Sinusoidal Applications in Radians

In Grade 11 you built sinusoidal models for Ferris wheels, tides, and temperatures in degrees. Scientists and engineers write the same models in radians, and that’s what you’ll do here. The steps are the same; only the formula for $k$ changes. Then you’ll use the model and its graph to answer real questions, like when the water in a harbour is deep enough for a boat.

Key ideas

Building a model in radians

For $y = a\sin\big(k(t - d)\big) + c$ or $y = a\cos\big(k(t - d)\big) + c$, with $t$ usually a time:

1. Amplitude: $a = \dfrac{\text{max} - \text{min}}{2}$
2. Axis: $c = \dfrac{\text{max} + \text{min}}{2}$
3. $k = \dfrac{2\pi}{\text{period}}$, with the period in the units of $t$ (seconds, hours, days, months, years).
4. Phase shift $d$: the time of a maximum (cosine), or of an upward crossing of the axis (sine). If the situation starts at a minimum, such as a rider boarding a Ferris wheel at the bottom, use a negative $a$ with cosine.

The only change from degrees is step 3: $2\pi$ instead of $360$. Then $k$ is in radians per unit of time. That’s the angular velocity of the turning wheel or the “turning” cycle.

Expanded and factored forms

Models are often written with $k$ multiplied out, like $h(t) = 4\cos\left(\dfrac{\pi t}{6} - \dfrac{\pi}{2}\right) + 5$. To read the phase shift, factor out $k$:

$$\frac{\pi t}{6} - \frac{\pi}{2} = \frac{\pi}{6}(t - 3)$$

so the phase shift is $3$ (units of time), not $\tfrac{\pi}{2}$.

Answering questions with a model

• Predict a value: substitute the time, with your calculator in radian mode.
• Find a maximum or minimum: it’s $c \pm \lvert a \rvert$, at times you can read from $d$ and the period.
• Find when a value occurs: set the model equal to the value and isolate the sine or cosine. If the result is a special value (like $\tfrac{1}{2}$ or $0$), use the exact values; otherwise, read the times from a graph, or use graphing technology to find where the curve meets a horizontal line. Solving trig equations in general comes in the next unit.
• Find for how long: find the times where the curve crosses a level, then use the graph to see which intervals are above or below it.

Real data never fits a model perfectly. Estimate the maximum, minimum, and period from the data, build the model, and check it against a few points.

Worked examples

Example 1: A Ferris wheel

A Ferris wheel has a diameter of $30$ m, its centre is $18$ m above the ground, and it turns once every $80$ s. A rider boards at the lowest point at $t = 0$.

• (a) Write a model for the rider’s height $h$, in metres, after $t$ seconds.
• (b) Find the height after $20$ s and after $30$ s.
• (c) When, during the first turn, is the rider $25.5$ m above the ground?

Solution.

(a) $a = 15$ (the radius), $c = 18$ (the centre), and $k = \dfrac{2\pi}{80} = \dfrac{\pi}{40}$. The rider starts at the minimum, so use a negative cosine:

$$h(t) = -15\cos\left(\frac{\pi t}{40}\right) + 18$$

(b) At $t = 20$, a quarter turn:

$$h(20) = -15\cos\frac{\pi}{2} + 18 = -15(0) + 18 = 18 \text{ m}$$

That’s level with the centre, which makes sense after a quarter turn. At $t = 30$:

$$h(30) = -15\cos\frac{3\pi}{4} + 18 = -15\left(-\frac{\sqrt{2}}{2}\right) + 18 \approx 28.61 \text{ m}$$

(c) Solve $-15\cos\left(\tfrac{\pi t}{40}\right) + 18 = 25.5$:

$$-15\cos\left(\frac{\pi t}{40}\right) = 7.5 \quad\Rightarrow\quad \cos\left(\frac{\pi t}{40}\right) = -\frac{1}{2}$$

During the first turn, $\tfrac{\pi t}{40}$ goes from $0$ to $2\pi$. Cosine is $-\tfrac{1}{2}$ at $\tfrac{2\pi}{3}$ and $\tfrac{4\pi}{3}$:

$$\frac{\pi t}{40} = \frac{2\pi}{3} \;\Rightarrow\; t = \frac{80}{3} \approx 26.7 \text{ s} \qquad\qquad \frac{\pi t}{40} = \frac{4\pi}{3} \;\Rightarrow\; t = \frac{160}{3} \approx 53.3 \text{ s}$$

The rider is at $25.5$ m on the way up at about $26.7$ s, and on the way down at about $53.3$ s.

Example 2: Is the harbour deep enough?

In a simplified model, the depth of water at a harbour entrance is

$$D(t) = 2.5\cos\left(\frac{\pi}{6}(t - 3)\right) + 4$$

where $D$ is in metres and $t$ is the number of hours after midnight. A fishing boat needs at least $5.25$ m of water. During which times of the day can it enter the harbour?

The depth is at least $5.25$ m in the shaded intervals.

Solution. First read the model: amplitude $2.5$ m, axis $4$ m, period $\dfrac{2\pi}{\pi/6} = 12$ hours, and high tide ($6.5$ m) at $t = 3$, 3:00 a.m. Low tide ($1.5$ m) is half a period later, at 9:00 a.m.

From the graph, the curve meets the line $D = 5.25$ at about $t = 1$, $5$, $13$, and $17$. Confirm with exact values:

$$2.5\cos\left(\frac{\pi}{6}(t - 3)\right) + 4 = 5.25 \quad\Rightarrow\quad \cos\left(\frac{\pi}{6}(t - 3)\right) = \frac{1}{2}$$

Cosine is $\tfrac{1}{2}$ at $\pm\tfrac{\pi}{3}$ (and every $2\pi$ after that). So $\tfrac{\pi}{6}(t - 3) = \pm\tfrac{\pi}{3}$, which gives $t - 3 = \pm 2$: $t = 1$ or $t = 5$. Adding the period of $12$ h gives $t = 13$ and $t = 17$.

Check: $D(1) = 2.5\cos\left(-\tfrac{\pi}{3}\right) + 4 = 2.5(0.5) + 4 = 5.25$. ✓

The depth is above the line around each high tide, so the boat can enter from 1:00 a.m. to 5:00 a.m. and from 1:00 p.m. to 5:00 p.m.

(Real tides along Canada’s coasts have a period closer to $12.4$ hours, so the times drift later each day.)

Example 3: A model from data

A mass bouncing on a spring is measured every $0.1$ s. Its displacement above its rest position (in cm; negative means below) is shown in the table and graph.

$t$ (s)	$0$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$
$y$ (cm)	$6.0$	$4.2$	$0.1$	$-4.3$	$-6.0$	$-4.2$	$0.0$	$4.3$	$6.0$

The data and the model $y = 6\cos\left(\tfrac{5\pi}{2}t\right)$.

Write a sinusoidal model, and use it to predict the displacement at $t = 1.3$ s.

Solution. The maximum is $6.0$ and the minimum is $-6.0$, so $a = 6$ and $c = 0$ (the rest position). The data go from one maximum ($t = 0$) to the next ($t = 0.8$), so the period is $0.8$ s and

$$k = \frac{2\pi}{0.8} = \frac{5\pi}{2}$$

The mass starts at a maximum, so use cosine with $d = 0$:

$$y = 6\cos\left(\frac{5\pi}{2}t\right)$$

Check: at $t = 0.3$, $y = 6\cos\tfrac{3\pi}{4} \approx -4.24$, close to the measured $-4.3$. ✓

At $t = 1.3$:

$$y = 6\cos\left(\frac{5\pi}{2}(1.3)\right) = 6\cos\frac{13\pi}{4} \approx -4.24 \text{ cm}$$

The mass will be about $4.2$ cm below its rest position. (Exactly: $\tfrac{13\pi}{4} = 3\pi + \tfrac{\pi}{4}$ has related angle $\tfrac{\pi}{4}$ in quadrant III, so the value is $6\left(-\tfrac{\sqrt{2}}{2}\right) = -3\sqrt{2}$.)

Example 4: Predators and prey

In a northern Canadian forest, the populations of snowshoe hares $H$ and lynx $L$ (which hunt the hares) are modelled by

$$H(t) = 30\,000 + 15\,000\sin\left(\frac{\pi t}{5}\right) \qquad\qquad L(t) = 2000 + 800\sin\left(\frac{\pi}{5}(t - 2)\right)$$

where $t$ is the time in years.

• (a) Find the period, maximum, and minimum of each population.
• (b) When does each population first reach its maximum? Explain the difference in the context.
• (c) Predict the hare population at $t = 7$.

Solution.

(a) Both have $k = \tfrac{\pi}{5}$, so both periods are $\dfrac{2\pi}{\pi/5} = 10$ years.

Hares: maximum $30\,000 + 15\,000 = 45\,000$, minimum $30\,000 - 15\,000 = 15\,000$.

Lynx: maximum $2000 + 800 = 2800$, minimum $2000 - 800 = 1200$.

(b) Sine reaches its maximum when its angle is $\tfrac{\pi}{2}$.

Hares: $\dfrac{\pi t}{5} = \dfrac{\pi}{2}$ gives $t = 2.5$ years.

Lynx: $\dfrac{\pi}{5}(t - 2) = \dfrac{\pi}{2}$ gives $t - 2 = 2.5$, so $t = 4.5$ years.

The lynx population peaks $2$ years after the hare population. When hares are plentiful, lynx have lots of food and their numbers grow. Then the many lynx eat more hares, the hare population falls, and later the lynx population falls too. The predator cycle lags behind the prey cycle.

(c)

$$H(7) = 30\,000 + 15\,000\sin\frac{7\pi}{5} \approx 30\,000 + 15\,000(-0.9511) \approx 15\,700$$

About $15\,700$ hares, close to the bottom of the cycle.

Common mistakes

Using 360 for the period. In radians, $k = \dfrac{2\pi}{\text{period}}$. A $12$-hour tide has $k = \tfrac{\pi}{6}$, not $30$.

Calculator in degree mode. Every model on this page needs radian mode. A quick test: $\cos\pi$ should give $-1$.

Reading the phase shift without factoring. In $\cos\left(\tfrac{\pi t}{6} - \tfrac{\pi}{2}\right)$ the phase shift is $3$, not $\tfrac{\pi}{2}$. Factor out $k$ first.

Rounding k too early. Keep $k$ as an exact expression like $\tfrac{2\pi}{365}$ on your calculator. Rounding it to $0.017$ can noticeably change predictions far from $t = 0$.

Finding only one time. In each cycle, a level is usually reached twice: once going up and once going down. Over several cycles, add the period to find more.

Ignoring the realistic domain. Times in a model usually start at $0$ and are limited by the question (for example, one day is $0 \le t \le 24$). Don’t report answers outside that range.

Practice

1. (Warm-up) Find $k$ for a period of (a) $24$ hours (b) $0.5$ s (c) $365$ days.

Solution

(a) $k = \dfrac{2\pi}{24} = \dfrac{\pi}{12}$ (b) $k = \dfrac{2\pi}{0.5} = 4\pi$ (c) $k = \dfrac{2\pi}{365}$

2. (Warm-up) The height of a buoy bobbing on waves is $h(t) = 3\sin\left(\dfrac{\pi t}{4}\right) + 10$, with $h$ in metres and $t$ in seconds. State the amplitude, period, axis, maximum height, and minimum height, with units.

Solution

Amplitude $3$ m, period $\dfrac{2\pi}{\pi/4} = 8$ s, axis $h = 10$ m, maximum $13$ m, minimum $7$ m.

3. (Warm-up) The displacement of a mass on a spring is $y(t) = -5\cos(2\pi t)$, in cm, after $t$ seconds. Where is the mass at $t = 0$, and how long does one bounce take?

Solution

At $t = 0$: $y = -5\cos 0 = -5$, so it’s $5$ cm below its rest position, at its lowest point.

Period: $\dfrac{2\pi}{2\pi} = 1$ s per bounce.

4. (Core) A Ferris wheel has a radius of $12$ m, its centre is $14$ m above the ground, and it turns once every $60$ s. A rider boards at the bottom at $t = 0$. Write a model in radians, and find the rider’s height after $10$ s and after $45$ s.

Solution

$a = 12$, $c = 14$, $k = \dfrac{2\pi}{60} = \dfrac{\pi}{30}$, starting at a minimum:

$$h(t) = -12\cos\left(\frac{\pi t}{30}\right) + 14$$

$h(10) = -12\cos\dfrac{\pi}{3} + 14 = -12\left(\dfrac{1}{2}\right) + 14 = 8$ m.

$h(45) = -12\cos\dfrac{3\pi}{2} + 14 = 0 + 14 = 14$ m (three-quarters of a turn, level with the centre).

5. (Core) In one Canadian city, the longest day of the year (day $172$) has $15.8$ hours of daylight, and the shortest has $8.6$ hours. Using a period of $365$ days, write a model for the hours of daylight on day $t$, and estimate the daylight on day $300$ (late October), to $1$ decimal place.

Solution

$a = \dfrac{15.8 - 8.6}{2} = 3.6$, $c = \dfrac{15.8 + 8.6}{2} = 12.2$, $k = \dfrac{2\pi}{365}$, with a maximum at day $172$:

$$D(t) = 3.6\cos\left(\frac{2\pi}{365}(t - 172)\right) + 12.2$$$$D(300) = 3.6\cos\left(\frac{2\pi(128)}{365}\right) + 12.2 \approx 10.1 \text{ hours}$$

6. (Core) At a harbour, high tide is $4.8$ m at 2:00 a.m., and the next low tide, $0.6$ m, is at 8:15 a.m. Write a model for the depth $t$ hours after midnight, and estimate the depth at noon, to $2$ decimal places.

Solution

$a = \dfrac{4.8 - 0.6}{2} = 2.1$ and $c = \dfrac{4.8 + 0.6}{2} = 2.7$. High to low tide is half a period, $6.25$ h, so the period is $12.5$ h and $k = \dfrac{2\pi}{12.5}$. The maximum is at $t = 2$:

$$D(t) = 2.1\cos\left(\frac{2\pi}{12.5}(t - 2)\right) + 2.7$$

At noon, $t = 12$:

$$D(12) = 2.1\cos\left(\frac{2\pi(10)}{12.5}\right) + 2.7 = 2.1\cos(1.6\pi) + 2.7 \approx 3.35 \text{ m}$$

7. (Core) Use the harbour model in Example 2, $D(t) = 2.5\cos\left(\dfrac{\pi}{6}(t - 3)\right) + 4$. During which times of the day ($0 \le t \le 24$) is the water less than $2.75$ m deep?

Solution

Find where the depth is exactly $2.75$ m:

$$2.5\cos\left(\frac{\pi}{6}(t - 3)\right) + 4 = 2.75 \quad\Rightarrow\quad \cos\left(\frac{\pi}{6}(t - 3)\right) = -\frac{1}{2}$$

Cosine is $-\tfrac{1}{2}$ at $\tfrac{2\pi}{3}$ and $\tfrac{4\pi}{3}$, so $\tfrac{\pi}{6}(t - 3) = \tfrac{2\pi}{3}$ or $\tfrac{4\pi}{3}$, giving $t - 3 = 4$ or $8$: $t = 7$ or $t = 11$. Adding the $12$ h period gives $t = 19$ and $t = 23$.

The depth is below $2.75$ m around each low tide ($t = 9$ and $t = 21$): from 7:00 a.m. to 11:00 a.m. and from 7:00 p.m. to 11:00 p.m.

Check: at low tide, $D(9) = 2.5\cos\pi + 4 = 1.5 \lt 2.75$. ✓

8. (Challenge) In a park, the numbers of rabbits $R$ and foxes $F$ are modelled by $R(t) = 1200 + 400\cos\left(\dfrac{\pi t}{6}\right)$ and $F(t) = 150 + 40\cos\left(\dfrac{\pi}{6}(t - 3)\right)$, where $t$ is in months.

• (a) Find the period of each model.
• (b) When, during the first year ($0 \le t \le 12$), is each population at its maximum?
• (c) How many foxes are there when the rabbit population is at its minimum? Explain what is happening in the park at that time.

Solution

(a) Both have $k = \tfrac{\pi}{6}$, so both periods are $\dfrac{2\pi}{\pi/6} = 12$ months.

(b) Cosine is at its maximum when its angle is $0$ or $2\pi$. Rabbits: $t = 0$ and $t = 12$ ($1600$ rabbits). Foxes: $t - 3 = 0$, so $t = 3$ ($190$ foxes). The fox maximum comes $3$ months after the rabbit maximum.

(c) The rabbit minimum is at $\tfrac{\pi t}{6} = \pi$, so $t = 6$ ($800$ rabbits). Then

$$F(6) = 150 + 40\cos\frac{\pi}{2} = 150 + 0 = 150 \text{ foxes}$$

The foxes are at their average number and falling: they peaked at month $3$, after the rabbits peaked, and now there are fewer rabbits to eat, so the fox population is shrinking toward its minimum at month $9$.

9. (Challenge) A Ferris wheel ride is modelled by $h(t) = 20\sin\left(\dfrac{\pi}{30}(t - 15)\right) + 22$, with $h$ in metres and $t$ in seconds.

• (a) Describe the wheel: its radius, the height of its centre, the time for one turn, and where the rider is at $t = 0$.
• (b) For how long during each turn is the rider more than $32$ m above the ground?

Solution

(a) Radius $20$ m, centre $22$ m above the ground, one turn every $\dfrac{2\pi}{\pi/30} = 60$ s. At $t = 0$:

$$h(0) = 20\sin\left(-\frac{\pi}{2}\right) + 22 = -20 + 22 = 2 \text{ m}$$

The rider is at the lowest point, $2$ m up, boarding from a platform. (This model is the same as $h(t) = -20\cos\left(\tfrac{\pi t}{30}\right) + 22$.)

(b) Solve $20\sin\left(\tfrac{\pi}{30}(t - 15)\right) + 22 = 32$:

$$\sin\left(\frac{\pi}{30}(t - 15)\right) = \frac{1}{2} \quad\Rightarrow\quad \frac{\pi}{30}(t - 15) = \frac{\pi}{6} \;\text{ or }\; \frac{5\pi}{6}$$

So $t - 15 = 5$ or $25$, giving $t = 20$ s and $t = 40$ s. Between these times the rider passes the top ($t = 30$, $h = 42$ m), so the rider is above $32$ m for $40 - 20 = 20$ s of each turn: one-third of the ride.