Builds sinusoidal models from measured data, computing amplitude, midline, period and phase shift from a maximum and a minimum. Extends to damped oscillation, where a decaying exponential multiplies the wave, and uses the models to answer questions by solving the equations of the previous section.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 7 — Trigonometric Identities and Equations
§7.6 Modeling with Trigonometric Functions, pp. 862-890
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 862-890 — the pages these objectives are drawn from
Warm-up
Two measurements determine more than they appear to.
Discussion prompt
A quantity peaks at 14 and troughs at 4. What can you say about the wave?
Hint: Where is its centre, and how far does it swing?
Answer:
The midline is halfway between them, at 9. That is the average value the quantity oscillates around.
The amplitude is half the gap, which is 5. That is how far it swings either side of the midline.
So two numbers give two of the four parameters, by nothing more than an average and a half-difference. The other two come from the times at which those values occurred, which is why a complete model needs four measurements rather than two.
Concept
A sinusoidal model has an amplitude, a midline, a period and a phase shift. Each is computed from measured data rather than fitted by trial.
\[ y=D+A\sin\!\left(\frac{2\pi}{P}(t-C)\right) \]
The vertical parameters come from the maximum and minimum values, and the horizontal ones from the times at which they occur. Nothing here requires a curve-fitting tool — four arithmetic steps produce the model.
Figure (svg): A labelled sinusoid showing the amplitude, midline, period and phase shift as measurable features of the graph
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 862-868
Section
Section 1
Concept
The midline is the average of the maximum and minimum, and the amplitude is half their difference. Both are arithmetic on two numbers.
It is worth checking the arithmetic by reconstructing: the midline plus the amplitude should give back the maximum, and the midline minus the amplitude the minimum. If it does not, one of the two was computed with the wrong operation.
Figure (svg): A labelled sinusoid showing the amplitude, midline, period and phase shift as measurable features of the graph
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 862-869
Picture it
Each one is a distance or a level that can be measured directly.
Figure (svg): A labelled sinusoid showing the amplitude, midline, period and phase shift as measurable features of the graph
The two dashed outer lines mark the extremes and the middle one the midline. The amplitude is the gap between the midline and either extreme.
Worked example
An average and a half-difference.
\[ \text{Temperatures range from } 52^\circ \text{ to } 78^\circ. \text{ Find the midline and amplitude.} \]
Add the extremes
Why: For the midline.
\[ 52 + 78 = 130 \]
Halve
Why: The average.
\[ \text{midline } = 65 \]
Subtract the extremes
Why: For the amplitude.
\[ 78 - 52 = 26 \]
Halve
Why: Half the swing.
\[ \text{amplitude } = 13 \]
Figure (svg): A worked ladder computing amplitude, midline, period and shift from a maximum of 14 at time 2 and a minimum of 4 at time 8
\[ D=65, \; A=13 \]
Verify: reconstruct the extremes
Why: Sixty-five plus thirteen is seventy-eight and sixty-five minus thirteen is fifty-two, recovering both given values. That reconstruction is the check that the sum and difference were not swapped, which is the only real risk here.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 863-866
Faded example
A quantity ranging from 3 to 11.
Fill in the blanks
D=\frac74=___, \quad A=\frac______=___
Why: The average gives the midline and half the difference gives the amplitude. Checking that seven plus four is eleven and seven minus four is three confirms both in one line.
Worked example
The parameters can be read out as easily as in.
\[ \text{For } y=9+5\sin(\ldots), \text{ find the maximum and minimum.} \]
Identify the midline
Why: The constant term.
\[ D = 9 \]
Identify the amplitude
Why: The coefficient's size.
\[ A = 5 \]
Add for the maximum
Why: The sine reaches 1.
\[ 9 + 5 = 14 \]
Subtract for the minimum
Why: The sine reaches -1.
\[ 9 - 5 = 4 \]
Figure (svg): The solution to Worked example work backwards from a model shown as a ladder of expressions, one row per legal move
\[ \text{max}=14, \; \text{min}=4 \]
Verify: check the sine's range
Why: The sine varies between negative one and one, so the whole expression varies between the midline minus the amplitude and the midline plus it. Nothing about the period or shift affects these values, which is why the vertical parameters can be read independently.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 866-869
Trap
\[ \text{max}=78 \;\Longrightarrow\; A=78 \]
Take the largest value as the amplitude
Why: The maximum is read directly as the amplitude with no reference to the minimum.
A model with amplitude 78 would swing between 0 and 156.
The amplitude is measured from the midline, not from zero. It is half the distance between the extremes.
Only when the midline happens to be zero do the two coincide, and in real data it almost never is.
Compute the midline first and the amplitude relative to it. The reconstruction check then catches any slip immediately.
Prediction
The midline is increased by 10 and nothing else changes.
Predict first
What happens to the amplitude?
Correct: Nothing; it is unchanged.
Why: The amplitude measures the swing about the midline, so moving the midline carries the whole wave without stretching it. Both extremes rise by 10 and their half-difference is untouched.
Sorting
Vertical measurements give vertical parameters.
Sort into buckets
Sort each piece of information.
Step zero
You are given a maximum and a minimum with their times.
Discussion prompt
What do you compute first, and why that order?
Hint: Which parameters need only the values?
Answer:
Compute the midline and amplitude first, since they need only the two values and no interpretation of the times.
Then reconstruct: midline plus amplitude should give the maximum back. That check costs a second and catches a swapped sum and difference before it propagates.
Only then turn to the times for the period and shift, which involve the genuinely error-prone step of remembering that a maximum to a minimum is half a period.
Section
Section 2
Concept
The period is the time for one full cycle, and the distance from a maximum to the next minimum is half of it. The shift is where the chosen wave form starts its cycle.
Choosing the cosine form when a maximum is given, and the sine form when a midline crossing is given, makes the shift a direct reading rather than a computation. That choice is free and worth making deliberately.
Figure (svg): A worked ladder computing amplitude, midline, period and shift from a maximum of 14 at time 2 and a minimum of 4 at time 8
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 869-875
Picture it
Each rung is one arithmetic step.
Figure (svg): A worked ladder computing amplitude, midline, period and shift from a maximum of 14 at time 2 and a minimum of 4 at time 8
The doubling in the period step is the one to watch. A maximum and the following minimum are half a cycle apart, not a whole one.
Worked example
Four measurements, four parameters.
\[ \text{A tide peaks at } 14 \text{ ft at } 2 \text{ h and troughs at } 4 \text{ ft at } 8 \text{ h. Model it.} \]
Compute the vertical parameters
Why: Average and half-difference.
\[ D = 9, A = 5 \]
Compute the period
Why: Peak to trough is half.
\[ 2(8 - 2) = 12 \]
Choose the cosine form
Why: A maximum is given.
Read the shift
Why: The time of the maximum.
\[ C = 2 \]
Figure (svg): A worked ladder computing amplitude, midline, period and shift from a maximum of 14 at time 2 and a minimum of 4 at time 8
\[ y=9+5\cos\!\left(\frac{2\pi}{12}(t-2)\right) \]
Verify: test at both given times
Why: At t equal to 2 the cosine's argument is zero, giving nine plus five, which is 14 — the stated maximum. At t equal to 8 the argument is pi, giving nine minus five, which is 4 — the stated minimum. Both data points are reproduced exactly.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 870-873
Prediction
A maximum occurs at hour 1 and the next minimum at hour 7.
Predict first
What is the period?
Correct: Twelve hours.
Why: A maximum to the following minimum is half a cycle, so the six-hour gap must be doubled. Forgetting the doubling halves every period and is the standard error when building a model from two extremes.
Worked example
When a midline crossing is what is given.
\[ \text{A quantity crosses its midline of } 20 \text{ rising at } t=3, \text{ with amplitude } 6 \text{ and period } 10. \text{ Model it.} \]
Choose the sine form
Why: It starts at a rising midline crossing.
Read the shift
Why: The time of that crossing.
\[ C = 3 \]
Write the coefficient
Why: Two pi over the period.
\[ 2 \pi / 10 \]
Assemble
Why: All four parameters.
\[ 20 + 6 \sin(2 \pi(t - 3) / 10) \]
Figure (svg): The solution to Worked example choose the sine form instead shown as a ladder of expressions, one row per legal move
\[ y=20+6\sin\!\left(\frac{2\pi}{10}(t-3)\right) \]
Verify: test the crossing
Why: At t equal to 3 the sine's argument is zero and the value is 20, the midline. And just after, the sine is increasing, so the quantity is rising — matching the stated condition. Choosing the form that matches the given event made the shift a direct reading.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 873-875
Error analysis
A student computes the period from a maximum at hour 2 and a minimum at hour 8.
Annotate
On: \( P=8-2=6 \)
This halves every period computed from an extreme-to-extreme measurement, which then makes the model wrong everywhere except at the two data points. Substituting both given times back into the finished model catches it reliably.
Matching
Choosing well makes the shift a direct reading.
Match the pairs
Why: Each wave form begins its cycle with a particular event, so choosing the form that matches what you were given turns the shift into a direct reading. Any of the four would work, but three of them require extra computation.
Faded example
A model with a period of eight.
Fill in the blanks
B=\frac84}=\frac______}
Why: The coefficient inside is two pi divided by the period, which simplifies here to pi over four. Getting this the wrong way up is a common slip, and checking that one period's worth of time gives an argument of two pi confirms it.
Explain it
Peak to trough is not a full cycle.
Discussion prompt
Explain to a classmate why the period is twice that gap.
Hint: What still has to happen after the trough?
Answer:
A cycle is complete only when the quantity returns to where it started. Reaching the minimum is halfway through — the wave has gone down but not come back.
The return trip from the minimum to the next maximum takes the same time as the descent, by the wave's symmetry. So the full cycle is twice the peak-to-trough gap.
A good explanation suggests the check that catches it: substitute both given times into the finished model. With the period halved, the second data point comes out wrong, which is immediate and unambiguous.
Section
Section 3
Concept
Once the model is built, questions about when a quantity reaches a value are exactly the trigonometric equations of the previous section.
Two solutions per cycle is not an artefact here — it corresponds to two physically distinct events, one while the quantity is rising and one while it is falling. Reporting only one answers half the question.
Figure (svg): A labelled sinusoid showing the amplitude, midline, period and phase shift as measurable features of the graph
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 875-881
Picture it
A horizontal line at the target value crosses twice per cycle.
Figure (svg): A labelled sinusoid showing the amplitude, midline, period and phase shift as measurable features of the graph
The two crossings are the rising one and the falling one, and both are genuine answers. Which is wanted depends on the question rather than on the mathematics.
Worked example
Set the model equal and solve.
\[ \text{With } y=9+5\cos\!\left(\frac{\pi}{6}(t-2)\right), \text{ when is } y=11.5 \text{ in the first 12 hours?} \]
Isolate the cosine
Why: Subtract and divide.
\[ \cos(...) = 0.5 \]
Apply the inverse
Why: The reference solution.
\[ \text{argument } = \frac{\pi}{3} \]
Take the second solution
Why: The cosine is positive in QIV too.
\[ \text{argument } = -\frac{\pi}{3} \]
Solve for t in each case
Why: Unwind the argument.
\[ t = 4\text{ and } t = 0 \]
Figure (svg): The solution to Worked example when does it reach a level shown as a ladder of expressions, one row per legal move
\[ t=0\text{ h and }t=4\text{ h} \]
Verify: check both against the peak
Why: The maximum is at hour 2, and the two answers sit symmetrically two hours either side of it — which is what a cosine's symmetry about its peak requires. Substituting either into the model gives nine plus five times one half, which is 11.5.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 876-879
Prediction
A level strictly between the midline and the maximum.
Predict first
How many times is it crossed per cycle?
Correct: Twice, once rising and once falling.
Why: A horizontal line strictly between the minimum and maximum meets one cycle of a sinusoid exactly twice. The two crossings are physically distinct events, which is why both must be reported.
Worked example
The answer is an interval between two crossings.
\[ \text{For the same model, for how long each cycle is } y>11.5? \]
Use the two crossing times
Why: From the previous example.
\[ t = 0\text{ and } t = 4 \]
Check between them
Why: At hour 2 the value is 14.
Compute the interval length
Why: The gap between crossings.
\[ 4\text{ hours} \]
State the result
Why: Per twelve-hour cycle.
\[ 4\text{ of every } 12\text{ hours} \]
Figure (svg): A labelled sinusoid showing the amplitude, midline, period and phase shift as measurable features of the graph
\[ 4\text{ h per cycle} \]
Verify: check the fraction is plausible
Why: Four hours out of twelve is a third of the cycle, and the threshold of 11.5 sits halfway between the midline of 9 and the maximum of 14. A third of the cycle above the halfway mark is right for a sinusoid, which spends more time near its extremes than a straight line would.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 879-881
Trap
\[ \text{the inverse gives one time, so that is the answer} \]
Report the calculator's single output
Why: The second crossing in the cycle is not sought.
Half the occurrences are missing, and any duration computed from one is meaningless.
A level is crossed twice per cycle, once rising and once falling. Both are real events at real times.
The second comes from the same symmetry used in §7.5 — reflect the argument and unwind. It is one extra line.
A duration question needs both, since the interval is defined by the two crossings. Reporting one makes the following question unanswerable.
Faded example
Solving for t after finding the argument.
Fill in the blanks
\frac24(t-2)=\frac______ \;\Longrightarrow\; t-2=___ \;\Longrightarrow\; t=___
Why: Multiplying both sides by six over pi isolates the shifted time, and adding the shift recovers t. Unwinding the argument is ordinary algebra once the trigonometric part has been resolved.
Sorting
Some questions want one event and some want an interval.
Sort into buckets
Sort each question.
Explain it to yourself
Solving a model equation gives two times per cycle.
Discussion prompt
Explain what the two answers mean physically.
Hint: What is the quantity doing at each?
Answer:
At one crossing the quantity is rising through the level and at the other it is falling through it. Both are moments when the value is exactly the target.
They are genuinely different events. For a tide, one is when the water reaches a depth on the way in and the other when it drops back through it on the way out.
So the two solutions are not a mathematical redundancy — they answer different practical questions, and which one is wanted depends on what the model is being used for. A duration question needs both, since the interval between them is the answer.
Section
Section 4
Concept
Multiplying a sinusoid by a decaying exponential produces an oscillation whose peaks shrink while its timing stays exactly the same.
The zeros being fixed is the clearest evidence that the period is untouched. A decaying exponential is never zero, so the product vanishes exactly where the sinusoid does — at the same times as in the undamped model.
Figure (svg): A damped oscillation, with the wave shrinking inside a decaying exponential envelope
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 881-887
Picture it
The dashed exponentials bound the wave.
Figure (svg): A damped oscillation, with the wave shrinking inside a decaying exponential envelope
Every crossing of the axis happens at the same time as it would without damping. Only the heights change, which is why damping is described as affecting amplitude rather than frequency.
Worked example
Successive peaks tell you which family applies.
\[ \text{Peaks measure } 8, 6, 4.5, 3.375. \text{ What kind of model is this?} \]
Compare successive peaks
Why: They shrink.
Look at the ratios
Why: Each is a fixed fraction of the last.
\[ 0.75\text{ each time} \]
Identify the pattern
Why: A constant ratio means exponential decay.
Conclude
Why: A damped sinusoid.
Figure (svg): A contrast between when a plain sinusoid model fits and when a damped model is needed
\[ \text{ratio}=0.75\text{ per cycle} \]
Verify: check a differences pattern instead
Why: The differences are 2, 1.5 and 1.125 — not constant, so the decay is not linear. The ratios are constant at 0.75, which is the signature of an exponential rather than a linear decline, and it is the ratio test that distinguishes the two.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 882-885
Prediction
A sinusoid is multiplied by a decaying exponential.
Predict first
What happens to its period?
Correct: Nothing; it is unchanged.
Why: The exponential factor is never zero, so the product vanishes exactly where the sinusoid does — at the same times as before. Only the heights of the peaks change, which is why damping affects amplitude and not frequency.
Worked example
The envelope and the wave are found separately.
\[ \text{A pendulum starts at } 8 \text{ cm, halves its swing every } 10 \text{ s, and has period } 2 \text{ s. Model it.} \]
Write the envelope
Why: Halving every ten seconds.
\[ 8(\frac{1}{2}) ^{\frac{t}{10}} \]
Write the sinusoid
Why: Period two seconds, starting at a maximum.
\[ \cos(\pi t) \]
Multiply them
Why: The damped model.
State the result
Why: Envelope times wave.
Figure (svg): The solution to Worked example build the damped model shown as a ladder of expressions, one row per legal move
\[ y=8\left(\tfrac{1}{2}\right)^{t/10}\cos(\pi t) \]
Verify: test at ten seconds
Why: At t equal to 10 the envelope is 4 and the cosine of ten pi is 1, so the value is 4 — half the starting swing, as required. And the period is still 2 seconds, since the cosine's argument gains two pi every two seconds regardless of the envelope.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 885-887
Error analysis
A student describes a damped pendulum.
Annotate
On: \( \text{as it damps, the swings get shorter, so the period increases} \)
This is why a pendulum clock keeps time as its swing decays. The confusion between a smaller swing and a slower one is natural but the model says clearly which changes and which does not.
Sorting
Look at successive peaks.
Sort into buckets
Sort each situation.
Faded example
An amplitude of 12 halving every 5 seconds.
Fill in the blanks
\text12=5\left(\frac______\right)^___}}
Why: The starting amplitude multiplies a half raised to the number of halving periods elapsed, which is t over five. That envelope then multiplies the sinusoid to give the damped model.
Explain it
A pendulum's swing shrinks but its clock stays accurate.
Discussion prompt
Explain this using the damped model.
Hint: What does the exponential factor affect?
Answer:
The model is an exponential times a sinusoid, and the two factors control different things. The exponential sets how far the pendulum swings.
The sinusoid sets when it crosses the centre, and the exponential cannot change that — it is never zero, so it cannot create or move a crossing.
So the timing is set entirely by the sinusoid and is unaffected by the decay. A clock counts crossings, not distances, which is exactly why it keeps time while the swing visibly shrinks. That separation of amplitude from timing is the single most useful idea in the section.
Section
Section 5
Concept
Look at successive peaks before choosing a model: equal heights call for a plain sinusoid, shrinking heights for a damped one, and growing heights for an amplified one.
The final test is not a formality. Substituting the original measurements back into the finished model catches a halved period, a swapped sum and difference, or a shift applied in the wrong direction — the three errors that account for most wrong models.
Figure (svg): A contrast between when a plain sinusoid model fits and when a damped model is needed
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 862-890
Picture it
Successive peak heights are the deciding evidence.
Figure (svg): A contrast between when a plain sinusoid model fits and when a damped model is needed
The question is answered by the data rather than by the context, though the context usually agrees: a driven system keeps its amplitude and a freely decaying one loses it.
Worked example
Substitute the data that built it.
\[ \text{Check } y=9+5\cos\!\left(\frac{\pi}{6}(t-2)\right) \text{ against a peak of } 14 \text{ at } t=2. \]
Substitute the time
Why: Into the argument.
\[ \text{argument } = 0 \]
Evaluate the cosine
Why: Of zero.
\[ 1 \]
Compute the value
Why: Midline plus amplitude.
\[ 9 + 5 = 14 \]
Compare
Why: It matches the data.
Figure (svg): A worked ladder computing amplitude, midline, period and shift from a maximum of 14 at time 2 and a minimum of 4 at time 8
\[ y(2)=14\;\checkmark \]
Verify: test the other data point too
Why: At t equal to 8 the argument is pi, the cosine is negative one, and the value is 4 — the stated minimum. Testing both points is what catches a halved period, since a wrong period reproduces the first point and fails the second.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 873-878
Prediction
Measured peaks are 10, 10, 10 and 10.
Predict first
Which model family fits?
Correct: A plain sinusoid.
Why: Equal successive peaks mean the amplitude is constant, so no envelope is needed. A damped model would require the peaks to shrink by a constant ratio, and there is no such shrinkage here.
Worked example
Which parameter is wrong is readable from which test fails.
\[ \text{A model matches the maximum but not the minimum. Which parameter is wrong?} \]
Note what matched
Why: The maximum was reproduced.
Note what failed
Why: The minimum's timing.
Identify the likely cause
Why: The period was not doubled.
Test the fix
Why: Double the period and recheck.
Figure (svg): The solution to Worked example diagnose a wrong model shown as a ladder of expressions, one row per legal move
\[ \text{period, not doubled} \]
Verify: confirm the diagnosis logically
Why: The vertical parameters affect both extremes equally, so a vertical error would fail both tests. Only a horizontal error can reproduce one time and miss the other, which narrows the diagnosis to the period or the shift before any recomputation.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 878-884
Trap
\[ \text{four parameters computed, so the model is finished} \]
Report the model without substituting anything back
Why: The arithmetic was straightforward, so it is assumed correct.
A halved period or a swapped sum passes unnoticed and the model is wrong everywhere between the data points.
Substitute both original measurements back. It takes two lines and confirms all four parameters at once.
The characteristic errors here — a halved period, a swapped sum and difference, a shift in the wrong direction — all fail this test loudly.
The test also diagnoses, since which point fails narrows down which parameter is wrong before any recomputation.
Sorting
What fails tells you what is wrong.
Sort into buckets
Sort each symptom.
Faded example
Testing a model at the time of its stated maximum.
Fill in the blanks
\text1=0, \;\cos 0=1, \;y=D+A\cdot___=\text___
Why: At the shift time the cosine's argument is zero and the cosine is 1, so the value is the midline plus the amplitude — the maximum. That is why a cosine form with the shift at the peak reproduces it automatically.
Explain it
The model is built from four numbers and tested against them.
Discussion prompt
Explain to a classmate why testing against the same data is not circular.
Hint: What could go wrong between the data and the model?
Answer:
The test does not check the data; it checks the arithmetic that turned the data into parameters. Those are different things.
A halved period, a swapped sum and difference, or a shift applied the wrong way all produce a model that no longer reproduces the numbers it came from. The test catches every one of them.
And it diagnoses as well as detects: a vertical error fails both points and a horizontal one usually fails just the second. A good explanation adds that two lines of substitution is a very cheap way to verify four separate computations.
Comparison
Fill the blanks from memory. Successive peaks decide which applies.
Comparison matrix
| plain sinusoid | damped sinusoid | |
|---|---|---|
| successive peaks | equal heights | shrink by a constant ratio |
| the period | constant | also constant, unaffected by damping |
| parameters needed | four | four plus a decay rate |
| typical context | tides, daylight, seasons | pendulums, springs, plucked strings |
The second row is the one people expect to differ and it does not. Damping changes how far, never how often.
Pattern
Five steps, and the last is what makes the first four trustworthy.
Step 3's doubling and step 5's check are the two places where models go right or wrong. The rest is arithmetic that rarely fails.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 869-875
Check
The vertical parameters.
Check your understanding
A quantity ranges from 20 to 60. What is its amplitude?
Answer: A
Why: The amplitude is half the difference between the extremes, which is half of forty. It measures the swing from the midline of 40, not from zero.
Check
The period.
Check your understanding
A maximum occurs at hour 3 and the next minimum at hour 9. What is the period?
Answer: A
Why: A maximum to the following minimum is half a cycle, so the six-hour gap must be doubled. Forgetting the doubling makes the model correct at the two data points and wrong everywhere between them.
Check
Damping.
Check your understanding
What does multiplying a sinusoid by a decaying exponential change?
Answer: A
Why: The exponential is never zero, so the product vanishes exactly where the sinusoid does — the zeros and hence the period are untouched. Only the heights of the peaks decay, which is why a pendulum clock keeps time as its swing shrinks.
Real world
Suspension design is a damped-oscillation problem with a target decay rate.
Discussion prompt
A car's suspension is designed so that a bump dies away quickly. What is the engineer choosing?
Hint: Which factor of the model controls how fast it settles?
Answer:
The decay rate of the exponential envelope, which is set by how much the damper resists motion. That is what the shock absorber provides.
Too little damping and the car keeps bouncing — the envelope decays slowly and several oscillations are felt. Too much and the suspension cannot respond to the next bump.
So the design is a deliberate choice of decay rate, balancing comfort against responsiveness. The mathematics separates the two concerns cleanly: the spring sets the period and the damper sets the decay, and each can be tuned without disturbing the other.
Commit first
State your confidence along with your answer.
Predict first
Why does damping leave the period unchanged?
Correct: The exponential is never zero, so it cannot move the sinusoid's zeros.
Why: The product vanishes exactly where the sinusoid does, since the other factor is never zero. Those crossings define the period, so it is untouched. This is why a pendulum clock keeps accurate time as its swing decays.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate computed a period as the gap from a maximum to the next minimum. Explain the error.
Hint: What has the wave not yet done?
Answer:
Reaching the minimum is halfway through a cycle. The wave has gone down but has not come back up to where it started.
A cycle is complete only on return to the starting value moving the same way, so the full period is twice that gap.
And there is a check that makes this self-correcting: substitute both data points into the finished model. With a halved period the first point still works and the second fails, which points straight at the period. A good explanation gives the check as well as the correction.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is where nearly all model-building errors occur, almost always through a period that was not doubled. The fourth is the idea that generalises furthest into physics and engineering.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Sketch a sinusoid and label all four parameters with the arithmetic that produces each from measured data. Beside it, build a complete model from a maximum of 14 at hour 2 and a minimum of 4 at hour 8, then substitute both times back to check. Underneath, sketch a damped oscillation and write one sentence on why its period matches the undamped one.
If your period step shows the doubling explicitly and your check reproduces both data points, the two places where models go wrong are both covered.
Recap
Five things, and the last protects the other four.
| if you remember one thing | it should be this |
|---|---|
| about the vertical parameters | average for the midline, half-difference for the amplitude |
| about the period | peak to trough is half a cycle, so double it |
| about damping | it changes how far, never how often |
| about finishing | substitute the data back; it detects and diagnoses |
Chapter 8 turns from waves to triangles, solving ones that have no right angle with the laws of sines and cosines — and then reaches the polar coordinate system, where the whole plane is described by an angle and a distance.
OpenStax, Precalculus, §7.6 Modeling with Trigonometric Functions §7.6, pp. 862-890 — everything on these slides traces back here
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