A depth that rises and falls
The depth of water at a harbor mouth changes through the day. A gauge records it every hour, and the readings make the graph below: the depth starts at 5 meters at midnight, climbs to 8 meters at 3 a.m., falls to 2 meters at 9 a.m., and is back at 5 meters at noon. Then the same pattern starts again.
A quantity that repeats the same rise and fall on a fixed cycle has the shape of a sine wave. To model it, read three numbers off the data: the level it swings about, how far it swings, and how long one cycle takes.
The depth d in meters, t hours after midnight. It rises to 8 m at t = 3 and falls to 2 m at t = 9, swinging about the dashed line d = 5.
The principal axis and the amplitude
The highest depth is 8 meters and the lowest is 2 meters. The principal axis, the level the water swings about, is halfway between them: (8 + 2) ÷ 2 = 5 meters.
The amplitude is how far the water rises above that level, which is half the gap between the highest and lowest values: (8 − 2) ÷ 2 = 3 meters. The water rises 3 meters above 5 and falls 3 meters below it.
So the model starts as d = 5 + 3 sin(bt)°, with the number b still to be found. A sine lies between −1 and 1, so 3 sin(bt)° lies between −3 and 3, and d lies between 5 − 3 = 2 and 5 + 3 = 8, as the data say.
The period sets the number in the bracket
One full cycle, from 5 meters up to 8, down to 2 and back to 5, takes 12 hours. A sine completes one cycle when its angle grows by 360°. So the angle bt must grow by 360° when t grows by 12: 12b = 360, and .
The model is d = 5 + 3 sin(30t)°, with t in hours after midnight. Every hour the angle inside the sine turns through 30°.
Check it against the data. At t = 3, the angle is 30 × 3 = 90°, and d = 5 + 3 sin 90° = 5 + 3 = 8. At t = 9, the angle is 270°, and d = 5 + 3 × (−1) = 2. At t = 0, 6 and 12 the sine is 0 and d = 5.
Using the model forward
Put a time in, and the model gives the depth. At 1 a.m., t = 1: d = 5 + 3 sin 30° = 5 + 3 × ½ = 6.5 meters.
At 4 a.m., t = 4: d = 5 + 3 sin 120°. The sine of 120° equals the sine of 180° − 120° = 60°, which is , so meters, to 2 decimal places.
Running the model backwards
Now ask the question the other way: at what times in the first 6 hours is the depth 6.5 meters? Set the model equal to 6.5: 5 + 3 sin(30t)° = 6.5.
Subtract 5: 3 sin(30t)° = 1.5. Divide by 3: sin(30t)° = ½. The angle 30t has a sine of ½ at 30° and at 180° − 30° = 150°. So 30t = 30, giving t = 1, or 30t = 150, giving t = 5.
The depth is 6.5 meters at 1 a.m., while the water is rising, and again at 5 a.m., on its way down. In the first 12 hours there are only those two times: the next angles with a sine of ½ are 390° and 510°, which give t = 13 and t = 17.
The dashed line d = 6.5 crosses the model at t = 1 and t = 5. Between those times the water is deeper than 6.5 m.
How long the water stays deep
A boat needs at least 6.5 meters of water to cross the harbor mouth. Between t = 1 and t = 5 the curve is above the line d = 6.5, so the boat can cross for 5 − 1 = 4 hours in each 12-hour cycle.
A larger boat needs 7 meters. Set 5 + 3 sin(30t)° = 7, so . The inverse sine gives 30t = 41.81°, to 2 decimal places, and the pairing gives 30t = 180° − 41.81° = 138.19°. Dividing by 30: t = 1.394 or t = 4.606, to 3 decimal places.
In hours and minutes, 0.394 of an hour is 0.394 × 60 = 23.6 minutes and 0.606 of an hour is 0.606 × 60 = 36.4 minutes. So the larger boat can cross from about 1:24 a.m. to about 4:36 a.m., a little over 3 hours.
Daylight through the year
The number of hours of daylight in a city changes through the year in the same way. In this city the longest day, in June, has 14.5 hours of daylight and the shortest, in December, has 9.5 hours. The cycle repeats every 12 months.
The principal axis is (14.5 + 9.5) ÷ 2 = 12 hours, the amplitude is (14.5 − 9.5) ÷ 2 = 2.5 hours, and the number in the bracket is 360 ÷ 12 = 30.
This time the cycle does not start on the principal axis at month 0. A sine wave rises through its principal axis a quarter of a cycle before its highest point. A quarter of 12 months is 3 months, and the highest point is at month 6, so the wave rises through 12 hours at month 3. Shifting the wave 3 months to the right gives D = 12 + 2.5 sin(30(m − 3))°, where m is the month number.
Check: in June, m = 6, the angle is 30 × 3 = 90°, and D = 12 + 2.5 = 14.5. In December, m = 12, the angle is 270°, and D = 12 − 2.5 = 9.5.
Measured hours of daylight, with the model D = 12 + 2.5 sin(30(m − 3))° drawn through them. The points lie close to the curve, which is what it means for the model to fit.
A wheel that starts at the bottom
A Ferris wheel has a radius of 10 meters, its center is 12 meters above the ground, and it turns once every 4 minutes. A rider boards at the lowest point, 12 − 10 = 2 meters up.
The principal axis is the height of the center, 12 meters, and the amplitude is the radius, 10 meters. One turn takes 4 minutes, so the number in the bracket is 360 ÷ 4 = 90.
A sine starts on its principal axis, but the rider starts at the lowest point. The cosine starts at its highest value, 1, so −cos starts at its lowest, −1. The model is h = 12 − 10 cos(90t)°, with t in minutes.
Check: at t = 0, h = 12 − 10 × 1 = 2, the boarding height. At t = 2, half a turn, h = 12 − 10 cos 180° = 12 + 10 = 22, the top of the wheel. At t = 1, h = 12 − 10 cos 90° = 12, level with the center.
the arm's height is carried across to the graph: sin θ is that height, so the wave is the circle unrolled, and it repeats every 360° because the arm does
Choose tan and turn the arm to 90°
An arm turns on a circle, and its height is carried across to a graph against the angle. Drag the arm around: the height traces one full sine wave per turn, which is why anything turning at a steady rate is modeled by a sine or a cosine.
The usual mistakes
Reading the amplitude as the highest value. For d = 5 + 3 sin(30t)°, the amplitude is 3, the rise above the principal axis. The greatest depth is 5 + 3 = 8 meters.
Reading the number in the bracket as the period. The 30 is how many degrees the angle turns in one hour. The period is 360 ÷ 30 = 12 hours.
Stopping at one time. The water passes any depth between the lowest and highest twice in each cycle, once rising and once falling, so the equation has two solutions in each cycle.
Starting every model with a sine. A sine model is at its principal axis and rising at the start. If the data start at a highest or lowest point, use cos or −cos, or shift the sine.
Worked example: The Depth of Water at a Harbor Mouth Through the Day
Question At a harbor mouth the depth of water, d meters, x hours after midnight is modeled by d = 3 sin (30x)° + 5 for 0 ≤ x ≤ 24. (a) What is the depth at high water, the depth at low water, and the mean depth about which the water rises and falls? (b) How long is it from one high water to the next, and at what time is the first high water after midnight?
1.The sine of any angle lies between −1 and 1, so 3 sin (30x)° lies between −3 and 3. The amplitude is 3 m: the water rises 3 m above its middle level and falls 3 m below it.
The sine lies between −1 and 1, so the water rises and falls 3 m about its middle level: the amplitude is 3 m. 2.The principal axis is d = 5, from the + 5 in the model. (a) The mean depth is 5 m, the depth at high water is 5 + 3 = 8 m and the depth at low water is 5 − 3 = 2 m.
(a) The principal axis is d = 5: the mean depth is 5 m, high water 8 m, low water 2 m. 3.The sine repeats every 360°, and 30x grows by 360 when x grows by 36030 = 12. The period is 12 hours, the time from one high water to the next.
The angle 30x grows by 360 in 36030 = 12 hours: the period is 12 hours. 4.High water is where sin (30x)° = 1, first when 30x = 90, so x = 3. (b) It is 12 hours from one high water to the next, and the first high water is at 3 a.m. Check: at x = 3, d = 3 sin 90° + 5 = 3 + 5 = 8 m, and the next high water is at x = 3 + 12 = 15, which is 3 p.m.
(b) High water first comes where 30x = 90, at x = 3, which is 3 a.m.; the next is at 3 p.m.
Answer: (a) 8 m at high water, 2 m at low water, and a mean depth of 5 m; (b) 12 hours, and the first high water is at 3 a.m.
Common mistakes
- Reading the amplitude 3 as the depth at high water. The amplitude is how far the water rises ABOVE the mean level, so the depth at high water is 5 + 3 = 8 m.
- Taking the period as 30 hours, from the 30 in the model. The 30 is how many degrees the angle turns in one hour, and a full cycle of 360° takes 360 ÷ 30 = 12 hours.