B. Trigonometric Functions · Radian Measure and Trigonometric Functions

Solving Problems with Sinusoidal Models: Times, Intervals and Rates of Change

B2.7About 75 minutes

Learning goals

I can find when a sinusoidal model reaches a given value, using inverse trigonometric functions in radian mode and the two-solutions idea.
I can find how long a model stays above or below a given value in each cycle.
I can calculate the average rate of change of a sinusoidal model over an interval and interpret it with units.
I can estimate an instantaneous rate of change and explain where a sinusoidal model changes fastest and slowest.

Words to know

average rate of change
The change in output divided by the change in input over an interval: [f(b) − f(a)] ÷ (b − a). It is the slope of the secant line.

secant line
A line through two points on a graph.

instantaneous rate of change
The rate of change at one instant: the slope of the tangent line. It can be estimated with an average rate over a very short interval.

tangent line
A line that touches a curve at one point and has the same direction as the curve there.

inverse cosine (cos⁻¹)
The calculator function that gives an angle (in radian mode, a number from 0 to π) whose cosine is a given value.

Questions models can answer

Once you have a model, you can ask much more than "what is the value at time t?" Today you will answer:

When does the model reach a certain value?
For how long is it above or below a value?
How fast is it changing?

Remember from Grade 11: in Sinusoidal Functions you solved model equations using inverse ratios and the fact that a ratio has two angles in each cycle. The same idea works in radians. Set your calculator to radian mode: sin⁻¹ gives a number between −π/2 and π/2, and cos⁻¹ gives a number between 0 and π.

Worked example

Example 1: When is the rider at 32 m?

For the Ferris wheel h(t) = −20 cos(πt/40) + 22 (t in s, h in m), find the times in the first rotation when the rider is 32 m above the ground, and how long each rotation the rider spends above 32 m.

Set up: −20 cos(πt/40) + 22 = 32, so −20 cos(πt/40) = 10 and cos(πt/40) = −1/2.
Let θ = πt/40. In one rotation, 0 ≤ t ≤ 80, so 0 ≤ θ ≤ 2π.
cos θ = −1/2 in quadrants II and III with β = π/3: θ = 2π/3 or 4π/3.
Solve for t: t = 40θ/π. t = 40(2/3) = 80/3 ≈ 26.7 s and t = 40(4/3) = 160/3 ≈ 53.3 s.
Between these times the rider is higher than 32 m (the top, 42 m, comes at t = 40, between them).

At t ≈ 26.7 s and t ≈ 53.3 s. Above 32 m for 160/3 − 80/3 = 80/3 ≈ 26.7 s each rotation, one third of the ride.

The line h = 32 meets one cycle twice. The shaded band, from about 26.7 s to 53.3 s, is when the rider is higher than 32 m.

Worked example

Example 2: A non-special value

Using D(t) = 3.2 sin((2π/365)(t − 80)) + 12.2 for Toronto, find the days of the year when there are more than 14 hours of daylight.

Solve 3.2 sin θ + 12.2 = 14, where θ = (2π/365)(t − 80). Then sin θ = 1.8/3.2 = 0.5625.
Radian mode: θ = sin⁻¹(0.5625) ≈ 0.5974. The other solution in the cycle is π − 0.5974 ≈ 2.5442.
Convert back: t = 80 + 365θ/(2π). t ≈ 80 + 34.7 = 114.7 and t ≈ 80 + 147.8 = 227.8.
Between those days, sin θ > 0.5625, so the daylight is more than 14 h.

From about day 115 (April 25) to day 228 (August 16), roughly 113 days, according to the model.

Rates of change

The average rate of change of a function f between t = a and t = b is

[f(b) − f(a)] ÷ (b − a)

It is the slope of the secant line through the two points, with units such as metres per second. For a sinusoidal model, the rate keeps changing:

it is 0 at a maximum or minimum (the graph is momentarily flat);
it is largest in size where the graph crosses its axis (the graph is steepest there).

The instantaneous rate of change at one moment is the slope of the tangent line. You can estimate it with an average rate over a very short interval, such as from t = 20 to t = 20.01. You will study this idea in depth in the unit Rates of Change, and in Calculus and Vectors (MCV4U) you will find it exactly with derivatives.

Worked example

Example 3: Average rates on the Ferris wheel

For h(t) = −20 cos(πt/40) + 22, find the average rate of change of height from t = 0 to t = 10 s and from t = 10 to t = 20 s.

h(0) = 2, h(10) = 22 − 10√2 ≈ 7.858, h(20) = −20 cos(π/2) + 22 = 22.
From 0 to 10: (7.858 − 2) ÷ 10 ≈ 0.586 m/s.
From 10 to 20: (22 − 7.858) ÷ 10 ≈ 1.414 m/s.

About 0.59 m/s then 1.41 m/s. The rider rises slowly at first and faster as the seat swings toward the side of the wheel.

Worked example

Example 4: Estimating an instantaneous rate

Estimate how fast the rider's height is changing at t = 20 s, using short intervals.

Use [20, 20.1]: h(20.1) ≈ 22.15708, so the rate ≈ (22.15708 − 22) ÷ 0.1 ≈ 1.5708 m/s.
Use [20, 20.01]: the rate ≈ 1.570796 m/s.
The estimates settle on about 1.5708, which is π/2.

About 1.57 m/s (it is exactly π/2 m/s, which calculus confirms). This is the fastest the height ever changes: at t = 20 the rider is level with the centre.

Two secant lines (dashed) show the average rates over 0 to 10 s and 10 to 20 s. The dotted tangent line at t = 20 is steeper still: the height changes fastest when the rider is level with the centre.

Solving model equations

When you solve, always (1) isolate the sine or cosine, (2) find all solutions of the angle θ in the cycle (two, usually), (3) convert each θ back to t, and (4) add multiples of the period for later cycles. Check one answer by substituting it into the model.

Just for fun

On a Ferris wheel, your height changes fastest when you are level with the centre, and for an instant at the very top it does not change at all. So the top really is the best place for a photo.

Hands-on activity

Track the daylight

You need: a weather app, a newspaper, or a website that gives today's sunrise and sunset times for where you live (your local weather forecast usually shows them), a calculator and paper.

Write down today's sunrise and sunset times. Compute today's hours of daylight (sunset minus sunrise), in hours and minutes, then in decimal hours.
Look up (or record a week from now) the sunrise and sunset for a date 7 days later.
Compute the average rate of change of daylight in minutes per day over that week.
Use the Toronto model D(t) = 3.2 sin((2π/365)(t − 80)) + 12.2 to compute the same rate (find the day numbers first). Compare.
Is daylight increasing or decreasing right now? Where on the curve is your date: near a maximum, a minimum, or the axis?

What to look for: near the March or September equinox the rate is largest; the model gives about 0.055 h/day, about 3.3 minutes per day. Near June 20 or December 21 the rate is close to 0. Places farther north than Toronto have bigger swings (a larger amplitude), so their rates are larger. Your local numbers will differ from the Toronto model if you live elsewhere.

Your turn

For h(t) = −20 cos(πt/40) + 22, find the two times in the first rotation when the rider is at 12 m.

For how long in each rotation is the rider above 12 m?

Find the average rate of change of h from t = 0 to t = 40 s.

Find the average rate of change of h from t = 40 to t = 80 s.

At which times in the first rotation (0 ≤ t ≤ 80) is the instantaneous rate of change of h equal to 0?

Solve 3 sin(πt/6) + 10 = 11.5 for 0 ≤ t ≤ 12.

Estimate the rate of change of h at t = 20 s using the interval [20, 20.1].

More practice: warm-up

Write the formula for the average rate of change of f from x = a to x = b.

Solve sin θ = 0.5 for 0 ≤ θ ≤ 2π.

Solve cos θ = −1 for 0 ≤ θ ≤ 2π.

What is the instantaneous rate of change at a maximum of a sinusoidal model?

What are the units of the rate of change of height (m) with respect to time (s)?

More practice: on your own

For D(t) = 3.2 sin((2π/365)(t − 80)) + 12.2, on which days is the daylight exactly 12.2 h?

For the tide model h(t) = 2.5 cos((2π/12.4)(t − 3)) + 4, when does the depth first fall to 3 m after the high tide at t = 3?

For the same tide model, find the average rate of change of depth from t = 3 to t = 6.1.

For the spring d(t) = −6 cos(5πt/2) (cm), find the average velocity from t = 0 to t = 0.2 s.

Estimate the rate of change of daylight at day 80 using the interval [80, 81]. Give the answer in minutes per day.

Challenge

For h(t) = −20 cos(πt/40) + 22, what fraction of each rotation does the rider spend above 35 m?

Compute the average rate of change of y = sin x over [0, 0.1] and [0, 0.01] (radian mode). What value is it approaching, and what would change if x were in degrees?

Using the Toronto daylight model, for how many days of the year is there more than 15 hours of daylight?

Check your understanding

Quiz

Which expression gives the average rate of change of f from x = a to x = b?

What are the solutions of 2 sin x = 1 for 0 ≤ x ≤ 2π?

Where is the instantaneous rate of change of a sinusoidal model largest in size?

For h(t) = −20 cos(πt/40) + 22, when is the rider first 22 m above the ground?

What is the average rate of change of y = sin x over [0, π]?

A model with minimum 2 and maximum 42 is set equal to 30. How many solutions are there in one full cycle?

For parents and teachers

Big idea (B2.7): sinusoidal models answer real questions: when a value is reached (solve with inverse trig in radian mode, two answers per cycle), how long a value is exceeded (subtract the times), and how fast things change (average rate = slope of a secant; instantaneous rate = slope of a tangent, estimated with a tiny interval).

This lesson connects to the unit Rates of Change, and to Calculus and Vectors (MCV4U), where the exact rates come from derivatives.

How to help: The daylight activity is a great family project: check sunrise and sunset once a week for a month and graph the hours of daylight together.

Common difficulties:

Finding only one solution per cycle.
Forgetting to convert the angle θ back to time t.
Using degree mode with sin⁻¹ or cos⁻¹.
Rounding intermediate values too early when computing rates over short intervals (keep all calculator digits).