D. Trigonometric Functions · Sinusoidal Functions

Solving Problems with Sinusoidal Models

D3.5 D3.2About 75 minutes

Learning goals

I can answer questions about a sinusoidal model by reading its graph.
I can solve a sinusoidal model for the input, using inverse trigonometric ratios and the two-angles idea.
I can find how long, or how many times, a model is above or below a given value.
I can pose my own problems about a sinusoidal situation and solve them.

Words to know

solve a model
Find the input value(s) that give a required output, for example the times when a rider is 32 m high.

inverse cosine / inverse sine
cos⁻¹ and sin⁻¹ on a calculator give one angle with a given ratio; the second angle in each cycle comes from the related acute angle.

interval
A range of input values, such as 20 < t < 40, during which something is true.

pose a problem
Write your own question about a situation, one that can be answered with the model.

Asking the model questions

A model is only useful if it answers questions. Some questions go forward: "How high is the rider after 50 s?" (substitute and calculate). Others go backward: "When is the rider higher than 32 m?" or "On which days are there more than 14 hours of daylight?" For backward questions you can read a graph (drawn by hand or with technology) or solve algebraically using inverse ratios, just as you found two angles with the same ratio in Trigonometric Ratios and Laws.

Solving algebraically: the plan

Set the model equal to the target value.
Isolate the sine or cosine: subtract c, then divide by a.
Find the angle(s) θ in one cycle (0° to 360°) with that ratio: one from the calculator (cos⁻¹ or sin⁻¹) and the second from the related acute angle (CAST).
Set the inside, k(t − d), equal to each angle and solve for t.
Add whole periods to find later solutions, and keep only the ones in the domain.
Answer in context (with units, and clock times or dates if needed).

Worked example

Example 1: When is the rider above 32 m?

For the Ferris wheel h = −20 cos(6t) + 22 (60 s per turn), find when the rider is above 32 m during the first turn, and for how long.

Set equal: −20 cos(6t) + 22 = 32.
Isolate: −20 cos(6t) = 10, so cos(6t) = −0.5.
cos θ = −0.5 for θ in 0° to 360°: related angle 60°, cosine negative in quadrants 2 and 3: θ = 120° and 240°.
6t = 120° gives t = 20 s; 6t = 240° gives t = 40 s.
The rider passes 32 m going up at 20 s and going down at 40 s, and is above 32 m in between: 20 < t < 40.
That is 20 s of every 60 s turn: one-third of the ride. (Later turns repeat: 80 < t < 100, and so on.)

Above 32 m from 20 s to 40 s: 20 s per turn.

Reading the same answer from the graph: the curve is above the line h = 32 between 20 s and 40 s, and again between 80 s and 100 s.

Worked example

Example 2: Long summer days

The daylight model for a southern Ontario city is D(n) = 3.2 sin(0.986(n − 80)) + 12.2. On which days of the year are there more than 14 hours of daylight?

Set equal: 3.2 sin(0.986(n − 80)) + 12.2 = 14.
Isolate: sin(0.986(n − 80)) = 1.8 ÷ 3.2 = 0.5625.
sin⁻¹(0.5625) ≈ 34.2°. Sine is also positive in quadrant 2: 180° − 34.2° = 145.8°.
0.986(n − 80) = 34.2 gives n − 80 ≈ 34.7, so n ≈ 114.7. 0.986(n − 80) = 145.8 gives n − 80 ≈ 147.8, so n ≈ 227.8.
Between these, the graph is above 14 h (it peaks near day 171). Whole days with more than 14 h: day 115 to day 227.
In a non-leap year, day 115 is April 25 and day 227 is August 15: about 113 days.

From about April 25 to August 15.

The daylight model is above 14 h between day 114.7 and day 227.8, the shaded part of the year.

Worked example

Example 3: Can the boat leave?

Sample tide model for a harbour on the Bay of Fundy: h = 6 cos(29.0(t − 4.5)) + 7 (metres, t hours after midnight; k = 360 ÷ 12.4). A fishing boat needs at least 4 m of water to leave the harbour. During what time in the morning and early afternoon is the water too shallow?

Set equal: 6 cos(29.0(t − 4.5)) + 7 = 4, so cos(29.0(t − 4.5)) = −0.5.
θ = 120° or 240°.
29.0(t − 4.5) = 120 gives t − 4.5 ≈ 4.13, so t ≈ 8.63 h ≈ 8:38 a.m.
29.0(t − 4.5) = 240 gives t − 4.5 ≈ 8.27, so t ≈ 12.77 h ≈ 12:46 p.m.
Between these times the tide is at its low point (low tide is at about 10:42 a.m.), so the water is below 4 m from about 8:38 a.m. to 12:46 p.m. The boat should leave before 8:38 a.m. or wait until after 12:46 p.m.

Too shallow from about 8:38 a.m. to 12:46 p.m.

Two solutions per cycle, and check with a sketch

Inside each cycle, a horizontal line between the minimum and the maximum crosses a sinusoidal graph twice. Your calculator gives only one angle, so always find the second one with the related acute angle. Then make a quick sketch to decide whether the answer is the interval between the two times or outside them.

Worked example

Example 4: Posing and solving your own problem

Pose a question about the pendulum model d = 12 cos(180t) (cm, seconds) that needs the model to answer, and solve it.

A question: "During the first swing cycle, for how long is the pendulum more than 6 cm to the right of its resting point?"
Solve 12 cos(180t) = 6: cos(180t) = 0.5, so 180t = 60° or 300°: t = 1/3 s or t = 5/3 s.
The pendulum starts at 12 cm (more than 6 cm) and is more than 6 cm to the right for 0 ≤ t < 1/3 and again for 5/3 < t ≤ 2.
Total: 1/3 + 1/3 = 2/3 s of each 2 s cycle.

About 0.67 s of every 2 s cycle.

Just for fun

An old proverb says, "Time and tide wait for no one." A sinusoidal model cannot make the tide wait either, but it can give you a good estimate of how long you have before your boat is sitting on the mud.

Hands-on activity

Your own sinusoidal question

You need: one model from this unit (the paper-plate Ferris wheel, the daylight model, your circle-walk model or a model you made from data), paper, a calculator and a free graphing tool if you have one.

Write down your model and what each variable means (with units).
Pose three questions about it: one forward question ("What is the value when…?"), one backward question ("When is it equal to…?"), and one "how long" or "how many times" question.
Solve each question algebraically.
Check each answer by reading a graph (sketch it by hand or use technology).
Swap questions with a family member or friend: can they understand and answer your questions?

What to look for: good questions name the units and the time period ("during the first turn", "in one year"). Answers should state both solutions in a cycle where needed and be written in a full sentence, for example: "The rider is above 32 m from 20 s to 40 s, which is 20 s per turn."

Your turn

For h = −20 cos(6t) + 22, when is the rider first at a height of 12 m? When is the second time in the first turn?

For the same wheel, how long during each turn is the rider below 12 m?

How many times does the rider reach the top (42 m) in the first 3 minutes after boarding at t = 0?

Use D(n) = 3.2 sin(0.986(n − 80)) + 12.2 to find the hours of daylight on day 200.

For the pendulum d = 12 cos(180t), when is d = 6 cm for the first time?

For T = 14 cos(30(m − 7)) + 8, in which months is the average temperature above 15 °C?

Pose a question about the tide model h = 6 cos(29.0(t − 4.5)) + 7 that involves a time, and answer it.

More practice: warm-up

For h = −20 cos(6t) + 22, find h(0).

For h = −20 cos(6t) + 22, find h(30).

Find all θ from 0° to 360° with cos θ = 0.5.

Find all θ from 0° to 360° with sin θ = −0.5.

What is the maximum of T = 14 cos(30(m − 7)) + 8, and in which month?

More practice: on your own

For the larger wheel h = −25 cos(4t) + 27 (90 s per turn), when is the rider first 40 m high, and how long per turn is the rider above 40 m?

Tide model h = 3 cos(29.0(t − 1)) + 5 (t hours after midnight). Between midnight and 4:00 p.m., when is the water at least 6.5 m deep?

A spring's height is h = 5 cos(450t) + 35 (cm, seconds). When does the weight first reach 32.5 cm?

For the outlet model V = 170 sin(21.6t) (t in ms), when does the voltage first reach 120 V?

Use the symmetry of the daylight model (maximum at day 171.25) to find the other day of the year with the same daylight as day 100.

Challenge

A ride on the wheel h = −20 cos(6t) + 22 lasts 3 full turns. What total time is the rider above 32 m, and what fraction of the ride is that?

Using Example 2, about how many days a year have more than 14 h of daylight? Then find how many days have less than 10.4 h (by symmetry, without a new calculation).

A cargo ship needs at least 9 m of water to dock. Using the sample tide model h = 6 cos(29.0(t − 4.5)) + 7, find every time interval in the 24 h day when it can dock.

Check your understanding

Quiz

For h = −20 cos(6t) + 22, when is the rider first at the top?

Which angles from 0° to 360° have cos θ = −0.5?

For h = −20 cos(6t) + 22, how many seconds of each turn is the rider above 32 m?

For d = 12 cos(180t), when does the pendulum first pass its resting point (d = 0)?

For T = 14 cos(30(m − 7)) + 8, which months have an average temperature above 15 °C?

Why must you look for a second angle after using cos⁻¹ on a calculator?

For parents and teachers

Big idea: D3.5 asks students to pose and solve problems based on applications of sinusoidal functions, using a given graph or one made from a table or equation. Students can answer by reading graphs or algebraically using inverse ratios and the "two angles per cycle" idea from the unit Trigonometric Ratios and Laws.

How to help: Ask your teen to explain a solution aloud from beginning to end, including the sketch. Ask "Why two answers?" and "Is the answer between them or outside them?"

Common difficulties:

Forgetting the second angle in the cycle.
Dividing by k before subtracting d, or forgetting to add d back. Solve k(t − d) = θ step by step: t − d = θ/k, then t = θ/k + d.
Converting decimal hours to clock times: 0.63 h = 0.63 × 60 ≈ 38 min.
Giving an answer without context. A complete answer is a sentence with units and times or dates.
Calculator in radian mode.