IBDP MAI : Topic 1 Number and algebra - AHL 1.12 Complex numbers AI HL Paper 3

(a) Recognition that period = 669, \( b = \frac{2\pi}{669} \) OR \( b = 0.00939190… \), \( b \approx 0.00939 \)

Detailed Solution: The function \( R(t) = a \sin(bt) + c \) models sunrise time over a Martian year (669 days).

The period of \( \sin(bt) \) is \( \frac{2\pi}{b} \), and since the graph completes one cycle in 669 days, \( \frac{2\pi}{b} = 669 \). Solve for \( b \): \( b = \frac{2\pi}{669} \approx \frac{6.283185307}{669} \approx 0.009391904… \), which rounds to 0.00939 (to 5 d.p.), as required.

(b) Length of day = \( 24\frac{2}{3} \) hours,

\( \frac{2\pi}{24\frac{2}{3}} = 0.255 \) radians (= 0.254723…, \(\frac{3\pi}{37}\), 14.5945…°)

Detailed Solution: Mars rotates \( 2\pi \) radians in one day (24 hours 40 minutes = \( 24 + \frac{40}{60} = 24\frac{2}{3} = \frac{74}{3} \) hours).

Angular speed = \( \frac{2\pi}{\frac{74}{3}} \)

= \( \frac{2\pi \cdot 3}{74} \) 

= \( \frac{6\pi}{74}\) 

= \( \frac{3\pi}{37} \approx \frac{9.42477796}{37} \approx 0.254723… \) radians per hour, approximately 0.255 (to 3 d.p.). \)

(c) (i) Substitution of either value of \( \delta \) into equation, correct use of arccos,

\( \cos \omega = 0.839 \tan(-0.440) \),

\( \omega = 1.97684… \approx 1.98 \)
(ii) \( \delta = 0.440 \),

\( \omega = 1.16 (1.16474…) \)

Detailed Solution: Given \( \cos \omega = 0.839 \tan \delta \),

\( \delta \) ranges from \(-0.440\) to 0.440,

and \( 0 \leq \omega \leq \pi \).

(i) Max \( \omega \) when \( \tan \delta \) is minimized (most negative):

\( \delta = -0.440 \),

\( \tan(-0.440) \approx -0.47591 \),

\( \cos \omega = 0.839 \cdot (-0.47591) \approx -0.39929 \),

\( \omega = \arccos(-0.39929) \approx 1.97684… \approx 1.98 \) (3 s.f.).

(ii) Min \( \omega \) when \( \tan \delta \) is maximized:

\( \delta = 0.440 \), \( \tan 0.440 \approx 0.47591 \),

\( \cos \omega = 0.839 \cdot 0.47591 \approx 0.39929 \),

\( \omega = \arccos(0.39929) \approx 1.16474… \approx 1.16 \).

(d) (i) \( R_{\text{max}} = \frac{1.97684…}{0.254723…} = 7.76 \) hours (7.76075…)
(ii) \( R_{\text{min}} = \frac{1.16474…}{0.254723…} = 4.57 \) hours (4.57258…)

Detailed Solution: \( R(t) \) is time from day start to sunrise,

\( \omega \) is the rotation angle,

and from (b), Mars rotates 0.254723 rad/hour.

(i) Max \( \omega = 1.97684 \),

time = \( \frac{1.97684}{0.254723} \approx 7.76075… \approx 7.76 \) hours.

(ii) Min \( \omega = 1.16474 \),

time = \( \frac{1.16474}{0.254723} \approx 4.57258… \approx 4.57 \) hours.

(e) \( a = \frac{7.76075… – 4.57258…}{2} \approx 1.59408… \approx 1.6 \) (correct to 2 s.f.)

Detailed Solution: For \( R(t) = a \sin(bt) + c \), amplitude \( a \) is half the range of \( R(t) \).

From (d), max \( R = 7.76075 \),

min \( R = 4.57258 \),

range = \( 7.76075 – 4.57258 = 3.18817 \),

\( a = \frac{3.18817}{2} \approx 1.59408… \approx 1.6 \) (2 s.f.).

(f) EITHER \( c = \frac{7.76075… + 4.57258…}{2} \)

OR \( c = 4.57258… + 1.59408… \)

OR \( c = 7.76075… – 1.59408… \),

THEN \( c = 6.17 (6.16666…) \)

Detailed Solution: \( c \) is the midline of \( R(t) \),

average of max and min:

\( c = \frac{7.76075 + 4.57258}{2} \)

\( = \frac{12.33333}{2} \approx 6.16666… \approx 6.17 \).

Alternatively,

\( c = R_{\text{min}} + a = 4.57258 + 1.59408 \approx 6.16666 \),

or \( c = R_{\text{max}} – a = 7.76075 – 1.59408 \approx 6.16666 \).

(g) \( d = 18.65 – 6.16666… = 12.5 (12.4833…) \)

Detailed Solution: \( L(t) = S(t) – R(t) \),

where \( S(t) = 1.5 \sin(0.00939t + 2.83) + 18.65 \),

\( R(t) = 1.6 \sin(0.00939t) + 6.16666 \).

Then,

\( L(t) = [1.5 \sin(0.00939t + 2.83) + 18.65] – [1.6 \sin(0.00939t) + 6.16666] \)

\( = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) + (18.65 – 6.16666) \).

So, \( d = 18.65 – 6.16666 \approx 12.48334 \approx 12.5 \).

(h) (i) \( z_1 = 1.5 e^{i(0.00939t + 2.83)} \),

\( z_2 = 1.6 e^{i(0.00939t)} \)
(ii) EITHER

\( z_1 – z_2 = 1.5 e^{i(0.00939t + 2.83)} – 1.6 e^{i(0.00939t)} \)

\( = e^{i(0.00939t)} (1.5 e^{i 2.83} – 1.6) \)

\( = e^{i(0.00939t)} (3.06249… e^{i 2.99089…}) \),

OR \( p = 3.06249… \),

\( r = 2.99086… \),

THEN \( L(t) = 3.06 \sin(0.00939t + 2.99) + 12.5 \) (or \( 3.06248… \sin(0.00939t + 2.99086…) + 12.4833… \))
(iii) Shortest time = \( 12.4833… – 3.06249… = 9.42 \) hours (9.420843…)

Detailed Solution: (i) \( f(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) = \text{Im}(z_1 – z_2) \),

So \( z_1 = 1.5 e^{i(0.00939t + 2.83)} \),

\( z_2 = 1.6 e^{i(0.00939t)} \) (constant moduli 1.5, 1.6).

(ii) \( z_1 – z_2 = e^{i(0.00939t)} (1.5 e^{i 2.83} – 1.6) \),

Compute \( 1.5 e^{i 2.83} – 1.6 \),

where \( e^{i 2.83} = \cos 2.83 + i \sin 2.83 \),

Magnitude \( \sqrt{(1.5 \cos 2.83 – 1.6)^2 + (1.5 \sin 2.83)^2} \approx 3.06249 \),

Phase \( \arctan\left(\frac{1.5 \sin 2.83}{1.5 \cos 2.83 – 1.6}\right) + \pi \approx 2.99086 \),

So \( L(t) = 3.06 \sin(0.00939t + 2.99) + 12.5 \).

(iii) Min \( L(t) = d – p = 12.4833 – 3.06249 \approx 9.42084 \approx 9.42 \) hours.