A Ferris wheel has a diameter of $20$ m and its centre is $12$ m above the ground. It turns once every $40$ s, and a rider boards at the bottom at $t = 0$. Write the rider's height $h(t)$ in metres.

$h(t) = -10 \cos\left(\frac{\pi}{20} t\right) + 12$

The depth of water at a dock is $d(t) = 3 \cos\left(\frac{\pi}{6} t\right) + 5$ metres, $t$ hours after high tide. Find the depth at high and low tide, the time between high tides, and the depth $4$ hours after high tide.

High tide $8$ m, low tide $2$ m, $12$ hours apart; $3.5$ m after $4$ hours

A mass on a spring bounces between $4$ cm and $16$ cm above a table, making $3$ full bounces every second. At $t = 0$ it is at its highest point. Write its height $h(t)$ and find its height at $t = 0.1$ s.

$h(t) = 6 \cos(6\pi t) + 10$; about $8.1$ cm at $t = 0.1$ s

For the Ferris wheel in the first example, $h(t) = -10 \cos\left(\frac{\pi}{20} t\right) + 12$, when is the rider first $17$ m above the ground?

At $t = \frac{40}{3} \approx 13.3$ s

A Ferris wheel $30$ m across has its centre $17$ m above the ground and turns once every $60$ s. A rider boards at the bottom at $t = 0$. For how many seconds of each turn is the rider more than $24.5$ m above the ground?

$20$ seconds of each turn, from $t = 20$ s to $t = 40$ s

Grade 12 · MHF4U · Advanced Functions

MHF4U: Sinusoidal Word Problems

Anything that repeats on a regular cycle can be modelled with $y = a \cos(k(t - d)) + c$ or the sine version: a rider on a Ferris wheel, the depth of water at a harbour, a mass bouncing on a spring. A word problem gives you the cycle in words, and your job is to turn those words into the four numbers.

How It Works

Start with the highest and lowest values. The axis is halfway between them, $c = \frac{\max + \min}{2}$ , and the amplitude is half the gap, $|a| = \frac{\max - \min}{2}$ . For a Ferris wheel the axis is the height of the centre and the amplitude is the radius.

Then the period. The problem gives it as the time for one full cycle: one rotation, high tide to high tide, one bounce. Then $k = \frac{2\pi}{\text{period}}$ . A problem that gives a frequency, say 3 cycles per second, means a period of $\frac{1}{3}$ of a second.

Then where the cycle starts. Choose the function that starts where the problem does. Cosine starts at a maximum, so it fits a rider at the top or water at high tide. $-\cos$ starts at a minimum, which fits a rider boarding at the bottom. Sine starts on the axis going up. If the start is not at $t = 0$ , that time is $d$ .

Then answer the question. A question that gives a time and asks for a value is a substitution: put $t$ in and evaluate, in radians. A question that gives a value and asks for a time is a trig equation: isolate the cosine, find the first angle, then use the period to find the others. Questions like "how long is the rider above 25 m" ask for the length of the interval between two such times.

Check the model against the story before you use it: at $t = 0$ it should give the starting height, and half a period later it should give the opposite extreme.

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 $20$ m and its centre is $12$ m above the ground. It turns once every $40$ s, and a rider boards at the bottom at $t = 0$ . Write the rider's height $h(t)$ in metres.
Show solution
1. The radius is $10$ , so $a = 10$ . The centre is the axis: $c = 12$ .
2. Period $40$ s, so $k = \frac{2\pi}{40} = \frac{\pi}{20}$ .
3. The rider starts at the bottom, a minimum, so use $-\cos$ with no shift: $h(t) = -10 \cos\left(\frac{\pi}{20} t\right) + 12$ .
4. Check: $h(0) = -10 + 12 = 2$ , the bottom, and $h(20) = 10 + 12 = 22$ , the top.
Answer: $h(t) = -10 \cos\left(\frac{\pi}{20} t\right) + 12$
Easy
The depth of water at a dock is $d(t) = 3 \cos\left(\frac{\pi}{6} t\right) + 5$ metres, $t$ hours after high tide. Find the depth at high and low tide, the time between high tides, and the depth $4$ hours after high tide.
Show solution
1. High tide is $5 + 3 = 8$ m and low tide is $5 - 3 = 2$ m.
2. Period $\frac{2\pi}{\pi/6} = 12$ hours between high tides.
3. $d(4) = 3 \cos\left(\frac{2\pi}{3}\right) + 5 = 3\left(-\frac{1}{2}\right) + 5 = 3.5$ .
Answer: High tide $8$ m, low tide $2$ m, $12$ hours apart; $3.5$ m after $4$ hours
Medium
A mass on a spring bounces between $4$ cm and $16$ cm above a table, making $3$ full bounces every second. At $t = 0$ it is at its highest point. Write its height $h(t)$ and find its height at $t = 0.1$ s.
Show solution
1. $c = \frac{16 + 4}{2} = 10$ and $a = \frac{16 - 4}{2} = 6$ .
2. $3$ bounces per second is a period of $\frac{1}{3}$ s, so $k = \frac{2\pi}{1/3} = 6\pi$ .
3. It starts at a maximum, so use cosine: $h(t) = 6 \cos(6\pi t) + 10$ .
4. $h(0.1) = 6 \cos(0.6\pi) + 10 \approx 6(-0.309) + 10 \approx 8.15$ .
Answer: $h(t) = 6 \cos(6\pi t) + 10$ ; about $8.1$ cm at $t = 0.1$ s
Medium
For the Ferris wheel in the first example, $h(t) = -10 \cos\left(\frac{\pi}{20} t\right) + 12$ , when is the rider first $17$ m above the ground?
Show solution
1. Set $-10 \cos\left(\frac{\pi}{20} t\right) + 12 = 17$ , so $\cos\left(\frac{\pi}{20} t\right) = -\frac{1}{2}$ .
2. The first positive angle with cosine $-\frac{1}{2}$ is $\frac{2\pi}{3}$ .
3. $\frac{\pi}{20} t = \frac{2\pi}{3}$ gives $t = \frac{40}{3} \approx 13.3$ .
Answer: At $t = \frac{40}{3} \approx 13.3$ s
Test level
A Ferris wheel $30$ m across has its centre $17$ m above the ground and turns once every $60$ s. A rider boards at the bottom at $t = 0$ . For how many seconds of each turn is the rider more than $24.5$ m above the ground?
Show solution
1. $a = 15$ , $c = 17$ , $k = \frac{2\pi}{60} = \frac{\pi}{30}$ , starting at the bottom: $h(t) = -15 \cos\left(\frac{\pi}{30} t\right) + 17$ .
2. $h(t) > 24.5$ means $-15 \cos\left(\frac{\pi}{30} t\right) > 7.5$ , so $\cos\left(\frac{\pi}{30} t\right) < -\frac{1}{2}$ .
3. On one turn, $0 \le \frac{\pi}{30} t \le 2\pi$ , the cosine is below $-\frac{1}{2}$ for angles between $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ .
4. So $20 < t < 40$ : the rider is above $24.5$ m from $20$ s to $40$ s.
Answer: $20$ seconds of each turn, from $t = 20$ s to $t = 40$ s

Learn It on Martian Lab

Martian Lab teaches this in the Sinusoidal Functions lesson of its MHF4U Trigonometric Functions unit. Watch it, then take the Trigonometric Functions practice tests to see if it stuck.

Watch the Sinusoidal Functions lesson free

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Free Trigonometric Functions practice test, with solutions

More Trigonometric Functions topics: Solving Trig Equations · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Proving Trig Identities · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

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