Rates Depend on Both the Model and Its Units
A sinusoidal model can describe a quantity that rises and falls repeatedly: a water level, a temperature, or the position of an oscillating object. Differentiating the model gives its instantaneous rate of change. To interpret that rate correctly, it is not enough to calculate a derivative: we must also track the units of the quantity and time, and understand how the model’s parameters affect the result.
For instance, a height measured in centimetres and modeled as a function of time in hours has a rate measured in centimetres per hour. The numerical value of that rate depends on the model’s amplitude and angular frequency, as well as on the phase at the time of interest.
A Sinusoidal Model and Its Parameters
Consider the model
where \(y(t)\) is the modeled quantity at time \(t\). We use \(A>0\) for the amplitude, \(\omega>0\) for the angular frequency, \(B\) for the vertical offset, and \(\phi\) for the phase. The angle \(\omega t+\phi\) is measured in radians. If \(y\) has units \(U\) and \(t\) has units \(T\), then \(A\) and \(B\) have units \(U\), while \(\omega\) has units radians per \(T\). The phase \(\phi\) is measured in radians.
Using the derivative of sine and the chain rule, the rate for this model is
The factor \(\omega\) appears because the angle inside the sine changes with time at rate \(\omega\). It also gives the product \(A\omega\) units \(U/T\), as a rate must have.
The parameters play different roles. The amplitude \(A\) scales the size of the oscillation in \(y\); the angular frequency \(\omega\) describes how quickly the phase changes with time. Both scale the rate through their product. The offset \(B\) shifts the model up or down without changing its rate. The phase \(\phi\) shifts when the model rises and falls, so it affects the rate at a particular time without changing the maximum possible rate.
Why the Rate Has Units of Quantity per Time
The units of a derivative follow from its definition as a limit of average rates. For a nonzero time change \(h\), the difference quotient is
The numerator is a difference between two values of \(y\), so it has units \(U\). The denominator has units \(T\). Thus every difference quotient has units \(U/T\). If the limit exists, taking that limit changes the numerical value but not the units. Therefore \(y'(t)\) has units \(U/T\).
For the sinusoidal model, the same conclusion is visible in its derivative: \(A\) has units \(U\), \(\omega\) has units radians per \(T\), and the cosine factor is dimensionless. Radians are retained in the description of angular frequency to make its meaning clear; the cosine itself contributes no units.
How the Parameters Bound the Rate
Since the cosine of any real angle lies between \(-1\) and \(1\), the rate satisfies
Consequently, the greatest possible magnitude of the instantaneous rate is \(A\omega\). This bound is attained when the cosine factor is \(1\) or \(-1\), which occurs during a complete cycle. The sign of the derivative indicates direction: a positive rate means \(y\) is increasing at that instant, while a negative rate means it is decreasing.
This result also explains why changing the amplitude and changing the frequency can both make rates larger. Multiplying \(A\) by a factor multiplies every rate by that factor. Multiplying \(\omega\) by a factor does the same, provided the time unit remains fixed. Changing only \(B\) does not affect the rate.
Worked Rates in Applied Models
Worked Example: Rate of Change of a Modeled Water Height
A water height \(h(t)\) is measured in centimetres and time \(t\) in hours. Suppose
Find the instantaneous rate at \(t=2\) hours, and state its units.
Here the amplitude is \(0.80\) cm, the offset is \(4.2\) cm, and the angular frequency is \(\pi/6\) radians per hour. Differentiate with respect to \(t\):
The coefficient is \(0.80(\pi/6)=0.4\pi/3=2\pi/15\), approximately \(0.418879\) cm per hour. At \(t=2\), the phase is
Using \(\cos(0.647198)\approx0.7978\), we obtain
To three decimal places, the height is increasing at \(0.334\) cm per hour at that time. Substituting the phase and cosine approximation into the derivative gives \(0.418879(0.7978)\approx0.3342\), confirming the reported rate. Its units are height divided by time, not centimetres alone.
Worked Example: Velocity from an Oscillating Position
An object’s position \(x(t)\) is measured in metres, with \(t\) in seconds. Let
Find its velocity at \(t=2\) seconds. The amplitude is \(3\) m, and the angular frequency is \(0.5\) radians per second. Differentiating gives
At \(t=2\), the phase is \(0.5(2)+1.2=2.2\) radians. Since \(\cos(2.2)\approx-0.588501\),
To three decimal places, the velocity is \(-0.883\) m/s. The negative sign means the position is decreasing at this instant. Substitution into the derivative verifies the calculation: \(1.5\cos(2.2)\approx-0.882752\) m/s. The offset \(12\) m does not appear in the velocity because a constant shift in position has zero derivative.
Worked Example: Comparing Amplitude and Frequency Effects
Consider a temperature model, with temperature in degrees Celsius and time \(t\) in minutes:
Compare its rate at \(t=0\) with the rate for a second model that has a different offset, amplitude, and angular frequency:
For the first model, differentiation gives
At \(t=0\), the phase is \(0.70\) radians, so using \(\cos(0.70)\approx0.764842\) gives \(T_1'(0)\approx0.60(0.764842)=0.458905\) degrees Celsius per minute.
For the second model,
Its phase at \(t=0\) is also \(0.70\) radians. Therefore \(T_2'(0)\approx1.50(0.764842)=1.147263\) degrees Celsius per minute. The ratio of the rates is \(1.147263/0.458905=2.5\), matching the ratio of the products \(A\omega\): \(1.50/0.60=2.5\). The offset changes from \(18\) to \(25\), but does not affect either rate. Substitution of the shared phase and each model’s coefficient into its derivative confirms both values.
Keep the Time Unit Consistent
A parameter’s numerical value is tied to the time unit used in the model. For example, an angular frequency of \(\pi/3\) radians per hour is not \(\pi/3\) radians per minute. To express that same frequency per minute, divide by \(60\):
Suppose time is entered in minutes in a model whose frequency is still written as \(\pi/3\) radians per hour. Substituting \(t=30\) directly into \((\pi/3)t\) treats \(30\) as hours, not minutes. The phase would be miscalculated. Convert \(30\) minutes to \(0.5\) hours before using the hourly model, or rewrite the frequency as \(\pi/180\) radians per minute and use \(t=30\). Either consistent choice gives the same phase.
Interpreting a Rate Before Reporting It
When using a sinusoidal rate in an applied setting, check three things. First, identify which quantity is changing and which variable represents time. Second, confirm that the angular frequency and time use consistent units. Third, interpret both the size and sign of the derivative: the size measures how quickly the output changes, while the sign indicates whether it is increasing or decreasing.
The amplitude and angular frequency determine the largest possible rate through their product \(A\omega\). The phase determines where the model is in its cycle at the chosen time, and therefore where the rate lies between its positive and negative bounds. The offset sets the model’s baseline but contributes nothing to the instantaneous rate.
Check Your Understanding
Answer each question using the model and units stated.
- If \(y(t)=6+2\sin(0.4t+0.3)\), with \(y\) in metres and \(t\) in seconds, what are the units of \(y'(t)\)?
- For \(y(t)=B+A\sin(\omega t+\phi)\), which parameters determine the maximum possible magnitude of the rate?
- Why does changing \(B\) leave the rate unchanged?
- If a model’s output is measured in degrees Celsius and time in minutes, what units should be used for its derivative?
- A frequency is given in radians per hour, but time is entered in minutes. What must be changed or converted to use the model consistently?
