Periodic Function Modeling - IB Mathematics Applications and Interpretation

1. The amplitude of the pendulum's swing decreases over time. To the nearest hundredth of a second, how long does it take for the amplitude to be halved?
   Answer: 13.86 seconds The amplitude of the oscillation is given by the function A(t) = $15e^{-0.05t}$. The initial amplitude (at t=0) is 15 cm. We need to find the time t when the amplitude is halved, i.e., A(t) = 7.5. $$15e^{-0.05t} = 7.5 \implies e^{-0.05t} = 0.5$$ Taking the natural logarithm of both sides: $$-0.05t = \ln(0.5) \implies t = \frac{\ln(0.5)}{-0.05} \approx 13.86 \text{ seconds.}$$
2. During the first two full revolutions, what is the total amount of time that the rider is above 40.5 metres?
   Answer: 26.67 s When you encounter a Ferris wheel problem involving trigonometric functions, you're dealing with periodic motion where you need to find when the function exceeds a certain threshold value. To find when the rider is above 40.5 meters, set up the inequality: $25\sin\left(\frac{\pi}{20}(t-10)\right) + 28 > 40.5$ Solving this: $25\sin\left(\frac{\pi}{20}(t-10)\right) > 12.5$, which gives $\sin\left(\frac{\pi}{20}(t-10)\right) > 0.5$ Since $\sin(\theta) > 0.5$ when $\frac{\pi}{6} < \theta < \frac{5\pi}{6}$, you have: $\frac{\pi}{6} < \frac{\pi}{20}(t-10) < \frac{5\pi}{6}$ Solving for t: $\frac{10}{3} < t-10 < \frac{50}{3}$, so $\frac{40}{3} < t < \frac{80}{3}$ This means the rider is above 40.5m from $t = \frac{40}{3}$ to $t = \frac{80}{3}$ seconds in each revolution. The duration per revolution is $\frac{80}{3} - \frac{40}{3} = \frac{40}{3}$ seconds. Since the period is $\frac{2\pi}{\pi/20} = 40$ seconds, two full revolutions take 80 seconds. The pattern repeats in the second revolution, so the total time above 40.5m is $2 \times \frac{40}{3} = \frac{80}{3} = 26.67$ seconds. Answer D (26.67s) is correct. Answer A (13.33s) gives only one revolution's worth. Answer B (20.00s) might come from incorrect trigonometric calculations. Answer C (40.00s) represents the full period duration, not the time above the threshold. Remember: in periodic motion problems, always identify the period first, then solve for one cycle and multiply by the number of cycles requested.
3. Which of the following functions best models the water depth, d(t)?
   Answer: $d(t) = 4\cos\left(\frac{\pi}{6}(t-9)\right) + 8$ The vertical shift (mean depth) is D = (12+4)/2 = 8 m. The amplitude is A = (12-4)/2 = 4 m. The time from a minimum to a maximum is half a period, so T/2 = 9 - 3 = 6 hours. The period is T = 12 hours. The angular frequency is B = 2π/12 = π/6. Using a cosine model, d(t) = Acos(B(t-C)) + D, the phase shift C corresponds to the time of a maximum. Since the maximum is at t=9, C=9. Thus, the model is d(t) = 4cos(π/6(t-9)) + 8.
4. Which of the following functions correctly models the moon's distance from the planet?
   Answer: $d(t) = 100\sin\left(\frac{\pi}{10}t\right) + 250$ The mean distance (vertical shift) is D = (350+150)/2 = 250. The amplitude is A = (350-150)/2 = 100. The period is T=20 days, so B = 2π/20 = π/10. The model is of the form d(t) = 100f(π/10(t-C)) + 250. The condition 'at t=0, moon is at mean distance' means the sinusoidal term is zero. This suggests a sine function with C=0. The condition 'moving towards its maximum distance' means the function is increasing at t=0. For d(t) = 100sin(π/10 t) + 250, d(0)=250 and the derivative d'(0) > 0, so it is increasing. This matches the conditions.
5. Which of the following correctly interprets the quantity D - A?
   Answer: The minimum height of the buoy. In the model h(t) = A\cos(B(t-C)) + D, D represents the vertical shift or the mean height. A represents the amplitude, which is the maximum displacement from the mean height. The maximum height is D + A. The minimum height occurs when the cosine term is -1, so h_min = A(-1) + D = D - A.
6. What is the maximum speed of the particle?
   Answer: 3.6 m/s When you encounter a displacement function and need to find maximum speed, remember that speed is the magnitude of velocity, and velocity is the derivative of displacement with respect to time. Given the displacement function $x(t) = 1.2\sin(3t - 0.5)$, you need to find the velocity by taking the derivative. Using the chain rule: $v(t) = \frac{dx}{dt} = 1.2 \cos(3t - 0.5) \cdot 3 = 3.6\cos(3t - 0.5)$ The maximum speed occurs when the absolute value of velocity is greatest. Since cosine oscillates between -1 and 1, the velocity ranges from -3.6 to +3.6 m/s. Therefore, the maximum speed is $|3.6| = 3.6$ m/s. Looking at the wrong answers: Choice A (1.2 m/s) represents the amplitude of the displacement function, not the speed. This is a common misconception—students often confuse the amplitude of position with maximum speed. Choice B (3.0 m/s) is the angular frequency coefficient from the original function, which appears in the derivative but isn't the final answer. Choice C (4.2 m/s) might result from incorrectly adding the amplitude and frequency (1.2 + 3.0), but this has no mathematical basis. The correct answer is D (3.6 m/s). Study tip: For sinusoidal motion, maximum speed always equals amplitude × angular frequency. If displacement is $A\sin(\omega t + \phi)$, then maximum speed is $A\omega$. Here, $A = 1.2$ and $\omega = 3$, so maximum speed is $1.2 \times 3 = 3.6$ m/s.
7. What is the fundamental period of the combined sound wave S(t)?
   Answer: $\frac{1}{110}$ s The period of the first component, $5\sin(220\pi t)$, is $T_1 = \frac{2\pi}{220\pi} = \frac{1}{110}$ s. The period of the second component, $3\sin(440\pi t)$, is $T_2 = \frac{2\pi}{440\pi} = \frac{1}{220}$ s. The period of the combined wave is the least common multiple (LCM) of the individual periods. LCM$(\frac{1}{110}, \frac{1}{220}) = \frac{1}{110}$ s. The combined pattern repeats every $\frac{1}{110}$ seconds.
8. During one full revolution, for how long is a passenger at least 32 metres above the ground?
   Answer: 10.0 seconds We need to solve the inequality h(t) ≥ 32. $$-20 \cos\left(\frac{\pi}{15} t\right) + 22 \ge 32 \implies -20 \cos\left(\frac{\pi}{15} t\right) \ge 10 \implies \cos\left(\frac{\pi}{15} t\right) \le -0.5$$ Using a GDC or inverse trigonometry, the interval for the argument $\frac{\pi}{15} t$ within $[0, 2\pi]$ is $[2\pi/3, 4\pi/3]$. Solving for t: $$\frac{2\pi}{3} \le \frac{\pi}{15} t \le \frac{4\pi}{3} \implies 10 \le t \le 20$$ The duration is 20 - 10 = 10 seconds.
9. What is the approximate range of the number of daylight hours in this city over one year?
   Answer: [8.7, 15.7] hours The function is a cosine wave with amplitude A=3.5 and vertical shift D=12.2. The maximum value is D + A = 12.2 + 3.5 = 15.7 hours. The minimum value is D - A = 12.2 - 3.5 = 8.7 hours. Therefore, the range of daylight hours is [8.7, 15.7].
10. Which function best models the displacement d(t) in cm from the equilibrium position at time t seconds? (Note: $e^{-0.1625} \approx 0.85$)
    Answer: $d(t) = 10e^{-0.1625t} \cos(\pi t)$ The initial displacement is +10 cm from equilibrium, so a cosine model is appropriate, with d(0)=10. The period is T=2s, so the angular frequency is B = 2π/T = π. The amplitude decays exponentially. If the amplitude is multiplied by 0.85 each second, the amplitude function is $A(t) = 10(0.85)^t$. This is equivalent to $A(t) = 10(e^{\ln(0.85)})^t = 10e^{\ln(0.85)t}$. Since $\ln(0.85) \approx -0.1625$, the model is $d(t) = 10e^{-0.1625t} \cos(\pi t)$.
11. For how many hours during a 24-hour period is the temperature above 31°C?
    Answer: 8 hours We need to solve the inequality T(t) > 31: $$-12\cos\left(\frac{\pi}{12}t\right) + 25 > 31 \implies -12\cos\left(\frac{\pi}{12}t\right) > 6 \implies \cos\left(\frac{\pi}{12}t\right) < -0.5$$ The argument $\frac{\pi}{12}t$ must be in the interval $(\frac{2\pi}{3}, \frac{4\pi}{3})$. Solving for t: $$\frac{2\pi}{3} < \frac{\pi}{12}t < \frac{4\pi}{3} \implies 8 < t < 16$$ The duration is 16 - 8 = 8 hours.
12. What is the predicted revenue, in dollars, for the 27th month?
    Answer: $132,500 Substitute t=27 into the revenue function: $$R(27) = 2.5(27) + 50 + 15\sin\left(\frac{\pi}{6} \cdot 27\right)$$ $$R(27) = 67.5 + 50 + 15\sin\left(\frac{9\pi}{2}\right)$$ Since $\sin(x) = \sin(x - 2k\pi)$, $\sin(\frac{9\pi}{2}) = \sin(\frac{\pi}{2} + 4\pi) = \sin(\frac{\pi}{2}) = 1$. $$R(27) = 117.5 + 15(1) = 132.5$$ The revenue is in thousands of dollars, so the predicted revenue is $132,500.
13. Find the first time after midnight (t=0) that the height of the tide is exactly 8 metres.
    Answer: 2.23 hours The time between high and low tide is half a period, so T/2 = 6 hours, which means the period T = 12 hours. B = 2π/T = 2π/12 = π/6. The maximum height is D+A = 6.5+A = 9, so the amplitude A=2.5. The model is H(t) = 2.5cos(π/6(t-4)) + 6.5. We solve H(t) = 8: $$2.5\cos(\frac{\pi}{6}(t-4)) + 6.5 = 8 \implies \cos(\frac{\pi}{6}(t-4)) = 0.6$$ Let u = π/6(t-4). Then u = arccos(0.6) ≈ ±0.9273 + 2kπ. For the first positive t, we test u = -0.9273: $\frac{\pi}{6}(t-4) = -0.9273 \implies t-4 \approx -1.771 \implies t \approx 2.229$. The other solution (from u = +0.9273) gives t ≈ 5.771. The first time is t ≈ 2.23 hours.
14. What is the frequency of the beats produced?
    Answer: 4 Hz The beat frequency is the absolute difference between the two individual frequencies. Given $f_1 = 254$ Hz and $f_2 = 258$ Hz, the beat frequency is $f_{beat} = |f_2 - f_1| = |258 - 254| = 4$ Hz. This means the sound will get louder and softer 4 times per second.
15. What is the first positive time t, in seconds, at which the voltage is exactly +100 V?
    Answer: 0.00458 s Set V(t) = 100 and solve for t: $$170 \cos(120\pi t - \frac{\pi}{4}) = 100 \implies \cos(120\pi t - \frac{\pi}{4}) = \frac{10}{17}$$ Let $u = 120\pi t - \frac{\pi}{4}$. The principal value is $u = \arccos(\frac{10}{17}) \approx 0.9416$ radians. The general solutions for u are $\pm 0.9416 + 2n\pi$. We test the two cases for the first positive t. Case 1: $120\pi t - \frac{\pi}{4} = 0.9416 \implies 120\pi t \approx 1.7270 \implies t \approx 0.00458$. Case 2: $120\pi t - \frac{\pi}{4} = -0.9416 \implies 120\pi t \approx -0.1562 \implies t \approx -0.00041$. The first positive time is approximately 0.00458 s.
16. According to the models, what is the approximate fox population when the rabbit population is at its first peak?
    Answer: 506 The rabbit population, R(t), is at its first peak when the sine term is 1. This occurs when $\frac{\pi}{6}t = \frac{\pi}{2}$, which gives t = 3 months. To find the fox population at this time, we substitute t=3 into the function F(t): $$F(3) = 150 \sin\left(\frac{\pi}{6}(3-1.5)\right) + 400 = 150 \sin\left(\frac{\pi}{4}\right) + 400$$ $$F(3) = 150\left(\frac{\sqrt{2}}{2}\right) + 400 \approx 106.07 + 400 \approx 506$$
17. What is the average value of the current over the first positive half-cycle of the function?
    Answer: 9.55 A When you encounter AC circuit problems involving sinusoidal functions, you need to distinguish between instantaneous values and average values over specific intervals. The key insight is that while a complete sine cycle averages to zero, a half-cycle has a non-zero average. To find the average value of $I(t) = 15\sin(100\pi t)$ over the first positive half-cycle, you need to determine when this half-cycle occurs and then integrate. The function completes one full cycle when $100\pi t = 2\pi$, so $t = \frac{1}{50}$ seconds. The first positive half-cycle runs from $t = 0$ to $t = \frac{1}{100}$ seconds. The average value formula gives us: $\text{Average} = \frac{1}{\frac{1}{100} - 0} \int_0^{1/100} 15\sin(100\pi t) \, dt$ Evaluating this integral: $100 \cdot 15 \int_0^{1/100} \sin(100\pi t) \, dt = 1500 \left[-\frac{\cos(100\pi t)}{100\pi}\right]_0^{1/100} = \frac{15}{\pi}[\cos(0) - \cos(\pi)] = \frac{30}{\pi} \approx 9.55$ A. Choice A (0 A) would be correct for a complete cycle, not a half-cycle. Choice B (7.5 A) incorrectly assumes the average is half the amplitude. Choice C (15 A) confuses the average with the maximum amplitude. Choice D is correct at 9.55 A. Remember: For sinusoidal functions, the average over a half-cycle is $\frac{2A}{\pi}$ where A is the amplitude—a useful formula to memorize for AC circuit problems.
18. Which statement accurately describes the relationship between the peak sales months in the two cities?
    Answer: Peak sales in the South occur 6 months before peak sales in the North. Peak sales occur when the cosine term is 1. For the North, this is when $\frac{\pi}{6}(t-7) = 0$, so t=7 (July). For the South, this is when $\frac{\pi}{6}(t-1) = 0$, so t=1 (January). The difference in peak times is 7 - 1 = 6 months. Since January (t=1) comes before July (t=7), the peak sales in the South occur 6 months before the peak in the North, reflecting the opposite seasons.

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