Sinusoidal Modeling - Math 3

1. The water level in a tidal pool varies sinusoidally. At 6 AM, the water is 2.3 feet deep and rising. At 12:30 PM (6.5 hours later), the water reaches its maximum depth of 5.1 feet. What will be the approximate water depth at 3 PM?
   Answer: 4.2 feet, since the tide is still relatively high in mid-afternoon When you encounter sinusoidal motion problems, you need to establish the function's key parameters: amplitude, period, phase shift, and vertical shift. This tidal pool follows a predictable wave pattern that repeats over time. Given that the water is 2.3 feet at 6 AM (rising) and reaches maximum depth of 5.1 feet at 12:30 PM, you can determine the function structure. The amplitude is $\frac{5.1 - \text{minimum}}{2}$, and since the water was rising at 6 AM, this point occurs partway up the sinusoidal curve. The maximum occurs 6.5 hours later, indicating this is roughly a quarter-cycle from the rising midpoint to the peak. For a typical tidal pattern, if maximum depth is 5.1 feet and the water was at 2.3 feet while rising, the minimum depth would be approximately 1.3 feet (making the midline around 3.2 feet). By 3 PM (9 hours after 6 AM, or 2.5 hours after the maximum), the tide would be declining but still relatively high, placing the water level around 4.2 feet. Option A (3.8 feet) underestimates how high the tide remains in mid-afternoon. Option B (1.7 feet) incorrectly assumes 3 PM is near the daily minimum, which would occur much later. Option C (2.9 feet) overestimates the decline rate from the maximum. Remember: sinusoidal functions change gradually near their peaks and troughs. The steepest changes occur at the midpoints, so water levels remain relatively high for several hours after reaching maximum depth.
2. The angular position of a pendulum is given by $\theta(t) = 0.15\cos(3.14t + 0.52)$ radians, where $t$ is in seconds. At $t = 2$ seconds, the pendulum is moving in which direction and at what angular speed?
   Answer: Moving clockwise at 0.31 rad/s, as determined by the negative derivative value When you encounter a question about the motion of an oscillating object like a pendulum, you need to find the angular velocity by taking the derivative of the position function. The sign of this velocity tells you the direction of motion. Given $\theta(t) = 0.15\cos(3.14t + 0.52)$, the angular velocity is found by differentiating: $\omega(t) = \frac{d\theta}{dt} = -0.15 \times 3.14 \times \sin(3.14t + 0.52) = -0.471\sin(3.14t + 0.52)$ At $t = 2$ seconds: $\omega(2) = -0.471\sin(3.14 \times 2 + 0.52) = -0.471\sin(6.80)$ Since $\sin(6.80) \approx 0.657$, we get $\omega(2) = -0.471 \times 0.657 \approx -0.31$ rad/s. The negative sign indicates clockwise motion, and the magnitude gives us the angular speed of 0.31 rad/s, confirming answer C. Answer A incorrectly attempts to determine direction from position rather than velocity. Answer B makes a sign error in the derivative calculation, leading to a positive velocity and wrong direction. Answer D misinterprets the phase shift (0.52) as indicating direction, which is incorrect—direction comes solely from the sign of the derivative. Remember: for oscillatory motion problems, always differentiate the position function to find velocity. The sign of the velocity (positive or negative) determines the direction of motion, while its magnitude gives you the speed.
3. A child's swing reaches a maximum height of 4 feet above ground at the back of its arc and 1.5 feet at the front. The swing completes one full back-and-forth cycle every 3 seconds. If we define $t = 0$ when the swing is at its highest point moving forward, which function models the swing's height $h(t)$?
   Answer: $h(t) = 1.25\cos\left(\frac{2\pi t}{3}\right) + 2.75$, with maximum height occurring at $t = 0$ When modeling periodic motion like a swing, you need to identify three key components: amplitude (how far from center), period (time for one complete cycle), and vertical shift (average height). The choice between sine and cosine depends on what's happening at your starting point. First, let's find these values. The swing goes from 1.5 feet (minimum) to 4 feet (maximum), so the amplitude is $\frac{4-1.5}{2} = 1.25$ feet, and the vertical shift (average height) is $\frac{4+1.5}{2} = 2.75$ feet. With a 3-second period, the frequency term is $\frac{2\pi}{3}$. The crucial detail is that $t = 0$ occurs "when the swing is at its highest point." Since cosine starts at its maximum value when there's no phase shift, and the swing starts at maximum height, we need $h(t) = 1.25\cos\left(\frac{2\pi t}{3}\right) + 2.75$. This gives us $h(0) = 1.25(1) + 2.75 = 4$ feet, which matches our maximum height. Choice A uses sine with a phase shift, which is unnecessarily complicated when cosine naturally starts at maximum. Choice B has the right structure but uses $-1.25\cos$, which would start at the minimum height (1.5 feet) instead of maximum. Choice D uses sine without phase shift, starting the swing at average height rather than maximum height. Remember: cosine starts at maximum, sine starts at zero (average). Choose your trigonometric function based on where your periodic motion begins in its cycle.
4. A sound engineer models the air pressure variation of a musical note as $P(t) = 0.02\sin(880\pi t)$, where $P$ is pressure deviation in pascals and $t$ is time in seconds. If the human ear can detect pressure changes as small as 0.01 pascals, what fraction of each cycle is the sound audible?
   Answer: $\frac{2}{3}$ The sound is audible when $|P(t)| \geq 0.01$, so $|0.02\sin(880\pi t)| \geq 0.01$, which means $|\sin(880\pi t)| \geq 0.5$. This occurs when $\sin(880\pi t) \geq 0.5$ or $\sin(880\pi t) \leq -0.5$. In one period of sine, $|\sin(x)| \geq 0.5$ occurs for $x \in [\pi/6, 5\pi/6] \cup [7\pi/6, 11\pi/6]$, which spans $\frac{4\pi/6 + 4\pi/6}{2\pi} = \frac{2\pi/3}{\pi} = \frac{2}{3}$ of each cycle.
5. A Ferris wheel has a diameter of 80 feet and its center is 50 feet above the ground. It completes one full rotation every 12 minutes. If a passenger boards at the lowest point and the wheel begins turning counterclockwise, which function best models the passenger's height above ground after $t$ minutes?
   Answer: $h(t) = -40\cos\left(\frac{\pi t}{6}\right) + 50$ The amplitude is 40 (radius), vertical shift is 50 (center height), and period gives $\omega = \frac{2\pi}{12} = \frac{\pi}{6}$. Since the passenger starts at the lowest point (10 feet), we need a function that equals 10 when $t = 0$. Only $h(t) = -40\cos\left(\frac{\pi t}{6}\right) + 50$ satisfies this: $h(0) = -40(1) + 50 = 10$. Choice A gives $h(0) = 50$, choice C gives $h(0) = 90$, and choice D gives $h(0) = 50$.
6. A Ferris wheel has a radius of 25 meters and its center is 30 meters above the ground. The wheel completes one revolution every 4 minutes. If a passenger boards at the lowest point and the wheel starts moving, what function models the passenger's height $h(t)$ above ground after $t$ minutes?
   Answer: $h(t) = -25\cos\left(\frac{\pi t}{2}\right) + 30$ At $t = 0$, the passenger is at the lowest point, which is $30 - 25 = 5$ meters above ground. The period is 4 minutes, so $B = \frac{2\pi}{4} = \frac{\pi}{2}$. Since we start at the minimum and use a cosine function, we need $h(t) = -25\cos\left(\frac{\pi t}{2}\right) + 30$. This gives $h(0) = -25(1) + 30 = 5$ meters (correct starting position) and $h(2) = -25(-1) + 30 = 55$ meters (maximum height).
7. The number of hours of daylight in Anchorage, Alaska varies sinusoidally throughout the year, with a maximum of 19.5 hours on June 21st (day 172) and a minimum of 5.5 hours on December 21st (day 355). Which equation best models $D(d)$, the hours of daylight on day $d$ of the year?
   Answer: $D(d) = 7\cos\left(\frac{2\pi(d-172)}{365}\right) + 12.5$ Amplitude = (19.5 - 5.5)/2 = 7, vertical shift = (19.5 + 5.5)/2 = 12.5, period = 365 days so $\omega = \frac{2\pi}{365}$. Since maximum occurs at day 172, we use cosine with phase shift: $D(d) = 7\cos\left(\frac{2\pi(d-172)}{365}\right) + 12.5$. Choice B uses sine with wrong phase shift. Choice C has wrong amplitude (14 instead of 7). Choice D has wrong period (366 instead of 365).
8. A population of seasonal workers in a resort town varies according to $P(t) = 2400\cos\left(\frac{\pi(t-6)}{6}\right) + 3600$, where $t$ represents the month (January = 1, February = 2, etc.). The town needs to maintain infrastructure for at least 4500 people. During how many months of the year does the population exceed this threshold?
   Answer: 6 months, covering the extended tourist season from April through September We need $P(t) > 4500$, so $2400\cos\left(\frac{\pi(t-6)}{6}\right) + 3600 > 4500$. This gives $\cos\left(\frac{\pi(t-6)}{6}\right) > \frac{900}{2400} = 0.375$. The cosine is greater than 0.375 when $\frac{\pi(t-6)}{6}$ is between $-\arccos(0.375)$ and $+\arccos(0.375)$, approximately $\pm 1.159$ radians. Solving gives approximately $t$ between 3.79 and 8.21, which corresponds to 6 months (April through September). Choice A misses the shoulder months, C undercounts, and D is too restrictive.
9. A tuning fork produces a sound wave with frequency 440 Hz. The air pressure variation can be modeled by $P(t) = 0.02\sin(880\pi t) + 14.7$, where $P$ is in psi and $t$ is in seconds. What does the coefficient 0.02 represent in this physical context?
   Answer: The maximum deviation in air pressure from atmospheric pressure, measured in psi In the sinusoidal model $P(t) = 0.02\sin(880\pi t) + 14.7$, the coefficient 0.02 is the amplitude, representing the maximum deviation from the average pressure (14.7 psi). Choice B incorrectly relates amplitude to frequency calculation. Choice C confuses amplitude with the derivative (rate of change). Choice D incorrectly describes amplitude as total variation rather than maximum deviation from center.

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