Show that $b\approx 0.00939$.

Find the angle through which Mars rotates on its axis each hour.

Show that the maximum value of $\omega =1.98$, correct to three significant figures.

Find the minimum value of $\omega$.

the maximum value of $R\left(t\right)$.

the minimum value of $R\left(t\right)$.

Hence show that $a=1.6$, correct to two significant figures.

Find the value of $c$.

Find the value of $d$.

Write down $z_1$ and $z_2$ in exponential form, with a constant modulus.

Hence or otherwise find an equation for $L$ in the form $L\left(t\right)=p \sin (qt+r)+d$, where $p, q, r, d\in \mathrm{ℝ}$.

Find, in hours, the shortest time from sunrise to sunset at point $\text{A}$ that is predicted by this model.

recognition that period $=669$                  (M1)

$b=\frac{2\mathrm{\pi }}{669}$  OR  $b=0.00939190\dots$                A1

Note: Award A1 for a correct expression leading to the given value or for a correct value of $b$ to 4 sf or greater accuracy.

$b\approx 0.00939$             AG

[2 marks]

length of day$=24\frac{2}{3}$ hours                (A1)

Note: Award A1 for $\frac{2}{3}, 0.666\dots , 0.\overline{6}$ or $0.667$.

$\frac{2\mathrm{\pi }}{24{\displaystyle \frac{2}{3}}}$               (M1)

Note: Accept $\left(\frac{360}{24{\displaystyle \frac{2}{3}}}\right)$.

$=0.255$ radians $\left(0.254723\dots , \frac{3\mathrm{\pi }}{37}, 14.5945\dots °\right)$               A1

[3 marks]

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).

$\cos \omega =0.839 \tan \left(-0.440\right)$               A1

$\omega =1.97684\dots$

$\approx 1.98$               AG

Note: For substitution of $1.98$ award M0A0.

[3 marks]

$\delta =0.440$

$\omega =1.16 (1.16474\dots )$           A1

[1 mark]

$R_{\text{max}}=\frac{1.97684\dots }{0.25472\dots }$           (M1)

$=7.76 \text{hours} (7.76075\dots )$           A1

Note: Accept $7.70$ from use of $1.98$.

[2 marks]

$R_{\text{min}}=\frac{1.16474\dots }{0.25472\dots }$

$=4.57 \text{hours} (4.57258\dots )$           A1

Note: Accept $4.55$ and $4.56$ from use of rounded values.

[1 mark]

$a=\frac{7.76075\dots -4.57258\dots }{2}$           M1

$\approx 1.59408\dots$           A1

Note: Award M1 for substituting their values into a correct expression. Award A1 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 $2$ sf)          AG

[2 marks]

EITHER

$c=\frac{7.76075\dots +4.57258\dots }{2} \left(=\frac{12.333\dots }{2}\right)$           (M1)

OR

$c=4.57258\dots +1.59408\dots$  or  $c=7.76075\dots -1.59408\dots$

THEN

$=6.17 \left(6.16666\dots \right)$           A1

Note: Accept $6.16$ from use of rounded values. Follow through on their answers to part (d) and $1.6$.


[2 marks]

$d=18.65-6.16666\dots$           (M1)

$=12.5 \left(12.4833\dots \right)$           A1

Note: Follow through for $18.65$ minus their answer to part (f).


[2 marks]

at least one expression in the form $r{\text{e}}^{g\left(t\right)\text{i}}$           (M1)

$z_1=1.5{\text{e}}^{\left(0.00939t+2.83\right)\text{i}}\mathrm{,} {\mathrm{z}}_2=1.6{\text{e}}^{\left(0.00939t\right)\text{i}}$           A1A1

[3 marks]

EITHER

$z_1-z_2=1.5{\text{e}}^{\left(0.00939t+2.83\right)\text{i}}-1.6{\text{e}}^{\left(0.00939t\right)\text{i}}$

$={\text{e}}^{0.00939t\text{i}}\left(1.5{\text{e}}^{2.83\text{i}}-1.6\right)$               (M1)

$={\text{e}}^{0.00939t\text{i}}\left(3.06249\dots {\text{e}}^{2.99086\dots \text{i}}\right)$            (A1)(A1)

OR

graph of $L$ or $f$

$p=3.06249...$            (A1)

$r=-0.150729...$  OR  $r=2.99086...$            (M1)(A1)

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


THEN

$L\left(t\right)=3.06 \sin (0.00939t+2.99)+12.5$                 A1

$\left(L\left(t\right)=3.06248\dots \sin (0.00939t+2.99086\dots )+12.4833\dots \right)$

Note: Accept equivalent forms, e.g. $L\left(t\right)=3.06 \sin (0.00939t-0.151)+12.5$.
Follow through on their answer to part (g) replacing $12.5$.

 
[4 marks]

shortest time between sunrise and sunset

$12.4833\dots -3.06249\dots$               (M1)

$=9.42$ hours  $\left(9.420843\dots \right)$                 A1

Note: Accept $9.44$ from use of $3$ sf values.

[2 marks]