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]