Build and check a daylight model

Explain it like I am five

Picture a year of daylight on a fridge calendar: short winter days, long summer days, repeating. A model is a formula following that rise and fall. Here t counts months (t = 1 is January) and D(t), read "D of t", counts daylight hours.

The notes' table (60° latitude, far north) peaks at 18.72 hours and bottoms at 5.88. A = $18.72 − 5.88 over 2$ = 6.42 and B = $18.72 + 5.88 over 2$ = 12.3. It repeats every 12 months: ω = $2 π over 12$ = $π over 6$ . A positive sine starts where the wave crosses its midline going up: April, t = 4, the rising month nearest 12.3 hours. So the start is h = 4 and φ = ω × h = $4 π over 6$ = $2 π over 3$ , giving D(t) = 6.42 sin( $π over 6$ t − $2 π over 3$ ) + 12.3.

Check July, t = 7: the inside is $7 π over 6$ − $4 π over 6$ = $π over 2$ , so sine is 1 and D = 6.42 + 12.3 = 18.72, a match. December, t = 12, gives about 6.74 against the table's 5.88: a model follows the pattern and can miss single months.

In plain words

Think of drawing a smooth line through a year's calendar of daylight measurements. Winter days are short, summer days are long, and the pattern returns the next year. The measurements are Data: recorded values. A Sinusoidal model is a sine or cosine rule chosen to approximate their repeating rise and fall. The month number is the input, and the number of daylight hours is the output. Here Latitude describes how far north or south a location is; the notes give observations at 60°. You use the observed top and bottom to set the wave's size and center, then choose a practical start. A smooth approximation can miss individual dots.

Dots show the repeated monthly observations; the smooth wave is the instructor's approximation, so some dots lie above or below it.

Reminder

Decimal arithmetic. 18.72 − 5.88 = 12.84; half is 6.42. Their sum is 24.60; half is 12.30.
Month input. D(7) means the model output for July because t = 7 labels July.
Positive sine start. At t = 4 the model's inside is 0, so its output is B = 12.3 and it rises.
Fraction addition. $7 π over 6$ − $2 π over 3$ = $7 π over 6$ − $4 π over 6$ = $π over 2$ .
Special sine value. sin( $π over 3$ ) = $square root of 3 over 2$ comes from the 30°, 60°, 90° triangle.

Why it works. The observed extremes give a reasonable height span and center. Annual repetition gives T = 12, fixing the inside speed. A rising middle crossing can anchor positive sine, but sampled months do not usually land exactly on that crossing. The notes choose April as an approximate rising anchor, so the resulting formula is a model, not exact interpolation. Checking another month exposes the difference between a predicted output and an observed value and keeps the approximation honest.

RuleRule: for a fitted positive-A sine wave, A = , B = , ω = , φ = ωh. Choose h near a rising middle crossing and compare predictions with observations.

The same idea, five ways

Say it

A sinusoidal model is a repeating sine or cosine estimate of measurements.

Write it

Choose a smooth periodic wave to approximate the observed daylight data.

In math

D(t) = A sin(ωt − φ) + B
D(t) = A cos(ωt − φ) + B
D(t + 12) = D(t) for this model
Graph words: observed dots compared with a smooth wave.

Trace a smooth seasonal guide through a calendar's recorded daylight heights.

See it

Dots show the repeated monthly observations; the smooth wave is the instructor's approximation, so some dots lie above or below it.

The same idea, other ways

As a smooth guide through dots

The dots are measurements. The curve is an approximation chosen from a fixed family, so it may pass above or below a dot.

Dots show the repeated monthly observations; the smooth wave is the instructor's approximation, so some dots lie above or below it.

As the graph method with units

A = 6.42 is in hours; T = 12 and h = 4 are in months. The inside $π over 6$ t − $2 π over 3$ is an angle in radians.

.1Read the table and repeat it for two years

The same observed month returns twelve input units later in this plotted repeating pattern. Second-year observations are assumed repeats for this exercise, not new measurements.

Rule: January is t = 1 and the next January is t = 13.
Rule: repeat the plotted observations by adding 12 to the input.

Dots show the repeated monthly observations; the smooth wave is the instructor's approximation, so some dots lie above or below it.

The same idea, five ways

Say it

Read the table and repeat it for two years

Write it

The same observed month returns twelve input units later in this plotted repeating pattern. Second-year observations are assumed repeats for this exercise, not new measurements.

In math

Rule: January is t = 1 and the next January is t = 13.
Rule: repeat the plotted observations by adding 12 to the input.

Copy one year's calendar onto the next year's dates.

See it

Dots show the repeated monthly observations; the smooth wave is the instructor's approximation, so some dots lie above or below it.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: January in the second year is t = 12.

t = 12 already means first-year December.

✓ Instead: The next January is t = 13.

Tips and tricks

Tip: The next January is t = 13.

.2Choose height, center and timing

Half-span and average set the heights. The annual cycle sets speed. Positive sine needs a rising middle crossing; positive cosine needs a peak.

Rule: A = 6.42, B = 12.3, T = 12.
Rule: positive cosine can use h = 7 at the July peak.

Read two adjacent matching points for the width and the two extreme heights for the center and size.

The same idea, five ways

Say it

Choose height, center and timing

Write it

Half-span and average set the heights. The annual cycle sets speed. Positive sine needs a rising middle crossing; positive cosine needs a peak.

In math

Rule: A = 6.42, B = 12.3, T = 12.
Rule: positive cosine can use h = 7 at the July peak.

Read the wheel's size and center first, then choose a point on its timetable.

See it

Read two adjacent matching points for the width and the two extreme heights for the center and size.

Worked exampleUse the July peak as a cosine start

Write the equivalent positive cosine daylight approximation from the same extrema and the July peak at t = 7. This asks for another start description of the same chosen smooth wave.

Read two adjacent matching points for the width and the two extreme heights for the center and size.

Keep A = 6.42, B = 12.3, T = 12 and ω = $π over 6$ .The size, center and repeat width stay the same.
Choose h = 7 for positive cosine because July is the observed peak.Positive cosine reaches its maximum at inside input 0.
φ = ωh = $7 π over 6$ . Plug back: $π over 6$ × 7 − $7 π over 6$ = 0.Convert the chosen peak time into an inside offset.
Write D(t) = 6.42 cos( $π over 6$ t − $7 π over 6$ ) + 12.3.The new start is one quarter period after the sine start 4.

Answer

D(t) = 6.42 cos(t − ) + 12.3

Check Both representations peak at t = 7, trough at t = 1 and 13, and cross the middle rising at t = 4. Their five cycle stages agree because the starts differ by = 3 months.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: October is closer to the center, so use it as a positive-sine start.

October occurs while daylight is falling. Positive sine starts by rising.

✓ Instead: Use the rising April anchor approximately, or use cosine at July.

Tips and tricks

Tip: Use the rising April anchor approximately, or use cosine at July.

.3Check prediction against observation

Predicting means evaluate the fitted rule. Checking the fit means compare that prediction with the recorded value in the table. Substitution means replacing t with the chosen month number.

Rule: a model prediction need not equal the observed output.

Dots show the repeated monthly observations; the smooth wave is the instructor's approximation, so some dots lie above or below it.

The same idea, five ways

Say it

Put the month into the model, then compare the predicted hours with the recorded hours.

Write it

Predicting means evaluate the fitted rule. Checking the fit means compare that prediction with the recorded value in the table.

In math

D(6) = 12.3 + 3.21 $square root of 3$ ≈ 17.8599
Observed June daylight = 18.28
18.28 − D(6) ≈ 0.4201 hour

Compare a predicted timetable with the arrival times you actually recorded.

See it

Dots show the repeated monthly observations; the smooth wave is the instructor's approximation, so some dots lie above or below it.

Worked examplePredict June and compare with the observation

Use the instructor's sine model to estimate daylight at t = 6. This asks for the model output for June, then how it differs from the source table.

Dots show the repeated monthly observations; the smooth wave is the instructor's approximation, so some dots lie above or below it.

Substitute t = 6: inside = $π over 6$ × 6 − $2 π over 3$ = π − $2 π over 3$ = $π over 3$ .Evaluation means put the known time into every t-slot.
sin( $π over 3$ ) = $square root of 3 over 2$ .A radian angle $π over 3$ is 60°, whose sine comes from the special triangle.
D(6) = 6.42 × $square root of 3 over 2$ + 12.3 = 12.3 + 3.21 $square root of 3$ .Multiply the numerical coefficient by one half.
D(6) ≈ 17.8599 hours, rounded to four decimal places.The radical is exact, while its decimal approximation is rounded.
The June column gives 18.28. Using the exact prediction, observed minus modeled = 18.28 − (12.3 + 3.21 $square root of 3$ ) ≈ 0.4201 hour, rounded to four decimal places.Compare predicted output with observed output, keeping rounding at the end.

Answer

Exact model output: 12.3 + 3.21 $square root of 3$ hours
Rounded model output: 17.8599 hours
Observed June daylight: 18.28 hours

Check The prediction is below the model's July maximum 18.72 and above its center 12.3, consistent with the rising part of the wave.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: The predicted June output must be exactly 18.28 because that is the table entry.

The table is measured data, and the simple sine model was chosen from approximate features.

✓ Instead: Report both the model output and the observed output.

Tips and tricks

Tip: Report both the model output and the observed output.

Strategy: step by step

Define the time input and the output units.
Draw the table and plot the observed points.
Read the observed maximum, minimum and repeating interval.
Choose a suitable sine crossing or cosine peak.
Assemble the model and check several months.
Repeat the observation pattern by T for the requested longer plot.

Strategy

Strategy: fit and check a repeating data model

Are the values rising at the chosen middle-height sample?

YesIt can approximate a positive-sine start.

NoUse a rising sample instead, or choose a cosine peak.

↓

Does the model miss a measured point?

YesReport the difference; a fitted model can be approximate.

NoCheck another point before claiming an exact match.

Define t and the measured output.
Use extrema for A and B and repetition for T.
Pick a reference stage with the correct direction.
Compute φ from h, then compare the formula with several observed dots.

Worked exampleModel rung 1: one exact repeating table

A synthetic repeating table is shown. The first zero crossing is upward, and the next matching crossing is at t = 4. Write a sine model. This asks for a rule that follows the recorded rise and fall.

The output crosses 0 upward at 0 and repeats that stage at 4.

The synthetic measurements follow one repeat every 4 input units.

Read the high output 1 and low output −1. A = $1 − (−1) over 2$ = 1; B = $1 + (−1) over 2$ = 0.Half the span gives height distance; half the sum gives center.
T = 4 − 0 = 4, so ω = $2 π over 4$ = $π over 2$ .Adjacent upward crossings give the full repeat width.
Choose positive sine at h = 0. φ = ωh = 0. Plug back: $π over 2$ × 0 − 0 = 0.This puts the inside start at the recorded rising center crossing.
Write y = sin( $π over 2$ t). Read the columns under 1 and 3: the inside angles are $π over 2$ and $3 π over 2$ , giving 1 and −1.The quarter-turn values reproduce the two recorded extremes.

Answer

y = sin( $π over 2$ t)
Amplitude = 1
Period = 4
Midline: y = 0

Check The columns under 0 and 4 both give 0, and the rule rises immediately after both. All five listed outputs match the table.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Model rung 1: one exact repeating table

A synthetic repeating table is shown. The first zero crossing is upward, and the next matching crossing is at t = 4. Write a sine model. This asks for a rule that follows the recorded rise and fall.

The output crosses 0 upward at 0 and repeats that stage at 4.

The synthetic measurements follow one repeat every 4 input units.

Read the high output 1 and low output −1. A = $1 − (−1) over 2$ = 1; B = $1 + (−1) over 2$ = 0.Half the span gives height distance; half the sum gives center.
T = 4 − 0 = 4, so ω = $2 π over 4$ = $π over 2$ .Adjacent upward crossings give the full repeat width.
Choose positive sine at h = 0. φ = ωh = 0. Plug back: $π over 2$ × 0 − 0 = 0.This puts the inside start at the recorded rising center crossing.
Write y = sin( $π over 2$ t). Read the columns under 1 and 3: the inside angles are $π over 2$ and $3 π over 2$ , giving 1 and −1.The quarter-turn values reproduce the two recorded extremes.

Answer

y = sin( $π over 2$ t)
Amplitude = 1
Period = 4
Midline: y = 0

Check The columns under 0 and 4 both give 0, and the rule rises immediately after both. All five listed outputs match the table.

Rung 2Model rung 2: a raised repeating height

The synthetic table shows a peak at t = 0, a low at t = 5, and the next peak at t = 10. Write a positive cosine model. This asks for its center, size and repeating timing.

The peak columns at 0 and 10 enclose one whole repeat.

A center height 12 and amplitude 5 place the extremes at 7 and 17.

A = $17 − 7 over 2$ = $10 over 2$ = 5; B = $17 + 7 over 2$ = $24 over 2$ = 12.The largest and smallest outputs in the picture set the amplitude and center.
T = 10 − 0 = 10; ω = $2 π over 10$ = $π over 5$ .Peak to next peak measures a complete repeat.
Choose cosine with h = 0; φ = ωh = 0. At t = 0 the inside is 0.Positive cosine starts at the chosen peak.
Write y = 5 cos( $π over 5$ t) + 12. At t = 5, the inside is π, so y = 5(−1) + 12 = 7.The independent half-period check reproduces the recorded minimum.

Answer

y = 5 cos( $π over 5$ t) + 12
Amplitude = 5
Period = 10
Midline: y = 12

Check At t = 10 the inside is 2π and the output is 17 again. At t = it is and the output is 12, agreeing with the middle-height column.

Rung 3Model rung 3: recover a shifted repeat

The synthetic table has consecutive peaks at t = 6 and t = 22 and a trough at t = 14. Write a cosine model and explain its shift. This asks you to convert the recorded peak time into the inside offset.

The peaks at 6 and 22 are sixteen units apart; the trough halfway between is at 14.

The chosen cosine peak is at 6; the wave repeats that peak at 22.

A = $19 − 3 over 2$ = 8; B = $19 + 3 over 2$ = 11.Half the extreme separation sets size and their average sets center.
T = 22 − 6 = 16; ω = $2 π over 16$ = $π over 8$ .The consecutive peak columns show a repeat distance, not the absolute position 22.
Choose h = 6 for positive cosine. Make the inside zero there: $π over 8$ × 6 − φ = 0, so φ = $6 π over 8$ = $3 π over 4$ .Solving finds the inside subtraction that places the model peak at the recorded time.
Plug back: $π over 8$ × 6 − $3 π over 4$ = $3 π over 4$ − $3 π over 4$ = 0.The candidate offset feeds cosine its peak input at t = 6.
Write y = 8 cos( $π over 8$ t − $3 π over 4$ ) + 11. The selected shift is 6 units right.The four controls now reproduce the recorded height, width, peak position and center.

Answer

y = 8 cos( $π over 8$ t − $3 π over 4$ ) + 11
Amplitude = 8
Period = 16
Phase shift = 6 units right
Midline: y = 11

Check At t = 14 the inside is − = π, so y = 3. At t = 22 it is 2π and y = 19. At t = 10 it is and y = 11, so both extrema and an intermediate column agree.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: April 13.27 is the value closest to 12.3 in the entire table.

Its difference is 0.97; October 11.55 differs by only 0.75. October is on the falling side.

✓ Instead: April is the closest sampled value on the rising part, so h = 4 is an approximate rising anchor.

✗ Not this: Counterexample: this model's lowest month must be December because the data's lowest month is December.

The chosen sine start is 4 and its trough is three quarters of a period later, at 13, equivalent to January at 1.

✓ Instead: The model's minimum occurs at January; the observed minimum is December. State the approximation.

Tips and tricks

Tip: Attach units to amplitude and period; they measure different directions.
Tip: Check a high point and at least one middle or low point.
Tip: Keep π exact until evaluating a rounded numerical prediction.

Trap. Do not confuse an observed height with a model prediction. The instructor's formula approximates the table and does not exactly place every observed extreme in its recorded month.

Keep in mind

A model approximates the data: D(12) ≈ 6.74 hours while the table shows 5.88, and that gap does not mean you made a mistake.
Reduce φ after multiplying: with ω = $π over 6$ and h = 4, φ = $4 π over 6$ = $2 π over 3$ .
Keep the units straight: t is in months, D is in hours, and the period is 12 because the daylight pattern repeats every 12 months.

Memory hookHalf the gap is A, half the sum is B, the repeat length T gives ω = , and the rising midline crossing gives h.

Flash cards: say the answer out loud, then flip

What is a sinusoidal model?

A sine or cosine formula chosen to follow repeating data approximately.

Why is the period 12 in the daylight model?

t counts months, and the daylight pattern repeats every 12 months.

Daylight peaks at 16 hours and bottoms at 8 hours. Find A and B.

A = $16 − 8 over 2$ = 4
B = $16 + 8 over 2$ = 12

A yearly model (T = 12) crosses its midline going up at t = 3. Find ω and φ for a sine model.

ω = $2 π over 12$ = $π over 6$
φ = ω × h = $π over 6$ × 3 = $π over 2$

Use D(t) = 6.42 sin(t − ) + 12.3 to predict April, t = 4.

The inside is − = 0, so D = 12.3 hours. The table shows 13.27.

October has 11.55 hours, close to 12.3. Can it be the start of a positive sine model?

No. Daylight is falling in October, and a positive sine starts on a rising crossing, so use April.