Math Core

A seat on a Ferris wheel, the water level at a pier, the number of daylight hours in a city: each rises and falls in a regular cycle. Sine and cosine are the natural tools for describing them. In this lesson you'll turn a description of a repeating situation into an equation you can use to make predictions.

From a story to four numbers

A sinusoidal model has the form

y = A\cos\big(B(t - C)\big) + D \qquad \text{or} \qquad y = A\sin\big(B(t - C)\big) + D,

where $t$ is usually time. You already know what each letter does. In a word problem, you find them from the highest value, the lowest value, and the timing of the cycle.

Building a sinusoidal model

Midline: $D = \dfrac{\text{max} + \text{min}}{2}$ , the average level.
Amplitude: $|A| = \dfrac{\text{max} - \text{min}}{2}$ , how far it swings from average.
Period: the time for one full cycle. Then $B = \dfrac{2\pi}{\text{period}}$ .
Starting point: pick the function that starts where your data does.
- Starts at a maximum: use $A\cos$ , with $C$ = time of the maximum.
- Starts at a minimum: use $-A\cos$ (negative coefficient), with $C$ = time of the minimum.
- Starts at the midline, rising: use $A\sin$ , with $C$ = that time.

There is always more than one correct equation for the same situation, because you can start the cycle at a different point. Cosine models starting at a maximum or minimum are usually the easiest, because maxima and minima are the easiest moments to identify.

A Ferris wheel

Worked example: Height on a Ferris wheel

A Ferris wheel has a diameter of $40$ meters, and its center is $25$ meters above the ground. It makes one full revolution every $8$ minutes. You board at the lowest point at time $t = 0$ . Write a model for your height $h$ (in meters) after $t$ minutes, and find your height at $t = 2$ and $t = 4$ .

Solution.

The radius is $20$ m, so the highest point is $25 + 20 = 45$ m and the lowest is $25 - 20 = 5$ m.
Midline: $D = \dfrac{45 + 5}{2} = 25$ . Amplitude: $\dfrac{45 - 5}{2} = 20$ .
Period $8$ minutes, so $B = \dfrac{2\pi}{8} = \dfrac{\pi}{4}$ .
You start at the minimum at $t = 0$ , so use a negative cosine with $C = 0$ .

h(t) = -20\cos\left(\frac{\pi}{4}t\right) + 25

At $t = 2$ : $h(2) = -20\cos\left(\dfrac{\pi}{2}\right) + 25 = 0 + 25 = 25$ m. A quarter turn in, you are level with the center.

At $t = 4$ : $h(4) = -20\cos(\pi) + 25 = 20 + 25 = 45$ m. Halfway around, you are at the top.

Height (m) against time (min). The rider starts at 5 m, peaks at 45 m after 4 minutes, and repeats every 8 minutes. The midline y = 25 is dashed.Open in grapher →

Tides

Worked example: Water depth at a pier

At a pier, high tide of $10$ feet occurs at $2$ a.m., and the next low tide of $2$ feet occurs at $8$ a.m. Model the depth $d$ in feet $t$ hours after midnight, and predict the depth at $4$ a.m.

Solution.

Midline $D = \dfrac{10 + 2}{2} = 6$ . Amplitude $\dfrac{10 - 2}{2} = 4$ .
High tide to low tide is half a cycle: $8 - 2 = 6$ hours. So the period is $12$ hours and $B = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$ .
A maximum occurs at $t = 2$ , so use cosine with $C = 2$ .

d(t) = 4\cos\left(\frac{\pi}{6}(t - 2)\right) + 6

At $4$ a.m., $t = 4$ : $d(4) = 4\cos\left(\dfrac{\pi}{6} \cdot 2\right) + 6 = 4\cos\left(\dfrac{\pi}{3}\right) + 6 = 4 \cdot \dfrac{1}{2} + 6 = 8$ feet.

Common mistake

The time from a maximum to the next minimum is half a period, not a full period. Doubling it is one of the most common steps students forget. A full period goes from one maximum to the next maximum.

Using a model to answer "when" questions

Sometimes you are asked how long a quantity stays above or below a level. Set up an inequality with the model and use the unit circle.

Worked example: Time spent near the top

For the Ferris wheel $h(t) = -20\cos\left(\dfrac{\pi}{4}t\right) + 25$ , during the first revolution ( $0 \le t \le 8$ ), how many minutes is the rider at least $35$ meters high?

Solution. Solve $h(t) \ge 35$ :

\begin{aligned} -20\cos\left(\tfrac{\pi}{4}t\right) + 25 &\ge 35 \\ -20\cos\left(\tfrac{\pi}{4}t\right) &\ge 10 \\ \cos\left(\tfrac{\pi}{4}t\right) &\le -\tfrac{1}{2} \end{aligned}

(Dividing by $-20$ flips the inequality.) On one cycle, $\cos\theta \le -\dfrac{1}{2}$ when $\dfrac{2\pi}{3} \le \theta \le \dfrac{4\pi}{3}$ . With $\theta = \dfrac{\pi}{4}t$ , multiply by $\dfrac{4}{\pi}$ : $\dfrac{8}{3} \le t \le \dfrac{16}{3}$ . That is $\dfrac{16}{3} - \dfrac{8}{3} = \dfrac{8}{3} \approx 2.67$ minutes.

Daylight over a year

Worked example: Hours of daylight

In one city, the longest day of the year has $15$ hours of daylight (day $172$ ) and the shortest has $9$ hours. Model the hours of daylight $L$ on day $t$ of the year, using a period of $365$ days.

Solution. Midline $D = \dfrac{15 + 9}{2} = 12$ , amplitude $\dfrac{15 - 9}{2} = 3$ , $B = \dfrac{2\pi}{365}$ , and a maximum at $t = 172$ :

L(t) = 3\cos\left(\frac{2\pi}{365}(t - 172)\right) + 12.

The model predicts about $12$ hours of daylight a quarter year from the longest day, near $t = 172 + 91.25 \approx 263$ , which is close to the autumn equinox.

Tip

Check a model at a moment you know. In the tide model, $t = 8$ gives $4\cos(\pi) + 6 = 2$ , the low tide. If a check fails, the most likely culprits are the period and the sign of $A$ .

Practice

Practice 1

Over one day, the temperature in a city varies sinusoidally between a low of $58^\circ$ F and a high of $82^\circ$ F. What is the amplitude of the model, in degrees?