Grade by Grade
D. Characteristics of Functions · Combining Functions and Modelling

Real-World Combinations: Damped Motion, Tones and Profit

D2.2 D3.3About 75 minutes
Learning goals
  • I can recognize real situations modelled by sums, differences, products and quotients of functions.
  • I can model a damped oscillation as the product of an exponential function and a trigonometric function.
  • I can explain why a phone keypad tone is the sum of two trigonometric functions.
  • I can solve problems about combined models graphically, and check them algebraically when possible.

Words to know

damped oscillation
Back-and-forth motion whose size shrinks over time, such as a swing or a spring coming to rest. A common model is y = a·bᵗ·cos(kt) with 0 < b < 1.
amplitude envelope
The curves y = ±a·bᵗ that a damped oscillation stays between. They show how the size of the swings shrinks.
dual-tone multi-frequency (DTMF)
The system that makes phone keypad sounds: each key plays two pure tones at once, one for its row and one for its column.
break-even point
A quantity where revenue equals cost, so profit P = R − C is zero.
profit
Revenue minus cost: P(x) = R(x) − C(x).

A swing that slowly stops

Give a little cousin one push on a swing in the park and walk away. The swing goes back and forth (a periodic motion), but each swing is a bit smaller than the last (an exponential decrease, caused by air resistance and friction). The motion is a product:

(size that shrinks exponentially) × (back-and-forth trigonometric motion).

In this lesson you will meet four real situations, each built by combining functions you already know. For each one, ask: which simple functions are combined, and how?

SituationCombinationModel
Profit of a small businessdifferenceP(x) = R(x) − C(x)
A hot drink cooling in a roomsumT(t) = (room temperature) + (exponential decay)
A swing or spring coming to restproductd(t) = a·bᵗ·cos(kt)
A phone keypad tonesump(t) = sin(2πf₁t) + sin(2πf₂t)
Average cost per itemquotientA(n) = C(n)/n
Five combined models. The last one appeared in the lesson Quotients of Functions.

1. Damped oscillation: a product

A weight bounces on a spring. Its displacement from rest, in centimetres, after t seconds is modelled by

d(t) = 12(0.9)ᵗ cos(πt).

  • The factor cos(πt) makes it go back and forth with period 2π/π = 2 s.
  • The factor 12(0.9)ᵗ shrinks by 10% each second. It is the amplitude.
  • Because −1 ≤ cos(πt) ≤ 1, the graph stays between y = 12(0.9)ᵗ and y = −12(0.9)ᵗ, the envelope, and touches it whenever cos(πt) = ±1 (at t = 0, 1, 2, ...).

This is the same idea as the envelope of y = x sin x in the lesson Products of Functions, but now the envelope shrinks.

2468101214161820-12-9-6-336912t ≈ 13.2envelope y = 12(0.9)ᵗy = 3d (cm)t (s)
The solid curve d(t) = 12(0.9)ᵗ cos(πt) bounces between the dashed envelope curves. The upper envelope crosses the dotted line y = 3 at t ≈ 13.2 s; after that, every swing is smaller than 3 cm.
Worked example

Example 1: When do the bounces become small?

For d(t) = 12(0.9)ᵗ cos(πt), find d(1) and d(2), and find when the amplitude first drops below 3 cm. Solve graphically, then check with logarithms.

  1. d(1) = 12(0.9)¹ cos π = 12(0.9)(−1) = −10.8 cm (10.8 cm below rest).
  2. d(2) = 12(0.81) cos 2π = 9.72 cm.
  3. The amplitude is the envelope 12(0.9)ᵗ. Graphically: draw y = 3 and find where it meets the upper envelope. The graph shows about t ≈ 13.
  4. Check algebraically: 12(0.9)ᵗ = 3, so 0.9ᵗ = 0.25, and t = log 0.25 / log 0.9 ≈ (−0.6021)/(−0.04576) ≈ 13.16. (You solved equations like this in the sibling unit Exponential and Logarithmic Functions.)

d(1) = −10.8 cm, d(2) = 9.72 cm. The amplitude drops below 3 cm after about 13.2 s.

2. Phone keypad tones: a sum

Press a key on a phone keypad and you hear one sound, but it is really two pure tones played together. The keypad is a grid. Each row has a low frequency and each column has a high frequency:

  • Rows: 697 Hz (1 2 3), 770 Hz (4 5 6), 852 Hz (7 8 9), 941 Hz (* 0 #)
  • Columns: 1209 Hz (1 4 7 *), 1336 Hz (2 5 8 0), 1477 Hz (3 6 9 #)

So key 5 (row 2, column 2) is modelled by

p(t) = sin(2π · 770t) + sin(2π · 1336t), with t in seconds.

A frequency of 770 Hz means 770 cycles per second, so sin(2π · 770t) has period 1/770 s. The sum is not a sinusoid: its shape is irregular, and its height reaches nearly 2 when both tones peak together. Equipment at the other end of the line detects the two frequencies and knows which key you pressed.

1−1t21−1−2tsin(2π · 770t) (dashed) and sin(2π · 1336t) (dotted)0.002 s0.004 s0.006 sTheir sum: the sound of key 5
Top: the two pure tones of key 5. Bottom: their sum, which is what the speaker actually plays. The sum is not a sine curve; where the tones line up it is taller, and where they oppose each other it nearly cancels.
Worked example

Example 2: Reading a keypad tone

(a) Write the model for key 9. (b) For key 5, find p(0) and explain why |p(t)| ≤ 2. (c) Which key has the model sin(2π · 941t) + sin(2π · 1336t)?

  1. (a) Key 9 is row 3 (852 Hz), column 3 (1477 Hz): p(t) = sin(2π · 852t) + sin(2π · 1477t).
  2. (b) p(0) = sin 0 + sin 0 = 0. Each sine is between −1 and 1, so their sum is between −2 and 2.
  3. (c) 941 Hz is the bottom row (* 0 #) and 1336 Hz is the middle column (2 5 8 0). The key in both is 0.

(a) sin(2π · 852t) + sin(2π · 1477t) (b) p(0) = 0, and the sum of two values in [−1, 1] is in [−2, 2] (c) key 0.

Just for fun

Funny but true: the name dual-tone multi-frequency means exactly what the math says, two tones (a sum of two sine functions) chosen from several frequencies. Every phone number you dial is really a little song of sums of sines.

3. Profit: a difference

A family maple syrup farm sells bottles of syrup at a market. To sell more bottles, they lower the price: at x bottles, the price is 25 − 0.05x dollars each. So the revenue is a product, R(x) = x(25 − 0.05x) = 25x − 0.05x². Their cost is $1000 for the market stall and equipment plus $5 per bottle: C(x) = 1000 + 5x. The profit is the difference

P(x) = R(x) − C(x) = −0.05x² + 20x − 1000.

Graphically, the profit is the vertical gap between the revenue curve and the cost line. The farm breaks even where the graphs of R and C intersect, which are exactly the zeros of P (as you saw in the lesson Sums and Differences of Functions).

100200300400-10001000200030004000x$(200, 1000)R(x) = 25x − 0.05x²C(x) = 1000 + 5xP(x) = R(x) − C(x)Break-even whereR and C meet:x ≈ 59 and x ≈ 341Maximum profit $1000at x = 200 bottles
Revenue (dashed) minus cost (dotted) gives profit (solid). The break-even points, where R and C cross, line up with the zeros of P. Profit is greatest halfway between them.
Worked example

Example 3: Break-even and maximum profit

For P(x) = −0.05x² + 20x − 1000, find the break-even quantities and the maximum profit.

  1. Break-even: P(x) = 0. Multiply by −20: x² − 400x + 20 000 = 0.
  2. Quadratic formula: x = [400 ± √(160 000 − 80 000)]/2 = [400 ± √80 000]/2 = 200 ± 141.4.
  3. So x ≈ 58.6 and x ≈ 341.4. Since bottles are whole numbers, the farm makes a profit from 59 to 341 bottles.
  4. Maximum: halfway between the zeros, x = 200. P(200) = −0.05(40 000) + 4000 − 1000 = $1000.
  5. Graphical check: the vertical gap between R and C is largest at x = 200, where R(200) = 3000 and C(200) = 2000.

Profit from 59 to 341 bottles; maximum profit $1000 at 200 bottles (price $15 each).

4. A hot drink cooling: a sum

A mug of hot chocolate at 90°C sits in a 20°C kitchen on a cold January day. A good model is

T(t) = 20 + 70(0.95)ᵗ, t in minutes.

It is the sum of a constant function (the room temperature, 20) and a decaying exponential (the extra heat, which starts at 70 and loses 5% each minute). The horizontal asymptote is T = 20: the drink never gets colder than the room.

Worked example

Example 4: When is it safe to drink?

Using T(t) = 20 + 70(0.95)ᵗ, find when the hot chocolate has cooled to 60°C.

  1. Graphically: draw T(t) and the line T = 60, and read the intersection. A table also works: T(10) ≈ 61.9 and T(11) ≈ 59.8, so it is between 10 and 11 minutes.
  2. Algebraically: 70(0.95)ᵗ = 40, so 0.95ᵗ = 4/7.
  3. t = log(4/7) / log 0.95 ≈ (−0.2430)/(−0.02228) ≈ 10.9.

After about 10.9 minutes.

Hands-on activity

A homemade damped pendulum

You need: about 1 m of string, a small heavy object to tie on the end (a set of keys or a large metal washer), tape, a ruler or measuring tape, and a helper.

  1. Tape the top of the string firmly to the top of a doorframe or the edge of a sturdy table so the keys hang freely.
  2. Put the ruler on a chair or box just under the keys, lined up with the swing.
  3. Pull the keys 20 cm to one side and let go (do not push). Your helper counts swings.
  4. Every 10 complete swings, estimate how far (in cm) the keys reach to the side. Record the amplitude A at 0, 10, 20, 30 and 40 swings.
  5. Divide each amplitude by the one before it. Are the ratios about the same?
  6. Tape a small piece of paper to the keys (more air resistance) and repeat.

What to look for: the amplitudes shrink, and the ratios are roughly constant (for example 20, 16, 13, 10.5, 8.5 cm gives ratios near 0.8). A constant ratio means the envelope is exponential, so the motion fits d = A₀ rⁿ cos(...). With the paper attached, the ratio is smaller: the swings die out faster. Your numbers will differ; that is expected for a real experiment.

Your turn

1

A spring's displacement is d(t) = 8(0.85)ᵗ sin(2πt) cm. What is the period, and what is the envelope?

2

For d(t) in the previous question, find d(0.25) and d(1.25).

3

Write the keypad model for key 1 and for key #.

4

R(x) = 40x − 0.1x² and C(x) = 2000 + 10x. Write P(x) and find the maximum profit.

5

A cup of tea cools by T(t) = 22 + 73(0.92)ᵗ (t in minutes). What is the starting temperature and the room temperature?

6

Name the combination in each: (a) total cost = fixed cost + variable cost (b) area of a rectangle whose sides both change with time (c) kilometres per litre.

More practice: warm-up

1

What is the envelope of y = 5(0.7)ᵗ cos t?

2

Is a damped oscillation a sum, difference, product or quotient?

3

How many pure tones make up one keypad sound?

4

P(x) = R(x) − C(x). What is P at a break-even point?

5

T(t) = 18 + 60(0.9)ᵗ. What value does T approach in the long run?

More practice: on your own

1

A ski shop's monthly sales (in thousands of dollars) are modelled by S(t) = 200 + 5t + 40 sin(πt/6), t in months. Name the two parts of the sum and find S(3) and S(9).

2

For d(t) = 10(0.8)ᵗ cos(2πt), find when the amplitude drops below 1 cm.

3

Find the break-even points for R(x) = 30x − 0.02x² and C(x) = 4000 + 10x.

4

Hot chocolate follows T(t) = 20 + 70(0.95)ᵗ. Find T(20) and T(30) and explain why the temperature drops more slowly later.

5

Explain why y = sin(2π · 770t) + sin(2π · 1336t) can be close to 0 even when neither tone is 0.

Challenge

1

A pendulum's displacement is d(t) = 20(0.8)ᵗ sin(2πt) cm. How many complete oscillations happen before the amplitude first drops below 1 cm?

2

Two mugs cool in a 20°C room: A(t) = 20 + 70(0.95)ᵗ and B(t) = 20 + 50(0.9)ᵗ. Show that A − B is positive for all t ≥ 0, and describe what that means.

3

Design a damped model for a diving board that starts 15 cm above rest, bounces with period 0.5 s, and loses 40% of its amplitude every second. Then find when the amplitude falls below 1 cm.

Check your understanding

Quiz

1.

Which combination models a pendulum that swings with decreasing size?

2.

What is the upper envelope of y = 6(0.75)ᵗ sin(3t)?

3.

Key 8 on a phone keypad is in the row with 852 Hz and the column with 1336 Hz. Which model fits?

4.

Revenue and cost graphs cross at x = 40 and x = 160. What are these points for the profit function?

5.

A drink cools by T(t) = 21 + 64(0.9)ᵗ. Which statement is true?

6.

The amplitude of d(t) = 12(0.9)ᵗ cos(πt) falls below 3 cm after about how many seconds?

For parents and teachers

Big idea: D2.2 asks students to recognize real situations that combine functions (the curriculum's own examples are a damped pendulum, modelled by a trigonometric function times an exponential, and phone keypad tones, modelled by a sum of two trigonometric functions) and to solve related problems graphically. This lesson also starts D3.3 (solving problems by reasoning with function models).

How to help: for any formula, ask "Which simple functions are inside it, and are they added, multiplied or divided?" Then ask what each part means in the story (the 20 in the cooling model is the room temperature, and so on).

Common difficulties:

  • Treating the envelope as the graph itself, or forgetting the negative envelope.
  • Adding the two keypad frequencies into one tone.
  • Mixing up radians and degrees in sin(πt).
  • Rounding the log values too early (keep calculator values until the last step).

Safety: in the activity, hang only light objects and keep the string clear of faces; an adult should check that the tape holds.