A synthetic level follows the complete stipulated graph H(t)=2+3cos(πt/4), with t in hours and H in meters. First read its landmarks: at times 0,2,4,6,8 the heights are 5,2,−1,2,5. A negative height means below the chosen reference plane, not a negative distance traveled. One cycle takes eight hours; the distance from the midline 2 to either extreme is 3 meters.
This is the graph of the stated sinusoid, not an arbitrary interpolating curve through five observations. The supplied formula supports statements between the marked points.
Predict
The height increases from t=4 to t=8. Does its rate of increase keep growing over that whole interval? Predict which one-hour rise is larger: from 4 to 5, or from 7 to 8. Then predict whether a positive height implies a positive rate.
The readings at t=5 and t=7 are 2−3√2/2 and 2+3√2/2. Consequently the rises over [4,5] and [7,8] are both 3−3√2/2≈0.87868 meters. The middle rises over [5,6] and [6,7] are both 3√2/2≈2.12132 meters. These are one-hour average rates with units m/h. The graph is rising throughout, but it first becomes steeper and then less steep. At t=1 it has positive height while nearby forward intervals have negative average rate. Level, direction and changing rate answer different questions.
Experiment
Construct a new graph from this description: midline 1 meter, amplitude 2 meters, period 6 hours, and a maximum at t=2. Mark a complete cycle beginning at 2, using quarter-period steps. Write one cosine equation and explain why the phase choice matches a maximum rather than a midline crossing.
Check the construction
One equation is J(t)=1+2cos(π(t−2)/3). The five cycle landmarks are (2,3),(3.5,1),(5,−1),(6.5,1),(8,3). Connect them with the specified sinusoidal shape, then repeat after six hours. A negative coefficient with an adjusted half-period phase can describe the same graph; amplitude remains the nonnegative size 2.
Return to H. Calculate symmetric secant rates centered at t=5 and t=7, using the same small positive half-width h, then compare their signs and magnitudes. Do not label either finite secant an instantaneous derivative. Using cosine addition gives
[H(t+h)−H(t−h)]/(2h)=−3sin(πt/4)sin(πh/4)/h.
For equal sufficiently small h, the two centers have the same positive secant rate by symmetry. At center 6 the positive rate is larger. Changing h changes the interval evidence; the formula, rather than a handful of rounded values, justifies the exact symmetry.
Understand
For a nonconstant sinusoid d+Acos(2π(t−s)/P), choose A>0 and P>0. The maximum and minimum are d+A and d−A, so their average gives d and half their difference gives A. Consecutive maxima are P apart, and s is a maximum time. A graph with only a partial branch may leave the period ambiguous; equivalent choices s+kP describe the same function. Stating a phase convention avoids mistaking equivalent equations for different models.
Concavity describes chord position, not output sign. A concave-down graph lies on or above its chords; a concave-up graph lies on or below them. For H and any symmetric interval with midpoint m and half-width h, cosine addition gives
H(m)−[H(m−h)+H(m+h)]/2=3cos(πm/4)[1−cos(πh/4)].
The bracket is nonnegative. On the portions (0,2) and (6,8), the cosine is positive, so midpoint values lie above their chords: the graph is concave down. On (2,6), the cosine is negative and midpoint values lie below: concave up. For the concavity claim, keep both endpoints and the entire interval inside the same listed concavity region. To pass from midpoints to every chord fraction, repeatedly bisect each interval. The inequalities hold at all fractions with denominator a power of two; continuity supplies the remaining fractions by limits. This is a school-level chord argument, not a hidden derivative calculation. At t=2 and t=6 the shape changes concavity, although the function is perfectly defined there.
Unit-circle cosine decreases over angles 0 to π and increases over π to 2π. Thus H decreases on (0,4) and increases on (4,8). Combining direction with concavity explains four distinct pieces: falling faster, falling slower, rising faster and rising slower. A vertical translation changes the sign of some heights but none of these direction or chord comparisons.
Apply
For K(t)=−1+2sin(πt/3) on 0≤t≤6, construct the complete graph and label each extremum and midline crossing. Identify intervals where it is increasing and where it is concave down. Find one interval where its value is negative even though it is increasing.
Check direction and shape separately
The landmarks are (0,−1),(1.5,1),(3,−1),(4.5,−3),(6,−1). It increases on (0,1.5) and (4.5,6), decreases on (1.5,4.5), is concave down on (0,3) and concave up on (3,6). Near t=0 it rises from −1; for example throughout 0<t<0.5 it is negative and increasing. These statements use the stipulated sine law and its unit-circle behavior, not only the five plotted points.
If those landmarks were measurements without a sinusoidal-family assumption, many continuous curves could pass through them with different between-point behavior. A construction task must specify the family or acknowledge that freedom. Record which part of a graph comes from measured evidence and which part follows from a chosen mathematical model.
Continue
Keep your equation together with the graph, units, domain, phase convention and the intervals supporting each shape claim. Do not turn a finite secant into an exact instantaneous rate, and do not infer increasing rate merely from increasing output. Next, fit competing periods and test unused predictions. A clean sinusoidal formula can describe an ideal signal while a noisy observation set requires a separate model-selection and validation argument.
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Practice and reasoning
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A stipulated sinusoidal graph G has consecutive maxima(1,7),(7,7), minimum(4,−1), and midline crossings(2.5,3),(5.5,3). Construct a complete cycle with these landmarks and recover a positive-amplitude cosine rule. Identify midline, amplitude, period and the meaning of the horizontal shift.
Structure your argument and preserve revisions
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For H(t)=5+2 cos(πt/2), compute unit-interval secants on [0,1],[1,2],[2,3],[3,4]. Then use half-unit intervals[2,2.5],[2.5,3],[3,3.5],[3.5,4] to distinguish increasing output from increasing rate.
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For q(t)=−2+4 sin t on 0≤t≤2π, identify open intervals of increasing/decreasing behavior and concave-up/concave-down shape without derivatives. Justify the concavity using the midpoint identity q(m)−[q(m−h)+q(m+h)]/2=4 sin(m)(1−cos h) when all three inputs belong to one half-cycle.
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Explain why 3+2 cos(2t−π) and 3−2 cos(2t) describe the same full graph. State its period, range and maxima. Why would reporting a negative amplitude be a confusing description of its size?
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A synthetic oscillating level follows L(t)=4−3 cos(πt/3) meters on 0≤t≤9 hours. Construct its graph over the whole window and list every maximum/minimum and rising/falling interval.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
At t=0,1,2,3,4 a measured signal is 3,1,−1,1,3. A learner claims those readings prove y=1+2 cos(πt/2) at every time. Give another continuous function with the same five readings but a different value at 0.5, and explain what new evidence could distinguish them.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
