Precalculus · F.TF.B.5

Tides, daylight and sound

The water in a harbor rises and falls twice a day, the hours of daylight grow and shrink over a year, and a pure tone pushes the air back and forth hundreds of times a second. How do a few facts about such a cycle become a formula that predicts it?

Where you are 3 in Modeling things that repeat

Trigonometry7

Trigonometric functions and their graphs8

Modeling things that repeat3

Height on a turning wheel Simple harmonic motion

Tides, daylight and sound

When you reach for it

Reach for this move when a quantity rises and falls on a steady cycle and you know four facts about it: its highest value, its lowest value, how long one cycle lasts, and when it peaks. The depth of water at a harbor, the hours of daylight over a year and the pressure of the air carrying a pure tone all behave this way. A cosine with those four facts built in describes the whole cycle, and then it answers two kinds of question: what the value is at a given time, and when the value reaches a given level.

The move puts together four pieces you already have: the Amplitude and reflection, the midline, the period from the value of b and the phase shift. Writing an equation from a wave reads the same four numbers off a picture, and Height on a turning wheel builds them from a wheel; here they arrive as facts in words. The harbor and the town in this lesson are made up, and their numbers are illustrative, not measurements of a real place.

The procedure

Building the model. Call the highest value M, the lowest value m, the period P, and a time at which the quantity peaks c. Then the model is

y = A\cos\bigl(b(t - c)\bigr) + k

with

\begin{aligned} A &= \frac{M - m}{2}, \quad k = \frac{M + m}{2}, \\ b &= \frac{2\pi}{P} \end{aligned}

Each piece is forced. The cosine of an angle is largest, 1, when the angle is 0; here the angle is b(t - c), which is 0 exactly when t = c, so the model peaks at time c. That is why a cosine is the natural choice when a peak time is known: the peak time goes straight in as the phase shift, and A stays positive. As \cos runs between -1 and 1, the output runs between k - A and k + A, and the formulas give

\begin{aligned} k + A &= \tfrac{M + m}{2} + \tfrac{M - m}{2} = M \\ k - A &= \tfrac{M + m}{2} - \tfrac{M - m}{2} = m \end{aligned}

Finally the angle grows by 2\pi, one full turn of cosine, when t grows by \displaystyle \frac{2\pi}{b} = P, so one cycle lasts P.

For sound the cycle is usually given as a frequency f, the number of cycles per second, measured in hertz (\text{Hz}). Then \displaystyle P = \frac{1}{f} and b = 2\pi f.

Predicting. To predict a value, substitute the time for t and evaluate, with the calculator in radian mode, since b is in radians per unit of time.

Finding when a level is crossed. To find when the quantity equals a level L, write \theta = b(t - c) and solve

A\cos\theta + k = L

so that

\cos\theta = \frac{L - k}{A}

A cosine lies between -1 and 1, so there is a solution only when m \le L \le M. Let \displaystyle \alpha = \arccos\frac{L - k}{A}, the one angle in [0, \pi] given by the inverse cosine. Cosine is even, so -\alpha works too, and on the cycle -\pi \le \theta \le \pi around a peak these are the only two solutions. Every solution repeats when \theta grows by 2\pi, that is when t grows by P, so the crossings are

t = c \pm \frac{\alpha}{b} + nP

for n any integer.

On [0, \pi] the cosine is decreasing, so \cos\theta > \cos\alpha exactly when |\theta| < \alpha. The quantity is above L between the two crossings around each peak, for a time of \displaystyle \frac{2\alpha}{b} in every cycle. The exception is a level at the very top or bottom: when L = M the angle \alpha is 0 and when L = m it is \pi, the two crossings merge, and the level is touched only once per cycle. This is the method of Solving trigonometric equations, applied to a model.

Worked examples

Example 1

Nergui’s fjord tide

At a harbor in a made-up fjord, high tide is 4.6 meters at 3:00 a.m., low tide is 0.8 meters, and one tide cycle lasts 12.4 hours. With t in hours after midnight, write a model for the depth h and find the times of that day’s low tides and its second high tide.

Solution

The highest and lowest depths give the amplitude and the midline:

\begin{aligned} A &= \frac{4.6 - 0.8}{2} = 1.9 \\ k &= \frac{4.6 + 0.8}{2} = 2.7 \end{aligned}

The period is \displaystyle 12.4 = \frac{62}{5} hours, so \displaystyle b = 2\pi \cdot \frac{5}{62} = \frac{5\pi}{31}. A peak is at t = 3, so c = 3:

h(t) = 1.9\cos\left(\frac{5\pi}{31}(t - 3)\right) + 2.7

The cosine reaches -1 when its angle is \pi, half a turn after the peak, so low tide comes half a period, 6.2 hours, after high tide: at t = 9.2, which is 9:12 a.m., since 0.2 hours is 12 minutes. The next high tide is one period after the first, at t = 15.4, 3:24 p.m., and the second low tide at t = 21.6, 9:36 p.m. Check: h(9.2) = 1.9\cos\pi + 2.7 = 0.8, the low tide.

Example 2

Depth at noon

Use Nergui’s model to predict the depth at noon.

Solution

Noon is t = 12, so the angle is \displaystyle \frac{5\pi}{31} \cdot 9 = \frac{45\pi}{31} \approx 4.560 radians, and

\begin{aligned} h(12) &= 1.9\cos\frac{45\pi}{31} + 2.7 \\ &\approx 1.9(-0.1514) + 2.7 \\ &\approx 2.41 \end{aligned}

The depth at noon is about 2.41 meters. That fits the cycle: low tide was at 9:12 a.m., and the water returns to the midline 2.7 a quarter period, 3.1 hours, later, at 12:18 p.m. So at noon the water is rising and should be just below 2.7 meters, as it is.

Example 3

When the boat can leave

Nergui’s boat needs at least 3.5 meters of water to leave the harbor. Between midnight and the next midnight, when may it leave?

Solution

Set the model equal to 3.5:

\begin{aligned} 1.9\cos\theta + 2.7 &= 3.5 \\ \cos\theta &= \frac{0.8}{1.9} = \frac{8}{19} \end{aligned}

The level 3.5 lies between 0.8 and 4.6, so there are solutions. The inverse cosine gives \displaystyle \alpha = \arccos\frac{8}{19} \approx 1.1362 radians, and dividing by b turns that angle into hours:

\frac{\alpha}{b} \approx 1.1362 \cdot \frac{31}{5\pi} \approx 2.242

The water is above 3.5 meters within 2.242 hours on either side of each high tide. Around the high tide at t = 3 that is from t \approx 0.758 to t \approx 5.242, about 12:45 a.m. to 5:15 a.m. Around the high tide at t = 15.4 it is from t \approx 13.158 to t \approx 17.642, about 1:09 p.m. to 5:39 p.m. The next high tide, at t = 27.8, opens its window after midnight, so it does not count. Each window lasts 2 \cdot 2.242 \approx 4.48 hours.

Figure 1Nergui’s tide model. The water is at least 3.5 meters deep while the curve is above the dashed line, a window of about 4.5 hours around each high tide.

Example 4

Solongo’s daylight

In a made-up town the longest day has 15.2 hours of daylight, on day 172 of the year (June 21 in a year that is not a leap year); the shortest has 9.2 hours; the cycle is 365 days. Write a model for the hours of daylight D on day t, predict day 80, and find the days with at least 13.7 hours of daylight.

Solution

Here \displaystyle A = \frac{15.2 - 9.2}{2} = 3, \displaystyle k = \frac{15.2 + 9.2}{2} = 12.2, \displaystyle b = \frac{2\pi}{365} and c = 172:

D(t) = 3\cos\left(\frac{2\pi}{365}(t - 172)\right) + 12.2

On day 80 the angle is \displaystyle \frac{2\pi}{365}(-92) \approx -1.5837, whose cosine is about -0.0129, so D(80) \approx 12.16 hours. That is near the midline, as it should be: day 80 is 92 days before the peak, close to a quarter of a year.

For the level, 3\cos\theta + 12.2 = 13.7 gives \displaystyle \cos\theta = \frac{1}{2}, so \displaystyle \alpha = \frac{\pi}{3} exactly, and \displaystyle \frac{\alpha}{b} = \frac{\pi}{3} \cdot \frac{365}{2\pi} = \frac{365}{6} \approx 60.83 days. The daylight is at least 13.7 hours for t from 172 - 60.83 = 111.17 to 172 + 60.83 = 232.83: the whole days 112 through 232, which are April 22 through August 20. Since 2\alpha is \displaystyle \frac{2\pi}{3}, a third of a full turn, that stretch is a third of the year.

Figure 2Solongo’s daylight model. The days with at least 13.7 hours of daylight lie between the two crossings, exactly a third of the year.

Example 5

Ebere’s tone

Ebere plays a pure tone of 250 hertz. The air pressure at her ear changes from its normal value by at most 0.02 pascal either way, and the change is at a peak at t = 0, with t in seconds. Write a model, count the cycles in 0.1 second, and find the first two times the change is -0.01 pascal.

Solution

The highest value is 0.02 and the lowest -0.02, so A = 0.02 and k = 0. The period is \displaystyle P = \frac{1}{250} = 0.004 second and b = 2\pi \cdot 250 = 500\pi. With c = 0,

p(t) = 0.02\cos(500\pi t)

In 0.1 second there are 250 \cdot 0.1 = 25 cycles. For the level, 0.02\cos\theta = -0.01 gives \displaystyle \cos\theta = -\frac{1}{2}, so \displaystyle \alpha = \frac{2\pi}{3}. Here \theta = 500\pi t. The first time is \displaystyle \theta = \frac{2\pi}{3}, so \displaystyle t = \frac{2\pi/3}{500\pi} = \frac{1}{750} second. The partner solution -\alpha is behind t = 0, but one turn later it gives \displaystyle \theta = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}, so \displaystyle t = \frac{1}{375} second. These are about 1.33 and 2.67 milliseconds.

Figure 3Ebere’s tone, with time in milliseconds. Each cycle lasts 4 milliseconds, and the pressure change is -0.01 pascal twice in every cycle.

Boundaries

The model assumes a perfectly steady cycle: the same high, the same low and the same period every time. A real tide or a real year of daylight is only close to this, so a model is trusted near the facts it was built from, and a small error in P grows with every cycle.
It needs M > m and P > 0. The time c may be any peak: c + nP for any integer n gives the same function.
A level L is reached only when m \le L \le M. Strictly between them it is crossed twice per cycle; at M or m only once.
The inverse cosine returns one angle, in [0, \pi]. The second crossing comes from -\alpha, and the rest from adding multiples of P to t.
t and P must be in the same unit, and the angle b(t - c) is in radians.

Where this goes next

Simple harmonic motion uses the same model for a spring or a pendulum, with time in seconds and the cycle often given as a frequency, as for the tone here. Every model in Modeling things that repeat answers its questions the same two ways: substitute a time to predict, or solve a trigonometric equation to find when a level is crossed.

Exercises

Building a model

A second harbor in the same made-up fjord has high tide 3.9 meters at 5:00 a.m. and low tide 1.1 meters, with a 12.4 hour cycle. Write a model h(t), t in hours after midnight, and give the time of the first low tide.
In another made-up town the longest day has 16.4 hours of daylight, on day 172, and the shortest has 8.2 hours, with a 365 day cycle. Write a model D(t) and find the day of the shortest day.
A pure tone of 200 hertz changes the air pressure by at most 0.05 pascal either way, with a peak at t = 0. Write a model p(t), t in seconds.
A tone has a period of 2.5 milliseconds. Find its frequency in hertz and the value of b when t is in seconds.
A tide has low tide 0.6 meters at 1:00 a.m. and high tide 3.4 meters, with a 12.4 hour cycle. Write a model using a peak time, and show it equals \displaystyle -1.4\cos\left(\frac{5\pi}{31}(t - 1)\right) + 2.

Predicting

Use Nergui’s model to predict the depth at 6:00 a.m.
Use Nergui’s model to predict the depth at 8:00 p.m.
Use Solongo’s model to predict the hours of daylight on day 1.
Use Ebere’s model to find the pressure change at t = 0.0005 second.
Give the next three high tides after 3:00 a.m. in Nergui’s harbor. How much later is the first high tide each day than the day before?

Crossing a level

In Nergui’s harbor, between midnight and noon, when is the depth exactly 2.7 meters?
In Nergui’s harbor, between 3:00 a.m. and 3:24 p.m., when is the depth less than 1.5 meters?
In Solongo’s town, which days have fewer than 10.7 hours of daylight?
For Ebere’s tone, during the first cycle 0 \le t \le 0.004, when is the pressure change at least 0.01 pascal, and for what fraction of the cycle?
For the harbor in Exercise 1, between midnight and noon, when is the depth at least 3.2 meters?

Reasoning

Nergui asks when the depth reaches 5.0 meters. Explain why the equation has no solution and what that says about the model.
Explain why a level strictly between the lowest and highest values is reached exactly twice in each cycle, while the highest value is reached only once.
Suppose the true tide cycle is 12.42 hours, not 12.4. After 60 cycles, how far off is the model’s time of high tide? What does this say about long forecasts?
Without a calculator, explain why the days with at least 13.7 hours of daylight in Solongo’s town make up exactly a third of the year.
Explain why the time per cycle that a model spends above a level L is \displaystyle \frac{2\alpha}{b}, and why it does not depend on c.

Answers

\displaystyle A = \frac{3.9 - 1.1}{2} = 1.4, \displaystyle k = \frac{3.9 + 1.1}{2} = 2.5, c = 5: \displaystyle h(t) = 1.4\cos\left(\frac{5\pi}{31}(t - 5)\right) + 2.5. Low tide is half a period later, t = 5 + 6.2 = 11.2, at 11:12 a.m.
\displaystyle A = \frac{16.4 - 8.2}{2} = 4.1, \displaystyle k = \frac{16.4 + 8.2}{2} = 12.3: \displaystyle D(t) = 4.1\cos\left(\frac{2\pi}{365}(t - 172)\right) + 12.3. The shortest day is half a cycle after the longest, at t = 172 + 182.5 = 354.5, around day 354 or 355.
\displaystyle P = \frac{1}{200} = 0.005 second and b = 2\pi \cdot 200 = 400\pi, so p(t) = 0.05\cos(400\pi t).
P = 0.0025 second, so \displaystyle f = \frac{1}{0.0025} = 400 hertz and b = 2\pi \cdot 400 = 800\pi.
A = 1.4 and k = 2; the peak is half a period after the low, at t = 7.2, so \displaystyle h(t) = 1.4\cos\left(\frac{5\pi}{31}(t - 7.2)\right) + 2. Since \displaystyle \frac{5\pi}{31} \cdot 6.2 = \pi, the angle is \displaystyle \frac{5\pi}{31}(t - 1) - \pi, and \cos(\theta - \pi) = -\cos\theta turns the model into \displaystyle -1.4\cos\left(\frac{5\pi}{31}(t - 1)\right) + 2.
The angle is \displaystyle \frac{5\pi}{31} \cdot 3 = \frac{15\pi}{31} \approx 1.520, with cosine about 0.0507, so h(6) \approx 2.80 meters, just above the midline and falling.
The angle is \displaystyle \frac{5\pi}{31} \cdot 17 = \frac{85\pi}{31} \approx 8.614, with cosine about -0.689, so h(20) \approx 1.39 meters, falling toward the 9:36 p.m. low tide.
The angle is \displaystyle \frac{2\pi}{365}(-171) \approx -2.944, with cosine about -0.981, so D(1) \approx 9.26 hours.
The angle is \displaystyle 500\pi \cdot 0.0005 = \frac{\pi}{4}, so \displaystyle p = 0.02 \cdot \frac{\sqrt{2}}{2}, which is \displaystyle 0.01\sqrt{2} \approx 0.0141 pascal.
Adding 12.4 hours each time: t = 15.4 (3:24 p.m.), t = 27.8 (3:48 a.m. the next day) and t = 40.2 (4:12 p.m. the next day). Two cycles take 24.8 hours, so the first high tide comes 0.8 hour, 48 minutes, later each day.
\cos\theta = 0, so \displaystyle \alpha = \frac{\pi}{2} and \displaystyle \frac{\alpha}{b} = 3.1. The crossings t = 3 \pm 3.1 + 12.4n are \dots, -0.1, 6.1, 12.3, \dots, so the only one is t = 6.1, 6:06 a.m.
\displaystyle \cos\theta = \frac{1.5 - 2.7}{1.9} = -\frac{12}{19}, so \alpha \approx 2.2541 and \displaystyle \frac{\alpha}{b} \approx 4.449. The depth is below 1.5 meters when |\theta| > \alpha, from t \approx 3 + 4.449 = 7.449 to t \approx 15.4 - 4.449 = 10.951: about 7:27 a.m. to 10:57 a.m.
\displaystyle \cos\theta = \frac{10.7 - 12.2}{3} = -\frac{1}{2}, so \displaystyle \alpha = \frac{2\pi}{3} and \displaystyle \frac{\alpha}{b} = \frac{365}{3} \approx 121.67. Daylight is above 10.7 hours from t \approx 50.33 to t \approx 293.67, so fewer than 10.7 hours on days 1 through 50 and 294 through 365.
\displaystyle \cos(500\pi t) \ge \frac{1}{2} with 500\pi t in [0, 2\pi] means \displaystyle 500\pi t \le \frac{\pi}{3} or \displaystyle 500\pi t \ge \frac{5\pi}{3}: from t = 0 to \displaystyle \frac{1}{1500} second and from \displaystyle \frac{1}{300} to \displaystyle \frac{1}{250} second. That is \displaystyle \frac{2}{1500} = \frac{1}{750} second of \displaystyle \frac{1}{250}, one third of the cycle.
\displaystyle \cos\theta = \frac{3.2 - 2.5}{1.4} = \frac{1}{2}, so \displaystyle \alpha = \frac{\pi}{3} and \displaystyle \frac{\alpha}{b} = \frac{\pi}{3} \cdot \frac{31}{5\pi} = \frac{31}{15} \approx 2.067 hours. From t \approx 2.933 to t \approx 7.067: about 2:56 a.m. to 7:04 a.m.
It would need \displaystyle \cos\theta = \frac{5.0 - 2.7}{1.9} = \frac{23}{19}, which is more than 1, and no angle has a cosine above 1. The model never goes above k + A = 4.6 meters, the high tide it was built from.
One cycle is \theta running over an interval of length 2\pi, say -\pi < \theta \le \pi. For a value strictly between -1 and 1, \alpha is strictly between 0 and \pi, and \alpha and -\alpha are two different angles in that interval. For the value 1, \alpha = 0 = -\alpha, a single angle, so the peak is reached once.
Each cycle the model’s high tide is 0.02 hour early, so after 60 cycles it is 60 \cdot 0.02 = 1.2 hours, 72 minutes, off. A small error in P adds up, so the model is good for days and poor for months.
The level gives \displaystyle \cos\theta = \frac{1}{2}, so \displaystyle \alpha = \frac{\pi}{3}. The daylight is at least 13.7 hours for \displaystyle -\frac{\pi}{3} \le \theta \le \frac{\pi}{3}, a stretch of \displaystyle \frac{2\pi}{3} out of each 2\pi, which is one third of the cycle, and the cycle is the year.
The quantity is above L while -\alpha < \theta < \alpha, an angle stretch of 2\alpha. Since \theta = b(t - c), the angle grows by b for every unit of time, so that stretch takes \displaystyle \frac{2\alpha}{b} units. Changing c slides the whole graph sideways without changing its shape, so every window moves but none changes length.