Home / IBDP MAI :Topic 2:Functions : AHL 2.9-Sinusoidal models.Exam Style Questions Paper 3

IBDP MAI :Topic 2:Functions : AHL 2.9-Sinusoidal models.Exam Style Questions Paper 3

Question

A suitable site for the landing of a spacecraft on the planet Mars is identified at a point, A. The shortest time from sunrise to sunset at point

A must be found.

Radians should be used throughout this question. All values given in the question should be treated as exact.
Mars completes a full orbit of the Sun in 669 Martian days, which is one Martian year.

On day $t$, where $t \in \mathbb{Z}$, the length of time, in hours, from the start of the Martian day until sunrise at point $\mathrm{A}$ can be modelled by a function, $R(t)$, where
$$
R(t)=a \sin (b t)+c, t \in \mathbb{R} .
$$

The graph of $R$ is shown for one Martian year.

Mars completes a full rotation on its axis in 24 hours and 40 minutes.

The time of sunrise on Mars depends on the angle, $\delta$, at which it tilts towards the Sun. During a Martian year, $\delta$ varies from -0.440 to 0.440 radians.

The angle, $\omega$, through which Mars rotates on its axis from the start of a Martian day to the moment of sunrise, at point A, is given by $\cos \omega=0.839 \tan \delta, 0 \leq \omega \leq \pi$.

Use your answers to parts (b) and (c) to find

Let $S(t)$ be the length of time, in hours, from the start of the Martian day until sunset at point A on day $t . S(t)$ can be modelled by the function
$$
S(t)=1.5 \sin (0.00939 t+2.83)+18.65 .
$$

The length of time between sunrise and sunset at point $\mathrm{A}, L(t)$, can be modelled by the function
$$
L(t)=1.5 \sin (0.00939 t+2.83)-1.6 \sin (0.00939 t)+d .
$$

Let $f(t)=1.5 \sin (0.00939 t+2.83)-1.6 \sin (0.00939 t)$ and hence $L(t)=f(t)+d$.
$f(t)$ can be written in the form $\operatorname{Im}\left(z_1-z_2\right)$, where $z_1$ and $z_2$ are complex functions of $t$.

a. Show that $b \approx 0.00939$.
b. Find the angle through which Mars rotates on its axis each hour.
c.i. Show that the maximum value of $\omega=1.98$, correct to three significant figures.
c.ii.Find the minimum value of $\omega$.
d.i.the maximum value of $R(t)$.
d.ii.the minimum value of $R(t)$.
e. Hence show that $a=1.6$, correct to two significant figures.
f. Find the value of $c$.
g. Find the value of $d$.
h.i. Write down $z_1$ and $z_2$ in exponential form, with a constant modulus.
h.ii.Hence or otherwise find an equation for $L$ in the form $L(t)=p \sin (q t+r)+d$, where $p, q, r, d \in \mathbb{R}$.
h.iiiFind, in hours, the shortest time from sunrise to sunset at point $\mathrm{A}$ that is predicted by this model.

▶️Answer/Explanation

a. recognition that period $=669$
(M1)
$$
b=\frac{2 \pi}{669} \text { OR } b=0.00939190 \ldots
$$
A1
Note: Award $\boldsymbol{A 1}$ for a correct expression leading to the given value or for a correct value of $b$ to 4 sf or greater accuracy.
$$
b \approx 0.00939
$$
$A G$
[2 marks]
b. length of day $=24 \frac{2}{3}$ hours
(A1)
Note: Award $\boldsymbol{A 1}$ for $\frac{2}{3}, 0.666 \ldots, 0 . \overline{6}$ or 0.667 .
$\frac{2 \pi}{24 \frac{2}{3}}$
(M1)

Note: Accept $\left(\frac{360}{24 \frac{2}{3}}\right)$.
$=0.255$ radians $\left(0.254723 \ldots, \frac{3 \pi}{37}, 14.5945 \ldots{ }^{\circ}\right)$
A1
[3 marks]
c.i. substitution of either value of $\delta$ into equation
(M1)
correct use of arccos to find a value for $\omega$
(M1)
Note: Both (M1) lines may be seen in either part (c)(i) or part (c)(ii).
$$
\begin{aligned}
& \cos \omega=0.839 \tan (-0.440) \\
& \omega=1.97684 \ldots \\
& \approx 1.98 \quad \text { AG }
\end{aligned}
$$
A1
$A G$
Note: For substitution of 1.98 award MOAO.

c.ii. $\delta=0.440$
$\omega=1.16$ (1.16474 ..)
A1
[1 mark]
d.i. $R_{\max }=\frac{1.97684 \ldots}{0.25472 \ldots}$
(M1)
$=7.76$ hours $(7.76075 \ldots)$
A1
Note: Accept 7.70 from use of 1.98 .
[2 marks]
d.ii $R_{\min }=\frac{1.16474 \ldots}{0.25472 \ldots}$
$=4.57$ hours $(4.57258 \ldots)$
A1
Note: Accept 4.55 and 4.56 from use of rounded values.
[1 mark]
e. $a=\frac{7.76075 \ldots-4.57258 \ldots}{2} \quad$ M1
$$
\approx 1.59408 \text {.. }
$$
A1
Note: Award $\boldsymbol{M 1}$ for substituting their values into a correct expression. Award $\boldsymbol{A 1}$ for a correct value of $a$ from their expression which has at least 3 significant figures and rounds correctly to 1.6 .
$\approx 1.6$ (correct to $2 \mathrm{sf}$ )
AG

f. EITHER
$$
c=\frac{7.76075 \ldots+4.57258 \ldots}{2}\left(=\frac{12.333 \ldots}{2}\right) \quad \text { (M1) }
$$

OR
$c=4.57258 \ldots+1.59408 \ldots$ or $c=7.76075 \ldots-1.59408 \ldots$

THEN
$=6.17(6.16666 \ldots)$
A1

Note: Accept 6.16 from use of rounded values. Follow through on their answers to part (d) and 1.6.
[2 marks]
g. $d=18.65-6.16666 \ldots$
(M1)
$=12.5(12.4833 \ldots)$
A1
Note: Follow through for 18.65 minus their answer to part (f).
[2 marks]

h.i. at least one expression in the form $r \mathrm{e}^{g(t) \mathrm{i}}$
(M1)
$z_1=1.5 \mathrm{e}^{(0.00939 t+2.83) \mathrm{i}}, \mathrm{z}_2=1.6 \mathrm{e}^{(0.00939 t) \mathrm{i}}$
A1A1
[3 marks]
h.ii.EITHER
$$
\begin{aligned}
& z_1-z_2=1.5 \mathrm{e}^{(0.00939 t+2.83) \mathrm{i}}-1.6 \mathrm{e}^{(0.00939 t) \mathrm{i}} \\
& =\mathrm{e}^{0.00939 t \mathrm{i}}\left(1.5 \mathrm{e}^{2.83 \mathrm{i}}-1.6\right) \quad \text { (M1) } \\
& =\mathrm{e}^{0.00939 t \mathrm{i}}\left(3.06249 \ldots \mathrm{e}^{2.99086 \ldots \mathrm{i})}\right.
\end{aligned}
$$
(A1)(A1)

OR
graph of $L$ or $f$
$p=3.06249 \ldots$
(A1)
$r=-0.150729 \ldots$ OR $r=2.99086 \ldots$
(M1)(A1)

Note: The $p$ and $r$ variables (or equivalent) must be seen.

THEN
$$
\begin{aligned}
& L(t)=3.06 \sin (0.00939 t+2.99)+12.5 \quad \text { A1 } \\
& (L(t)=3.06248 \ldots \sin (0.00939 t+2.99086 \ldots)+12.4833 \ldots)
\end{aligned}
$$
A1
Note: Accept equivalent forms, e.g. $L(t)=3.06 \sin (0.00939 t-0.151)+12.5$.
Follow through on their answer to part $(\mathrm{g})$ replacing 12.5.
[4 marks]
h.iiishortest time between sunrise and sunset
$12.4833 \ldots-3.06249 \ldots$
(M1)
$=9.42$ hours $(9.420843 \ldots$ )
A1
Note: Accept 9.44 from use of $3 \mathrm{sf}$ values.
[2 marks]

 

Question

An estate manager is responsible for stocking a small lake with fish. He begins by introducing 1000 fish into the lake and monitors their population growth to determine the likely carrying capacity of the lake.

After one year an accurate assessment of the number of fish in the lake is taken and it is found to be 1200 .
Let $N$ be the number of fish $t$ years after the fish have been introduced to the lake.
Initially it is assumed that the rate of increase of $N$ will be constant.

When $t=8$ the estate manager again decides to estimate the number of fish in the lake. To do this he first catches 300 fish and marks them, so they can be recognized if caught again. These fish are then released back into the lake. A few days later he catches another 300 fish, releasing each fish after it has been checked, and finds 45 of them are marked.

Let $X$ be the number of marked fish caught in the second sample, where $X$ is considered to be distributed as $\mathrm{B}(n, p)$. Assume the number of fish in the lake is 2000 .

The estate manager decides that he needs bounds for the total number of fish in the lake.

The estate manager feels confident that the proportion of marked fish in the lake will be within 1.5 standard deviations of the proportion of marked fish in the sample and decides these will form the upper and lower bounds of his estimate.

The estate manager now believes the population of fish will follow the logistic model $N(t)=\frac{L}{1+C e^{-k t}}$ where $L$ is the carrying capacity and $C, k>0$. The estate manager would like to know if the population of fish in the lake will eventually reach 5000.
a. Use this model to predict the number of fish in the lake when $t=8$.
b. Assuming the proportion of marked fish in the second sample is equal to the proportion of marked fish in the lake, show that the estate manager will estimate there are now 2000 fish in the lake.
c.i. Write down the value of $n$ and the value of $p$.
c.ii.State an assumption that is being made for $X$ to be considered as following a binomial distribution.
d.i. Show that an estimate for $\operatorname{Var}(X)$ is 38.25 .
d.iiHence show that the variance of the proportion of marked fish in the sample, $\operatorname{Var}\left(\frac{X}{300}\right)$, is 0.000425 .
e.i. Taking the value for the variance given in (d) (ii) as a good approximation for the true variance, find the upper and lower bounds for the proportion of marked fish in the lake.
e.ii.Hence find upper and lower bounds for the number of fish in the lake when $t=8$.
f. Given this result, comment on the validity of the linear model used in part (a).
g.i. Assuming a carrying capacity of 5000 use the given values of $N(0)$ and $N(1)$ to calculate the parameters $C$ and $k$.
g.ii.Use these parameters to calculate the value of $N(8)$ predicted by this model.
h. Comment on the likelihood of the fish population reaching 5000 .

▶️Answer/Explanation

a.$$
\begin{aligned}
& N(8)=1000+200 \times 8 \quad \text { M1 } \\
& =2600 \quad \text { A1 }
\end{aligned}
$$
A1
[2 marks]
b.
$$
\begin{array}{ll}
\frac{45}{300}=\frac{300}{N} & \text { M1A1 } \\
N=2000 & \text { AG }
\end{array}
$$
[2 marks]
c.i. $n=300, p=\frac{300}{2000}=0.15 \quad$ A1A1
[2 marks]
c.ii.Any valid reason for example:
R1
Marked fish are randomly distributed, so $p$ constant.
Each fish caught is independent of previous fish caught
[1 mark]
$$
\begin{aligned}
& \text { d.i. } \operatorname{Var}(X)=n p(1-p) \text { M1 } \\
&=300 \times \frac{300}{2000} \times \frac{1700}{2000} \text { A1 } \\
&=38.25 \quad \text { AG }
\end{aligned}
$$
[2 marks]
d.ii. $\operatorname{Var}\left(\frac{X}{300}\right)=\frac{\operatorname{Var}(X)}{300^2} \quad$ M1A1
$=0.000425 \quad$ AG
[2 marks]

e.i. $0.15 \pm 1.5 \sqrt{0.000425}$
(M1)
0.181 and 0.119
A1
[2 marks]
e.i. $\frac{300}{N}=0.181 \ldots, \frac{300}{N}=0.119 \ldots \quad$ M1
Lower bound 1658 upper bound 2519
A1
[2 marks]
f. Linear model prediction falls outside this range so unlikely to be a good model
R1A1
[2 marks]
$$
\begin{aligned}
& \text { g.i. } 1000=\frac{5000}{1+C} \quad \text { M1 } \\
& C=4 \quad \text { A1 } \\
& 1200=\frac{5000}{1+4 e^{-k}} \quad \text { M1 } \\
& e^{-k}=\frac{3800}{4 \times 1200} \quad \text { (M1) } \\
& k=-\ln (0.7916 \ldots)=0.2336 \ldots
\end{aligned}
$$
A1
[5 marks]
$$
\text { g.ii. } N(8)=\frac{5000}{1+4 e^{-0.2336 \times 8}}=3090 \quad \text { M1A1 }
$$

Note: Accept any answer that rounds to 3000 .
[2 marks]
h. This is much higher than the calculated upper bound for $N(8)$ so the rate of growth of the fish is unlikely to be sufficient to reach a carrying capacity of 5000 .
M1R1

Question

In this question you will explore possible models for the spread of an infectious disease

An infectious disease has begun spreading in a country. The National Disease Control Centre (NDCC) has compiled the following data after receiving alerts from hospitals.

 A graph of  n  against  d text  is shown below. 

The NDCC want to find a model to predict the total number of people infected, so they can plan for medicine and hospital facilities. After looking at the data, they think an exponential function in the form $n=a b^d$ could be used as a model.

Use your answer to part (a) to predict

The NDCC want to verify the accuracy of these predictions. They decide to perform a $\chi^2$ goodness of fit test.

The predictions given by the model for the first five days are shown in the table.

In fact, the first day when the total number of people infected is greater than 1000 is day 14 , when a total of 1015 people are infected.

Based on this new data, the NDCC decide to try a logistic model in the form $n=\frac{L}{1+c e^{-k d}}$.

Use the data from days $1-5$, together with day 14 , to find the value of
a. Use an exponential regression to find the value of $a$ and of $b$, correct to 4 decimal places.
b.i.the number of new people infected on day 6 .
b.ii.the day when the total number of people infected will be greater than 1000 .
c. Use your answer to part (a) to show that the model predicts 16.7 people will be infected on the first day.
d.i. Explain why the number of degrees of freedom is 2.
d.iiPerform a $\chi^2$ goodness of fit test at the $5 \%$ significance level. You should clearly state your hypotheses, the p-value, and your conclusion.
e. Give two reasons why the prediction in part (b)(ii) might be lower than 14.
f.i. $L$.
f.ii. $c$.
f.iii.k.
g. Hence predict the total number of people infected by this disease after several months.
h. Use the logistic model to find the day when the rate of increase of people infected is greatest.

▶️Answer/Explanation

a. $a=9.7782, b=1.7125 \quad$ M1A1A1
[3 marks]
b.i. $n(6)=247$
A1
number of new people infected $=247-140=107$ M1A1
[3 marks]
b.ii.use of graph or table M1
day 9 A1
[2 marks]
c. $9.7782(1.7125)^1 \quad \mathbf{M 1}$
$=16.7$ people $A G$
[1 mark]
d.i. 2 parameters $(a$ and $b$ ) were estimated from the data. $\quad \boldsymbol{R 1}$
$v=5-1-2 \quad M 1$
$=2 \quad A G$
[2 marks]
d.ii. $H_0$ : data is modeled by $n(d)=9.7782(1.7125)^d$ and $H_1$ : data is not modeled by $n(d)=9.7782(1.7125)^d$
A1
p-value $=0.893 \quad \boldsymbol{A 2}$
Since $0.893>0.05 \quad \boldsymbol{R 1}$
Insufficient evidence to reject $H_0$. So data is modeled by $n(d)=9.7782(1.7125)^d \quad$ A1
[5 marks]
e. vaccine or medicine might slow down rate of infection $\boldsymbol{R 1}$
People become more aware of disease and take precautions to avoid infection
R1
Accept other valid reasons

f.i. 1060
M1A1
[2 marks]
f.ii. 108
A1
[1 mark]
f.iii. 0.560
A1
[1 mark]
g. As $d \rightarrow \infty \quad$ M1
$$
n \rightarrow 1060
$$

A1
[2 marks]
h. sketch of $\frac{\mathrm{d} n}{\mathrm{~d} d}$ or solve $\frac{\mathrm{d}^2 n}{\mathrm{~d} d^2}=0 \quad$ M1
$$
d=8.36
$$
A1
Day 8 A1
[3 marks]

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