P3 4.4 - Modelling growth and decay | Edexcel A-Level International Mathematics

Use exponential functions to model quantities that grow or decay, and differentiate them to find instantaneous rates. Interpret initial values and long-term predictions, then use observations and physical limits to test and improve a model.

Initial values and exponential rates

A quantity is falling, but the fall is becoming slower. How can its rate keep changing while one simple equation describes it? An exponential decay model makes the instantaneous rate proportional to the amount remaining.

For example, $M=80e^{-0.12t}$ gives $M=80$ when $t=0$ , and its derivative is $dM/dt=-9.6e^{-0.12t}=-0.12M$ . The rate is negative, and its magnitude becomes smaller as $M$ decreases.

Initial value and exponential rate

In $Q=Ae^{kt}$ , for time $t\ge0$ , the initial value is $Q(0)=A$ . If $A>0$ , the model grows when $k>0$ , decays when $k<0$ and is constant when $k=0$ . Its derivative is

\frac{dQ}{dt}=kAe^{kt}=kQ.

The constant $k$ has units reciprocal to the time unit. The derivative has units of quantity per unit time.

“Initially” means substitute $t=0$ , not $t=1$ . A quantity decreasing at a rate of $r$ units per minute has signed derivative $-r$ , even though its rate of decrease has positive magnitude $r$ .

Worked example 1: growth and its rate

A model for a mass is $M=80e^{0.12t}$ , where $M$ is in grams and $t\ge0$ is in minutes. Find its initial mass and its instantaneous rate of change at $t=5$ .

Plan. Use the original model for mass and its derivative for rate.

\begin{aligned} M(0)&=80e^0=80 &&\text{initial means }t=0\\ \frac{dM}{dt}&=9.6e^{0.12t} &&\text{multiply by the input derivative}\\ M'(5)&=9.6e^{0.6}\approx17.5 &&\text{evaluate the rate}. \end{aligned}

Answer. Initial mass $80\,\mathrm g$ ; rate $17.5\,\mathrm{g\,min^{-1}}$ to 3 significant figures, increasing.

Check. $M(5)=80e^{0.6}$ , and $0.12M(5)=9.6e^{0.6}$ , giving the same rate.

Change only the sign of the exponent. The initial mass stays fixed, while the direction of change reverses.

Worked example 2: decay and its rate

For $M=80e^{-0.12t}$ , with the same units, find its initial mass and its instantaneous rate of change at $t=5$ .

Plan. Keep the negative input derivative when differentiating.

\begin{aligned} M(0)&=80,\\ \frac{dM}{dt}&=-9.6e^{-0.12t},\\ M'(5)&=-9.6e^{-0.6}\approx-5.27. \end{aligned}

Answer. Initial mass $80\,\mathrm g$ ; signed rate $-5.27\,\mathrm{g\,min^{-1}}$ , a decrease of $5.27\,\mathrm{g\,min^{-1}}$ .

Check. After five minutes the mass is below its initial value. Its rate magnitude is below the initial $9.6\,\mathrm{g\,min^{-1}}$ , consistent with slowing decay.

The exponential factor is positive; a negative derivative here comes from the multiplier. To find a doubling or halving time, set the model equal to twice or half its initial value and take natural logarithms. For $Q=Ae^{kt}$ with $k>0$ , $e^{kt}=2$ gives doubling time $t=\ln2/k$ . For $Q=Ae^{-kt}$ with $k>0$ , $e^{-kt}=1/2$ gives the same expression for the halving time. The coefficient $A>0$ cancels, so the time depends on the relative rate rather than the starting amount.

Derivative of a to x

A model with a base such as $1.04$ can be differentiated by rewriting it with base $e$ .

Since $a=e^{\ln a}$ for $a>0$ ,

a^x=e^{x\ln a}.

The input $x\ln a$ has derivative $\ln a$ , because $a$ is constant. The chain rule for this linear input therefore gives

\frac{d}{dx}a^x=(\ln a)e^{x\ln a}=a^x\ln a.

Derivative of an exponential with any positive base

For a constant $a>0$ ,

\frac{d}{dx}(a^x)=a^x\ln a.

In particular, $\frac{d}{dt}(Ab^t)=Ab^t\ln b$ for constants $A$ and $b>0$ . The booklet gives $e^{x\ln a}=a^x$ ; this derivative is not printed there: learn it. For $a=1$ , the derivative is zero because $\ln1=0$ .

For $Q=Ab^t$ , the factor $b$ compares values one time unit apart: $Q(t+1)=bQ(t)$ . The instantaneous relative rate is $Q'/Q=\ln b$ . Thus growth of $4\%$ over a whole time unit, $b=1.04$ , does not mean an instantaneous relative rate exactly equal to $0.04$ .

Worked example 3: a base above one

A quantity is modelled by $N=200(1.04)^t$ , where $t$ is in days. Find $dN/dt$ and its initial value.

Plan. Multiply the original exponential by the natural logarithm of its base.

\begin{aligned} \frac{dN}{dt}&=200(1.04)^t\ln1.04,\\ N'(0)&=200\ln1.04\approx7.84. \end{aligned}

Answer. $dN/dt=200(1.04)^t\ln1.04$ ; initial rate $7.84$ units per day to 3 significant figures.

Check. $\ln1.04>0$ , so the rate is positive. The increase over the first whole day is $8$ units; a slightly smaller initial instantaneous rate is consistent with accelerating growth.

Change only the base from $1.04$ to $0.96$ . The logarithm changes sign.

Worked example 4: a base below one

For $N=200(0.96)^t$ , with the same units, find $dN/dt$ and its initial value.

Plan. Use the same formula and retain the negative value of $\ln0.96$ .

\begin{aligned} \frac{dN}{dt}&=200(0.96)^t\ln0.96,\\ N'(0)&=200\ln0.96\approx-8.16. \end{aligned}

Answer. $dN/dt=200(0.96)^t\ln0.96$ ; initial signed rate $-8.16$ units per day.

Check. Over the first whole day the loss is $8$ units. The initial rate magnitude is slightly larger because the decline slows as the quantity falls.

Use $\ln b$ , rather than $\log_{10}b$ , when differentiating $b^t$ .

Long-term behaviour

A model can describe the first few observations well and still make an unsuitable prediction far into the future.

For $k>0$ , the term $e^{-kt}$ tends to zero as $t$ increases. Thus $Ae^{-kt}$ tends to zero, while $C+Ae^{-kt}$ tends to $C$ . The latter has horizontal asymptote $Q=C$ . For finite $t$ , the exponential term is never zero, so the asymptote is approached without being reached.

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P3 4.4 - Modelling growth and decay

Initial values and exponential rates

Derivative of a to x

Long-term behaviour

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