IBDP MAI : Topic 2 Function - AHL 2.7 Composite functions f∘g AI HL Paper 3
Question
A suitable landing site for a spacecraft on Mars is identified at point A. This question examines the shortest duration from sunrise to sunset at point A. Radians should be used throughout, and all values given are exact.
Mars completes a full orbit of the Sun in 669 Martian days, one Martian year. On day \( t \), where \( t \in \mathbb{Z} \), the time from the start of the Martian day to sunrise at point A, in hours, is modelled by \( R(t) = a \sin(bt) + c \), \( t \in \mathbb{R} \). 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 sunrise time depends on the angle \( \delta \), ranging from \(-0.440\) to \(0.440\) radians, with \( \cos \omega = 0.839 \tan \delta \), \( 0 \leq \omega \leq \pi \).
(a) Show that \( b \approx 0.00939 \). [2]
(b) Find the angle through which Mars rotates on its axis each hour. [3]
(c)
(i) Show that the maximum value of \( \omega = 1.98 \), correct to three significant figures. [2]
(ii) Find the minimum value of \( \omega \). [2]
(i) Show that the maximum value of \( \omega = 1.98 \), correct to three significant figures. [2]
(ii) Find the minimum value of \( \omega \). [2]
(d) Using your answers to parts (b) and (c), find:
(i) the maximum value of \( R(t) \). [2]
(ii) the minimum value of \( R(t) \). [1]
(i) the maximum value of \( R(t) \). [2]
(ii) the minimum value of \( R(t) \). [1]
(e) Hence show that \( a = 1.6 \), correct to two significant figures. [2]
(f) Find the value of \( c \). [2]
The time from sunrise to sunset is given by \( L(t) = S(t) – R(t) \), where \( S(t) = 1.5 \sin(0.00939t + 2.83) + 18.65 \).
(g) Find the value of \( d \). [2]
(h)
(i) Write down \( z_1 \) and \( z_2 \) in exponential form, with a constant modulus. [2]
(ii) Hence or otherwise find an equation for \( L(t) \) in the form \( L(t) = p \sin(qt + r) + d \), where \( p, q, r, d \in \mathbb{R} \). [4]
(iii) Find, in hours, the shortest time from sunrise to sunset at point A predicted by this model. [2]
(i) Write down \( z_1 \) and \( z_2 \) in exponential form, with a constant modulus. [2]
(ii) Hence or otherwise find an equation for \( L(t) \) in the form \( L(t) = p \sin(qt + r) + d \), where \( p, q, r, d \in \mathbb{R} \). [4]
(iii) Find, in hours, the shortest time from sunrise to sunset at point A predicted by this model. [2]
▶️ Answer/Explanation
Markscheme
(a)
The period of \( R(t) = a \sin(bt) + c \) is 669 days. M1
Since \( \sin(bt) \) has period \( \dfrac{2\pi}{b} \), set \( \dfrac{2\pi}{b} = 669 \).
Solve: \( b = \dfrac{2\pi}{669} \approx \dfrac{6.283185307}{669} \approx 0.009391904 \approx 0.00939 \) (to 5 d.p.). A1
[2 marks]
The period of \( R(t) = a \sin(bt) + c \) is 669 days. M1
Since \( \sin(bt) \) has period \( \dfrac{2\pi}{b} \), set \( \dfrac{2\pi}{b} = 669 \).
Solve: \( b = \dfrac{2\pi}{669} \approx \dfrac{6.283185307}{669} \approx 0.009391904 \approx 0.00939 \) (to 5 d.p.). A1
[2 marks]
(b)
Mars rotates \( 2\pi \) radians in one day (24 hours 40 minutes = \( 24 + \dfrac{40}{60} = \dfrac{74}{3} \) hours). M1
Angular speed: \( \dfrac{2\pi}{\dfrac{74}{3}} = \dfrac{2\pi \times 3}{74} = \dfrac{6\pi}{74} = \dfrac{3\pi}{37} \). A1
\( \dfrac{3\pi}{37} \approx \dfrac{9.42477796}{37} \approx 0.254723 \approx 0.255 \) radians per hour. A1
[3 marks]
Mars rotates \( 2\pi \) radians in one day (24 hours 40 minutes = \( 24 + \dfrac{40}{60} = \dfrac{74}{3} \) hours). M1
Angular speed: \( \dfrac{2\pi}{\dfrac{74}{3}} = \dfrac{2\pi \times 3}{74} = \dfrac{6\pi}{74} = \dfrac{3\pi}{37} \). A1
\( \dfrac{3\pi}{37} \approx \dfrac{9.42477796}{37} \approx 0.254723 \approx 0.255 \) radians per hour. A1
[3 marks]
(c)(i)
Given \( \cos \omega = 0.839 \tan \delta \), with \( \delta = -0.440 \): M1
\( \tan(-0.440) \approx -0.47591 \), so \( \cos \omega = 0.839 \times (-0.47591) \approx -0.39929 \).
\( \omega = \arccos(-0.39929) \approx 1.97684 \approx 1.98 \) (3 s.f.). A1
[2 marks]
Given \( \cos \omega = 0.839 \tan \delta \), with \( \delta = -0.440 \): M1
\( \tan(-0.440) \approx -0.47591 \), so \( \cos \omega = 0.839 \times (-0.47591) \approx -0.39929 \).
\( \omega = \arccos(-0.39929) \approx 1.97684 \approx 1.98 \) (3 s.f.). A1
[2 marks]
(c)(ii)
For \( \delta = 0.440 \): M1
\( \tan 0.440 \approx 0.47591 \), so \( \cos \omega = 0.839 \times 0.47591 \approx 0.39929 \).
\( \omega = \arccos(0.39929) \approx 1.16474 \approx 1.16 \). A1
[2 marks]
For \( \delta = 0.440 \): M1
\( \tan 0.440 \approx 0.47591 \), so \( \cos \omega = 0.839 \times 0.47591 \approx 0.39929 \).
\( \omega = \arccos(0.39929) \approx 1.16474 \approx 1.16 \). A1
[2 marks]
(d)(i)
Using \( \omega = 1.97684 \) from (c)(i) and angular speed 0.254723 from (b): M1
\( R_{\text{max}} = \dfrac{1.97684}{0.254723} \approx 7.76075 \approx 7.76 \) hours. A1
[2 marks]
Using \( \omega = 1.97684 \) from (c)(i) and angular speed 0.254723 from (b): M1
\( R_{\text{max}} = \dfrac{1.97684}{0.254723} \approx 7.76075 \approx 7.76 \) hours. A1
[2 marks]
(d)(ii)
Using \( \omega = 1.16474 \) from (c)(ii):
\( R_{\text{min}} = \dfrac{1.16474}{0.254723} \approx 4.57258 \approx 4.57 \) hours. A1
[1 mark]
Using \( \omega = 1.16474 \) from (c)(ii):
\( R_{\text{min}} = \dfrac{1.16474}{0.254723} \approx 4.57258 \approx 4.57 \) hours. A1
[1 mark]
(e)
For \( R(t) = a \sin(bt) + c \), amplitude \( a = \dfrac{R_{\text{max}} – R_{\text{min}}}{2} \). M1
\( a = \dfrac{7.76075 – 4.57258}{2} = \dfrac{3.18817}{2} \approx 1.59408 \approx 1.6 \) (2 s.f.). A1
[2 marks]
For \( R(t) = a \sin(bt) + c \), amplitude \( a = \dfrac{R_{\text{max}} – R_{\text{min}}}{2} \). M1
\( a = \dfrac{7.76075 – 4.57258}{2} = \dfrac{3.18817}{2} \approx 1.59408 \approx 1.6 \) (2 s.f.). A1
[2 marks]
(f)
\( c \) is the midline: \( c = \dfrac{R_{\text{max}} + R_{\text{min}}}{2} \). M1
\( c = \dfrac{7.76075 + 4.57258}{2} = \dfrac{12.33333}{2} \approx 6.16666 \approx 6.17 \). A1
[2 marks]
\( c \) is the midline: \( c = \dfrac{R_{\text{max}} + R_{\text{min}}}{2} \). M1
\( c = \dfrac{7.76075 + 4.57258}{2} = \dfrac{12.33333}{2} \approx 6.16666 \approx 6.17 \). A1
[2 marks]
(g)
Given \( L(t) = S(t) – R(t) \), where \( S(t) = 1.5 \sin(0.00939t + 2.83) + 18.65 \), \( R(t) = 1.6 \sin(0.00939t) + 6.16666 \). M1
\( L(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) + (18.65 – 6.16666) \).
\( d = 18.65 – 6.16666 \approx 12.48334 \approx 12.5 \). A1
[2 marks]
Given \( L(t) = S(t) – R(t) \), where \( S(t) = 1.5 \sin(0.00939t + 2.83) + 18.65 \), \( R(t) = 1.6 \sin(0.00939t) + 6.16666 \). M1
\( L(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) + (18.65 – 6.16666) \).
\( d = 18.65 – 6.16666 \approx 12.48334 \approx 12.5 \). A1
[2 marks]
(h)(i)
For \( f(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) = \text{Im}(z_1 – z_2) \):
\( z_1 = 1.5 e^{i(0.00939t + 2.83)} \). A1
\( z_2 = 1.6 e^{i(0.00939t)} \). A1
[2 marks]
For \( f(t) = 1.5 \sin(0.00939t + 2.83) – 1.6 \sin(0.00939t) = \text{Im}(z_1 – z_2) \):
\( z_1 = 1.5 e^{i(0.00939t + 2.83)} \). A1
\( z_2 = 1.6 e^{i(0.00939t)} \). A1
[2 marks]
(h)(ii)
Compute \( z_1 – z_2 = e^{i(0.00939t)} (1.5 e^{i 2.83} – 1.6) \). M1
\( 1.5 e^{i 2.83} = 1.5 (\cos 2.83 + i \sin 2.83) \). A1
Magnitude: \( \sqrt{(1.5 \cos 2.83 – 1.6)^2 + (1.5 \sin 2.83)^2} \approx 3.06249 \). A1
Phase: \( \arctan\left(\dfrac{1.5 \sin 2.83}{1.5 \cos 2.83 – 1.6}\right) + \pi \approx 2.99086 \).
Thus, \( L(t) = 3.06 \sin(0.00939t + 2.99) + 12.5 \). A1
[4 marks]
Compute \( z_1 – z_2 = e^{i(0.00939t)} (1.5 e^{i 2.83} – 1.6) \). M1
\( 1.5 e^{i 2.83} = 1.5 (\cos 2.83 + i \sin 2.83) \). A1
Magnitude: \( \sqrt{(1.5 \cos 2.83 – 1.6)^2 + (1.5 \sin 2.83)^2} \approx 3.06249 \). A1
Phase: \( \arctan\left(\dfrac{1.5 \sin 2.83}{1.5 \cos 2.83 – 1.6}\right) + \pi \approx 2.99086 \).
Thus, \( L(t) = 3.06 \sin(0.00939t + 2.99) + 12.5 \). A1
[4 marks]
(h)(iii)
Shortest \( L(t) = d – p \). M1
\( 12.4833 – 3.06249 \approx 9.42084 \approx 9.42 \) hours. A1
[2 marks]
Shortest \( L(t) = d – p \). M1
\( 12.4833 – 3.06249 \approx 9.42084 \approx 9.42 \) hours. A1
[2 marks]
Total Marks: 27