From a Periodic Model to a Rate at One Time
A sinusoidal model describes a quantity that rises and falls repeatedly. In an application, it is often not enough to know the model’s value at a particular time. You may also need to know whether the quantity is increasing or decreasing at that instant, and how quickly.
The derivative provides that instantaneous rate. To find the rate at a specified time, first write the derivative as a function of time. Then substitute the specified time into that derivative and evaluate the trigonometric expression that remains. The value of the original model and its rate at that time are different quantities, so it is important to evaluate the derivative—not the model—when the question asks for a rate.
We will use sinusoidal models whose trigonometric arguments are in radians, as required for the standard derivative formulas for sine and cosine.
The Derivative of a Sinusoidal Model
Consider a sinusoidal model with constant parameters \(C\), \(A\), \(\omega\), and \(\phi\), where \(t\) is the input time:
Here \(C\) shifts the model vertically, \(A\) sets its signed amplitude, \(\omega\) controls how quickly its phase changes with time, and \(\phi\) is a phase shift. These parameters are fixed when the model is differentiated. The derivative rule for a sine with an affine input gives
The constant vertical shift \(C\) contributes zero to the derivative. The factor \(\omega\) appears because the input to the cosine’s derivative is \(\omega t+\phi\), not just \(t\). For a cosine model, the corresponding formula is
The phase \(\omega t_0+\phi\) must be calculated before evaluating its sine or cosine. Keeping that phase in parentheses helps avoid substituting the time into only part of the argument.
Why Substitution Gives the Rate at That Time
The derivative \(y'(t)\) is a function: for each input \(t\) where the derivative exists, it gives the instantaneous rate of change of \(y\) at that input. This follows directly from the definition of the derivative at a particular time \(t_0\):
provided the limit exists. In words, the derivative at \(t_0\) is the limiting change in the model’s value per unit time over shorter intervals beginning at \(t_0\). For the sine model, the derivative rule gives a formula for that rate at any input \(t\). Setting \(t=t_0\) in that formula therefore gives the rate at the requested time:
This explains the order of the calculation: differentiate to obtain a formula for rates, then evaluate that formula at the specified time. Substituting \(t_0\) into \(y(t)\) instead answers a different question: it gives the model’s value at that time.
A Reliable Evaluation Process
Record the model’s constants and the input value \(t_0\) at which the rate is required.
Use the appropriate sine or cosine derivative rule to obtain a rate formula in \(t\).
Calculate the complete trigonometric argument using \(t=t_0\), with parentheses around the full phase.
Use the phase in radians, retain sufficient calculator precision, and round only the final rate.
Confirm that you evaluated the derivative, and interpret a positive or negative rate as increasing or decreasing at that instant.
Worked Evaluations
Worked Example: Rate of a Sine Model at a Given Time
Suppose a height model is \(H(t)=14+6\sin(0.8t-0.5)\), with time \(t\) measured in seconds. Find the rate at \(t=2\).
First differentiate. The derivative of the constant \(14\) is zero, and the sine term contributes the frequency factor \(0.8\):
Now substitute \(t=2\) into the derivative’s phase:
Thus the required rate is
Using \(\cos(1.1)\approx 0.4535961214\),
Rounded to six decimal places, the rate is \(2.177261\) height units per second. Check: substituting \(t=2\) into the derivative gives the phase \(1.1\) and the same product \(4.8\cos(1.1)\). The positive result indicates that the modeled height is increasing at that time.
Worked Example: A Decreasing Cosine Model
Let \(R(t)=50-8\cos(0.3t+0.2)\). Find the rate at \(t=10\).
Differentiate the cosine model, including the negative sign in the derivative of cosine:
Substitute \(t=10\) into the phase:
Evaluate the derivative at that time:
Using \(\sin(3.2)\approx -0.0583741434\),
Rounded to six decimal places, \(R'(10)\approx -0.140098\) model units per unit of time. Check: the derivative is \(2.4\sin(0.3t+0.2)\); at \(t=10\), its phase is \(3.2\), whose sine is negative, so the evaluated rate must be negative. This agrees with the computed result and means the modeled quantity is decreasing at \(t=10\).
Worked Example: Substituting into a Shifted Time Input
Consider \(D(t)=100+4\sin(1.1(t-0.5)+0.25)\). Find the rate at \(t=1.5\). The shifted input \(t-0.5\) is part of the phase and must be kept intact during substitution.
Differentiate using the coefficient of \(t\) inside the phase, which is \(1.1\):
At \(t=1.5\), first calculate the shifted time and then the full phase:
Therefore,
Using \(\cos(1.35)\approx 0.2190066871\),
Rounded to six decimal places, the rate is \(0.963629\) model units per unit of time. Check: evaluating the derivative formula at \(t=1.5\) gives \(4.4\cos(1.35)\), exactly the expression used in the calculation. Its positive value indicates an increasing model at that time.
Common Pitfall: Evaluating the Model Instead of Its Derivative
A question asking for a rate at a time does not ask for the model’s value at that time. These calculations can look similar because both involve substituting the same time into a trigonometric expression, but they have different formulas and answer different questions.
For example, in the first worked example, the model value at \(t=2\) is
whereas the rate is
A wrong calculation for the rate would be to report \(14+6\sin(1.1)\). That is a height, not a rate. The correction is to differentiate first and then substitute: \(H'(t)=4.8\cos(0.8t-0.5)\), so \(H'(2)=4.8\cos(1.1)\approx 2.177261\). Checking whether the final expression came from the derivative formula is a simple way to catch this mistake.
What the Evaluated Rate Tells You
The derivative at one time is a signed rate. A positive value means the modeled quantity is increasing at that instant; a negative value means it is decreasing. A value near zero means the instantaneous rate is small, though the model’s value itself need not be near zero. The rate is determined by the derivative formula’s coefficient and by the trigonometric value at the phase reached at that time.
When evaluating a rate, keep the calculation organized: write the derivative, substitute into its complete phase, evaluate the trigonometric function, and multiply. This makes it easier to distinguish the model’s level from its motion and to see how the chosen time affects the answer.
Check Your Understanding
For each question, distinguish the derivative formula from the original model before evaluating a rate.
- For \(f(t)=9+3\sin(0.5t+0.4)\), write \(f'(t)\), then set up the expression for the rate at \(t=2\). What phase must be evaluated?
- For \(g(t)=20-5\cos(0.2t-0.1)\), write \(g'(t)\). At \(t=5\), is the rate positive or negative?
- In a model \(p(t)=C+A\sin(\omega t+\phi)\), which expression gives the model value at \(t_0\), and which gives its rate there?
- Why must the complete phase be evaluated after substituting the specified time?
- If a computed rate is negative, what does that say about the model at that instant?
