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D3.4 · Predict how changing conditions changes a periodic model

Learn to predict how changing conditions changes a periodic model through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Trigonometric Functions

MCR3U – D3.4: Sinusoidal Functions and Real-World Change

Many real-world quantities repeat in a regular, wave-like pattern. The height of a tide, the temperature across seasons, the position of a rotating blade — all of these rise and fall over equal time intervals. In MCR3U you have already seen that a sinusoidal function can model this kind of behaviour. This lesson focuses on a key skill: if something in the real situation changes — the wave gets taller, the cycle speeds up, the whole pattern shifts up or down — how does the equation change, and what can you predict about the new behaviour? Working through this lesson carefully will let you answer those questions confidently.

What you will learn

Prerequisite Bridge: The Four Parameters of a Sinusoidal Model

Before predicting what happens when conditions change, you need to be comfortable reading a sinusoidal equation. The standard form used in this course is y=asin⁡(bx)+dy = a\sin(bx) + d, where xx is typically time (in hours, days, months, etc.) and yy is the quantity being modelled.
Each letter controls one geometric feature of the wave. The parameter aa is the amplitude — it tells you how far the wave reaches above and below its middle value. The parameter bb controls the period, which is the length of one complete cycle: period=360°b\text{period} = \frac{360°}{b} when xx is in degrees. The parameter dd is the vertical shift (also called the midline or equilibrium value) — it slides the entire wave up or down. A horizontal shift (phase shift) can also appear, but this lesson focuses on aa, bb, and dd because those are the ones most directly connected to changing real-world conditions.
Quick memory check: if a=3a = 3, b=2b = 2, and d=5d = 5, then the wave oscillates between a minimum of 5−3=25 - 3 = 2 and a maximum of 5+3=85 + 3 = 8, and one full cycle takes 360°2=180°\frac{360°}{2} = 180° (or 180 days if xx is in days).
Period=360°b\text{Period} = \frac{360°}{b}

What 'Changing Conditions' Means in a Real-World Model

When a real-world situation is updated, one or more of the four parameters must change to reflect the new reality. The key skill is matching the real-world change to the correct parameter.
If the quantity swings higher and lower than before — for example, coastal tides become more extreme because of a storm — the amplitude |a| increases. The midline stays the same, but the peaks go higher and the troughs go lower.
If the cycle speeds up or slows down — for example, a fan blade spins faster — the period changes. A shorter period means more cycles per unit of time, so bb increases (because b=360°periodb = \frac{360°}{\text{period}}). A longer period means bb decreases.
If the whole pattern shifts upward or downward — for example, average daily temperature rises by several degrees across all seasons — the midline dd changes. Both the maximum and the minimum move by the same amount; the amplitude does not change.
The table in this lesson summarises these connections. Use it as a reference when you read a scenario and need to decide which parameter to update.
b=360°periodb = \frac{360°}{\text{period}}

How to Write the Revised Equation and Make Predictions

Once you know which parameter changes and by how much, writing the new equation is straightforward: substitute the new value into y=asin⁡(bx)+dy = a\sin(bx) + d and leave every other parameter alone.
After writing the new equation, you can predict the new maximum, minimum, and period directly from the parameters. Maximum =d+∣a∣= d + |a|. Minimum =d−∣a∣= d - |a|. Period =360°b= \frac{360°}{b}. These three formulas do all the prediction work.
A useful habit: always check that your updated maximum and minimum make sense in the real-world context. If you are modelling the height of water in metres and your new minimum is negative, ask yourself whether negative height is physically possible — and if not, re-read the scenario to find your error.
When two conditions change at once, handle them one parameter at a time. Change aa for a new amplitude, then change dd for a new midline, for example. The parameters are independent, so the order does not matter as long as you update each one correctly.
max=d+∣a∣,min=d−∣a∣\text{max} = d + |a|, \text{min} = d - |a|

Putting It All Together: Reading a Scenario Step by Step

When you see a word problem about changing conditions, follow a four-step approach. Step 1: Identify the original model and label each parameter. Step 2: Read the change described in the scenario and decide which parameter it affects. Step 3: Calculate the new parameter value. Step 4: Write the updated equation and state your predictions.
This process keeps your work organised and makes it easy to check. The two worked examples below apply this approach to two different real-world contexts — one involving tides and one involving temperature — so you can see how the same reasoning transfers across topics.
One more point worth noting: you do not need to redraw or re-sketch the whole graph to answer prediction questions, although doing so is a great way to double-check. If you can read the parameters, you can predict the key features without a graph.

Matching Real-World Changes to Equation Parameters

Real-World ChangeParameter AffectedEffect on EquationEffect on Graph
Swings become larger (more extreme highs and lows)Amplitude |a||a| increasesWave gets taller
Swings become smaller (highs and lows closer together)Amplitude |a||a| decreasesWave gets flatter
Cycle speeds up (more cycles in same time)Period → bbbb increasesWave is horizontally compressed
Cycle slows down (fewer cycles in same time)Period → bbbb decreasesWave is horizontally stretched
Entire pattern shifts upward (higher average)Midline dddd increasesWave moves up
Entire pattern shifts downward (lower average)Midline dddd decreasesWave moves down

Worked example

Example 1 – Tidal Height: Amplitude and Midline Both Change

A harbour's tidal height (in metres above sea level) is modelled by h=3sin⁡(30t)+5h = 3\sin(30t) + 5, where tt is time in hours and 30t30t is measured in degrees. Due to seasonal conditions, the tides become more extreme so that the amplitude increases by 1.5 m, and rising sea levels shift the entire tidal pattern up by 0.5 m. Write the new equation and predict the new maximum height, minimum height, and period.
  1. Identify the original parameters
    Read the original equation h=3sin⁡(30t)+5h = 3\sin(30t) + 5. The amplitude is a=3a = 3, the value of bb is 3030, and the midline is d=5d = 5. Original maximum =5+3=8= 5 + 3 = 8 m; original minimum =5−3=2= 5 - 3 = 2 m.
    a=3,b=30,d=5a = 3, b = 30, d = 5
  2. Update the amplitude
    The amplitude increases by 1.5 m, so the new amplitude is 3+1.5=4.53 + 1.5 = 4.5. Only aa changes here; bb and dd stay the same for this step.
    anew=3+1.5=4.5a_{\text{new}} = 3 + 1.5 = 4.5
  3. Update the midline
    Rising sea levels shift the whole pattern up by 0.5 m, so the new midline is 5+0.5=5.55 + 0.5 = 5.5. Only dd changes here.
    dnew=5+0.5=5.5d_{\text{new}} = 5 + 0.5 = 5.5
  4. Write the new equation
    Replace aa with 4.54.5 and dd with 5.55.5 in the model. The value of bb does not change because the speed of the tidal cycle has not changed.
    h=4.5sin⁡(30t)+5.5h = 4.5\sin(30t) + 5.5
  5. Predict the new maximum and minimum
    Use the formulas: maximum =d+∣a∣= d + |a| and minimum =d−∣a∣= d - |a|, with the updated values.
    max=5.5+4.5=10 m,min=5.5−4.5=1 m\text{max} = 5.5 + 4.5 = 10 \text{ m}, \text{min} = 5.5 - 4.5 = 1 \text{ m}
  6. Find the period
    Because b=30b = 30 is unchanged, the period is 360°30=12\frac{360°}{30} = 12 hours. One complete tidal cycle still takes 12 hours.
    period=360°30=12 h\text{period} = \frac{360°}{30} = 12 \text{ h}
Answer: New equation: h=4.5sin⁡(30t)+5.5h = 4.5\sin(30t) + 5.5. New maximum: 10 m. New minimum: 1 m. Period: 12 hours (unchanged).
Check: Original max was 8 m; the amplitude grew by 1.5 and the midline rose by 0.5, so the new max should be 8+1.5+0.5=108 + 1.5 + 0.5 = 10 m. ✓ Original min was 2 m; the amplitude drop pulls it down by 1.5 but the midline rise pushes it up by 0.5, giving 2−1.5+0.5=12 - 1.5 + 0.5 = 1 m. ✓

Worked example

Example 2 – Seasonal Temperature: Period Changes

A researcher studying a planet in a computer simulation models its surface temperature (in degrees Celsius) with T=12sin⁡(1.5x)+20T = 12\sin(1.5x) + 20, where xx is time in days and 1.5x1.5x is in degrees. The simulation is adjusted so that the planet's year (one complete temperature cycle) shortens from 240 days to 180 days, while the temperature range and average remain the same. Write the new equation and state the new maximum, minimum, and period.
  1. Identify the original parameters
    From T=12sin⁡(1.5x)+20T = 12\sin(1.5x) + 20: amplitude a=12a = 12, b=1.5b = 1.5, midline d=20d = 20. Verify the original period: 360°1.5=240\frac{360°}{1.5} = 240 days. This matches the scenario.
    a=12,b=1.5,d=20,period=240 daysa = 12, b = 1.5, d = 20, \text{period} = 240 \text{ days}
  2. Identify what changes and what stays the same
    The temperature range (amplitude) and average (midline) are unchanged, so a=12a = 12 and d=20d = 20 stay the same. Only the period changes, from 240 days to 180 days. A shorter period means bb must increase.
    new period=180 days\text{new period} = 180 \text{ days}
  3. Calculate the new value of b
    Use the period formula rearranged for bb: b=360°periodb = \frac{360°}{\text{period}}. Substitute the new period of 180 days.
    bnew=360°180=2b_{\text{new}} = \frac{360°}{180} = 2
  4. Write the new equation
    Replace b=1.5b = 1.5 with b=2b = 2. Keep a=12a = 12 and d=20d = 20 unchanged.
    T=12sin⁡(2x)+20T = 12\sin(2x) + 20
  5. Predict the new maximum and minimum
    Since aa and dd did not change, the maximum and minimum are the same as before.
    max=20+12=32°C,min=20−12=8°C\text{max} = 20 + 12 = 32°\text{C}, \text{min} = 20 - 12 = 8°\text{C}
  6. Confirm the new period
    Check by computing the period from the new equation: 360°2=180\frac{360°}{2} = 180 days, which matches the updated scenario.
    period=360°2=180 days\text{period} = \frac{360°}{2} = 180 \text{ days}
Answer: New equation: T=12sin⁡(2x)+20T = 12\sin(2x) + 20. Maximum temperature: CAD 32°C. Minimum temperature: CAD 8°C. Period: 180 days.
Check: Increasing bb from 1.5 to 2 should compress the cycle: 360°2=180\frac{360°}{2} = 180 days < 360°1.5=240\frac{360°}{1.5} = 240 days. ✓ The amplitude and midline were not to change, so max and min stay at 32°C and 8°C. ✓

Common mistakes and how to avoid them

Changing the amplitude when the midline shifts. For example, if the average temperature rises by 3°C, some students incorrectly increase aa by 3 instead of dd.
Correction: A shift in the overall average moves the midline dd. The amplitude |a| only changes if the range between the highest and lowest values changes.
Confusing a shorter period with a smaller value of bb. Students sometimes reason that 'shorter means smaller' and decrease bb.
Correction: Because b=360°periodb = \frac{360°}{\text{period}}, a shorter period produces a larger bb. They are inversely related: as one goes down, the other goes up.
Recalculating the maximum as just |a| instead of d+∣a∣d + |a|.
Correction: The maximum value of the function is d+∣a∣d + |a|, not |a| alone. The midline dd sets the centre, and |a| measures the distance above and below that centre.
Changing bb when the problem says the cycle speeds up, but also accidentally changing aa or dd at the same time.
Correction: Each parameter is independent. Change only the parameter linked to the stated condition. Re-read the problem to confirm which features of the wave are described as staying the same.
Forgetting to check whether the new minimum makes physical sense (e.g., a negative water height when the context says the harbour never runs dry).
Correction: After writing the new equation, always evaluate the minimum =d−∣a∣= d - |a| and ask whether that value is reasonable given the real-world context. A physically impossible answer signals an error somewhere.

Lesson summary

Check your understanding

Question 1

A Ferris wheel's height is modelled by h=10sin⁡(6t)+12h = 10\sin(6t) + 12, where tt is in seconds and 6t6t is in degrees. The wheel is replaced with a taller one so the height swings 4 m more above and below the centre, but the centre height and rotation speed stay the same. What is the new equation?
  1. h=10sin⁡(6t)+16h = 10\sin(6t) + 16
  2. h=14sin⁡(6t)+12h = 14\sin(6t) + 12
  3. h=10sin⁡(10t)+12h = 10\sin(10t) + 12
  4. h=14sin⁡(10t)+16h = 14\sin(10t) + 16
Show answer and explanation
h=14sin⁡(6t)+12h = 14\sin(6t) + 12
A larger swing means the amplitude increases. The original amplitude is 10; adding 4 gives a new amplitude of 14. The centre (midline d=12d = 12) and rotation speed (b=6b = 6) are both unchanged. The new equation is h=14sin⁡(6t)+12h = 14\sin(6t) + 12.

Question 2

For the model T=8sin⁡(2x)+15T = 8\sin(2x) + 15 (with 2x2x in degrees), what is the period?
  1. CAD 2°
  2. CAD 45°
  3. CAD 180°
  4. CAD 360°
Show answer and explanation
CAD 180°
The period equals 360°b\frac{360°}{b}. Here b=2b = 2, so the period =360°2=180°= \frac{360°}{2} = 180°. Option A confuses bb with the period. Option D is the period only when b=1b = 1.

Question 3

A lake's water level (in metres) follows w=2sin⁡(30t)+6w = 2\sin(30t) + 6, where tt is in months and 30t30t is in degrees. Drought conditions lower the average water level by 1.5 m but do not change the size of seasonal swings or their timing. What are the new maximum and minimum water levels?
  1. Maximum 7.5 m, minimum 2.5 m
  2. Maximum 8 m, minimum 4 m
  3. Maximum 6.5 m, minimum 2.5 m
  4. Maximum 6.5 m, minimum 3.5 m
Show answer and explanation
Maximum 6.5 m, minimum 2.5 m
The drought lowers the midline by 1.5 m: new d=6−1.5=4.5d = 6 - 1.5 = 4.5. Amplitude stays at ∣a∣=2|a| = 2. New maximum =4.5+2=6.5= 4.5 + 2 = 6.5 m. New minimum =4.5−2=2.5= 4.5 - 2 = 2.5 m.

Question 4

A periodic model has a period of 90 days. A change in conditions makes the cycle complete in 60 days instead. If the original equation contained b=4b = 4, what is the new value of bb?
  1. b=3b = 3
  2. b=4b = 4
  3. b=6b = 6
  4. b=8b = 8
Show answer and explanation
b=6b = 6
Use b=360°periodb = \frac{360°}{\text{period}}. With the new period of 60 days: b=360°60=6b = \frac{360°}{60} = 6. A shorter period always gives a larger bb because they are inversely related. You can also verify the original: 360°4=90\frac{360°}{4} = 90 days. ✓

Key terms

Sinusoidal function
A function whose graph has a smooth, repeating wave shape, modelled in this course by y=asin⁡(bx)+dy = a\sin(bx) + d.
Amplitude
The distance from the midline to the highest (or lowest) point of the wave. It equals |a| in the model y=asin⁡(bx)+dy = a\sin(bx) + d.
Period
The horizontal length of one complete cycle of the wave. Calculated as 360°b\frac{360°}{b} when xx is in degree measure.
Midline
The horizontal line that runs through the middle of the wave, halfway between the maximum and minimum. It is the value of dd in y=asin⁡(bx)+dy = a\sin(bx) + d.
Maximum value
The highest output value of the function. For y=asin⁡(bx)+dy = a\sin(bx) + d, it equals d+∣a∣d + |a|.
Minimum value
The lowest output value of the function. For y=asin⁡(bx)+dy = a\sin(bx) + d, it equals d−∣a∣d - |a|.
Parameter
A constant in an equation whose value shapes the graph. In y=asin⁡(bx)+dy = a\sin(bx) + d, the parameters are aa, bb, and dd.
Periodic model
An equation or function used to represent a real-world quantity that repeats in regular cycles over time.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation D3.4. It is a study resource, not an official curriculum publication.

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