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Sinusoidal models and their rates of change

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Rates from periodic models

A periodic model has an amplitude, an equilibrium level and a period. Differentiate the model with respect to time before interpreting an instantaneous rate

\[y(t)=A\sin\left(k\pi t\right)+E\quad\Rightarrow\quad y'(t)=Ak\pi\cos\left(k\pi t\right)\]
  1. 1Read amplitude, equilibrium and period For a nonconstant model y(t) = A sin(kπt) + E with k > 0, the amplitude is |A|, the central level is E and the period is 2/k. \(\displaystyle y(t)=3\sin\left(\frac{1}{3}\pi t\right)+10\quad\Rightarrow\quad A=3,\ E=10,\ P=\frac{2}{1/3}=6\)
  2. 2Include the inner coefficient Differentiating the inner expression multiplies the amplitude by k pi \(\displaystyle \frac{d}{dt}\left(A\sin\left(k\pi t\right)+E\right)=Ak\pi\cos\left(k\pi t\right)\)
  3. 3Substitute time into the phase Calculate the angle inside sine or cosine before evaluating the trigonometric value \(\displaystyle t=3\quad\Rightarrow\quad \frac{1}{3}\pi(3)=\pi\)
  4. 4Interpret the derivative sign A positive derivative means the modelled quantity is increasing, and a negative derivative means it is decreasing \(\displaystyle y'(3)=\pi\cos\left(\pi\right)=-\pi<0\quad\Rightarrow\quad y\text{ is decreasing}\)
Year 12, VCE Units 3 & 4, Chapter 5, 4 difficulty levels, Calculator allowedBuild a worksheet

Rates from periodic models

A periodic model has an amplitude, an equilibrium level and a period. Differentiate the model with respect to time before interpreting an instantaneous rate

\[y(t)=A\sin\left(k\pi t\right)+E\quad\Rightarrow\quad y'(t)=Ak\pi\cos\left(k\pi t\right)\]
  1. Read amplitude, equilibrium and period For a nonconstant model y(t) = A sin(kπt) + E with k > 0, the amplitude is |A|, the central level is E and the period is 2/k. \(\displaystyle y(t)=3\sin\left(\frac{1}{3}\pi t\right)+10\quad\Rightarrow\quad A=3,\ E=10,\ P=\frac{2}{1/3}=6\)
  2. Include the inner coefficient Differentiating the inner expression multiplies the amplitude by k pi \(\displaystyle \frac{d}{dt}\left(A\sin\left(k\pi t\right)+E\right)=Ak\pi\cos\left(k\pi t\right)\)
  3. Substitute time into the phase Calculate the angle inside sine or cosine before evaluating the trigonometric value \(\displaystyle t=3\quad\Rightarrow\quad \frac{1}{3}\pi(3)=\pi\)
  4. Interpret the derivative sign A positive derivative means the modelled quantity is increasing, and a negative derivative means it is decreasing \(\displaystyle y'(3)=\pi\cos\left(\pi\right)=-\pi<0\quad\Rightarrow\quad y\text{ is decreasing}\)

Worked example

A height in metres is modelled by the expression shown, where time is in hours. Find the instantaneous rate at two hours

\[h(t)=2\cos\left(\frac{1}{4}\pi t\right)+8\]
  1. \(\displaystyle P=\frac{2}{1/4}=8\text{ h}\)The coefficient k=1/4 gives a period of 8 hours
  2. \(\displaystyle h'(t)=-2\times\frac{1}{4}\pi\sin\left(\frac{1}{4}\pi t\right)=-\frac{1}{2}\pi\sin\left(\frac{1}{4}\pi t\right)\)Cosine differentiates to negative sine and the inner coefficient 1/4 is included
  3. \(\displaystyle \frac{1}{4}\pi(2)=\frac{1}{2}\pi,\quad \sin\left(\frac{1}{2}\pi\right)=1\)Substitute t=2 into the phase and evaluate the sine value
  4. \(\displaystyle h'(2)=-\frac{1}{2}\pi\text{ m/h}\)The instantaneous height is decreasing at a rate of pi/2 metres per hour

Common mistakes

Missing the inner coefficient
✕\(\displaystyle y'(t)=3\pi\cos\left(\frac{1}{3}\pi t\right)\)✓\(\displaystyle y'(t)=\pi\cos\left(\frac{1}{3}\pi t\right)\)
The chain rule multiplies by the inner coefficient 1/3, so the amplitude 3 becomes a coefficient of 1.
Swapping amplitude and equilibrium
✕\(\displaystyle A=10,\quad E=3\)✓\(\displaystyle A=3,\quad E=10\)
The amplitude is the size of the oscillating term, while the added constant gives the equilibrium level
Misreading the period
✕\(\displaystyle P=\frac{1}{3}\)✓\(\displaystyle P=\frac{2}{1/3}=6\)
For a positive coefficient k in the angle k pi t, the period is 2 divided by k, not k itself

Difficulty levels

introRead amplitude, equilibrium level and period from a periodic model
standardEvaluate a periodic model or its derivative at a specified time
challengeFind a time from a rate condition or compare the rate over an interval
examSolve a mixed periodic modelling question involving phase, rates and intervals

Questions about this topic

How is the period found from a sine model?+

For a nonconstant sine model, the period is two divided by the magnitude of the nonzero phase coefficient. A constant model has no least positive period

Why does the derivative include the inner coefficient?+

The chain rule multiplies the derivative of the outer sine or cosine by the derivative of its inner angle

What does the equilibrium level mean?+

It is the added constant in the sine model, around which the periodic quantity oscillates

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 5: Differentiation

In this chapter

Evaluate limits and test continuityGradient of a tangent from first principlesDifferentiate power functions and evaluate a derivativeExpand and factor polynomial expressionsPascal rows and binomial coefficient structure5A Limits, first principles and the power rule: Simplify and combine algebraic fractionsEvaluate a polynomial or recover a coefficient parameterSubstitute values into a formula5A Limits, first principles and the power rule: Solve a linear equationTest or construct a smooth piecewise joinRead and construct a gradient function from a function graphAnalyse a cubic graph or determine its ruleEvaluate and determine function rulesEquation of a tangent or normal lineRecover a polynomial rule from derivative and tangent conditionsFind the gradient of a line through two pointsFind the equation of a line from its graph or from its intercept and slopeForm an equation of a parallel or perpendicular lineSolve a pair of simultaneous linear equationsDetermine parameters in a polynomial identity5C Tangents, normals and derivative conditions: Solve a linear equationDifferentiate natural exponential functions and solve simple derivative conditionsExponential growth and decay rate models5D Exponential derivatives and rate models: Simplify and combine algebraic fractions5D Exponential derivatives and rate models: Round to decimal places and significant figuresDifferentiate sine, cosine and tangent functions and solve simple derivative conditionsGeometric optimisation with a trigonometric modelSinusoidal models and their rates of changeSolve a practical right-angled-triangle problemPerimeter and area of standard shapes5E Trigonometric differentiation and rates: Round to decimal places and significant figuresApproximate a cubic root using Newton's method