Find the equation of the regression line of $h$ on $t$.

Interpret the meaning of parameter $a$ in the context of the model.

Suggest why Eva’s use of the linear regression equation in this way could be unreliable.

Find the equation of the least squares quadratic regression curve.

Find the value of $k$.

Hence, write down a suitable domain for Eva’s function $h\left(t\right)=p{t}^2+qt+r$.

Show that $\frac{dh}{dt}=-R^2\sqrt{70 560h}$.

By solving the differential equation $\frac{dh}{dt}=-R^2\sqrt{70 560h}$, show that the general solution is given by $h=17 640{\left(c-R^{2}t\right)}^2$, where $c\in \mathrm{ℝ}$.

Use the general solution from part (d) and the initial condition $h\left(0\right)=3.2$ to predict the value of $T$.

Find this new height.

Show that $\frac{dH}{dt}\approx 0.2514-0.009873t-0.1405\sqrt{H}$, where $0\le t\le T$.

Use Euler’s method with a step length of $0.5$ minutes to estimate the maximum value of $H$.

$h\left(t\right)=-0.134t+3.1$           A1A1

Note: Award A1 for an equation in $h$ and $t$ and A1 for the coefficient $-0.134$ and constant $3.1$.

[2 marks]

EITHER

the rate of change of height (of water in metres per minute)           A1

Note: Accept “rate of decrease” or “rate of increase” in place of “rate of change”.

OR

the (average) amount that the height (of the water) decreases each minute           A1

[1 mark]

EITHER

unreliable to use $h$ on $t$ equation to estimate $t$        A1

OR

unreliable to extrapolate from original data        A1

OR

rate of change (of height) might not remain constant (as the water drains out)      A1

[1 mark]

$h\left(t\right)=0.002t^2-0.174t+3.2$        A1

[1 mark]

$0.002t^2-0.174t+3.2=0$         (M1)

$26.4 \left(26.4046\dots \right)$        A1

[2 marks]

EITHER

$\left(0\le \right) t\le 26.4 \left(t\le 26.4046\dots \right)$           A1

OR

$\left(0\le \right) t\le 20$ (due to range of original data / interpolation)           A1

[1 mark]

$V=\mathrm{\pi }{\left(1\right)}^{2}h$               (A1)

EITHER

$\frac{dV}{d\text{t}}=\mathrm{\pi }\frac{dh}{d\text{t}}$             M1

OR

attempt to use chain rule             M1

$\frac{dh}{dt}=\frac{dh}{dV}\times \frac{dV}{dt}$

THEN

$\frac{dh}{dt}=\frac{1}{\mathrm{\pi }}\times -\mathrm{\pi }{R}^2\sqrt{70 560h}$           A1

$\frac{dh}{dt}=-R^2\sqrt{70 560h}$           AG

[3 marks]

attempt to separate variables             M1

$\int \frac{1}{\sqrt{70 560h}}dh=\int -R^{2}dt$           A1

$\frac{2\sqrt{h}}{\sqrt{70 560}}=-R^{2}t+c$           A1A1

Note: Award A1 for each correct side of the equation.

$\sqrt{h}=\frac{\sqrt{70 560}}{2}\left(c-R^{2}t\right)$           A1

Note: Award the final A1 for any correct intermediate step that clearly leads to the given equation.

$h=17 640{\left(c-R^{2}t\right)}^2$          AG

[5 marks]

$t=0 \Rightarrow 3.2=17640c^2$               (M1)

$c=0.0134687\dots$               (A1)

substituting $h=0$ and their non-zero value of $c$               (M1)

$T=\frac{c}{R^2}=\frac{0.0134687\dots }{0.{023}^2}$

$=25.5$ (minutes)  $\left(25.4606\dots \right)$           A1

[4 marks]

$h=0 \Rightarrow c=R^{2}t$

$c=0.{023}^2\times 15 \left(=0.007935\right)$              (A1)

$t=0 \Rightarrow h=17640{\left(0.{023}^2\times 15\right)}^2$               (M1)

$h=1.11$ (metres)  $\left(1.11068\dots \right)$           A1

[3 marks]

let $h$ be the height of water in the highest container from parts (d) and (e) we get

$\frac{dh}{dt}=-35280R^2\left(0.0134687\dots -R^{2}t\right)$               (M1)(A1)

so $\frac{dH}{dt}=35280R^2\left(0.0135-R^{2}t\right)-R^2\sqrt{70560H}$               M1A1

$\left(\frac{dH}{dt}=18.6631\dots \left(0.0134687\dots -0.000529t\right)-0.000529\sqrt{70560H}\right)$

$\left(\frac{dH}{dt}=0.251367\dots -0.0987279\dots -0.140518\dots \sqrt{H}\right)$

$\frac{dH}{dt}\approx 0.2514-0.009873t-0.1405\sqrt{H}$           AG

[4 marks]

evidence of using Euler’s method correctly

e.g. $y_1=1.05545\dots$               (A1)

maximum value of $H=1.45$ (metres) (at $8.5$ minutes)             A2

($1.44678\dots$ metres)

[3 marks]

All parts were answered well. In part(a)(i) a few candidates lost a mark from either not writing an equation or not using the variables $h$ and $t$. In part (a)(ii) some candidates incorrectly stated it was the rate of change of water, instead of the rate of change of the height of the water. A few weaker candidates simply stated it is the gradient of the line. In part (a)(iii) some candidates incorrectly criticized the linear model, instead of addressing the question about why it could be unreliable to use the model to make a prediction about the future.

All parts were answered well. In part(a)(i) a few candidates lost a mark from either not writing an equation or not using the variables $h$ and $t$. In part (a)(ii) some candidates incorrectly stated it was the rate of change of water, instead of the rate of change of the height of the water. A few weaker candidates simply stated it is the gradient of the line. In part (a)(iii) some candidates incorrectly criticized the linear model, instead of addressing the question about why it could be unreliable to use the model to make a prediction about the future.

All parts were answered well. In part(a)(i) a few candidates lost a mark from either not writing an equation or not using the variables $h$ and $t$. In part (a)(ii) some candidates incorrectly stated it was the rate of change of water, instead of the rate of change of the height of the water. A few weaker candidates simply stated it is the gradient of the line. In part (a)(iii) some candidates incorrectly criticized the linear model, instead of addressing the question about why it could be unreliable to use the model to make a prediction about the future.

This question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for $k$, showing a lack of understanding of the context of the model.

This question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for $k$, showing a lack of understanding of the context of the model.

This question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for $k$, showing a lack of understanding of the context of the model.

Many candidates recognized the need to use related rates of change, but could not present coherent working to reach the given answer. Often candidates either did not appreciate the need to use the equation for the volume of a cylinder or did not simplify their equation using $\mathrm{r}=1$. Many candidates wrote nonsense arguments trying to cancel the factor of $\mathrm{\pi }$. In these long paper 3 questions, the purpose of “show that” parts is often to enable candidates to re-enter a question if they are unable to do a previous part.

Many candidates were able to correctly separate the variables, but many found the integral of $\frac{1}{\sqrt{h}}$ to be too difficult. A common incorrect approach was to use logarithms. A surprising number also incorrectly wrote $\int -R^2\mathrm{d}t=-\frac{R^3}{3}+c$, showing a lack of understanding of the difference between a parameter and a variable. Given that most questions in this course will be set in context, it is important that candidates learn to distinguish these differences.

Generally done well.

Many candidates found this question too difficult.

Part (g)(i) was often left blank and was the worst answered question on the paper. Part (g)(ii) was answered correctly by a number of candidates, who made use of the given answer from part (g)(i).

Part (g)(i) was often left blank and was the worst answered question on the paper. Part (g)(ii) was answered correctly by a number of candidates, who made use of the given answer from part (g)(i).

$\gamma MS$

$\delta S$

show that, at the equilibrium point, the value of the mackerel population density is $\frac{\delta }{\gamma }$;

find the value of the shark population density at the equilibrium point.

Toxic sewage is added to the Mediterranean Sea. Alessia claims this reduces the shark population growth rate and hence the value of $\gamma$ is halved. No other parameter changes.

Global warming increases the temperature of the Mediterranean Sea. Alessia claims that this promotes the mackerel population growth rate and hence the value of $\alpha$ is doubled. No other parameter changes.

Write down the differential equation for $M$ that models this situation.

Show that the expression for the mackerel population density after $t$ years is $M=M_0{\text{e}}^{\alpha t}$

Alessia estimates that the mackerel population density increases by a factor of three every two years. Show that $\alpha =0.549$ to three significant figures.

Write down expressions for $M_{n+1}$ and $S_{n+1}$ in terms of $M_n$ and $S_n$.

Use Euler’s method to find an estimate for the mackerel population density after one year.

Use Euler’s method to sketch the trajectory of the phase portrait, for $4\le M\le 7$ and $1.5\le S\le 3$, over the first nine years.

Using your phase portrait, or otherwise, determine whether the mackerel population density would be sufficient to support sustainable fishing during the first nine years.

State two reasons why Alessia’s conclusion, found in part (f)(ii), might not be valid.

population growth rate / birth rate of sharks (due to eating mackerel)            A1

[1 mark]

(net) death rate of sharks          A1

[1 mark]

$\gamma MS-\delta S=0$          A1

since $S\ne 0$          R1

Note: Accept $S>0$.

getting to given answer without further error by either cancelling or factorizing           A1

$M=\frac{\delta }{\gamma }$           AG

[3 marks]

$\frac{dM}{dt}=0$

$\alpha M-\beta MS=0$          (M1)

(since $M\ne 0$)  $S=\frac{\alpha }{\beta }$          A1

[2 marks]

$M_{eq}=\frac{\delta }{\gamma }\Rightarrow \frac{\delta }{{\displaystyle \frac{1}{2}}\gamma }=2M_{eq}$                      M1

Note: Accept equivalent in words.

Doubles          A1

Note: Do not accept “increases”.

[2 marks]

$M_{eq}=\frac{\delta }{\gamma }$ is not dependent on $\alpha$                      R1

Note: Award R0 for any contextual argument.

no change         A1

Note: Do not award R0A1.

[2 marks]

$\frac{dM}{dt}=\alpha M$                    A1

[1 mark]

$\int \frac{1}{M}dM=\int \alpha \text{d}t$                  M1

Note: Award M1 is for an attempt to separate variables. This means getting to the point $\int f\left(M\right) dM=\int g\left(t\right) \text{d}t$ where the integral can be seen or implied by further work.

$\ln \left|M\right|=\alpha t+c$                  A1

Note: Accept $\ln M$. Condone missing constant of integration for this mark.

$M=k{\text{e}}^{\alpha t}$

when $t=0, M_0=k$                 M1

Note: Award M1 for a clear attempt at using initial conditions to find a constant of integration. Only possible if the constant of integration exists. $t=0$ or “initially” or similar must be seen. Substitution may appear earlier, following the integration.

initial conditions and all other manipulations correct and clearly communicated to get to the final answer                   A1

$M=M_0{\text{e}}^{\alpha t}$                  AG

[4 marks]

$M=3M_0$ seen anywhere                (A1)

substituting $t=2, M=3M_0$ into equation $M=M_0{\text{e}}^{\alpha t}$                (M1)

$3M_0=M_0{\text{e}}^{2\alpha }$

$\alpha =\frac{1}{2}\ln 3$  OR  $0.549306\dots$                  A1

Note: The A1 requires either the exact answer or an answer to at least $4$ sf.

$\approx 0.549$                 AG

[3 marks]

an attempt to set up one recursive equation                (M1)

Note: Must include two given parameters and $M_n$ and $S_n$ and $M_{n+1}$ or $S_{n+1}$ for the (M1) to be awarded.

$M_{n+1}=M_n+0.1\left(0.549M_n-0.236M_n{S}_n\right)$                 A1

$S_{n+1}=S_n+0.1\left(0.244M_n{S}_n-1.39S_n\right)$                 A1

[3 marks]

EITHER
$6.12 (6.11609\dots )$             A2

OR
$6120 (6116.09\dots )$ (mackerel per ${\text{km}}^3$)             A2

[2 marks]

spiral or closed loop shape         A1

approximately $1.25$ rotations (can only be awarded if a spiral)         A1

correct shape, in approximately correct position (centred at approx. $(5.5, 2.5)$)         A1

Note: Award A0A0A0 for any plot of $S$ or $M$ against $t$.

[3 marks]

EITHER

approximate minimum is $(5.07223\dots ) 5.07$ (which is greater than $5$)           A1

OR

the line $M=5$ clearly labelled on their phase portrait           A1

THEN

(the density will not fall below $5000$) hence sufficient for sustainable fishing           A1


Note: Do not award A0A1. Only if the minimum point is labelled on the sketch then a statement here that “the mackerel population is always above $\mathit{5000}$” would be sufficient. Accept the value $5.07$ seen within a table of values.

[2 marks]

Any two from:                           A1A1

• Current values / parameters are only an estimate,

• The Euler method is only an approximate method / choosing $h=0.1$ might be too large.

• There might be random variation / the model has no stochastic component

• Conditions / parameters might change over the nine years,

• A discrete system is being approximated by a continuous system,

Allow any other sensible critique.

If a candidate identifies factors which the model ignores, award A1 per factor identified. These factors could include:

• Other predators

• Seasonality

• Temperature

• The effect of fishing

• Environmental catastrophe

• Migration

Note: Do not allow:
             “You cannot have $5.07$ mackerel”.
             It is only a model (as this is too vague).
             Some factors have been ignored (without specifically identifying the factors).
             Values do not always follow the equation / model. (as this is too vague).

[2 marks]

Use the trapezoidal rule to find an estimate for the area.

With reference to the shape of the graph, explain whether your answer to part (a)(i) will be an over-estimate or an underestimate of the area.

Use all the coordinates in the table to find the equation of the least squares cubic regression curve.

Write down the coefficient of determination.

Write down an expression for the area enclosed by the cubic function, the $x$-axis, the $y$-axis and the line $x=4.4$.

Find the value of this area.

Show that $\text{ln}y=qx+\text{ln}p$.

Hence explain how a straight line graph could be drawn using the coordinates in the table.

By finding the equation of a suitable regression line, show that $p=1.83$ and $q=0.986$.

Hence find the area enclosed by the exponential function, the $x$-axis, the $y$-axis and the line $x=4.4$.

Area $=\frac{1.1}{2}\left(2+2\left(5+15+47\right)+148\right)$         M1A1

Area = 156 units²          A1

[3 marks]

The graph is concave up,         R1

so the trapezoidal rule will give an overestimate.         A1

[2 marks]

$f\left(x\right)=3.88x^3-12.8x^2+14.1x+1.54$         M1A2

[3 marks]

$R^2=0.999$        A1

[1 mark]

Area $=\underset{0}{\overset{4.4}{\int }}\left(3.88x^3-12.8x^2+14.1x+1.54\right)dx$        A1A1

[2 marks]

Area = 145 units²    (Condone 143–145 units², using rounded values.)      A2

[2 marks]

$\text{ln}y=\text{ln}\left(p{\text{e}}^{qx}\right)$      M1

$\text{ln}y=\text{ln}p+\text{ln}\left({\text{e}}^{qx}\right)$      A1

$\text{ln}y=qx+\text{ln}p$      AG

[2 marks]

Plot $\text{ln}y$ against $p$.      R1

[1 mark]

Regression line is $\text{ln}y=0.986x+0.602$       M1A1

So $q=$ gradient = 0.986    R1

$p=e^{0.602}=1.83$       M1A1

[5 marks]

Area $=\underset{0}{\overset{4.4}{\int }}1.83e^{0.986x}dx=140$ units²     M1A1

[2 marks]

Find the equilibrium point for this system.

If initially $x=100$ and $y=50$ use Euler’s method with an time increment of 0.1 to find an approximation for the values of $x$ and $y$ when $t=1$.

Extend this method to conjecture the limit of the ratio $\frac{y}{x}$ as $t\to \mathrm{\infty }$.

Show how using the substitution $X=x-10\text{,}Y=y-20$ transforms the system of differential equations into $\begin{array}{c}\dot{X}=X+2Y\\ \dot{Y}=2X+Y\end{array}$.

Solve this system of equations by the eigenvalue method and hence find the general solution for $\left(\begin{array}{c}x\\ y\end{array}\right)$ of the original system.

Find the particular solution to the original system, given the initial conditions of part (b).

Hence find the exact values of $x$ and $y$ when $t=1$, giving the answers to 4 significant figures.

Use part (f) to find limit of the ratio $\frac{y}{x}$ as $t\to \mathrm{\infty }$.

With the initial conditions as given in part (b) state if the equilibrium point is stable or unstable.

If instead the initial conditions were given as $x=20$ and $y=10$, find the particular solution for $\left(\begin{array}{c}x\\ y\end{array}\right)$ of the original system, in this case.

With the initial conditions as given in part (j), determine if the equilibrium point is stable or unstable.

$\begin{array}{c}\dot{x}=0\Rightarrow x+2y-50=0\\ \dot{y}=0\Rightarrow 2x+y-40=0\end{array}$     $\Rightarrow x=10\text{,}y=20$      M1A1

[2 marks]

Using $\begin{array}{c}{x}_{n+1}=x_n+0.1\left(x_n+2y_n-50\right)\\ y_{n+1}=y_n+0.1\left(2x_n+y_n-40\right)\\ t_{n+1}=t_n+0.1\end{array}$

Gives $x\left(1\right)\simeq 848\text{,}y\left(1\right)\simeq 837\left(3sf\right)$      M1A1A1

[3 marks]

By extending the table, conjecture that $\underset{t\to \mathrm{\infty }}{\text{lim}}\frac{y}{x}=1$      M1A1

[2 marks]

$X=x-10\text{,}Y=y-20\Rightarrow \dot{X}=\dot{x}\text{,}\dot{Y}=\dot{y}$      R1

$\begin{array}{c}\dot{X}=\left(X+10\right)+2\left(Y+20\right)-50=X+2Y\\ \dot{Y}=2\left(X+10\right)+\left(Y+20\right)-40=2X+Y\end{array}$     M1A1AG

[3 marks]

     M1A1A1

$\lambda =-1$         an eigenvector is $\left(\begin{array}{c}1\\ -1\end{array}\right)$

$\lambda =3$           an eigenvector is $\left(\begin{array}{c}1\\ 1\end{array}\right)$     M1A1A1

$\left(\begin{array}{c}X\\ Y\end{array}\right)=A{e}^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+B{e}^{3t}\left(\begin{array}{c}1\\ 1\end{array}\right)\Rightarrow \left(\begin{array}{c}x\\ y\end{array}\right)=A{e}^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+B{e}^{3t}\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}10\\ 20\end{array}\right)$      A1A1

[8 marks]

$\begin{array}{c}100=A+B+10\\ 50=-A+B+20\end{array}\Rightarrow A=30\text{,}B=60$     M1

$\left(\begin{array}{c}x\\ y\end{array}\right)=30e^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+60e^{3t}\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}10\\ 20\end{array}\right)$      A1

[2 marks]

$x\left(1\right)=1226\text{,}y\left(1\right)=1214\left(4sf\right)$     A1A1

[2 marks]

Dominant term is $60e^{3t}\left(\begin{array}{c}1\\ 1\end{array}\right)$ so $\underset{t\to \mathrm{\infty }}{\text{lim}}\frac{y}{x}=1$    M1A1

[2 marks]

The equilibrium point is unstable.               R1

[1 mark]

$\begin{array}{c}20=A+B+10\\ 10=-A+B+20\end{array}\Rightarrow A=10\text{,}B=0$            M1

$\left(\begin{array}{c}x\\ y\end{array}\right)=10e^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+\left(\begin{array}{c}10\\ 20\end{array}\right)$             A1

[2 marks]

As $e^{-t}\to 0$ as $t\to \mathrm{\infty }$ the equilibrium point is stable.           R1A1

[2 marks]

Show that the matrix  has (sadly) only one eigenvalue.  Find this eigenvalue and an associated eigenvector.

Hence, verify that $\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t}$ is a solution to the above system.

Verify that $\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}t\\ -t+1\end{array}\right)e^{2t}$ is also a solution.

If initially at $t=0\text{,}x=20\text{,}y=10$ find the particular solution.

Find the values of $x$ and $y$ when $t=2$.

 Find the equation of this asymptote.

State the direction of the trajectory, including the quadrant it is in as it approaches this asymptote.

      M1A1

${\lambda }^2-4\lambda +4=0\Rightarrow {\left(\lambda -2\right)}^2=0$      A1A1

So only one solution    $\lambda =2$      AGA1

      M1

So an eigenvector is $\left(\begin{array}{c}1\\ -1\end{array}\right)$      A1

[7 marks]

So             M1A1A1

and $\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t}\Rightarrow \left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)=\left(\begin{array}{c}1\\ -1\end{array}\right)2e^{2t}$      M1A1

showing that $\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t}$ is a solution      AG

[5 marks]

$\left(\begin{array}{c}3x+y\\ -x+y\end{array}\right)=\left(\begin{array}{c}3t-t+1\\ -t-t+1\end{array}\right)e^{2t}=\left(\begin{array}{c}2t+1\\ -2t+1\end{array}\right)e^{2t}$          M1A1

$\left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)=\left(\begin{array}{c}{e}^{2t}+t2e^{2t}\\ -e^{2t}+(-t+1)2e^{2t}\end{array}\right)=\left(\begin{array}{c}2t+1\\ -2t+1\end{array}\right)e^{2t}$      M1A1A1

Verifying that $\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}t\\ -t+1\end{array}\right)e^{2t}$ is also a solution      AG

[5 marks]

Require $\left(\begin{array}{c}20\\ 10\end{array}\right)=A\left(\begin{array}{c}1\\ -1\end{array}\right)+B\left(\begin{array}{c}0\\ 1\end{array}\right)\Rightarrow A=20\text{,}B=30$           M1A1

$\left(\begin{array}{c}x\\ y\end{array}\right)=20\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t}+30\left(\begin{array}{c}t\\ -t+1\end{array}\right)e^{2t}$    A1

[3 marks]

$t=2\Rightarrow x=4370\text{,}y=-2730(3sf)$      A1A1

[2 marks]

As $t\to \mathrm{\infty }\text{,}x\simeq 30t{e}^{2t}\text{,}y\simeq -30t{e}^{2t}$      M1A1

so asymptote is $y=-x$      A1

[3 marks]

Will approach the asymptote in the 4th quadrant, moving away from the origin.      R1

[1 mark]

Find the equation of the regression line of $Q\left(t\right)$ on $t$.

Write down the value of $r$, Pearson’s product-moment correlation coefficient.

Explain why it would not be appropriate to conduct a hypothesis test on the value of $r$ found in (a)(ii).

Find the general solution of the differential equation $Q\text{'}\left(t\right)=\beta NQ\left(t\right)$.

Using the data in the table write down the equation for an appropriate non-linear regression model.

Write down the value of $R^2$ for this model.

Hence comment on the suitability of the model from (b)(ii) in comparison with the linear model found in part (a).

By considering large values of $t$ write down one criticism of the model found in (b)(ii).

Use your answer from part (b)(ii) to estimate the time taken for the number of infected computers to double.

Find in which city, X or Y, the computer virus is spreading more easily. Justify your answer using your results from part (b).

Determine the value of $a$ and of $b$. Give your answers correct to one decimal place.

Use linear regression to estimate the value of $k$ and of $L$.

The solution to the differential equation is given by

$Q\left(t\right)=\frac{L}{1+C{\text{e}}^{-kt}}$

where $C$ is a constant.

Using your answer to part (f)(i), estimate the percentage of computers in city X that are expected to have been infected by the virus over a long period of time.

$Q\left(t\right)=3090t-54000 \left(3094.27\dots t-54042.3\dots \right)$         A1A1

Note: Award at most A1A0 if answer is not an equation. Award A1A0 for an answer including either $x$ or $y$.

[2 marks]

$0.755 \left(0.754741\dots \right)$         A1

[1 mark]

$t$ is not a random variable OR it is not a (bivariate) normal distribution

OR data is not a sample from a population

OR data appears nonlinear

OR $r$ only measures linear correlation         R1

Note: Do not accept “$r$ is not large enough”.

[1 mark]

attempt to separate variables            (M1)

$\int \frac{1}{Q}dQ=\int \beta N dt$

$\ln \left|Q\right|=\beta Nt+c$           A1A1A1 

Note: Award A1 for LHS, A1 for $\beta Nt$, and A1 for $+c$.

Award full marks for $Q={\text{e}}^{\beta Nt+c}$  OR  $Q=A{\text{e}}^{\beta Nt}$.

Award M1A1A1A0 for $Q={\text{e}}^{\beta Nt}$

[4 marks]

attempt at exponential regression           (M1)

$Q=1.15{\text{e}}^{0.292t} \left(Q=1.14864\dots {\text{e}}^{0.292055\dots t}\right)$           A1

OR

attempt at exponential regression           (M1)

$Q=1.15\times 1.{34}^t \left(1.14864\dots \times 1.33917{\dots }^t\right)$           A1

Note: Condone answers involving $y$ or $x$. Condone absence of “$Q=$” Award M1A0 for an incorrect answer in correct format.

[2 marks]

$0.999 \left(0.999431\dots \right)$          A1

[1 mark]

comparing something to do with $R^2$ and something to do with $r$        M1

Note:   Examples of where the M1 should be awarded:

$R^2>r$
$R>r$
$0.999>0.755$
$0.999>0.{755}^2 \left(=0.563\right)$
The “correlation coefficient” in the exponential model is larger.
Model B has a larger $R^2$

Examples of where the M1 should not be awarded:

The exponential model shows better correlation (since not clear how it is being measured)
Model 2 has a better fit
Model 2 is more correlated

an unambiguous comparison between $R^2$ and $r^2$ or $R$ and $r$ leading to the conclusion that the model in part (b) is more suitable / better          A1

Note: Condone candidates claiming that $R$ is the “correlation coefficient” for the non-linear model.

[2 marks]

it suggests that there will be more infected computers than the entire population       R1

Note: Accept any response that recognizes unlimited growth. 

[1 mark]

$1.15{\text{e}}^{0.292t}=2.3$  OR  $1.15\times 1.{34}^t=2.3$  OR  $t=\frac{\ln 2}{0.292}$  OR using the model to find two specific times with values of $Q\left(t\right)$ which double          M1

$t=2.37$  (days)          A1

Note: Do not FT from a model which is not exponential. Award M0A0 for an answer of $2.13$ which comes from using $(10, 20)$ from the data or any other answer which finds a doubling time from figures given in the table.

[2 marks]

an attempt to calculate $\beta$ for city X          (M1)

$\beta =\frac{0.292055\dots }{2.6\times {10}^6}$  OR  $\beta =\frac{\ln 1.33917\dots }{2.6\times {10}^6}$

$=1.12328\dots \times {10}^{-7}$          A1

this is larger than $9.64\times {10}^{-8}$ so the virus spreads more easily in city X         R1

Note: It is possible to award M1A0R1.
Condone “so the virus spreads faster in city X” for the final R1.

[3 marks]

$a=38.3, b=3086.1$          A1A1

Note: Award A1A0 if values are correct but not to $1$ dp.

[2 marks]

$\frac{Q\text{'}}{Q}=0.42228-2.5561\times {10}^{-6}Q$          (A1)(A1)

Note: Award A1 for each coefficient seen – not necessarily in the equation. Do not penalize seeing in the context of $y$ and $x$.

identifying that the constant is $k$ OR that the gradient is $-\frac{k}{L}$          (M1)

therefore $k=0.422 \left(0.422228\dots \right)$          A1

$\frac{k}{L}=2.5561\times {10}^{-6}$

$L=165000 \left(165205\right)$          A1

Note: Accept a value of $L$ of $164843$ from use of $3$ sf value of $k$, or any other value from plausible pre-rounding.
Allow follow-through within the question part, from the equation of their line to the final two A1 marks.

[5 marks]

recognizing that their $L$ is the eventual number of infected        (M1)

$\frac{165205\dots }{2600000}=6.35\% \left(6.35403\dots \%\right)$          A1

Note: Accept any final answer consistent with their answer to part (f)(i) unless their $L$ is less than $120146$ in which case award at most M1A0.

[2 marks]

A significant minority were unable to attempt 1(a) which suggests poor preparation for the use of the GDC in this statistics-heavy course. Large numbers of candidates appeared to use $y, x, Q$ and $t$ interchangeably. Accurate use of notation is an important skill which needs to be developed.

1(a)(iii) was a question at the heart of the Applications and interpretations course. In modern statistics many of the calculations are done by a computer so the skill of the modern statistician lies in knowing which tests are appropriate and how to interpret the results. Very few candidates seemed familiar with the assumptions required for the use of the standard test on the correlation coefficient. Indeed, many candidates answered this by claiming that the value was either too large or too small to do a hypothesis test, indicating a major misunderstanding of the purpose of hypothesis tests.

1(b)(i) was done very poorly. It seems that perhaps adding parameters to the equation confused many candidates – if the equation had been $Q\text{'}\left(t\right)=5Q\left(t\right)$ many more would have successfully attempted this. However, the presence of parameters is a fundamental part of mathematical modelling so candidates should practise working with expressions involving them.

1(b)(ii) and (iii) were done relatively well, with many candidates using the data to recognize an exponential model was a good idea. Part (iv) was often communicated poorly. Many candidates might have done the right thing in their heads but just writing that the correlation was better did not show which figures were being compared. Many candidates who did write down the numbers made it clear that they were comparing an $R^2$ value with an $r$ value.

1(c) was not meant to be such a hard question. There is a standard formula for half-life which candidates were expected to adapt. However, large numbers of candidates conflated the data and the model, finding the time for one of the data points (which did not lie on the model curve) to double. Candidates also thought that the value of t found was equivalent to the doubling time, often giving answers of around 40 days which should have been obviously wrong.

1(d) was quite tough. Several candidates realized that $\beta$ was the required quantity to be compared but very few could calculate $\beta$ for city X using the given information.

1(e) was meant to be relatively straightforward but many candidates were unable to interpret the notation given to do the quite straightforward calculation.

1(f) was meant to be a more unusual problem-solving question getting candidates to think about ways of linearizing a non-linear problem. This proved too much for nearly all candidates.

Show that if the system has two distinct real eigenvalues then ${\left(a-d\right)}^2+4bc>0$.

Find two conditions that must be satisfied by $a$, $b$, $c$, $d$.

Explain why $b$ and $c$ must have opposite signs.

In the case when there is a pair of purely imaginary eigenvalues you are told that the solution will form an ellipse.  You are also told that the initial conditions are such that the ellipse is large enough that it will cross both the positive and negative $x$ axes and the positive and negative $y$ axes.

By considering the differential equations at these four crossing point investigate if the trajectory is in a clockwise or anticlockwise direction round the ellipse. Give your decision in terms of $b$ and $c$. Using part (b) (ii) show that your conclusions are consistent.

The characteristic equation is given by

      M1A1A1

$\lambda =\frac{a+d\pm \sqrt{{\left(a+d\right)}^2-4\left(ad-bc\right)}}{2}$

For two distinct real roots require ${\left(a+d\right)}^2-4\left(ad-bc\right)>0$       R1

$\Rightarrow a^2+2ad+d^2-4ad+4bc>0\Rightarrow a^2-2ad+d^2+4bc>0$      A1A1

$\Rightarrow {\left(a-d\right)}^2+4bc>0$        AG

[6 marks]

Using the working from part (a) (or using the characteristic equation) for a pair of purely imaginary eigenvalues require

$a+d=0$ and     R1A1A1

$\Rightarrow d=-a$ and       A1A1

[5 marks]

so $b$ and $c$ must have opposite signs      M1AG

[1 mark]

When crossing the $x$ axes, $y=0$ so $\dot{y}=cx$      M1A1

When crossing the positive $x$ axes, $\dot{y}$ has the sign of $c$.       A1

When crossing the negative $x$ axes, $\dot{y}$ has the sign of $-c$.       A1

Hence if $c$ is positive the trajectory is anticlockwise and if $c$ is negative the trajectory is clockwise.       R1R1

When crossing the $y$ axes, $x=0$ so $\dot{x}=by$      M1A1

When crossing the positive $y$ axis, $\dot{x}$ has the sign of $b$.       A1

When crossing the negative $y$ axes, $\dot{x}$ has the sign of $-b$.       A1

Hence if $b$ is positive the trajectory is clockwise and if $b$ is negative the trajectory is anticlockwise.       R1R1

Since by (b)(ii), $b$ and $c$ have opposite signs the above conditions agree with each other.       R1

[13 marks]

Find the value of $\underset{4}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x^3}\text{d}x$.

Illustrate graphically the inequality .

Hence write down a lower bound for $\sum _{n=4}^{\mathrm{\infty }}\frac{1}{n^3}$.

Find an upper bound for $\sum _{n=4}^{\mathrm{\infty }}\frac{1}{n^3}$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\underset{4}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x^3}\text{d}x=\underset{R\to \mathrm{\infty }}{\text{lim}}\underset{4}{\overset{R}{\int }}\frac{1}{x^3}\text{d}x$      (A1)

Note: The above A1 for using a limit can be awarded at any stage.
Condone the use of $\underset{x\to \mathrm{\infty }}{\text{lim}}$.

Do not award this mark to candidates who use $\mathrm{\infty }$ as the upper limit throughout.

= $\underset{R\to \mathrm{\infty }}{\text{lim}}{\left[-\frac{1}{2}{x}^{-2}\right]}_4^R\left(={\left[-\frac{1}{2}{x}^{-2}\right]}_4^{\mathrm{\infty }}\right)$     M1

$=\underset{R\to \mathrm{\infty }}{\text{lim}}\left(-\frac{1}{2}\left(R^{-2}-4^{-2}\right)\right)$

$=\frac{1}{32}$     A1

[3 marks]

      A1A1A1A1

A1 for the curve
A1 for rectangles starting at $x=4$
A1 for at least three upper rectangles
A1 for at least three lower rectangles

Note: Award A0A1 for two upper rectangles and two lower rectangles.

sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so

      AG

[4 marks]

a lower bound is $\frac{1}{32}$     A1

Note: Allow FT from part (a).

[1 mark]

METHOD 1

    (M1)

$\frac{1}{64}+\sum _{n=5}^{\mathrm{\infty }}\frac{1}{n^3}=\frac{1}{32}+\frac{1}{64}$     (M1)

, an upper bound      A1

Note: Allow FT from part (a).

METHOD 2

changing the lower limit in the inequality in part (b) gives

     (A1)

     (M1)

, an upper bound     A1

Note: Condone candidates who do not use a limit.

[3 marks]

Find the equilibrium population of brown squirrels suggested by this model.

Explain why the population of squirrels is increasing for values of $x$ less than this value.

Verify that $x=800$, $y=600$ is an equilibrium point.

Find the other three equilibrium points.

By using separation of variables, show that the general solution of $\frac{\text{d}x}{\text{d}t}=2x$ is $x=A{\text{e}}^{2t}$.

Write down the general solution of $\frac{\text{d}y}{\text{d}t}=3y$.

If both populations contain 10 squirrels at $t=0$ use the solutions to parts (c) (i) and (ii) to estimate the number of black and brown squirrels when $t=0.2$. Give your answers to the nearest whole numbers.

Write down the expressions for $x_{n+1}$ and $y_{n+1}$ that the conservationists will use.

Given that the initial populations are $x=100$, $y=100$, find the populations of each species of squirrel when $t=1$.

Use further iterations of Euler’s method to find the long-term population for each species of squirrel from these initial values.

Use the same method to find the long-term populations of squirrels when the initial populations are $x=400$, $y=100$.

Use Euler’s method with step length 0.2 to sketch, on the same axes, the approximate trajectories for the populations with the following initial populations.

(i)      $x=1000$, $y=1500$

(ii)    $x=1500$, $y=1000$

Given that the equilibrium point at (800, 600) is a saddle point, sketch the phase portrait for $x$ ≥ 0 , $y$ ≥ 0 on the same axes used in part (e).

2000        (M1)A1

[2 marks]

because the value of $\frac{\text{d}x}{\text{d}t}$ is positive (for $x>0$)        R1

[1 mark]

substitute $x=800$, $y=600$ into both equations      M1

both equations equal 0       A1

hence an equilibrium point       AG

[3 marks]

$x=0$, $y=0$      A1

$x=2000$, $y=0$, $x=0$, $y=3000$       M1A1A1

Note: Award M1 for an attempt at solving the system provided some values of $x$ and $y$ are found.

[4 marks]

$\int \frac{1}{x}\text{d}x=\int 2\text{d}t$      M1

$\text{ln}x=2t+c$         A1A1

Note: Award A1 for RHS, A1 for LHS.

$x={\text{e}}^c{\text{e}}^{2t}$        M1

$x=A{\text{e}}^{2t}$  (where $A={\text{e}}^c$)       AG

[4 marks]

$y=B{\text{e}}^{3t}$        A1

Note: Allow any letter for the constant term, including $A$.

[1 mark]

$x=15$, $y=18$       (M1)A1