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Times and thresholds in trigonometric models

Find first and repeated event times, durations above a threshold and restricted-window predictions in trigonometric models. Check units, physical bounds and reciprocal-model assumptions.

Before you startTrigonometric modelling, periodic equations and the R formula

01 / Translate an event into an equation or inequality

An equality time and a duration answer different questions.

A height equal to a level gives an equation. A height above that level gives an inequality and usually a collection of time intervals. Define the observation window before reporting a first event or a total duration.

h(t) = 6 − 4cos(πt/6)
Full-cycle range: 2 ≤ h ≤ 10 m
Period: 12 seconds

The manual model lets you change both the level and the observation window. The written methods remain available while you explore.

When is the height above a level?Explore

Threshold (m)Observation window (s)

h(t) = 6 − 4cos(πt/6), t in seconds; angles in radians.

Equality times in [0, 12]: 4, 8 s.

Time strictly above 8 m: 4.0000 s out of 12 s.

First equality in this window: 4.0000 s.

The pale band is the observation window. Gold points mark equality; green segments show where the height is strictly above the threshold. At an extreme, equality can be a touch rather than a crossing. Readouts are rounded to four decimal places.

02 / Find every equality time in one cycle

Transform the whole time interval into an angle interval.

When does h(t) = 6 − 4cos(πt/6) equal 8 m during 0 ≤ t < 12?Worked example

cos(πt/6) = −1/2

Rearrange the original height equation.

Let θ = πt/6, so 0 ≤ θ < 2π

The input angle is not the time.

θ = 2π/3 or 4π/3

Both cosine branches are required.

t = 4 s or 8 s

Multiply each angle by 6/π.

h(4) = h(8) = 8 m

Check in the original model.

03 / Extend the event times by the model period

Choose the first event relative to the stated starting time.

h(t) = 8 at t = 4 + 12n or t = 8 + 12n, n ∈ ℤ

Find the first time at or after t = 9 s when the height equals 8 m.Worked example

Events near this window: 4, 8, 16, 20, …

Add the 12-second period to both event families.

The first admissible event is t = 16 s

The earlier root 8 is outside the requested window.

Waiting time from t = 9 is 7 s

An absolute event time and an elapsed waiting time are different answers.

If the question says “strictly after” a time that is itself an event, exclude that event. If it says “at or after”, include it.

04 / Use the graph sign to find time above a threshold

Subtract boundary times only after locating the correct interval.

For the same height model, how long is h(t) > 8 during one 12-second cycle?Worked example

−4cos(πt/6) > 2

Subtract 6.

cos(πt/6) < −1/2

Dividing by −4 reverses the inequality.

2π/3 < πt/6 < 4π/3

This is the part of the cycle around the cosine minimum.

4 < t < 8

Convert the endpoints back to seconds.

Duration = 8 − 4 = 4 s

This is one third of the 12-second cycle.

Watch the above-threshold interval appear

Pause, replay or seek freely. The notes explain the same idea and stay in view.

05 / Intersect each event interval with the actual observation window

A partial cycle need not have the full-cycle proportion.

Find the duration above 8 m within 3 ≤ t ≤ 9.Worked example

The full-cycle above-threshold interval is (4, 8)

This interval lies entirely inside [3, 9].

Duration = 4 s out of a 6-second observation

The observed proportion is 2/3, not the full-cycle proportion 1/3.

Now use the window 5 ≤ t ≤ 17.Worked example

Relevant above-threshold intervals: (4, 8) and (16, 20)

Extend by the 12-second period.

Intersections: [5, 8) and (16, 17]

Clip the intervals to the observation window.

Duration = (8 − 5) + (17 − 16) = 4 s

Add lengths of the disjoint pieces.

06 / Check the range before taking an inverse trig function

Some levels are never reached; an extreme may only be touched.

For a model with range [2, 10], level 11 is never reached and the height is never above 11. Level 1 is also never reached, but the height is always above 1. These are different outcomes despite both equality equations having no roots.

At level 10: isolated equality at each maximum; no time strictly above.
At level 2: equality at each minimum; strictly above at all other times.

Changing > to ≥ adds the isolated equality times. It changes interval membership, but not the total duration in a continuous time model. If the model were constant at the threshold over a whole interval, this conclusion would no longer hold.

07 / Handle a shifted input and keep the units visible

Solve for the angle, then undo both scale and shift.

A model is y(t) = 5 + 2cos(π(t − 1)/4). Find when it equals 6 during 0 ≤ t ≤ 8 s.Worked example

cos(π(t − 1)/4) = 1/2

The input range is [−π/4, 7π/4].

Input angles: π/3 and 5π/3

The root −π/3 is below the lower input bound.

t = 1 + 4/3 = 7/3 s, or 1 + 20/3 = 23/3 s

Undo the shift after dividing by the angular coefficient.

These are approximately 2.3333 s and 7.6667 s

Both lie in the time window.

If times are requested in minutes, convert only after solving consistently in seconds, or rewrite the entire model using a minute input first.

08 / Check positivity before reversing bounds in a reciprocal model

A zero denominator makes the expression undefined.

A positive repeating quantity is E(t) = 5 + 3sin t. Find the full-model range of 16/[E(t)]².Worked example

2 ≤ E(t) ≤ 8

The entire range is positive.

4 ≤ [E(t)]² ≤ 64

Squaring preserves this ordering for positive values.

1/4 ≤ 16/[E(t)]² ≤ 4

Taking a positive reciprocal reverses the endpoints.

The maximum occurs at the smallest E; the minimum at the largest E

Check which input values actually occur if the time window is restricted.

If E could pass through zero, there would be an excluded input and no finite global upper bound near that zero. Do not use the same endpoint shortcut across a sign change.

09 / State when an inverse-square time relation is valid

A varying power during a process needs a different calculation.

Suppose a device has power P = kE², with a positive constant k. If E is held constant during an experiment and the required energy is Q, the duration is τ = Q/P = Q/(kE²). The assumptions are part of the model.

Compare two constant-field experiments requiring the same energy.Worked example

Double E while keeping k and Q fixed

Power becomes four times as large.

A field changes during heating. Why is dividing the required energy by the initial power generally insufficient?

Hint

Power is the rate of energy transfer, and it changes.

Worked solution

The initial power does not describe the entire interval. Total energy must be accumulated from the varying power over time; using Q/P assumes constant power for the whole process.

11 / Recap

Solve events and measure intervals as separate steps.

Check the model range and state the time window.
Find all angle branches, then undo scale and phase.
Repeat event families by the period and filter the window.
Locate inequality intervals and add their clipped lengths.
Keep units, positivity restrictions and physical assumptions explicit.

Section 1 of 11 · Translate an event into an equation or inequality