5.07 Sinusoidal models
Many phenomena in the world around us change periodically, such as ocean tides, pendulums, springs, rotors, wheels, and even certain animal populations. Scientists observe this back-and-forth movement and collect data so they can model the situation using an equation or a graph. They then can use this information to answer certain questions about the situation. We can apply our knowledge of trigonometric functions to investigate and solve real-world problems.
Strategy
There will generally be four components to any problem involving the application of trigonometric functions.
| 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 need to find them in order to understand the situation and answer the questions. The situation is usually given first, but apart from that the order and format can vary from problem to problem.
Worked examples
Example 1
The depth of the water in metres at a certain pier is given by the equation $d\left(t\right)=2+\frac{1}{2}\sin\left(30t\right)$d(t)=2+12sin(30t), where $t$t is the number of hours after $10$10 a.m.
(a) How deep will the water be at high tide? How deep will it be at low tide?
(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?
(c) How many hours are there between two successive high tides?
(d) How many of these tide cycles are there in a $24$24 hour day?
(e) Sketch a graph of the depth of the water as a function of $t$t.
(f) There is a rock shelf near the pier that is only safe to visit when the depth of the water is $1.75$1.75 m or less. If Dakota wants to visit the rock shelf in the evening, between what times should she go?
Think: In this particular problem we've been given the situation with a corresponding equation, and we are to answer questions relating to it as well as construct a graph representing the situation.
Do:
(a) Recall that the sine function $\sin\left(nt\right)$sin(nt), no matter what the value of $n$n is, is always between $1$1 and $-1$−1. This means that $\frac{1}{2}\sin\left(nt\right)$12sin(nt) will always be between $\frac{1}{2}$12 and $-\frac{1}{2}$−12.
Hence, the maximum value of $d\left(t\right)$d(t) is $2+\frac{1}{2}$2+12 and the minimum value is $2-\frac{1}{2}$2−12. So the water is $2.5$2.5 m deep at high tide, and $1.5$1.5 m deep at low tide.
(b) 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$2 m. This means that mid-tide occurs whenever the $\frac{1}{2}\sin\left(30t\right)$12sin(30t) component is equal to zero in our function $d\left(t\right)=2+\frac{1}{2}\sin\left(30t\right)$d(t)=2+12sin(30t).
We want to find any times from $10$10 a.m. when this is the case - in other words, any values from $t=0$t=0 at which $\frac{1}{2}\sin\left(30t\right)=0$12sin(30t)=0.
Recall that $\sin x=0$sinx=0 at $x=0,180,360,540,\ldots$x=0,180,360,540,… etc. Hence, $\frac{1}{2}\sin\left(30t\right)=0$12sin(30t)=0 at $30t=0,180,360,540,\ldots$30t=0,180,360,540,… etc.
Since we want to find two values of $t$t, we solve for $30t=0$30t=0 and $30t=180$30t=180 only. This gives $t=0$t=0 and $t=6$t=6, which correspond to $10$10 a.m. and $4$4 p.m. respectively as the times when the first two mid tides occur.
(c) This question is really asking us what the period of the function is. Recall that the period $T$T is the amount of time it takes for the function to complete one full cycle, and is given by $T=\frac{360}{n}$T=360n.
In this problem, $n=\frac{180}{6}$n=1806, so we have a period $T=\frac{360}{\frac{180}{6}}$T=3601806 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) If our period is $12$12 hours, this means a cycle takes $12$12 hours. So there are going to be $2$2 full cycles in a $24$24 hour day.
(e) Using all the information from the previous parts, we can come up with the graph below.
(f) 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. To find these times, we set $d\left(t\right)=1.75$d(t)=1.75 and solve for $t$t.
| $2+\frac{1}{2}\sin\left(30t\right)$2+12sin(30t) | $=$= | $1.75$1.75 |
| $\frac{1}{2}\sin\left(30t\right)$12sin(30t) | $=$= | $-0.25$−0.25 |
| $\sin\left(30t\right)$sin(30t) | $=$= | $-0.5$−0.5 |
Recall our exact value angles and quadrants to notice that $\sin x=-\frac{1}{2}$sinx=−12 for $x=210,330,570,690,\ldots$x=210,330,570,690,… etc. Hence,$30t=210$30t=210 and $30t=330$30t=330 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.
Example 2
A circular Ferris wheel is $120$120 metres in diameter and contains several carriages. Asher and Jessica enter a carriage at the bottom of the wheel and get off $24$24 minutes later after having gone around $8$8 times. When a carriage is at the bottom of the wheel, it is $1$1 metre off the ground. The height of a certain carriage on the Ferris wheel as it rotates can be modelled using a cosine function.
(a) What is the maximum and minimum height of Asher and Jessica's carriage?
(b) What is the period of the function $h\left(t\right)$h(t), the height of the carriage $t$t minutes after it has started moving?
(c) Sketch a graph of the function $h\left(t\right)$h(t).
(d) Find the equation for the function $h\left(t\right)$h(t).
(e) The ride suddenly stops $5$5 minutes and $30$30 seconds after Asher and Jessica got on. How high is their carriage above the ground when this happens?
(f) How long does it take the carriage to first reach a height of $91$91 m above the ground?
Think: This is an example of a problem where we've been given the details of the situation only. We will use these details to find an equation, construct a graph, and then answer further questions.
Do:
(a) We've been given the diameter of the wheel, so we know that there is a distance of $120$120 m from the lowest point on the wheel to the highest point. A carriage is at its lowest height off the ground at the bottom of the wheel, so the minimum height is $1$1 m. The greatest height off the ground is $120$120 m above this, so the maximum height is $1+120=121$1+120=121 m.
(b) The period is the amount of time it takes to complete one full cycle from the minimum height, to the maximum height, and back to the minimum height again. In other words, how long it takes the wheel to go around once.
We already know that it takes $24$24 minutes to go around $8$8 times. Therefore, it must take $24\div8=3$24÷8=3 minutes to go up and then down once.
(c) Using the above information, we can come up with the following graph.
(d) How does the above cosine graph compare to a normal $y=\cos x$y=cosx graph? Notice that it has been vertically dilated by a factor of $60$60. It is also upside down, which means we have a negative cosine graph.
Now, the graph has also been vertically translated upwards. How far has the equilibrium position through the middle of the graph moved upwards? It used to be at $0$0, now it is halfway between $1$1 and $121$121, at $61$61. Hence, we have an equation that looks like $h\left(t\right)=-60\cos\left(nt\right)+61$h(t)=−60cos(nt)+61 for some value of $n$n.
Recall that the period is given by $T=\frac{360}{n}$T=360n. We can rearrange to make $n$n the subject and get $n=\frac{360}{T}$n=360T. Substituting in our period $T=3$T=3, we get $n=\frac{360}{3}$n=3603.
Hence, our equation for the height function is $h\left(t\right)=-60\cos\left(\frac{360}{3}t\right)+61$h(t)=−60cos(3603t)+61.
(e) We want to find the height after $5$5 minutes and $30$30 seconds - that is, at $t=5\frac{1}{2}=11/2$t=512=11/2 minutes.
Substituting this value of $t$t into our equation gives:
| $h$h$($($\frac{11}{2}$112$)$) | $=$= | $-60\cos\left(\left(\frac{360}{3}\times\left(\frac{11}{2}\right)\right)\right)+61$−60cos((3603×(112)))+61 |
|
| $=$= | $-60\cos660^\circ+61$−60cos660°+61 |
|
|
| $=$= | $-60\cos300^\circ+61$−60cos300°+61 |
Recognising that $660=300$660=300 on the unit circle. |
|
| $=$= | $-60\times\frac{1}{2}+61$−60×12+61 |
Recalling the exact values for sine and cosine angles |
|
| $=$= | $-30+61$−30+61 |
|
|
| $=$= | $31$31 |
|
Hence, the carriage is $31$31 m above the ground when the ride stops.
(f) We want to find the first time $t$t when $h=91$h=91. To do this, we set our equation $-60\cos\left(120t\right)+61=91$−60cos(120t)+61=91 and solve for $t$t.
| $-60\cos\left(120t\right)+61$−60cos(120t)+61 | $=$= | $91$91 |
|
| $-60\cos120t$−60cos120t | $=$= | $30$30 |
|
| $\cos\left(120t\right)$cos(120t) | $=$= | $-\frac{1}{2}$−12 |
|
| $120t$120t | $=$= | $120,240,\ldots$120,240,… |
Recalling the quadrants and exact values for sine and cosine angles |
| $t$t | $=$= | $1,2,\ldots$1,2,… |
|
| $t$t | $=$= | $1$1 |
Choosing our first value of $t$t only |
Hence, the carriage first reaches a height of $91$91 m after $1$1 minute.
Practice questions
Question 1
The height above the ground of a rider on a Ferris wheel can be modelled by the function $h\left(t\right)=25\cos\left(3\left(t-60\right)\right)+30$h(t)=25cos(3(t−60))+30, where $h\left(t\right)$h(t) is the height in metres, $t$t is the time in seconds, and the angular velocity of $3^\circ$3° per second is the speed at which the wheel rotates.
Draw the graph of $y=h\left(t\right)$y=h(t).
Loading Graph...What is the maximum height of the rider?
What is the minimum height of the rider?
At what height is the rider after $85$85 seconds? Give your answer to two decimal places.
How long does it take the rider to complete one full revolution?