Determining Rules for Graphs of Circular Functions Revision Notes for VCE SSCE Mathematical Methods

Determining Rules for Graphs of Circular Functions

This section focuses on finding the rule for a graph when you know it represents a circular function. The general form for these functions is:

f(t) = A \sin(nt + \varepsilon) + b

where:

$A$ is the amplitude (distance from centreline to peak)
$n$ is the frequency parameter (related to the period)
$\varepsilon$ (epsilon) is the phase shift (horizontal translation)
$b$ is the vertical shift (height of centreline)

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Understanding each parameter in the general form is crucial for determining the complete rule from a graph or given information. Each parameter controls a specific aspect of the function's behavior.

Key formulas

Period formula

The period of a circular function is related to the frequency parameter $n$ by:

\text{Period} = \frac{2\pi}{n}

Rearranging this formula allows us to find $n$ when we know the period:

n = \frac{2\pi}{\text{period}}

Finding amplitude and centreline from range

When given the range $[\text{minimum}, \text{maximum}]$ of a circular function:

The amplitude is:

A = \frac{1}{2}(\text{maximum} - \text{minimum})

The centreline has equation:

y = b \text{ where } b = \frac{1}{2}(\text{maximum} + \text{minimum})

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Finding Key Parameters from Range

These formulas allow you to determine both the amplitude and vertical shift without needing to see the graph. The amplitude tells you how far the function oscillates from its centreline, while $b$ tells you where that centreline is positioned vertically.

Worked example: Finding A and n from amplitude and period

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Worked Example: Finding A and n from amplitude and period

A function has rule $f(t) = A \sin(nt)$ . The amplitude is $6$ and the period is $10$ . Find $A$ and $n$ , and sketch the graph of $y = f(t)$ for $0 \leq t \leq 10$ .

Solution:

Finding n from the period:

We know that:

\text{Period} = \frac{2\pi}{n} = 10

Rearranging to solve for $n$ :

n = \frac{2\pi}{10} = \frac{\pi}{5}

Finding A from the amplitude:

Since the amplitude equals $6$ , we know that A = 6.

The complete rule:

The function has rule:

f(t) = 6 \sin\left(\frac{\pi t}{5}\right)

Sketch of the graph:

The graph shows one complete cycle from $t = 0$ to $t = 10$ , with maximum value $6$ and minimum value $-6$ .

Worked example: Finding A, n, and b from a graph

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Worked Example: Finding A, n, and b from a graph

The graph shown is that of a function with rule $y = A \sin(nt) + b$ . Find $A$ , $n$ , and $b$ .

Solution:

Finding the amplitude:

The graph oscillates between a minimum of $2$ and a maximum of $6$ . The amplitude is the distance from the centreline to either extreme:

A = 2

Finding the period and n:

From the graph, we can see that one complete cycle occurs over an interval of length $6$ . Therefore, the period is 6.

Using the period formula:

\frac{2\pi}{n} = 6

Solving for $n$ :

n = \frac{2\pi}{6} = \frac{\pi}{3}

Finding the vertical shift:

The centreline runs horizontally through the middle of the oscillation. We can calculate its position:

b = \frac{\text{maximum} + \text{minimum}}{2} = \frac{6 + 2}{2} = 4

Therefore, the centreline has equation $y = 4$ and b = 4.

The complete rule:

y = 2 \sin\left(\frac{\pi t}{3}\right) + 4

Worked example: Finding A, n, and b from range and period

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Worked Example: Finding A, n, and b from range and period

A function with rule $y = A \sin(nt) + b$ has range $[-2, 4]$ and period $3$ . Find $A$ , $n$ , and $b$ .

Solution:

Finding the amplitude:

Using the range $[-2, 4]$ , where the minimum is $-2$ and the maximum is $4$ :

A = \frac{1}{2}(4 - (-2)) = \frac{1}{2}(6) = 3

Finding the centreline:

The centreline is positioned at:

b = \frac{4 + (-2)}{2} = \frac{2}{2} = 1

Therefore, b = 1 and the centreline has equation $y = 1$ .

Finding n from the period:

The period is $3$ , so:

\frac{2\pi}{n} = 3

n = \frac{2\pi}{3}

The complete rule:

y = 3 \sin\left(\frac{2\pi}{3} t\right) + 1

Worked example: Finding A, n, and ε with a phase shift

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Worked Example: Finding A, n, and ε with a phase shift

A function with rule $y = A \sin(nt + \varepsilon)$ has the following properties:

range = $[-2, 2]$
period = $6$
when $t = 4$ , $y = 0$

Find values for $A$ , $n$ , and $\varepsilon$ .

Solution:

Finding the amplitude:

Since the range is $[-2, 2]$ , the amplitude is:

A = \frac{1}{2}(2 - (-2)) = \frac{1}{2}(4) = 2

Finding n from the period:

Since the period is $6$ :

\frac{2\pi}{n} = 6

n = \frac{\pi}{3}

Setting up the equation:

We now know that:

y = 2 \sin\left(\frac{\pi}{3} t + \varepsilon\right)

Finding ε using the given point:

When $t = 4$ , we know that $y = 0$ . Substituting these values:

2 \sin\left(\frac{4\pi}{3} + \varepsilon\right) = 0

Dividing both sides by $2$ :

\sin\left(\frac{4\pi}{3} + \varepsilon\right) = 0

The sine function equals zero when its argument is $0$ , $\pm\pi$ , $\pm2\pi$ , and so on:

\frac{4\pi}{3} + \varepsilon = 0 \text{ or } \pm\pi \text{ or } \pm2\pi \text{ or } \ldots

Choosing the simplest solution:

We select the simplest solution, which is:

\frac{4\pi}{3} + \varepsilon = 0

\varepsilon = -\frac{4\pi}{3}

The complete rule:

y = 2 \sin\left(\frac{\pi}{3} t - \frac{4\pi}{3}\right)

This rule satisfies all three given properties.

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Common Mistakes to Avoid

When determining rules for circular functions, watch out for these common errors:

Forgetting that period = $\frac{2\pi}{n}$ , so $n = \frac{2\pi}{\text{period}}$
Using negative values for amplitude. The amplitude is always positive: use $A = \frac{1}{2}(\text{max} - \text{min})$
Misidentifying the centreline. It's the horizontal line exactly halfway between the maximum and minimum values
When finding the phase shift $\varepsilon$ , remember to substitute the given point into your equation and solve for the simplest value
Not checking your final rule by verifying it satisfies all the given conditions

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Key Points to Remember:

The general form of a circular function is $f(t) = A \sin(nt + \varepsilon) + b$ , where $A$ is amplitude, $n$ relates to period, $\varepsilon$ is phase shift, and $b$ is vertical shift
The period and frequency parameter are related by: Period = $\frac{2\pi}{n}$ , therefore $n = \frac{2\pi}{\text{period}}$
Find amplitude from range using: $A = \frac{1}{2}(\text{maximum} - \text{minimum})$
Find the centreline from range using: $b = \frac{1}{2}(\text{maximum} + \text{minimum})$
When finding phase shift $\varepsilon$ , substitute a known point into the equation and choose the simplest solution from the infinite family of solutions

Determining Rules for Graphs of Circular Functions (VCE SSCE Mathematical Methods): Revision Notes