| SITUATION | Details of the real world situation being modelled. |
| EQUATION | The equation describing the situation. |
| GRAPH | The graph describing the situation. |
| QUESTIONS | Questions about specific features of the situation, such as the period, amplitude, certain values, maxima or minima. |
Sometimes the equation or the graph will be given, and sometimes you will be asked to find them.
Finding the equation of a model
Let's review the key features of a sine graph and how we might calculate the parameters given a graph or information.
| Feature | How to find |
|---|---|
| Amplitude, $a$a |
This is half the vertical distance between a maximum and a minimum, hence can be calculated as: $a=\frac{y_{max}-y_{min}}{2}$a=ymax−ymin2 |
| Period |
This is the time taken to repeat one cycle. This can be found as the horizontal distance between two successive maximums or minimums. For example: $Period,P=x_{max2}-x_{max}$Period,P=xmax2−xmax |
| $b$b |
The parameter $b$b impacts the period as it causes a horizontal dilation by a factor of $\frac{1}{b}$1b. It can be calculated as follows: $b=\frac{2\pi}{P}$b=2πP |
| Principal axis, $y=d$y=d |
The principal axis or mid-line can be found as the average of the $y$y-coordinates for a maximum and minimum. Hence, can be calculated using: $d=\frac{y_{max}+y_{min}}{2}$d=ymax+ymin2 |
| Phase shift, $c$c |
This is how far the model has been horizontally shifted. For a sine model this can be found visually as the $x$x-value where the graph crosses the mid-line between a minimum and maximum. Or calculated as the average of the $x$x-coordinates of a minimum followed by a maximum: (note order is important) $c=\frac{x_{min}+x_{max}}{2}$c=xmin+xmax2 For a cosine model this can be found as the $x$x-coordinate of the first maximum: $c=x_{max}$c=xmax |
The only difference between finding a sine model or a cosine model is the phase shift. We could in fact model all such graphs as sine models but if a graph starts at a maximum or minimum we can model using a cosine function and not require a phase shift.
Worked example
Example 1
Find the equation of the form $y=a\sin\left(b\left(x-c\right)\right)+d$y=asin(b(x−c))+d to fit the following model:
Think: Do we have enough information to find the model? In this case three points were labelled for us but if they were not we could label them or extract key information from a written question.
Do: Find or list each of the parameters. You may be able to easily read them from the graph or information given. In this case we will calculate each from the given points.
Amplitude:
| $a$a | $=$= | $\frac{y_{max}-y_{min}}{2}$ymax−ymin2 |
| $=$= | $\frac{2-\left(-4\right)}{2}$2−(−4)2 | |
| $=$= | $3$3 |
Period:
| $P$P | $=$= | $x_{max2}-x_{max}$xmax2−xmax |
| $=$= | $6.5-2.5$6.5−2.5 | |
| $=$= | $4$4 |
b:
| $b$b | $=$= | $\frac{2\pi}{P}$2πP |
| $=$= | $\frac{2\pi}{4}$2π4 | |
| $=$= | $\frac{\pi}{2}$π2 |
Principal axis:
| $d$d | $=$= | $\frac{y_{max}+y_{min}}{2}$ymax+ymin2 |
| $=$= | $\frac{2+\left(-4\right)}{2}$2+(−4)2 | |
| $=$= | $-1$−1 |
Phase shift:
| $c$c | $=$= | $\frac{x_{min}+x_{max}}{2}$xmin+xmax2 |
| $=$= | $\frac{0.5+2.5}{2}$0.5+2.52 | |
| $=$= | $1.5$1.5 |
Hence, the model that suits the graph is $y=3\sin\left(\frac{\pi}{2}\left(x-1.5\right)\right)-1$y=3sin(π2(x−1.5))−1
Reflect: Do the transformations fit the model shown? Check direction of shifts and period.
Practice question
Question 1
Determine the equation of the graphed function given that it is of the form $y=\sin\left(x-c\right)$y=sin(x−c), where $c$c is the least positive value.
Question 2
Determine the equation of the graph given that it is of the form $y=-\cos\left(x+c\right)-d$y=−cos(x+c)−d, where $c$c is the least positive value and $x$x is in radians.
Applications
For application questions, we will often be required to interpret key features in context. Think carefully about the situation, the units required in answers and if given answers make sense in the context. If the interpretation is the focus of the question make effective use of your calculator to graph the function and find answers.
Worked examples
Example 2
The depth of the water in metres at a certain pier is given by the equation $d\left(t\right)=2+\frac{1}{2}\sin\frac{\pi}{6}t$d(t)=2+12sinπ6t where $t$t is the number of hours after $10$10 a.m.
Think: In this particular problem we've been given the situation with a corresponding equation, and we have to answer questions relating to it and construct a graph. As we have been given the equation we could graph this using technology.
Do: If technology is available use your knowledge of the transformations and a suitable domain such as $0\le t\le24$0≤t≤24 representing a day, set up an appropriate viewing window and graph.
(a) How deep will the water be at high tide? How deep will it be at low tide?
Think: Recall that the sine function $\sin bt$sinbt, no matter what the value of $b$b is, is always between $1$1 and $-1$−1. This means that $\frac{1}{2}\sin bt$12sinbt will always be between $\frac{1}{2}$12 and $-\frac{1}{2}$−12.
Do: Hence, the maximum value of $d\left(t\right)$d(t) is $2+\frac{1}{2}$2+12 and the minimum value is $2+\left(-\frac{1}{2}\right)$2+(−12). So the water is $2.5$2.5m deep at high tide, and $1.5$1.5m deep at low tide.
Or find using technology.
(b) Mid tide occurs when the depth of the water is exactly halfway between high tide and low tide. From $10$10 a.m., what are the first two times when it is mid tide?
Think: Given that we now know the depth of the water at high tide and low tide, we can clearly see that the midpoint between the two values is $2$2m.
We want to find the first two times from $10$10 a.m. when $f(t)=2$f(t)=2, in other words, values from $t=0$t=0 at which $2+\frac{1}{2}\sin\frac{\pi}{6}t=2$2+12sinπ6t=2.
Do: We can do this using technology and graphing both sides of the equation as separate graphs and finding the $x$x-coordinates of points of intersection.
Algebraically we could solve this by rearranging the equation to $\sin\frac{\pi}{6}t=0$sinπ6t=0 and recalling that $\sin x=0$sinx=0 at $x=0,\pi,2\pi,3\pi$x=0,π,2π,3π$,$,... etc. Hence, $\sin\frac{\pi}{6}t=0$sinπ6t=0 at $\frac{\pi}{6}t=0,\pi,2\pi,3\pi$π6t=0,π,2π,3π$,$,...etc.
Since the question asked for two values of $t$t, we solve for $\frac{\pi}{6}t=0$π6t=0 and $\frac{\pi}{6}t=\pi$π6t=π only. This gives $t=0$t=0 and $t=6$t=6, which corresponds to $10$10 a.m. and $4$4 p.m. when the first two mid tides occur.
(c) How many hours are there between two successive high tides?
Think: This question is really asking us what the period of the function is. Recall that the period $P$P is the amount of time it takes for the function to complete one full cycle, and is given by $P=\frac{2\pi}{b}$P=2πb.
Do: In this problem, $b=\frac{\pi}{6}$b=π6, so we have a period $P=\frac{2\pi}{\frac{\pi}{6}}$P=2ππ6 which simplifies to $T=12$T=12.
Hence, the time between successive high tides is $12$12 hours. In fact, this is also the time between low tides, since one cycle will always take the same amount of time no matter where it starts from.
(d) How many of these tide cycles are there in a $24$24 hour day?
Think: We just found the period or cycle is 12 hours, how many will fit in 24 hours?
Do: $24\div12=2$24÷12=2, so there are going to be $2$2 full cycles in a $24$24 hour day.
(e) Graph the depth of the water as a function of $t$t.
Think: Using all the information from the previous parts, such as maximum and minimum depth, sketch the graph for an appropriate domain, such as one day.
Do:
(f) There is a rock shelf near the pier that is only safe to visit when the depth of the water is $1.75$1.75m or less. If Dakota wants to visit the rock shelf in the evening, between what times should she go?
Think: Looking at the graph above, we can see that there are certain times between $4$4 p.m. and $10$10 p.m. when the water will dip below $1.75$1.75, allowing for safe passage to the rocks.
Do: We can solve this using technology by graphing $y=1.75$y=1.75 and finding the $x$x-coordinates of the points of intersection.
To find these times algebraically, we set $d\left(t\right)=1.75$d(t)=1.75 and solve for $t$t.
| $2+\frac{1}{2}\sin\frac{\pi}{6}t$2+12sinπ6t | $=$= | $1.75$1.75 |
| $\frac{1}{2}\sin\frac{\pi}{6}t$12sinπ6t | $=$= | $-0.25$−0.25 |
| $\sin\frac{\pi}{6}t$sinπ6t | $=$= | $-0.5$−0.5 |
Recall our exact value angles and quadrants to notice that $\sin x=-\frac{1}{2}$sinx=−12 for $x=\frac{7\pi}{6},\frac{11\pi}{6},\frac{19\pi}{6},\frac{23\pi}{6}$x=7π6,11π6,19π6,23π6$,$,... etc. Hence,$\frac{\pi}{6}t=\frac{7\pi}{6}$π6t=7π6 and $\frac{\pi}{6}t=\frac{11\pi}{6}$π6t=11π6 for the times we are concerned with, giving $t=7$t=7 and $t=11$t=11.
So, if Dakota wants to visit the rock shelf, she will have to go between $5$5 p.m. and $9$9 p.m.
Reflect: How long do you think a model for tides such as the one above will be accurate? How does the timing between low and high tides change over time?
Example 3
The height of a point on a rotating disc is described by the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t−4))+8. The graph of the function is shown below.
Graph of the function $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t−4))+8.
(a) Find the initial height of the point.
Think: The initial point in time is the moment when $t=0$t=0 seconds.
Do: Substitute $t=0$t=0 into the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t−4))+8.
| $h$h | $=$= | $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5(t−4))+8 | |
| $=$= | $8\sin\left(\frac{\pi}{5}\left(0-4\right)\right)+8$8sin(π5(0−4))+8 | (Substitute the value of $t$t) | |
| $=$= | $8\sin\left(-\frac{4\pi}{5}\right)+8$8sin(−4π5)+8 | (Simplify the product) | |
| $=$= | $3.30$3.30 cm (2 d.p.) | (Evaluate the expression) |
The initial height of the point is $3.30$3.30 cm above the surface.
(b) Find the maximum height of the point.
Think: The sine function takes values between $-1$−1 and $1$1. The maximum height will correspond to the times at which $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)=1$sin(π5(t−4))=1.
Do: By substituting $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)=1$sin(π5(t−4))=1 into the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t−4))+8 we see that the maximum height is $h=8\times1+8=16$h=8×1+8=16 cm above the surface.
(c) Find the time when the point is first at a height of $12$12 cm.
Think: We have a value of $h$h, and we want to find the corresponding value of $t$t.
Do: Substitute $h=12$h=12 into the equation and rearrange to solve for $t$t.
| $h$h | $=$= | $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5(t−4))+8 | |
| $12$12 | $=$= | $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5(t−4))+8 | (Substitute the value of $h$h) |
| $4$4 | $=$= | $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)$8sin(π5(t−4)) | (Subtract $8$8 from both sides) |
| $\frac{1}{2}$12 | $=$= | $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)$sin(π5(t−4)) | (Divide both sides by $8$8) |
| $\frac{\pi}{6}$π6 | $=$= | $\frac{\pi}{5}\left(t-4\right)$π5(t−4) | (Take the inverse $\sin$sin of both sides) |
| $\frac{5}{6}$56 | $=$= | $t-4$t−4 | (Multiply both sides by $\frac{5}{\pi}$5π) |
| $t$t | $=$= | $4.83$4.83 seconds (2 d.p.) | (Add $4$4 to both sides) |
The point on the disk will first reach a height of $12$12 cm when it has rolled for $4.83$4.83 seconds.
Practice questions
question 3
The population (in thousands) of two different types of insects on an island can be modelled by the following functions: Butterflies: $f\left(t\right)=a+b\sin\left(mt\right)$f(t)=a+bsin(mt), Crickets: $g\left(t\right)=c-d\sin\left(kt\right)$g(t)=c−dsin(kt)
$t$t is the number of years from when the populations started being measured, and $a$a,$b$b,$c$c,$d$d,$m$m, and $k$k are positive constants. The graphs of $f$f and $g$g for the first $2$2 years are shown below.
State the function $f\left(t\right)$f(t) that models the population of Butterflies over $t$t years.
State the function $g\left(t\right)$g(t) that models the population of Crickets over $t$t years.
How many times over a $18$18 year period will the population of Crickets reach its maximum value?
How many years after the population of Crickets first starts to increase, does it reach the same population as the Butterflies?
Solve for $t$t, the number of years it takes for the population of Butterflies to first reach $200000$200000.
question 4
question 5
The change in voltage of overhead power lines over time can be modelled by a periodic function that oscillates between $-300$−300 and $300$300 kiloVolts with a frequency of $40$40 cycles per second.
Assuming that at $t=0$t=0 seconds, the voltage is $300$300 kV, which of the following is the most suitable model for voltage over time?
$V=A\cos nt$V=Acosnt
A$V=A\sin nt$V=Asinnt
B$V=A\sin t$V=Asint
C$V=A\cos t$V=Acost
DUsing the function $V=A\cos nt$V=Acosnt to model the voltage over time, solve for the value of $n$n.
Form an equation for the voltage $V$V as a function of time $t$t. You may assume the general form $V=A\cos nt$V=Acosnt.
Graph the voltage function over time.
Loading Graph...If the frequency were increased, how would that affect the graph of voltage over time?
It would increase the amplitude of the function and stretch the graph out vertically.
AIt would increase the period of the function and stretch the graph out horizontally.
BIt would decrease the amplitude of the function and compress the graph vertically.
CIt would decrease the period of the function and compress the graph horizontally.
D
question 6
Tobias is jumping on a trampoline. Victoria watches him bounce at a regular rate and wants to try to model his height over time. When Victoria starts her stopwatch, Tobias is at a minimum height of $30$30 cm below the trampoline frame. A moment later Victoria records Tobias reaching a maximum height of $50$50 cm above the trampoline frame. She uses the function $H\left(s\right)=a\sin\left(2\pi\left(s-c\right)\right)+d$H(s)=asin(2π(s−c))+d, where $H$H is the height in cm above the trampoline frame and $s$s is the time in seconds.
Find the value of $a$a, the amplitude of the function.
Find the value of $d$d.
Find the value of $H\left(0\right)$H(0).
Find the value of $c$c, if $0
At what time does Tobias first reach a height of $30$30 cm?