NZ Level 8 (NZC) Level 3 (NCEA) [In development] Applications of DE's
Lesson

The following examples show how differential equations can be applied to real world problems. There are many others of course, but the four examples discussed have a common theme. Each problem begins with the formulation of a rate of change $\frac{dy}{dx}$dydx. Nature often works this way - revealing itself through rates. These rates lead us to the models of nature - laws that seem to govern the behaviour of things - and the study of differential equations is the key to unlocking these laws.

##### Example 1

In a small country town in Victoria, Australia, the rate of increase of its population is proportional to the the number of people living there at any given time so that $\frac{dP}{dt}=kP$dPdt=kP

At the end of 2017 the population was recorded as $103000$103000. Show that the population growth model is given by $P=P_0e^{kt}$P=P0ekt. If the town is predicted to undergo a $4%$4% increase in 2018, find the growth rate $k$k and determine an estimate of the population in 2027. When will the population exceed $200000$200000.

Note the wording - the rate of growth is proportional to the population. This makes perfect sense - the more people around the larger the potential for population growth. For example a $10%$10% per year increase in a city like Sydney, Australia with a population of $5000000$5000000 means an extra $500000$500000 people per year. A $10%$10% growth in a smaller city means a smaller increase in numbers.

Hence we have the model $\frac{dP}{dt}=kP$dPdt=kP and this is a separable first order differential equation.

Thus:

 $\frac{dP}{dt}$dPdt​ $=$= $kP$kP $\int\frac{dP}{P}$∫dPP​ $=$= $k\int dt$k∫dt $\ln P$lnP $=$= $kt+C$kt+C $P$P $=$= $e^{kt+C}=e^C\times e^{kt}$ekt+C=eC×ekt $\therefore$∴  $P$P $=$= $Ae^{kt}$Aekt

Suppose we now set the year 2017 as our zero year, so that at $t=0$t=0, $P=103000$P=103000. Substituting these values into our model yields $$, and thus A=103000A=103000. Hence we can declare that the model for the country town is given by P=103000e^{kt}P=103000ekt. The town is expected to increase by 4%4% or 4,1204,120 people by the end of 2018. This means that for t=1t=1, P=103000+4120=107120P=103000+4120=107120. Putting this figure into our model we have that 107120=103000e^{k\times1}107120=103000ek×1 and this information will allow us to find kk.  107120107120 == 103000e^k103000ek \frac{107120}{103000}107120103000​ == e^kek 1.041.04 == e^kek \therefore∴ kk == \ln1.04ln1.04 == 0.03922071320.0392207132 Note carefully how the 4%4% growth rate relates to the constant kk. Therefore kk becomes fixed as 0.03922071320.0392207132 so that our model becomes; P=103000e^{0.0392207132t}P=103000e0.0392207132t or more correctly P=103000e^{(\ln1.04)t}P=103000e(ln1.04)t. At t=10t=10, in 2027 the population is expected to be given by P=P=103000e^{\ln1.04\times10}=152465P=P=103000eln1.04×10=152465 people. The time taken for the town's population to reach 200000200000 people is found by solving 200000=103000e^{(\ln1.04)t}200000=103000e(ln1.04)t as follows:  200000200000 == 103000e^{(\ln1.04)t}103000e(ln1.04)t 1.941747571.94174757 == e^{(\ln1.04)t}e(ln1.04)t (\ln1.04)t(ln1.04)t == \ln1.94174757ln1.94174757 \therefore∴ tt == \frac{\ln1.94174757}{\ln1.04}ln1.94174757ln1.04​ == 16.9216.92 So if we add 1717 years to 2017 we get the year 2034 when the population reaches 200000200000. ##### Example 2 The half life of Radium-266266 is 16001600 years. Given the rate of radioactive decay is proportional to the amount of Radium-266266 present at any given moment, estimate the time it takes for an amount to decay to 10%10% of its initial quantity. Along similar lines as Example 1, we can construct the decay model as \frac{dQ}{dt}=-kQdQdt=kQ and solve this to reveal the law of decay as Q=Q_0e^{-kt}Q=Q0ekt where Q_0Q0 is the initial amount of radio-active substance and kk is a constant peculiar to Radium-226226. A half-life is the time taken for a quantity of a radio-active substance to reduce by exactly one half. This means in the case of Radium-226226 a quantity QQ will decay to \frac{Q}{2}Q2 in 16001600 years and again to \frac{Q}{4}Q4 in a further 16001600 years, and again to \frac{Q}{8}Q8 in a further 16001600 years etc. In other words, in absolute terms, the amount of decay is not constant but falls to a level one half of its present value every 16001600 years. The question posed is straightforward. We first need to determine kk by setting$$.

Thus:

 $\frac{Q_0}{2}$Q0​2​ $=$= $Q_0e^{-k\times1600}$Q0​e−k×1600 $\frac{1}{2}$12​ $=$= $e^{-k\times1600}$e−k×1600 $-\ln2$−ln2 $=$= $-1600k$−1600k $\therefore$∴  $k$k $=$= $\frac{\ln2}{1600}$ln21600​ $=$= $0.000433217$0.000433217

So $k\approx0.000433217$k0.000433217 and all that we need to do is to solve for $t$t in the equation $10%Q_0=Q_0e^{-kt}$10%Q0=Q0ekt.

 $0.1Q_0$0.1Q0​ $=$= $Q_0e^{-0.000433217t}$Q0​e−0.000433217t $0.1$0.1 $=$= $e^{-0.000433217t}$e−0.000433217t $\ln0.1$ln0.1 $=$= $-0.000433217t$−0.000433217t $\therefore$∴ $t$t $=$= $\frac{\ln0.1}{-0.000433217}$ln0.1−0.000433217​ $=$= $5315$5315

Hence it will take approximately $5315$5315 years to reduce to $10%$10% of the original quantity.

##### Example 3

Sometimes population growth becomes hampered by certain constraints placed upon it. Here is one such idealised scenario where a modified model is suggested. In the year 1900, $50$50 rabbits were naively released onto a certain South Pacific Island. By the year 1950 the rabbit population had grown to an estimated $1500$1500 rabbits and, understandably, environmentalists thought at that point the rabbits would cause a serious threat to the island's bio-diversity.

They believe however that the rabbit population will eventually become self-limiting reaching an upper limit of $10000$10000, due largely to a shortage of a sustainable food supply.

In an effort to estimate future growth in the population, they devise a modified population model where the rate of growth is proportional to both the number $R$R present at any given time and the 'amount of room' left to expand as identified by an upper population limit $L$L for the island . That is to say, the 'room left' would be measured by the quantity $L-R$LR

The upper limit $L$L would be based on things like the size of the island and the amount of food resource available. In the model, $L$L would be like a ceiling number of rabbits - a "constant" hypothesised by the researchers themselves.

They propose that the rate of growth $\frac{dR}{dt}$dRdt would be proportional to the product of both $R$R and $(L-R)$(LR). They reason as follows:

In the beginning, $R$R would have been very small and thus $L-R$LR would have been large, but the size of the product $R(L-R)$R(LR) would be relatively small. In other words, the rate of growth, in the beginning, would be fairly moderate.

By the time $R$R became very large, so that $(L-R)$(LR) became small, the product $R(L-R)$R(LR) would again be relatively small and again the rate of growth would also be moderate.

Somewhere near the middle when $R$R and $(L-R)$(LR) were approximately the same size, the product would be relatively large, so the rate of growth would be at a maximum.

To help you think about this think about the product of two numbers $p$p and $100-p$100p. When $p$p is small, $100-p$100p will be large, and so the product will be moderate. Try $2$2 and $98$98 for example - the product is $196$196. Also when $p$p is large, $100-p$100p will be small, and again the product will be moderate. But when $p$p and $100-p$100p are about the same size, the product will be very large - try $p=50$p=50 and $100-p=50$100p=50.   If these products represented growth rates then it becomes clear that the rate of growth is maximised in the middle.

Hence a reasonable model for population growth in this contained space would be given by:

$\frac{dR}{dt}=kR(L-R)$dRdt=kR(LR)

Of course, it is only a model, and it does contain a hypothesised population limit $L$L

Can we solve this first order differential equation and use it to estimate future population growth on the island?

The equation is separable, so the solution starts:

 $\frac{dR}{dt}$dRdt​ $=$= $kR(L-R)$kR(L−R) $\int\frac{1}{R(L-R)}dR$∫1R(L−R)​dR $=$= $\int kdt$∫kdt

Using the method of partial fractions we set $\frac{1}{R(L-R)}=\frac{A}{R}+\frac{B}{L-R}$1R(LR)=AR+BLR and multiply through to form the equation:

$1=A(L-R)+BR$1=A(LR)+BR

This is an identity and as such is true for any substitution instance. For $R=0$R=0, we have $A=\frac{1}{L}$A=1L and for $R=L$R=L we have $B=\frac{1}{L}$B=1L. As we proceed with the solution it is important to remember that the population limit $L$L is a constant. Thus we continue with the solution:

 $\int\frac{\frac{1}{L}}{R}+\frac{\frac{1}{L}}{L-R}dR$∫1L​R​+1L​L−R​dR $=$= $\int kdt$∫kdt $\frac{1}{L}\int\frac{1}{R}+\frac{1}{L-R}dR$1L​∫1R​+1L−R​dR $=$= $\int kdt$∫kdt $\frac{1}{L}[(\ln R-\ln(L-R)]$1L​[(lnR−ln(L−R)] $=$= $kt+C$kt+C $\ln[\frac{R}{L-R}]$ln[RL−R​] $=$= $L(kt+C)$L(kt+C) $\frac{R}{L-R}$RL−R​ $=$= $e^{Lkt}\times e^{LC}$eLkt×eLC $\frac{R}{L-R}$RL−R​ $=$= $Ae^{bt}$Aebt

From here we work to isolate our function $R$R. We can simplify the working by reducing products and sums of constants to single letters. Note that $A$A replaced $e^{LC}$eLC and $b$b replaced $Lk$Lk in the above working.

Continuing:

 $R$R $=$= $LAe^{bt}-RAe^{bt}$LAebt−RAebt $R(1+Ae^{bt})$R(1+Aebt) $=$= $LAe^{bt}$LAebt $\therefore$∴  $R$R $=$= $\frac{LAe^{bt}}{1+Ae^{bt}}$LAebt1+Aebt​ $R$R $=$= $\frac{L}{1+qe^{-bt}}$L1+qe−bt​

Note that the last line was achieved by dividing the numerator and denominator by $Ae^{bt}$Aebt and in the process created $q$q to replace $\frac{1}{A}$1A.

From th information provided,  $50$50 rabbits were introduced to the island in 1900 (t=0). Assuming the island can carry no more than $L=10000$L=10000 rabbits, we have:

 $50$50 $=$= $\frac{10000}{1+q}$100001+q​ $50+50q$50+50q $=$= $10000$10000 $50q$50q $=$= $9950$9950 $\therefore$∴  $q$q $=$= $199$199 $\therefore$∴  $R$R $=$= $\frac{10000}{1+199e^{-bt}}$100001+199e−bt​

In 1950 ($t=50$t=50) they had estimated the population as $1500$1500 rabbits on the island. We can use that information to find the constant $b$b

 $1500$1500 $=$= $\frac{10000}{1+199e^{-50b}}$100001+199e−50b​ $1+199e^{-50b}$1+199e−50b $=$= $\frac{20}{3}$203​ $199e^{-50b}$199e−50b $=$= $\frac{17}{3}$173​ $e^{-50b}$e−50b $=$= $0.0284757$0.0284757 $-50b$−50b $=$= $-3.558704$−3.558704 $\therefore$∴  $b$b $=$= $0.0711741$0.0711741

Hence our complete model is given by $R=\frac{10000}{1+199e^{-0.0711741t}}$R=100001+199e0.0711741t

Here is the graph: Note the classic shape of the modified exponential growth curve - the $S$S shape or Sigmoid curve. As predicted it starts rising slowly, but then reaches a maximum rate of change around the year numbered $t=75$t=75, after which the gradient reduces in size as it nears the theoretical limit of 10000 rabbits.

Let's now predict the number of rabbits in 2020. That's $t=120$t=120 and from the graph, the population looks to be about $9600$9600. The model more accurately predicts $R=\frac{10000}{1+199e^{-0.0711741t}}=9626$R=100001+199e0.0711741t=9626 rabbits.

##### Example 4 (The Tractrix - an extension question)

The tractrix or the curve of pursuit gets its name from the Latin verb trahere meaning to pull or drag. We will examine this curve using the idealised scenario of a horse pulling along a stump.

Imagine a work horse harnessed by a rope  of length $a$a units which is attached to a heavy stump lying on the ground. At the beginning, the rope is taught and remains so for the duration. The horse begins to move at right angles to the direction of the rope, and of course, the stump moves with ground resistance along a curved path as shown here: Suppose the situation is superimposed onto the cartesian plane. Suppose we initially locate the stump at $S_0$S0 with coordinates $(a,0)$(a,0) and locate the horse at the origin. After any time $t$t, as the horse moves at right angles along the $y$y axis, the stump will shift to a variable point $S(x,y)$S(x,y).

The rope will always remain tangent to the "hitherto unknown" path $y(x)$y(x) of the stump as shown in the diagram, and hence the gradient of that tangent will be given by $\frac{dy}{dx}$dydx.

But there is another way we can measure the gradient of the rope.

Because the rope is of length $a$a, and the stump is at $(x,y)$(x,y), then we know by Pythagoras' Theorem, that the horse's rectangular displacement from the stump will be $\sqrt{a^2-x^2}$a2x2 The gradient of the rope (note that it is negative) is therefore given by:

$\frac{dy}{dx}=-\frac{\sqrt{a^2-x^2}}{x}$dydx=a2x2x.

This is a first order differential equation of the form $\frac{dy}{dx}=f(x)$dydx=f(x)

The derivative $\frac{dy}{dx}$dydx is algebraically difficult to determine, and goes beyond the level expected in most courses.   We will omit the details here and just state that the function $y(x)$y(x) is given by:

$y=a\ln[\frac{a+\sqrt{a^2-x^2}}{x}]-\sqrt{a^2-x^2}+C$y=aln[a+a2x2x]a2x2+C

Note that for $x=a,y=a\ln[\frac{a+\sqrt{a^2-a^2}}{a}]-\sqrt{a^2-a^2}+C=C$x=a,y=aln[a+a2a2a]a2a2+C=C, and since we started at the point $S(a,0)$S(a,0) we need $C$C to be equal to $0$0.

The curve for $a=10$a=10 is shown here: Imagine the horse beginning to pull the weight of the stump. Imagine how the stump would begin turning as if it were pursuing the horse. This is a wonderful example of how maths can be harnessed to explain physical phenomena.

### Outcomes

#### M8-12

Form differential equations and interpret the solutions

#### 91579

Apply integration methods in solving problems