F2, F7 - Exponential growth and decay models | AQA A-Level Mathematics

Use F2 to connect an exponential's gradient to its current value. Apply that proportional-rate idea in F7 models of continuous interest, population growth and decay, then test assumptions and refine models with a nonzero baseline.

Exponential gradients and proportional rates

If a growing population doubles, must it take longer to double again because it now needs more new organisms?

Imagine that each organism contributes the same average net growth rate. A population twice as large then adds organisms twice as quickly. The absolute rate changes, even while the rate per organism stays constant.

You already know

For $N=N_0e^{kt}$ , an interval of length $s$ multiplies the amount by $e^{ks}$ . Taking $\ln$ solves for an unknown time. These index and logarithm results are not supplied in the formula booklet.

The special property of $e^x$ is that its gradient equals its current value. More generally the gradient of $e^{kx}$ is $ke^{kx}$ . The notation $dy/dx$ means the instantaneous rate of change, or tangent gradient; use this exponential gradient result directly here. Later lessons develop differentiation methods and their derivations.

Exponential gradient and relative rate

For $y=Ae^{kx}$ ,

\frac{dy}{dx}=Ake^{kx}=ky

When $y\ne0$ , the constant ratio $\frac{1}{y}\frac{dy}{dx}=k$ is the relative rate. For positive $y$ , $k>0$ gives growth and $k<0$ gives decay.

Not in the formula booklet: learn it.

If time is in days, $k$ has units $\mathrm{day}^{-1}$ ; its product with time is dimensionless. An absolute rate has units of amount per day. A continuous relative rate of $0.08\,\mathrm{day}^{-1}$ gives a one-day percentage increase $100(e^{0.08}-1)\%\approx8.33\%$ , rather than exactly $8\%$ . This conversion is a consequence of the unsupplied exponential model.

Growth and decay with equal relative-rate magnitudes give changing absolute gradients

The growth curve steepens as more is present. The decay rate is negative, but its magnitude gets smaller as less remains. Adding a constant amount each day would instead give a straight line.

Worked example 1: a population's rate and doubling time

Question. A population is modelled by $N(t)=600e^{0.08t}$ , with $t\ge0$ in days. Find its initial growth rate, its rate after $5$ days, and its doubling time.

Plan. Multiply the population by its relative rate, then solve the equation for twice the starting amount.

\begin{aligned} \frac{dN}{dt}&=0.08N=48e^{0.08t} && \text{exponential gradient} \\ N'(0)&=48\qquad N'(5)=48e^{0.4}\approx71.6 && \text{rates in organisms per day} \\ 600e^{0.08T}&=1200 && \text{twice the starting amount} \\ T&=\frac{\ln2}{0.08}\approx8.66 && \text{divide and take }\ln \end{aligned}

Why this step? The amount $600$ cancels in the doubling equation, so the time depends on the relative rate, not the initial population.

Answer. Rates: initially $48$ and after $5$ days approximately $71.6$ organisms per day. Doubling time: approximately $8.66$ days.

Check. $e^{0.08T}=2$ . Over any further interval of length $T$ , the same factor $2$ applies again.

Change only the initial population to $1200$ . This doubles each absolute rate, but the time to multiply the population by $2$ should be unchanged.

Worked example 2: doubling the initial amount

Question. Repeat the rate and doubling-time calculations for $N(t)=1200e^{0.08t}$ , with time in days.

Plan. Use the same relative-rate multiplier and set the target to twice the new starting amount.

\begin{aligned} \frac{dN}{dt}&=96e^{0.08t} && 0.08\times1200=96 \\ N'(0)&=96\qquad N'(5)=96e^{0.4}\approx143 && \text{substitute the times} \\ 1200e^{0.08T}&=2400 && \text{new doubling target} \\ T&=\frac{\ln2}{0.08}\approx8.66 && \text{same amount ratio} \end{aligned}

Why this step? Doubling both the amount and its growth rate leaves the time to achieve a fixed multiplying factor unchanged.

Answer. Rates: $96$ and approximately $143$ organisms per day, respectively. The doubling time remains approximately $8.66$ days.

Check. Each unrounded rate is twice its value in example 1, while $2400/1200=1200/600=2$ .

The gradient rule works for negative exponents too. For $y=7e^{-2x}$ , the gradient is $-14e^{-2x}$ ; at $x=\ln2$ , it is $-14/4=-7/2$ . The negative gradient describes decrease, not a negative amount.

Watch out

Wrong: “The rate of $600e^{0.08t}$ is $0.08$ organisms per day.” The $0.08$ is a relative rate in $\mathrm{day}^{-1}$ . Multiply by the current population: the absolute rate is $48e^{0.08t}$ organisms per day.

Change only the initial population again. Use the rate's units to distinguish it from an amount.

Continuous compound interest

Continuous compounding models a balance increasing at a rate proportional to the balance at every instant.

For a nominal annual rate $r$ , compounding $n$ times per year gives $P_0(1+r/n)^{nt}$ after $t$ years. Increasing $n$ makes each interest addition smaller but more frequent; its limiting balance is $P_0e^{rt}$ . Neither of these compounding formulae is supplied in the formula booklet.

Continuous compound interest

P(t)=P_0e^{rt},\qquad t\ge0

Here $P_0$ is the initial balance, $r$ the continuous annual rate as a decimal, and $t$ time in years. Assume constant $r$ and no deposits, withdrawals or fees.

The one-year multiplier is $b=e^r$ , so the effective annual percentage increase is $100(e^r-1)\%$ . Conversely, $r=\ln b$ .

Not in the formula booklet: learn it.

A continuous rate of $4.5\%$ means $r=0.045\,\mathrm{year}^{-1}$ ; the effective annual increase is about $4.60\%$ . Use the same time units as the rate, and retain the exact rate until the final monetary rounding.

Worked example 3: a balance and a target time

Question. An initial £2500 earns continuous interest at rate $0.045\,\mathrm{year}^{-1}$ , with no other transactions. Find the balance after $6$ years to the nearest penny, and the time to reach £3500 to two decimal places.

Plan. Substitute the given time, then solve a separate exponential equation for the target balance.

\begin{aligned} P(6)&=2500e^{0.045(6)}\approx3274.91 && \text{balance in pounds} \\ 2500e^{0.045t}&=3500 && \text{target balance} \\ e^{0.045t}&=1.4 && \text{divide by }2500 \\ t&=\frac{\ln1.4}{0.045}\approx7.48 && \text{take }\ln\text{ and divide} \end{aligned}

Why this step? Divide by the original deposit before taking logarithms; the target is an amount ratio, not an amount to add to the exponent.

Answer. The balance is £3274.91, including £774.91 interest. The target is reached after approximately $7.48$ years.

Check. $2500e^{\ln1.4}=3500$ ; the target exceeds the six-year balance, and the target time exceeds $6$ .

Change only the target from £3500 to £4000. The account's model stays fixed, so compare the new time with the first target time.

Worked example 4: a higher target

Question. For the same initial £2500 and continuous rate $0.045\,\mathrm{year}^{-1}$ , find the time to reach £4000 to two decimal places.

Plan. Form the new balance ratio, then solve its logarithmic equation.

\begin{aligned} 2500e^{0.045t}&=4000 && \text{new target} \\ e^{0.045t}&=1.6 && \text{divide by initial balance} \\ 0.045t&=\ln1.6 && \text{inverse functions} \\ t&=\frac{\ln1.6}{0.045}\approx10.44 && \text{solve for time} \end{aligned}

Why this step? A target expressed as a multiplying factor fits the exponential model directly; adding a fixed amount each year would describe a different model.

Answer. Approximately $10.44$ years.

Check. $2500e^{\ln1.6}=4000$ . The higher target takes longer than $7.48$ years, as it should.

These are continuous times. Rounding up to the next integer would answer a different question about the first whole-year anniversary. For comparison, a £1800 deposit at $0.03\,\mathrm{year}^{-1}$ has balance $1800e^{0.12}\approx\text{£}2029.49$ after $4$ years.

Watch out

Wrong: $P(6)=2500e^{4.5(6)}$ . A percentage must be converted to a decimal: $4.5\%=0.045$ , so the exponent is $0.045(6)=0.27$ .

Change only the target again and preserve the unrounded logarithm until the end.

Decay constants and half-life

Exponential decay removes a fixed fraction over equal time intervals, so its modelled amount stays positive at every finite time.

A model $Q=Q_0e^{-\lambda t}$ uses a positive decay constant $\lambda$ . The rate is $-\lambda Q$ : the minus sign describes decrease. The half-life $H$ is the time for the amount to halve. To obtain its formula, divide $Q_0e^{-\lambda H}=Q_0/2$ by $Q_0$ , then take $\ln$ : $-\lambda H=-\ln2$ .

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Pure Mathematics

Mechanics

Statistics

F2, F7 - Exponential growth and decay models

Exponential gradients and proportional rates

Continuous compound interest

Decay constants and half-life

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