A Ferris wheel has a diameter of $40$ m, its centre is $22$ m above the ground, and it turns once every $2$ minutes. Find the maximum and minimum heights of a seat, the amplitude, the midline and $k$.

Maximum $42$ m, minimum $2$ m, amplitude $20$, midline $h = 22$, $k = 180$.

The water in a harbour is $2$ m deep at low tide and $10$ m deep at high tide, and high tides are $12.4$ hours apart. Find the midline, amplitude, period and $k$.

Midline $6$ m, amplitude $4$ m, period $12.4$ h, $k \approx 29.03$.

A rider boards the Ferris wheel from Example 1 at the bottom at $t = 0$. Write the height $h(t)$ in metres after $t$ minutes, and find the height after $40$ seconds.

$h(t) = -20\cos(180t) + 22$; $32$ m after $40$ seconds.

In the harbour from Example 2, high tide is at 3:00 a.m. Write a cosine model for the depth $D$ in metres, $t$ hours after midnight, and find the depth at noon.

$D(t) = 4\cos\left(\frac{360}{12.4}(t - 3)\right) + 6$; about $5.39$ m at noon.

A city gets a maximum of $15.5$ hours of daylight on day $172$ of the year and a minimum of $8.9$ hours half a year later. Using a period of $365$ days, write a model for the hours of daylight $D$ on day $t$. Find the daylight on day $100$, and the two days when there are $12.2$ hours.

$D(t) = 3.3\cos\left(\frac{360}{365}(t - 172)\right) + 12.2$; about $13.27$ hours on day $100$; $12.2$ hours around day $81$ and day $263$.

Grade 11 · MCR3U · Functions

MCR3U: Sinusoidal Word Problems

A sinusoidal word problem describes something that rises and falls the same way over and over: a seat on a Ferris wheel, the tide in a harbour, the hours of daylight over a year. The work is turning the story into four numbers, and the four numbers into an equation you can use.

How It Works

The model. Use $y = a\sin(k(x - d)) + c$ or $y = a\cos(k(x - d)) + c$ , working in degrees. Here $x$ is usually time, so read the units of the period, seconds, minutes, hours or days, and keep them.

Four numbers from the story. The midline is halfway between the extremes: $c = \frac{\max + \min}{2}$ . The amplitude is half the spread: $a = \frac{\max - \min}{2}$ . The period is the time for one full cycle, and $k = \frac{360^\circ}{\text{period}}$ . The phase shift $d$ places one point you know.

Pick the easiest start. If you know when a maximum happens, use cosine and set $d$ to that time. If the story starts at a minimum, use $-\cos$ , so a Ferris wheel rider who boards at the bottom gives $h(t) = -a\cos(kt) + c$ with no shift at all. If it starts on the midline going up, use sine.

Using the model. To find the value at a given time, substitute and evaluate. To find when it reaches a value, use the shape of the cycle: the graph crosses the midline a quarter period before and after each maximum, and by symmetry it reaches any other height at two times equally far from the peak.

Check it. Substitute the time of a known maximum or minimum back into your equation and make sure it gives the right value. Then answer in the units of the question, with a sensible rounding.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
A Ferris wheel has a diameter of $40$ m, its centre is $22$ m above the ground, and it turns once every $2$ minutes. Find the maximum and minimum heights of a seat, the amplitude, the midline and $k$ .
Show solution
1. The radius is $20$ m, so the maximum is $22 + 20 = 42$ m and the minimum is $22 - 20 = 2$ m.
2. Amplitude $a = \frac{42 - 2}{2} = 20$ and midline $c = \frac{42 + 2}{2} = 22$ .
3. The period is $2$ minutes, so $k = \frac{360}{2} = 180$ .
Answer: Maximum $42$ m, minimum $2$ m, amplitude $20$ , midline $h = 22$ , $k = 180$ .
Easy
The water in a harbour is $2$ m deep at low tide and $10$ m deep at high tide, and high tides are $12.4$ hours apart. Find the midline, amplitude, period and $k$ .
Show solution
1. Midline: $c = \frac{10 + 2}{2} = 6$ . Amplitude: $a = \frac{10 - 2}{2} = 4$ .
2. One high tide to the next is one cycle, so the period is $12.4$ hours and $k = \frac{360}{12.4} \approx 29.03$ .
Answer: Midline $6$ m, amplitude $4$ m, period $12.4$ h, $k \approx 29.03$ .
Medium
A rider boards the Ferris wheel from Example 1 at the bottom at $t = 0$ . Write the height $h(t)$ in metres after $t$ minutes, and find the height after $40$ seconds.
Show solution
1. The ride starts at the minimum, so use $-\cos$ with no shift: $h(t) = -20\cos(180t) + 22$ .
2. Check: $h(0) = -20 + 22 = 2$ and $h(1) = -20\cos 180^\circ + 22 = 42$ , the top after half a turn.
3. $40$ seconds is $\frac{2}{3}$ of a minute: $h\left(\frac{2}{3}\right) = -20\cos 120^\circ + 22 = -20\left(-\frac{1}{2}\right) + 22 = 32$ .
Answer: $h(t) = -20\cos(180t) + 22$ ; $32$ m after $40$ seconds.
Medium
In the harbour from Example 2, high tide is at 3:00 a.m. Write a cosine model for the depth $D$ in metres, $t$ hours after midnight, and find the depth at noon.
Show solution
1. High tide is a maximum, so use cosine with $d = 3$ : $D(t) = 4\cos\left(\frac{360}{12.4}(t - 3)\right) + 6$ .
2. Noon is $t = 12$ : $\frac{360}{12.4}(12 - 3) \approx 261.29^\circ$ .
3. $D(12) = 4\cos 261.29^\circ + 6 \approx 4(-0.1514) + 6 \approx 5.39$ .
Answer: $D(t) = 4\cos\left(\frac{360}{12.4}(t - 3)\right) + 6$ ; about $5.39$ m at noon.
Test level
A city gets a maximum of $15.5$ hours of daylight on day $172$ of the year and a minimum of $8.9$ hours half a year later. Using a period of $365$ days, write a model for the hours of daylight $D$ on day $t$ . Find the daylight on day $100$ , and the two days when there are $12.2$ hours.
Show solution
1. $c = \frac{15.5 + 8.9}{2} = 12.2$ , $a = \frac{15.5 - 8.9}{2} = 3.3$ and $k = \frac{360}{365}$ . The maximum is on day $172$ , so $D(t) = 3.3\cos\left(\frac{360}{365}(t - 172)\right) + 12.2$ .
2. Day $100$ : $\frac{360}{365}(100 - 172) \approx -71.01^\circ$ , so $D(100) \approx 3.3(0.3253) + 12.2 \approx 13.27$ .
3. $12.2$ hours is the midline, reached a quarter period, $\frac{365}{4} = 91.25$ days, before and after the maximum.
4. $172 - 91.25 = 80.75$ and $172 + 91.25 = 263.25$ .
Answer: $D(t) = 3.3\cos\left(\frac{360}{365}(t - 172)\right) + 12.2$ ; about $13.27$ hours on day $100$ ; $12.2$ hours around day $81$ and day $263$ .

Learn It on Martian Lab

Martian Lab teaches this in the Applications of Sinusoidal Functions lesson of its MCR3U Trigonometric Functions unit. Watch it, then take the Trigonometric Functions practice tests to see if it stuck.

Watch the Applications of Sinusoidal Functions lesson free

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More Trigonometric Functions topics: Transformations of Sine and Cosine

More MCR3U topics: Inverse Functions · Transformations of Functions · Sine Law and the Ambiguous Case · Domain, Range and Function Notation · Simplifying Rational Expressions · Operations with Rational Expressions · Max/Min and Vertex Form · Linear-Quadratic Systems · Quadratic Word Problems · Simplifying Radicals · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest · Annuities: Future and Present Value

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