=\mathrm e^u\mathrm e^v$ , so adding to an exponent multiplies the output. To solve $\mathrm e^u=v$ , with $v>0$ , use $u=\ln v$ . Keep unrounded calculator values until the final answer.

A surprising consequence is that doubling the initial population does not shorten its doubling time. Both populations multiply by exactly the same factor over every equal interval. We will calculate the opening model's doubling time below.

Exponential growth and decay

An exponential model $N(t)=N_0\mathrm e^{kt}$ , with $N_0>0$ , has a constant multiplicative change over equal time intervals. It grows when $k>0$ , decays when $k<0$ , and stays constant when $k=0$ . Use $t\ge0$ for time measured after a stated start, unless earlier times are justified.

Dividing two model values shows precisely which part stays constant.

Constant proportional change

For $N(t)=N_0\mathrm e^{kt}$ ,

\frac{N(t+h)}{N(t)}=\mathrm e^{kh}.

The percentage change over one time unit is $100(\mathrm e^k-1)\%$ ; for decay, the positive percentage decrease is $100(1-\mathrm e^k)\%$ . The known exponential gradient gives $\mathrm dN/\mathrm dt=kN$ .

Not in the formula sheet: learn it.

The exponent must be dimensionless: if time is in months, $k$ has units per month. The instantaneous proportional rate $k$ is usually different from the fractional change over an entire month.

Worked example 1: interpreting growth

Question. For $N=500\mathrm e^{0.12t}$ , with $t$ in months, find the initial population, the monthly percentage increase and the population after $3$ months.

Plan. Set $t=0$ , interpret the one-month multiplier, then evaluate the requested time.

\begin{aligned} N(0)&=500 &&\text{set the exponent to zero}\\ 100(\mathrm e^{0.12}-1)&\approx12.7\% &&\text{convert multiplier to increase}\\ N(3)&=500\mathrm e^{0.36}\approx716.665 &&\text{substitute the elapsed time} \end{aligned}

Why this step? A monthly multiplier $1.12749\ldots$ means an increase of $0.12749\ldots$ , not $1.12749\ldots$ , of the previous amount.

Answer. Initially $500$ ; monthly growth about $12.7\%$ ; about $717$ organisms after $3$ months.

Check. The three-month multiplier is $(\mathrm e^{0.12})^3=\mathrm e^{0.36}>1$ , consistent with growth.

Change only the sign of the rate constant. The amount still stays positive, but its multiplier is now less than $1$ .

Worked example 2: interpreting decay

Question. A count is modelled by $N=500\mathrm e^{-0.12t}$ , with $t$ in months. Find the monthly percentage decrease and the count after $3$ months.

Plan. Subtract the multiplier from $1$ to find the decrease, then evaluate the model.

\begin{aligned} q&=\mathrm e^{-0.12}\approx0.886920 &&\text{one-month multiplier}\\ 100(1-q)&\approx11.3\% &&\text{fraction lost as a percentage}\\ N(3)&=500\mathrm e^{-0.36}\approx348.838 &&\text{three months of decay} \end{aligned}

Why this step? A negative continuous rate does not mean subtracting $12\%$ each month; exponentiate it first.

Answer. Monthly decrease about $11.3\%$ ; a count of about $349$ after $3$ months.

Check. $0<N(3)<500$ , as a positive decay model requires.

Population counts may be non-integer model outputs. Round to whole individuals when interpreting them, while retaining the unrounded outputs in calculations.

Watch out

For $N=N_0\mathrm e^{0.12t}$ , calling the one-unit increase exactly $12\%$ confuses a continuous rate with a whole-interval change. The increase is $100(\mathrm e^{0.12}-1)\%$ .

Use the multiplier to interpret a different decay rate.

Fitting rates and finding threshold times

Two positive observations determine the parameters of a chosen exponential model. They do not establish that an exponential is the right choice; reserve further observations for checking it.

Fitting an exponential rate

For $N=N_0\mathrm e^{kt}$ , if $N(t_1)=N_1>0$ , $N(t_2)=N_2>0$ , and $t_2\ne t_1$ ,

\begin{aligned} \frac{N_2}{N_1}&=\mathrm e^{k(t_2-t_1)}\\ k&=\frac{\ln(N_2/N_1)}{t_2-t_1}\\ N_0&=N_1\mathrm e^{-kt_1} \end{aligned}

Not in the formula sheet: learn it.

Division cancels the unknown initial coefficient. Use elapsed time $t_2-t_1$ , not whichever time label is larger.

Worked example 3: fitting from an initial observation

Question. An amount is $300$ at $t=0$ and $450$ at $t=4$ days. Assuming $N=N_0\mathrm e^{kt}$ , determine the model and when it first reaches $600$ .

Plan. Use the first observation for $N_0$ , fit the rate from the ratio and solve the threshold equation.

\begin{aligned} N_0&=300 &&\text{the first time is zero}\\ \mathrm e^{4k}&=450/300=1.5 &&\text{use the second observation}\\ k&=\frac{\ln1.5}{4}\approx0.101366 &&\text{take logs}\\ \mathrm e^{kt}&=600/300=2 &&\text{set the target amount}\\ t&=\frac{\ln2}{k}\approx6.84 &&\text{solve in days} \end{aligned}

Why this step? The target equation compares the target with the initial amount; the fitted rate must be kept unrounded.

Answer. $N=300\mathrm e^{(\ln1.5)t/4}$ , reaching $600$ after about $6.84$ days.

Check. $N(6)\approx551<600$ and $N(7)\approx610>600$ . If only whole-day observations count, the first qualifying observation is day $7$ .

Now shift only the observation times by two days. The ratio determines the same rate, but the coefficient at $t=0$ is no longer the first observed amount.

Worked example 4: a different time origin

Question. The amount is $300$ at $t=2$ and $450$ at $t=6$ days. Determine the model and when it reaches $600$ .

Plan. Fit the rate over the four-day interval, recover the value at zero and solve from the known time $2$ .

\begin{aligned} k&=\frac{\ln(450/300)}{6-2}=\frac{\ln1.5}{4} &&\text{use elapsed time}\\ N_0&=300\mathrm e^{-2k}\approx244.949 &&\text{move back to time zero}\\ 600&=300\mathrm e^{k(t-2)} &&\text{measure time from the known value}\\ t&=2+\frac{\ln2}{k}\approx8.84 &&\text{return to the stated clock} \end{aligned}

Why this step? The coefficient of $\mathrm e^{kt}$ means the amount at time zero, not the amount in whichever observation happens to appear first.

Answer. $N=300\mathrm e^{k(t-2)}$ , with $k=\ln1.5/4$ ; the threshold occurs at about day $8.84$ .

Check. Substituting $t=2$ gives $300$ , and $t=6$ gives $300(1.5)=450$ .

A negative solution can describe a backward extension rather than a future event. For example, a growing model starting at $300$ does not fall to $200$ at a non-negative time. Check direction and domain before interpreting an answer.

Watch out

Do not round every threshold time upwards. Round up only when a discrete observation or completed-period rule asks for the first permitted integer time; otherwise report the continuous crossing at suitable accuracy.

Fit a decay model and find a later threshold.

Half-lives, doubling times and time constants

The initial amount cancels when you ask for a fixed multiplying factor. That is why equal half-lives remove equal fractions, rather than equal amounts.

Doubling time and half-life

For $N=N_0\mathrm e^{kt}$ , with $N_0>0$ ,

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

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

Not in the formula sheet: learn it.

For example, $160\mathrm e^{-(\ln2)t/6}$ has a half-life of $6$ hours: after $18$ hours, three halvings leave $160/8=20$ . This simple ratio check is often faster than a fresh logarithmic calculation.

Worked example 5: doubling with a continuous rate

Question. For $P=200\mathrm e^{0.08t}$ , with $t$ in years, determine its doubling time and its percentage increase over one year.

Plan. Set the ratio to $2$ , then separately calculate the annual multiplier.

\begin{aligned} 2&=\mathrm e^{0.08D} &&\text{doubling ratio}\\ D&=\frac{\ln2}{0.08}\approx8.66 &&\text{take logs}\\ 100(\mathrm e^{0.08}-1)&\approx8.33\% &&\text{annual increase} \end{aligned}

Why this step? A doubling time is determined by a ratio, so the starting amount $200$ cancels.

Answer. Doubling takes about $8.66$ years; the annual increase is about $8.33\%$ .

Check. $200\mathrm e^{0.08D}=200(2)=400$ , using the exact value of $D$ .

Change only the sign of the rate. We now seek halving, and both the time and the amount remain positive.

Worked example 6: halving with the opposite rate

Question. For $P=200\mathrm e^{-0.08t}$ , determine the half-life and the one-year percentage decrease.

Plan. Set the ratio to $1/2$ , then calculate the amount lost in one year.

\begin{aligned} \tfrac12&=\mathrm e^{-0.08H} &&\text{halving ratio}\\ H&=\frac{\ln(1/2)}{-0.08}=\frac{\ln2}{0.08}\approx8.66 &&\text{negative signs cancel}\\ 100(1-\mathrm e^{-0.08})&\approx7.69\% &&\text{annual decrease} \end{aligned}

Why this step? A half-life is a positive elapsed time; dividing $\ln2$ by a negative signed rate without adjusting the sign would give an impossible negative half-life.

Answer. Half-life about $8.66$ years; annual decrease about $7.69\%$ .

Check. After two half-lives the model gives $200/4=50$ .

A decay model can instead be written $N=N_0\mathrm e^{-t/\tau}$ , defining its time constant $\tau=1/|k|$ . At $t=\tau$ , the remaining fraction is $\mathrm e^{-1}\approx0.368$ , not $1/2$ ; its half-life is $\tau\ln2$ . These time-constant relations are not in the formula sheet: derive them from the exponent.

Watch out

A time constant and a half-life are different. Test the stated time in the exponent: $-1$ leaves $\mathrm e^{-1}$ , whereas $-\ln2$ leaves $1/2$ .

The opening culture's growth rate gives doubling time $\ln2/0.12\approx5.78$ months. Test the corresponding fixed-ratio reasoning with decay instead.

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Mechanics

1.06i - 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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