A. Rate of Change · Derivatives of Exponential and Sinusoidal Functions

Solving Problems with Sinusoidal Rates of Change

A3.5 A2.4About 75 minutes

Learning goals

I can differentiate sinusoidal models of real situations, such as Ferris wheels, tides, daylight and alternating current.
I can find and interpret the rate of change at a given time, with units.
I can find when the rate is greatest, least or zero, and connect these to the midline, maximums and minimums.
I can find the times when a sinusoidal rate reaches a given value by solving a trigonometric equation.

Words to know

sinusoidal model
A function h(t) = a sin(k(t − d)) + c or a cos(k(t − d)) + c used to describe a repeating situation.

midline
The horizontal line y = c halfway between the maximum and minimum. A sinusoid changes fastest as it crosses its midline.

maximum rate of change
For a sinusoid with amplitude |a| and k, the largest |rate| is |ak|.

slack water
The short time at high or low tide when the water level is barely changing.

alternating current (AC)
Electric current whose voltage varies sinusoidally, 60 cycles per second in Canadian household outlets.

Riding, rising and falling

In MHF4U you modelled periodic situations with sinusoidal functions and found approximate rates of change. Now you can find exact instantaneous rates with derivatives.

The strategy is the same roadmap as for exponential problems: model, differentiate (usually with the chain rule), substitute or solve, and interpret with units. Two facts make sinusoidal problems quicker:

The rate is zero at every maximum and minimum.
The rate has its greatest size, |ak|, every time the graph crosses its midline.

Worked example

Example 1: A Ferris wheel

A Ferris wheel has a radius of 15 m and its axle is 17 m above the ground. It turns once every 2 minutes. You board at the lowest point. Your height is h(t) = 17 − 15 cos(πt) metres after t minutes. (a) Find h′(t). (b) How fast are you rising at t = 1/3 min, and how high are you? (c) When are you rising fastest, and how fast?

(a) Chain rule: h′(t) = −15 × (−sin(πt)) × π = 15π sin(πt).
(b) h′(1/3) = 15π sin(π/3) = 15π(√3/2) ≈ 40.8 m/min. Height: h(1/3) = 17 − 15 cos(π/3) = 17 − 7.5 = 9.5 m.
(c) sin(πt) is largest (1) when πt = π/2, so t = 0.5 min. The rate is then 15π ≈ 47.1 m/min. At that moment h = 17 m: you are level with the axle.

(a) h′(t) = 15π sin(πt); (b) rising at about 40.8 m/min at a height of 9.5 m; (c) fastest at t = 0.5 min, 15π ≈ 47.1 m/min, as you pass axle level.

On the wheel, your seat climbs fastest as it passes the level of the axle. On the graph of height, that moment (t = 0.5 min) is where the curve crosses its midline, h = 17 m, with slope 15π ≈ 47.1 m/min.

Worked example

Example 2: Tides on the Bay of Fundy

The Bay of Fundy, between New Brunswick and Nova Scotia, has some of the highest tides in the world. At one harbour, the water depth is modelled by h(t) = 7 cos(2πt/12.4) + 8 metres, t hours after a high tide (the tide repeats about every 12.4 hours). (a) How fast is the water level changing 2 hours after high tide? (b) What is the fastest the water level changes, and when? (c) When, during the first falling tide, is the water dropping at 2 m per hour?

Let k = 2π/12.4 ≈ 0.5067. Then h′(t) = −7k sin(kt) ≈ −3.547 sin(kt).
(a) h′(2) ≈ −3.547 sin(1.0134) ≈ −3.01 m/h. The depth is h(2) ≈ 11.7 m.
(b) The greatest size is 7k ≈ 3.55 m/h, when sin(kt) = ±1: t = 3.1 h (falling, at mid-tide) and t = 9.3 h (rising).
(c) Solve −3.547 sin(kt) = −2: sin(kt) ≈ 0.5639. On the first falling tide, 0 < kt < π, so kt ≈ 0.599 or kt ≈ π − 0.599 ≈ 2.543.
Divide by k: t ≈ 1.18 h or t ≈ 5.02 h.

(a) Falling at about 3.0 m per hour; (b) 3.55 m per hour at mid-tide, t = 3.1 h (falling) and t = 9.3 h (rising); (c) about 1.2 h and 5.0 h after high tide.

The tide model: the depth changes fastest at mid-tide (h = 8 m), about 3.55 m per hour, and is momentarily still (slack water) at high and low tide.

Worked example

Example 3: Hours of daylight

A model for the hours of daylight in a city in southern Ontario is D(t) = 3.2 sin(2π(t − 80)/365) + 12.2, where t is the day of the year (t = 1 on January 1). Find D′(t) and the rate on the spring equinox (t = 80) and the summer solstice (t = 172), in minutes per day.

D′(t) = 3.2 cos(2π(t − 80)/365) × 2π/365 ≈ 0.05509 cos(2π(t − 80)/365) hours per day.
t = 80: cos 0 = 1, so D′ ≈ 0.05509 h/day × 60 ≈ 3.3 minutes per day.
t = 172: 2π(92)/365 ≈ 1.584 radians, cos ≈ −0.0128, so D′ ≈ −0.0007 h/day, about 0.04 minutes (2.6 seconds) per day.

In this model, days lengthen by about 3.3 minutes per day around the spring equinox, but barely change (almost 0) at the summer solstice. That matches what people notice: daylight changes fastest in March and September.

Worked example

Example 4: Household electricity

The voltage of a Canadian household outlet can be modelled by V(t) = 170 sin(120πt) volts, t in seconds (60 cycles per second). Find the greatest rate of change of voltage.

V′(t) = 170 cos(120πt) × 120π = 20 400π cos(120πt).
The greatest value of |cos| is 1, so the greatest rate is 20 400π ≈ 64 088 V/s.
It happens when V = 0, as the voltage crosses its midline.

About 64 000 volts per second, at the instants when the voltage passes through 0. (The peak of 170 V corresponds to the 120 V 'effective' voltage printed on appliances.)

Hands-on activity

Your community's daylight

You need: a computer or phone to look up sunrise and sunset times for your community (for example on a weather website or an almanac), a calculator, graph paper.

Look up sunrise and sunset for the 21st of each month this year (12 dates). Compute the hours of daylight for each (as a decimal, e.g. 9 h 15 min = 9.25 h).
Plot daylight against day of the year. Find the maximum, the minimum, the midline value c and the amplitude a.
Build a model D(t) = a sin(2π(t − 80)/365) + c (the spring equinox is near day 80).
Differentiate. Use your model to predict the change in daylight per day on March 21, June 21 and your birthday. Convert to minutes per day.
Check: look up sunrise and sunset for March 21 and March 22 and find the actual change.

What to look for: communities farther north have a bigger amplitude, so their daylight changes faster near the equinoxes. The model should predict a change near 0 on June 21 and the largest change in late March and late September. For most of southern Ontario, the March change is around 3 minutes per day; communities in Nunavut see much bigger changes.

Just for fun

Why did the Ferris wheel rider bring a calculus textbook? She wanted to know exactly when things would be looking up the fastest. (Answer: at axle level.)

Your turn

For h(t) = 17 − 15 cos(πt), find h′(1) and explain it.

For h(t) = 17 − 15 cos(πt), find h′(1.5).

A buoy bobs as y(t) = 0.6 sin(2t) metres, t in seconds. Find its greatest vertical speed.

For the tide model h(t) = 7 cos(2πt/12.4) + 8, find h′(6.2) and explain.

A mass on a spring has s(t) = 5 cos(4t) cm. Find v(t) and v(π/8).

For V(t) = 170 sin(120πt), find V′(0).

More practice: warm-up

Where on a sinusoidal graph is the rate of change 0?

Where is a sinusoid changing fastest?

What is the greatest rate of change of y = 4 sin(3t) + 10?

If h is in metres and t in minutes, what are the units of h′(t)?

Differentiate h(t) = 20 + 5 sin(πt/6).

More practice: on your own

A different Ferris wheel has h(t) = 25 − 20 cos(πt/3) metres, t in minutes. Find the radius, the period, and the greatest vertical speed.

For that wheel, find h′(1) and h(1).

The temperature in a town over a day is T(t) = −8 cos(πt/12) + 2 °C, t hours after midnight (a February day). When is the temperature rising fastest, and how fast?

For the daylight model D(t) = 3.2 sin(2π(t − 80)/365) + 12.2, find D′(105) in minutes per day.

For s(t) = 5 cos(4t), find all t in [0, π/2] where the speed is 10 cm/s.

Challenge

A weight on a spring moves as y(t) = 10e^(−0.1t) cos(2πt) cm (damped motion). Find y′(t) and y′(1).

For the Ferris wheel h(t) = 17 − 15 cos(πt), at what heights is your vertical speed half its maximum?

Show that the greatest rate of change of a daylight model D(t) = a sin(2π(t − 80)/365) + c is proportional to the amplitude a, and find a if daylight changes by at most 6 minutes per day.

Check your understanding

Quiz

h(t) = 17 − 15 cos(πt). What is h′(t)?

A sinusoidal model changes fastest:

What is the greatest rate of change of h(t) = 6 sin(πt/4) + 9?

For a tide model, slack water (rate 0) happens:

A buoy's height is y = 0.5 sin(3t) m. Its velocity at t = 0 is:

For the daylight model in this lesson, daylight changes most slowly:

For parents and teachers

Big idea (A3.5, A2.4): derivatives of sinusoidal models give exact rates for repeating situations. The rate is zero at maximums and minimums and greatest (|ak|) at the midline. Real contexts include Ferris wheels, Bay of Fundy tides, daylight hours and household electricity.

How to help: ask your teen to point on the graph to where the rate is 0 and where it is largest before calculating. Then check the calculation agrees. Make sure the calculator is in radian mode.

Common difficulties:

Forgetting the chain-rule factor k (for example 2π/12.4).
Answers in the wrong units (hours per day versus minutes per day).
Finding only one solution of a trigonometric equation when the context allows two (rising and falling).
Mixing up 'greatest rate' with 'greatest height'.