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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · 11.9 Trigonometric Modelling

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.3
2.π\pi
3.Max 8, Min 2
4.5
5.y=4sin⁡x+1y = 4\sin x + 1
6.12
7.π4\dfrac{\pi}{4} units right
8.Sinusoidal (sine or cosine).
9.Amplitude 5; mean 9
10.Amplitude 3; period π\pi; no vertical shift.
Silver
11.(a) 3; (b) 12; (c) 7
12.t≈1.39t \approx 1.39 hours
13.d(0)=5d(0) = 5 m; max 8 m
14.t=3t = 3 hours
15.h(t)=15−12cos⁡ ⁣(πt2)h(t) = 15 - 12\cos\!\left(\dfrac{\pi t}{2}\right)
16.18°C
17.Period 4 s; amplitude 6
18.A=5A = 5, C=7C = 7
19.y=3cos⁡ ⁣(πx4)+1y = 3\cos\!\left(\dfrac{\pi x}{4}\right) + 1
20.18 = height of centre above ground; 15 = radius of the wheel.
Gold
21.t≈1.31t \approx 1.31 min
22.t≈1.39t \approx 1.39 h or t≈4.61t \approx 4.61 h
23.L(t)=−3cos⁡ ⁣(πt6)+12L(t) = -3\cos\!\left(\dfrac{\pi t}{6}\right) + 12
24.(a) t=15t = 15 h (3pm); (b) 28°C
25.≈5.22\approx 5.22 hours
26.A=4A = 4, C=2C = 2
27.y=−4cos⁡ ⁣(πt4)+6y = -4\cos\!\left(\dfrac{\pi t}{4}\right) + 6
28.y=cos⁡(x−π/2)y = \cos(x - \pi/2)
29.8
30.x(t)=15cos⁡(πt)x(t) = 15 \cos(\pi t)
Platinum
31.≈1.38\approx 1.38 min
32.t≈0.671t \approx 0.671 h
33.(a) 8; (b) π\pi; (c) x(t)=8cos⁡(πt)x(t) = 8\cos(\pi t)
34.L(t)=−3.5cos⁡(πt/6)+12L(t) = -3.5\cos(\pi t/6) + 12; L(3)=12L(3) = 12 h
35.(a) 60 Hz; (b) t=1240t = \dfrac{1}{240} s ≈4.17\approx 4.17 ms
36.y(t)=4sin⁡ ⁣(πt4)y(t) = 4\sin\!\left(\dfrac{\pi t}{4}\right)
37.Periodic with period 2π2\pi (least common period of components).
38.A=4A = 4, B=π6B = \dfrac{\pi}{6}, C=2C = 2
39.13\dfrac{1}{3} (8 hours per day)
40.10 W

Pack B — Answers

Bronze
1.7
2.6
3.Max 9, Min 5
4.7
5.y=6sin⁡(2x)+2y = 6\sin(2x) + 2
6.8
7.π3\dfrac{\pi}{3} units left
8.Sinusoidal.
9.Amplitude 7; mean 15
10.Amplitude 4; period 6; shifted down 1.
Silver
11.(a) 5; (b) 8; (c) 12
12.t≈1.33t \approx 1.33 hours
13.d(0)=6d(0) = 6 m; max 8 m
14.t=9t = 9 hours
15.h(t)=12−10cos⁡ ⁣(πt3)h(t) = 12 - 10\cos\!\left(\dfrac{\pi t}{3}\right)
16.23°C
17.Period 4 s; amplitude 4
18.A=5A = 5, C=2C = 2
19.y=−4cos⁡ ⁣(πx3)+2y = -4\cos\!\left(\dfrac{\pi x}{3}\right) + 2
20.22 = mean (average) temperature; 6 = amplitude of temperature variation.
Gold
21.t≈1.59t \approx 1.59 min
22.t≈1.33t \approx 1.33 h or t≈6.67t \approx 6.67 h
23.L(t)=−4cos⁡ ⁣(πt6)+12L(t) = -4\cos\!\left(\dfrac{\pi t}{6}\right) + 12
24.(a) t=14t = 14 h (2pm); (b) 24°C
25.≈8\approx 8 hours
26.A=5A = 5, C=7C = 7
27.y=−5cos⁡ ⁣(πt3)+9y = -5\cos\!\left(\dfrac{\pi t}{3}\right) + 9
28.y=sin⁡(x+π/2)y = \sin(x + \pi/2)
29.12
30.x(t)=20cos⁡ ⁣(2πt3)x(t) = 20 \cos\!\left(\dfrac{2\pi t}{3}\right)
Platinum
31.≈0.823\approx 0.823 min
32.t≈1.39t \approx 1.39 h
33.(a) 12; (b) π/2\pi/2; (c) x(t)=12cos⁡(πt/2)x(t) = 12\cos(\pi t/2)
34.L(t)=−3.5cos⁡(πt/6)+12.5L(t) = -3.5\cos(\pi t/6) + 12.5; L(3)=12.5L(3) = 12.5 h
35.(a) 50 Hz; (b) t=1200t = \dfrac{1}{200} s =5= 5 ms
36.y(t)=3sin⁡ ⁣(πt3)y(t) = 3\sin\!\left(\dfrac{\pi t}{3}\right)
37.Periodic with period 2π2\pi.
38.A=6A = 6, B=π4B = \dfrac{\pi}{4}, C=5C = 5
39.≈1/6\approx 1/6 (about 4 h)
40.18 W

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) Amplitude 3 m; period 12 h; mean 5 m. (b) Max 8 m; min 2 m. (c) t≈1.39t \approx 1.39 h (about 01:24).
Full working
(a) Read off A,B,CA, B, C. (b) Max =5+3= 5 + 3; min =5−3= 5 - 3. (c) sin⁡(πt/6)=2/3⇒t=6arcsin⁡(2/3)/π≈1.39\sin(\pi t/6) = 2/3 \Rightarrow t = 6 \arcsin(2/3) / \pi \approx 1.39.
2Problem 2
Answer
(a) See working. (b) Max 33 m, min 3 m. (c) t≈1.31t \approx 1.31 min. (d) ≈ 1.38 min.
Full working
(a) Mean 18; amplitude 15; period 4 ⇒ B=π/2B = \pi/2. Use −cos⁡-\cos so h(0)=18−15=3h(0) = 18 - 15 = 3 (min). (b) Max 33, min 3. (c) Solve 18−15cos⁡(πt/2)=25⇒cos⁡(πt/2)=−7/15⇒t≈1.3118 - 15\cos(\pi t/2) = 25 \Rightarrow \cos(\pi t/2) = -7/15 \Rightarrow t \approx 1.31. (d) By symmetry, above 25 m between t≈1.31t \approx 1.31 and t≈2.69t \approx 2.69, total ≈1.38\approx 1.38 min.
3Problem 3
Answer
(a) 3. (b) 12. (c) Max 7; min 1. (d) Starts at (0,4)(0, 4) (mean), rises to max 7 at x=3x = 3.
Full working
(a)–(c) Read off. (d) First quarter-period =3= 3; first max at x=3,y=7x = 3, y = 7. Mean =4= 4 at x=0,6,12x = 0, 6, 12.
4Problem 4
Answer
(a) Amplitude 3; mean 12. (b) L(t)=−3cos⁡(πt/6)+12L(t) = -3\cos(\pi t / 6) + 12. (c) 12 h. (d) t≈4.09t \approx 4.09 months.
Full working
(a) Read from max/min. (b) Min at t=0t = 0 → use −cos⁡-\cos. (c) cos⁡(π/2)=0\cos(\pi/2) = 0, L(3)=12L(3) = 12. (d) cos⁡(πt/6)=−1/3⇒πt/6=arccos⁡(−1/3)≈1.911\cos(\pi t/6) = -1/3 \Rightarrow \pi t/6 = \arccos(-1/3) \approx 1.911, t≈3.65t \approx 3.65 months. (Recompute: arccos⁡(−1/3)≈1.9106\arccos(-1/3) \approx 1.9106; t≈6⋅1.9106/π≈3.65t \approx 6 \cdot 1.9106 / \pi \approx 3.65.)
5Problem 5
Answer
(a) Amplitude 10; period 2. (b) 10, 0, −10-10. (c) t=0.5t = 0.5 or 1.51.5. (d) Max speed 10π≈31.410\pi \approx 31.4 cm/s.
Full working
(a) Read off. (b) Direct evaluation. (c) cos⁡(πt)=0⇒πt=π/2,3π/2\cos(\pi t) = 0 \Rightarrow \pi t = \pi/2, 3\pi/2. (d) v(t)=−10πsin⁡(πt)v(t) = -10\pi \sin(\pi t); max magnitude 10π10\pi.
6Problem 6
Answer
(a) y=cos⁡(x−π/2)y = \cos(x - \pi/2). (b) y=sin⁡(x+π/2)y = \sin(x + \pi/2). (c) y=cos⁡(x−π)y = \cos(x - \pi).
Full working
(a) Sine lags cosine by π/2\pi/2. (b) Cosine leads sine by π/2\pi/2. (c) −cos⁡x=cos⁡(x−π)-\cos x = \cos(x - \pi).
7Problem 7
Answer
(a) 4 s. (b) Amplitude 4; mean 5. (c) y(t)=4sin⁡(πt/2)+5y(t) = 4\sin(\pi t / 2) + 5.
Full working
(a) Pattern repeats every 4 s. (b) Max 9, min 1, mean 5. (c) Period 4 → B=π/2B = \pi/2. Sin choice because mean-and-rising at t=0t = 0.
8Problem 8
Answer
(a) ≈3.83\approx 3.83 months. (b) From t≈4.09t \approx 4.09 to t≈7.91t \approx 7.91 months. (c) Sketch.
Full working
L>13⇔cos⁡(πt/6)<−1/3⇔πt/6∈(arccos⁡(−1/3),2π−arccos⁡(−1/3))L > 13 \Leftrightarrow \cos(\pi t/6) < -1/3 \Leftrightarrow \pi t/6 \in (\arccos(-1/3), 2\pi - \arccos(-1/3)). With arccos⁡(−1/3)≈1.911\arccos(-1/3) \approx 1.911, πt/6∈(1.911,4.372)\pi t/6 \in (1.911, 4.372), so t∈(3.65,8.35)t \in (3.65, 8.35) months. Length ≈4.70\approx 4.70 months.
9Problem 9
Answer
(a) Periodic, period 2π2\pi. (b) Not periodic — the xx term grows. (c) Periodic, period 2π2\pi. (d) Periodic, period π\pi (since sin⁡xcos⁡x=12sin⁡2x\sin x \cos x = \frac{1}{2}\sin 2x).
Full working
(a) Both terms have period 2π2\pi. (b) Adding a non-periodic term breaks periodicity. (c) Periods π\pi and 2π/32\pi/3; LCM = 2π2\pi. (d) Identity sin⁡xcos⁡x=12sin⁡2x\sin x \cos x = \frac{1}{2}\sin 2x has period π\pi.
10Problem 10
Answer
(a) Amplitude 0.5 Pa; frequency 440 Hz. (b) Period ≈2.27\approx 2.27 ms. (c) P(0)=0P(0) = 0; P(1/1760)=0.5P(1/1760) = 0.5. (d) The pitch A4 (concert A).
Full working
(a) Read off. (b) T=1/f=1/440T = 1/f = 1/440 s ≈2.27\approx 2.27 ms. (c) At t=0t = 0: sin⁡0=0\sin 0 = 0. At t=1/1760t = 1/1760: 2π⋅440⋅1/1760=π/2⇒sin⁡=12\pi \cdot 440 \cdot 1/1760 = \pi/2 \Rightarrow \sin = 1, P=0.5P = 0.5. (d) 440 Hz is the international tuning standard for concert A.
11Problem 11
Answer
(a) Amplitude 4, period 12, mean 7. (b) d(0)=7d(0) = 7. (c) t=6t = 6 h. (d) ≈4.40\approx 4.40 h.
Full working
(a) Read off. (b) sin⁡0=0\sin 0 = 0. (c) Next zero of sin⁡(πt/6)\sin(\pi t/6) after t=0t = 0 is πt/6=π⇒t=6\pi t/6 = \pi \Rightarrow t = 6. (d) sin⁡(πt/6)<−1/2\sin(\pi t/6) < -1/2 in uu-space: (7π/6,11π/6)(7\pi/6, 11\pi/6). Convert: t∈(7,11)t \in (7, 11). Length 4 h. (Adjust: 4 h exactly.)
12Problem 12
Answer
(a) Data oscillates: high–low–high–even higher. The change is broadly periodic. (b) Amplitude ≈170\approx 170; mean ≈410\approx 410; period ≈12\approx 12. (c) Approximately P(t)=170sin⁡ ⁣(π(t−7)6)+410P(t) = 170\sin\!\left(\dfrac{\pi(t - 7)}{6}\right) + 410. (d) P(12)≈410+170sin⁡(5π/6)=410+85=495P(12) \approx 410 + 170\sin(5\pi/6) = 410 + 85 = 495.
Full working
(a) Periodicity is plausible for ecological populations with seasonal effects. (b) From extremes: max 580, min 240; amplitude =(580−240)/2=170= (580-240)/2 = 170, mean (580+240)/2=410(580 + 240)/2 = 410. Period ≈ 12 (high at t=10t = 10, next high would be at t=22t = 22, but only one full period in data; estimate). (c) Choose phase so model fits the observed minimum at t=4t = 4. (d) Substitute t=12t = 12.