IBDP MAI : Topic 1 Number and algebra - AHL 1.13 Modulus–argument (polar) form AI HL Paper 3
Question
An appropriate location for a spacecraft landing on Mars is identified at point \( A \). The shortest duration from sunrise to sunset at point \( A \) needs to be calculated.
Use radians for all calculations in this question. Treat all given values as exact.
Mars orbits the Sun completely in 669 Martian days, equivalent to one Martian year.
On day \( t \), where \( t \in \mathbb{Z} \), the time in hours from the beginning of the Martian day to sunrise at point \( A \) is modeled by the function \( R(t) \), where
\( R(t) = a \sin(bt) + c, \quad t \in \mathbb{R}. \)
The graph of \( R \) is shown for one Martian year.
(a) Demonstrate that \( b \approx 0.00939 \). [2]
Mars completes one full rotation on its axis in 24 hours and 40 minutes.
(b) Determine the angle of rotation of Mars on its axis per hour. [3]
The sunrise time on Mars depends on the tilt angle, \( \delta \), relative to the Sun. Over a Martian year, \( \delta \) ranges from \(-0.440\) to \(0.440\) radians.
The angle \( \omega \), representing the rotation of Mars on its axis from the start of a Martian day to sunrise at point \( A \), is given by
\( \cos \omega = 0.839 \tan \delta, \quad 0 \leq \omega \leq \pi. \)
(c)(i) Prove that the maximum value of \( \omega = 1.98 \), correct to three significant figures. [3]
(c)(ii) Calculate the minimum value of \( \omega \). [1]
(d)(i) Using your answers from parts (b) and (c), compute the maximum value of \( R(t) \). [2]
(d)(ii) Compute the minimum value of \( R(t) \). [1]
(e) Show that \( a = 1.6 \), correct to two significant figures. [2]
(f) Determine the value of \( c \). [2]
Let \( S(t) \) represent the time in hours from the start of the Martian day to sunset at point \( A \) on day \( t \). The function is modeled as
\( S(t) = 1.5 \sin(0.00939t + 2.83) + 18.65. \)
The duration from sunrise to sunset at point \( A \), \( L(t) \), is modeled by
\( L(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) + d. \)
(g) Calculate the value of \( d \). [2]
Let \( f(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) \), so \( L(t) = f(t) + d \).
\( f(t) \) can be expressed as \( \mathrm{Im}(z_1 – z_2) \), where \( z_1 \) and \( z_2 \) are complex functions of \( t \).
(h)(i) Provide \( z_1 \) and \( z_2 \) in exponential form with constant modulus. [3]
(h)(ii) Derive an equation for \( L \) in the form \( L(t) = p \sin(qt + r) + d \), where \( p, q, r, d \in \mathbb{R} \). [4]
(h)(iii) Calculate, in hours, the shortest duration from sunrise to sunset at point \( A \) predicted by this model. [2]
▶️ Answer/Explanation
Markscheme
(a)
Recognize that the period of \( R(t) \) is 669 days (one Martian year).
For a sine function, period = \(\frac{2\pi}{b}\).
Thus, \(\frac{2\pi}{b} = 669\).
\( b = \frac{2\pi}{669} \approx 0.00939190\dots \approx 0.00939 \) (to 4 s.f.).
[2 marks]
Recognize that the period of \( R(t) \) is 669 days (one Martian year).
For a sine function, period = \(\frac{2\pi}{b}\).
Thus, \(\frac{2\pi}{b} = 669\).
\( b = \frac{2\pi}{669} \approx 0.00939190\dots \approx 0.00939 \) (to 4 s.f.).
[2 marks]
(b)
Length of a Martian day: \( 24 + \frac{40}{60} = 24 + \frac{2}{3} = \frac{74}{3} \) hours.
Mars rotates \( 2\pi \) radians in \(\frac{74}{3}\) hours.
Angle per hour: \(\frac{2\pi}{\frac{74}{3}} = \frac{2\pi \cdot 3}{74} \approx 0.254723 \approx 0.255\) radians.
[3 marks]
Length of a Martian day: \( 24 + \frac{40}{60} = 24 + \frac{2}{3} = \frac{74}{3} \) hours.
Mars rotates \( 2\pi \) radians in \(\frac{74}{3}\) hours.
Angle per hour: \(\frac{2\pi}{\frac{74}{3}} = \frac{2\pi \cdot 3}{74} \approx 0.254723 \approx 0.255\) radians.
[3 marks]
(c)(i)
Given: \(\cos \omega = 0.839 \tan \delta\), with \(\delta \in [-0.440, 0.440]\).
Maximum \(\omega\) occurs when \(\tan \delta\) is minimized (since \(\cos \omega\) is inversely related).
Minimum \(\tan \delta = \tan(-0.440) \approx -0.47524\).
\(\cos \omega = 0.839 \cdot (-0.47524) \approx -0.39873\).
\(\omega = \arccos(-0.39873) \approx 1.97684 \approx 1.98\) (to 3 s.f.).
[3 marks]
Given: \(\cos \omega = 0.839 \tan \delta\), with \(\delta \in [-0.440, 0.440]\).
Maximum \(\omega\) occurs when \(\tan \delta\) is minimized (since \(\cos \omega\) is inversely related).
Minimum \(\tan \delta = \tan(-0.440) \approx -0.47524\).
\(\cos \omega = 0.839 \cdot (-0.47524) \approx -0.39873\).
\(\omega = \arccos(-0.39873) \approx 1.97684 \approx 1.98\) (to 3 s.f.).
[3 marks]
(c)(ii)
Minimum \(\omega\) occurs when \(\tan \delta\) is maximized: \(\tan \delta = \tan(0.440) \approx 0.47524\).
\(\cos \omega = 0.839 \cdot 0.47524 \approx 0.39873\).
\(\omega = \arccos(0.39873) \approx 1.16474 \approx 1.16\).
[1 mark]
Minimum \(\omega\) occurs when \(\tan \delta\) is maximized: \(\tan \delta = \tan(0.440) \approx 0.47524\).
\(\cos \omega = 0.839 \cdot 0.47524 \approx 0.39873\).
\(\omega = \arccos(0.39873) \approx 1.16474 \approx 1.16\).
[1 mark]
(d)(i)
Maximum \( R(t) \): \(\omega_{\max} \approx 1.97684\), rotation rate \(\approx 0.254723\) radians/hour.
Time to rotate \(\omega_{\max}\): \(\frac{1.97684}{0.254723} \approx 7.76075 \approx 7.76\) hours.
[2 marks]
Maximum \( R(t) \): \(\omega_{\max} \approx 1.97684\), rotation rate \(\approx 0.254723\) radians/hour.
Time to rotate \(\omega_{\max}\): \(\frac{1.97684}{0.254723} \approx 7.76075 \approx 7.76\) hours.
[2 marks]
(d)(ii)
Minimum \( R(t) \): \(\omega_{\min} \approx 1.16474\).
Time to rotate \(\omega_{\min}\): \(\frac{1.16474}{0.254723} \approx 4.57258 \approx 4.57\) hours.
[1 mark]
Minimum \( R(t) \): \(\omega_{\min} \approx 1.16474\).
Time to rotate \(\omega_{\min}\): \(\frac{1.16474}{0.254723} \approx 4.57258 \approx 4.57\) hours.
[1 mark]
(e)
For \( R(t) = a \sin(bt) + c \), amplitude \( a = \frac{R_{\max} – R_{\min}}{2}\).
\( a = \frac{7.76075 – 4.57258}{2} \approx \frac{3.18817}{2} \approx 1.59408 \approx 1.6 \) (to 2 s.f.).
[2 marks]
For \( R(t) = a \sin(bt) + c \), amplitude \( a = \frac{R_{\max} – R_{\min}}{2}\).
\( a = \frac{7.76075 – 4.57258}{2} \approx \frac{3.18817}{2} \approx 1.59408 \approx 1.6 \) (to 2 s.f.).
[2 marks]
(f)
Midpoint \( c = \frac{R_{\max} + R_{\min}}{2}\).
\( c = \frac{7.76075 + 4.57258}{2} \approx \frac{12.33333}{2} \approx 6.16666 \approx 6.17\).
[2 marks]
Midpoint \( c = \frac{R_{\max} + R_{\min}}{2}\).
\( c = \frac{7.76075 + 4.57258}{2} \approx \frac{12.33333}{2} \approx 6.16666 \approx 6.17\).
[2 marks]
(g)
\( L(t) = S(t) – R(t) = [1.5 \sin(0.00939t + 2.83) + 18.65] – [1.6 \sin(0.00939t) + 6.16666]\).
\( L(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) + (18.65 – 6.16666)\).
\( d = 18.65 – 6.16666 \approx 12.4833 \approx 12.5\).
[2 marks]
\( L(t) = S(t) – R(t) = [1.5 \sin(0.00939t + 2.83) + 18.65] – [1.6 \sin(0.00939t) + 6.16666]\).
\( L(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) + (18.65 – 6.16666)\).
\( d = 18.65 – 6.16666 \approx 12.4833 \approx 12.5\).
[2 marks]
(h)(i)
For \( f(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) = \mathrm{Im}(z_1 – z_2)\):
\( z_1 = 1.5 e^{(0.00939t + 2.83)i}\), \( z_2 = 1.6 e^{(0.00939t)i}\).
[3 marks]
For \( f(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) = \mathrm{Im}(z_1 – z_2)\):
\( z_1 = 1.5 e^{(0.00939t + 2.83)i}\), \( z_2 = 1.6 e^{(0.00939t)i}\).
[3 marks]
(h)(ii)
Compute \( z_1 – z_2 = e^{0.00939ti} (1.5 e^{2.83i} – 1.6)\).
Convert \( 1.5 e^{2.83i} – 1.6 \) to polar form: magnitude \(\sqrt{(1.5 \cos 2.83 – 1.6)^2 + (1.5 \sin 2.83)^2} \approx 3.06249\).
Phase: \(\arctan\left(\frac{1.5 \sin 2.83}{1.5 \cos 2.83 – 1.6}\right) \approx 2.99086\).
Thus, \( f(t) = \mathrm{Im}(3.06249 e^{(0.00939t + 2.99086)i}) \approx 3.06 \sin(0.00939t + 2.99)\).
\( L(t) = 3.06 \sin(0.00939t + 2.99) + 12.5\).
[4 marks]
Compute \( z_1 – z_2 = e^{0.00939ti} (1.5 e^{2.83i} – 1.6)\).
Convert \( 1.5 e^{2.83i} – 1.6 \) to polar form: magnitude \(\sqrt{(1.5 \cos 2.83 – 1.6)^2 + (1.5 \sin 2.83)^2} \approx 3.06249\).
Phase: \(\arctan\left(\frac{1.5 \sin 2.83}{1.5 \cos 2.83 – 1.6}\right) \approx 2.99086\).
Thus, \( f(t) = \mathrm{Im}(3.06249 e^{(0.00939t + 2.99086)i}) \approx 3.06 \sin(0.00939t + 2.99)\).
\( L(t) = 3.06 \sin(0.00939t + 2.99) + 12.5\).
[4 marks]
(h)(iii)
Shortest time: \( L_{\min} = 12.5 – 3.06 \approx 12.4833 – 3.06249 \approx 9.42084 \approx 9.42\) hours.
[2 marks]
Shortest time: \( L_{\min} = 12.5 – 3.06 \approx 12.4833 – 3.06249 \approx 9.42084 \approx 9.42\) hours.
[2 marks]