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Trigonometric Derivatives · Tutorial 513 of 1495

Rates of Change in Periodic Height Models

Learn how a sinusoidal height model’s derivative describes vertical motion, and how its sign and zeros identify rising, falling, and turning points.

Intermediate 9 min read

What You'll Learn

  • Identify the mean height, amplitude, angular frequency, and phase in a periodic height model
  • Differentiate sinusoidal height models and attach the correct rate units
  • Interpret positive and negative derivatives as rising or falling height
  • Recognize when zero rates correspond to maxima or minima in a nonconstant sinusoidal model
  • Explain why equal heights can occur with opposite rates

Height Models Describe More Than Height

A periodic height model can describe the changing level of a tide, the vertical coordinate of a point on a rotating object, or another height that repeats over time. The model gives the height at each time. Its derivative gives a different kind of information: how quickly that height is changing at that time.

This distinction matters in practice. A particular height might occur while the water is rising or while it is falling. Knowing the height alone does not tell us which is happening; the rate of change does.

A Sinusoidal Height Model and Its Rate

Consider the model \(h(t)=D+A\sin(\omega t+\phi)\), where \(h(t)\) is height at time \(t\). We use \(A>0\) for amplitude, \(\omega>0\) for angular frequency, and \(\phi\) for phase. Angles in the model are measured in radians. The constant \(D\) is the midline, or mean height: the model oscillates between \(D-A\) and \(D+A\).

Definition: The instantaneous rate of change of height at time \(t\) is \(h'(t)\), the derivative of the height function with respect to time. For a sinusoidal model \(h(t)=D+A\sin(\omega t+\phi)\), the chain rule gives \(h'(t)=A\omega\cos(\omega t+\phi)\). If height is measured in metres and time in hours, for example, the rate is measured in metres per hour.

The factor \(A\omega\) sets the scale of the rate, while the cosine determines its sign and how it varies through the cycle. The period, or time for one full cycle, is \(P=2\pi/\omega\). A cosine model has the same structure: it can be rewritten as a sine model with a phase shift, or differentiated directly.

A positive rate means the height is increasing at that instant; a negative rate means it is decreasing. In an application, those usually correspond to rising and falling. A zero rate means that the instantaneous vertical motion is momentarily stationary. For a nonconstant sinusoidal model, its zero-rate points are turning points: the height is at a maximum or minimum. The height is not necessarily zero there.

Why the Sign Gives a Local Direction

The derivative is defined by the limiting value of a difference quotient. Its sign has a precise local interpretation. Suppose \(h'(t_0)>0\). By the definition of the limit, for sufficiently small nonzero \(\Delta t\), the quotient \(\frac{h(t_0+\Delta t)-h(t_0)}{\Delta t}\) is positive. In particular, take a sufficiently small positive \(\Delta t\). Since its denominator is positive, its numerator must also be positive, so \(h(t_0+\Delta t)>h(t_0)\). The height just after \(t_0\) is greater than the height at \(t_0\): it is rising locally.

If \(h'(t_0)<0\), the same reasoning gives a negative difference for sufficiently small positive \(\Delta t\), so the height is falling locally. This explains why the derivative’s sign describes direction rather than merely supplying a calculation. At a zero rate, this sign argument does not by itself decide what happens next. The sinusoidal formula provides more information: its zero-rate points are extrema, and the sign changes from positive to negative at a maximum or from negative to positive at a minimum.

Theorem: For \(h(t)=D+A\sin(\omega t+\phi)\) with \(A>0\) and \(\omega>0\), the rate is zero when \(\omega t+\phi=\frac{\pi}{2}+k\pi\), where \(k\) is an integer. When \(k\) is even the height is a maximum, \(D+A\); when \(k\) is odd it is a minimum, \(D-A\). This follows because the sine is \(1\) at the even-\(k\) phases and \(-1\) at the odd-\(k\) phases.

The conditions \(A>0\) and \(\omega>0\) matter: they describe a nonconstant repeating model. If \(A=0\), the height is constant and every rate is zero, with no alternating maxima and minima. Also, a zero rate does not mean the height itself is zero; it means the model has zero instantaneous change.

Reading the Rate Through a Cycle

For the sine model, the rate is \(A\omega\cos(\omega t+\phi)\). Consequently, the phase determines the direction of motion. When the cosine is positive, height is rising; when it is negative, height is falling. When the cosine is zero, the model is at a maximum or minimum. At the midline crossings, where the sine is zero, the rate has its largest magnitude, \(A\omega\), or its most negative value, \(-A\omega\).

This cycle-wide pattern is often more useful than a single rate value. It tells us, for instance, that a model’s height rises from its minimum to its maximum, then falls back to its minimum. The height changes direction at the extrema, not at the midline crossings.

Worked Example: Interpreting a Tide Model

Suppose a tide’s height above a reference level, in metres, is modeled by \(h(t)=2.4+0.6\cos\left(\frac{\pi t}{6}\right)\), where \(t\) is measured in hours. The midline is \(2.4\) metres, the amplitude is \(0.6\) metres, and the angular frequency is \(\frac{\pi}{6}\) radians per hour. Differentiate using the chain rule:

$$ h'(t)=0.6\left(-\sin\left(\frac{\pi t}{6}\right)\right)\frac{\pi}{6} =-\frac{\pi}{10}\sin\left(\frac{\pi t}{6}\right) \quad\text{metres per hour}. $$

At \(t=3\), the phase is \(\frac{\pi(3)}{6}=\frac{\pi}{2}\). Thus \(h(3)=2.4+0.6\cos(\frac{\pi}{2})=2.4\) metres, and \(h'(3)=-\frac{\pi}{10}\sin(\frac{\pi}{2})=-\frac{\pi}{10}\) metres per hour. The negative rate means the tide is falling. At \(t=9\), the phase is \(\frac{3\pi}{2}\), so \(h(9)=2.4\) metres and \(h'(9)=-\frac{\pi}{10}(-1)=\frac{\pi}{10}\) metres per hour: it is rising. At \(t=6\), the phase is \(\pi\), so \(h(6)=2.4+0.6(-1)=1.8\) metres and \(h'(6)=0\). This is the model’s minimum height. As a check, the model ranges from \(2.4-0.6=1.8\) to \(2.4+0.6=3.0\) metres, consistent with the stated minimum.

Turning Points and Rotating Heights

A rotating point has a height that repeats as it moves around its circle. If its vertical coordinate is modeled by a sinusoid, its rate describes the vertical component of its motion. At the top and bottom of the circular path, the vertical coordinate momentarily stops changing, even though the point continues moving along the circle. This is a useful reminder that a zero rate of height is not the same as being at rest in every direction.

Worked Example: A Point on a Rotating Object

A marker’s vertical coordinate is \(y(t)=1+4\sin\left(\frac{\pi t}{5}+\frac{\pi}{6}\right)\) centimetres, with \(t\) in seconds. Its rate is

$$ y'(t)=4\left(\frac{\pi}{5}\right)\cos\left(\frac{\pi t}{5}+\frac{\pi}{6}\right) =\frac{4\pi}{5}\cos\left(\frac{\pi t}{5}+\frac{\pi}{6}\right) \quad\text{centimetres per second}. $$

At \(t=0\), the phase is \(\frac{\pi}{6}\), so \(y(0)=1+4\left(\frac12\right)=3\) centimetres and \(y'(0)=\frac{4\pi}{5}\left(\frac{\sqrt3}{2}\right)=\frac{2\sqrt3\pi}{5}\) centimetres per second. The positive rate means the marker is moving upward vertically. When the phase reaches \(\frac{\pi}{2}\), the time satisfies \(\frac{\pi t}{5}+\frac{\pi}{6}=\frac{\pi}{2}\), giving \(t=\frac{5}{3}\) seconds. At that time, \(y=1+4(1)=5\) centimetres and \(y'=\frac{4\pi}{5}(0)=0\). This is the maximum height because the amplitude is \(4\) and the midline is \(1\), giving a maximum of \(1+4=5\) centimetres. The rate formula also checks directly: its cosine factor is zero at this phase.

The Same Height Can Have Different Rates

A height value alone cannot determine the direction of change. In a sinusoidal cycle, the same height is commonly reached once while rising and again while falling. The corresponding rates can have opposite signs, even though the heights are identical. This is why an interpretation of a periodic model should use both the height and the rate when direction matters.

Worked Example: Equal Heights, Opposite Directions

Let \(h(t)=5+2\sin\left(\frac{\pi t}{4}\right)\) metres, with \(t\) in minutes. Differentiation gives

$$ h'(t)=\frac{\pi}{2}\cos\left(\frac{\pi t}{4}\right) \quad\text{metres per minute}. $$

At \(t=\frac{2}{3}\) minutes, the phase is \(\frac{\pi}{6}\). Therefore \(h=5+2(\frac12)=6\) metres and \(h'=\frac{\pi}{2}(\frac{\sqrt3}{2})=\frac{\sqrt3\pi}{4}\) metres per minute. The height is rising. At \(t=\frac{10}{3}\) minutes, the phase is \(\frac{5\pi}{6}\). Then \(h=5+2(\frac12)=6\) metres again, but \(h'=\frac{\pi}{2}(-\frac{\sqrt3}{2})=-\frac{\sqrt3\pi}{4}\) metres per minute. The height is falling. Substituting both phase values into the original model verifies that the heights match; substituting them into the derivative verifies the opposite rates.

A Common Sign Mistake

When differentiating a cosine model, it is easy to lose the negative sign. For the tide model above, an incorrect calculation would be

$$ \frac{d}{dt}\left(2.4+0.6\cos\left(\frac{\pi t}{6}\right)\right) =\frac{\pi}{10}\sin\left(\frac{\pi t}{6}\right). $$

This is wrong because the derivative of \(\cos(u)\) is \(-\sin(u)\,u'\). The correct calculation is \(h'(t)=-\frac{\pi}{10}\sin(\frac{\pi t}{6})\). At \(t=3\), the incorrect expression gives a positive rate, contradicting the model’s movement from its midline toward its minimum. The corrected derivative gives \(-\frac{\pi}{10}\), which correctly indicates falling height. Checking the sign against the model’s position in the cycle is a useful way to catch this error.

Using Rates Responsibly

A derivative describes instantaneous change, not the total change over a long time interval. A rate of \(-0.1\) metres per hour means the height is decreasing at that instant; it does not by itself say how much lower the height will be several hours later. The sinusoidal model determines how the rate changes as time passes.

Units also help keep interpretations clear. If \(h\) is measured in metres and \(t\) in hours, then \(h'(t)\) is in metres per hour. The sign gives direction, while the magnitude gives the instantaneous speed of vertical change. At a zero rate, the vertical speed is momentarily zero; for a nonconstant sinusoid, that occurs at a highest or lowest point. A model’s phase is essential too: the same amplitude and period can describe different motion at a given clock time if the phase shift changes.

Check Your Understanding

Use the definitions and interpretations in this tutorial to answer the following questions.

  1. For \(h(t)=D+A\sin(\omega t+\phi)\), what does \(D\) represent, and what are the model’s minimum and maximum heights?
  2. In a height model, what does a negative derivative mean in context?
  3. Why does a zero rate occur at a maximum or minimum for a nonconstant sinusoidal height model?
  4. Can the same height occur once while rising and again while falling? Explain using the rate.
  5. If height is measured in feet and time in seconds, what units should the derivative have?

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