Sine and Cosine Applications
Now that we're familiar with the properties of sine and cosine equations and their graphs, we can apply these functions to real-world situations. 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 them using an equation or a graph. They then use this information to answer certain questions about the situation.
There will generally be four components to any problem involving the application of sine and cosine functions.
| SITUATION | Details of the real world situation being modeled. |
| 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. The situation is usually given first, but apart from that the order and format can vary from problem to problem.
Applying Sine Functions
Let's say we are given the following problem involving tides.
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.
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) Graph 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$1.75m or less. If Dakota wants to visit the rock shelf in the evening, between what times should she go?
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.
a) Recall that the sine function $\sin nt$sinnt, no matter what the value of $n$n is, is always between $1$1 and $-1$−1. This means that $\frac{1}{2}\sin nt$12sinnt 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+\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.
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$2m. This means that mid tide occurs whenever the $\frac{1}{2}\sin\frac{\pi}{6}t$12sinπ6t component is equal to zero in our function $d\left(t\right)=2+\frac{1}{2}\sin\frac{\pi}{6}t$d(t)=2+12sinπ6t.
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\frac{\pi}{6}t=0$12sinπ6t=0.
Recall that $\sin x=0$sinx=0 at $x=0,\pi,2\pi,3\pi$x=0,π,2π,3π$,$,... etc. Hence, $\frac{1}{2}\sin\frac{\pi}{6}t=0$12sinπ6t=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) 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 $T=\frac{2\pi}{n}$T=2πn.
In this problem, $n=\frac{\pi}{6}$n=π6, so we have a period $T=\frac{2\pi}{\frac{\pi}{6}}$T=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) 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\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.
Applying Cosine Functions
Suppose we are given the following problem involving Ferris wheels. The height of a certain carriage on a Ferris Wheel as it rotates can be modeled using the cosine function.
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.
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) Graph 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?
This is an example of a problem where we've been given the details of the situation only. We have to use this to find an equation, a graph, and answer further questions.
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 $h\left(t\right)=-60\cos nt+61$h(t)=−60cosnt+61 for some value of $n$n.
Recall that the period is given by $T=\frac{2\pi}{n}$T=2πn. We can rearrange to make $n$n the subject and get $n=\frac{2\pi}{T}$n=2πT. Substituting in our period $T=3$T=3, we get $n=\frac{2\pi}{3}$n=2π3.
Hence, our equation for the height function is $h\left(t\right)=-60\cos\frac{2\pi}{3}t+61$h(t)=−60cos2π3t+61.
e) We want to find the height after $5$5 minutes and $30$30 seconds, in other words, at $t=5\frac{1}{2}$t=512$=$=$\frac{11}{2}$112.
Substituting this value of $t$t into our equation gives:
| $h$h$($($\frac{11}{2}$112$)$) | $=$= | $-60\cos\left(\frac{2\pi}{3}\times\left(\frac{11}{2}\right)\right)+61$−60cos(2π3×(112))+61 | |
| $=$= | $-60\cos\frac{11\pi}{3}+61$−60cos11π3+61 | ||
| $=$= | $-60\cos\frac{5\pi}{3}+61$−60cos5π3+61 | Recalling that $\frac{11\pi}{3}=2\pi+\frac{5\pi}{6}$11π3=2π+5π6 and $\cos\left(2k\pi+\theta\right)=\cos\theta$cos(2kπ+θ)=cosθ for any integer $k$k. | |
| $=$= | $-60\times\frac{1}{2}+61$−60×12+61 | Recalling our 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 $-60\cos\frac{2\pi}{3}t+61=91$−60cos2π3t+61=91 and solve for $t$t.
| $-60\cos\frac{2\pi}{3}t+61$−60cos2π3t+61 | $=$= | $91$91 | |
|
$-60\cos\frac{2\pi}{3}t$−60cos2π3t |
$=$= | $30$30 | |
| $\cos\frac{2\pi}{3}t$cos2π3t | $=$= | $-\frac{1}{2}$−12 | |
| $\frac{2\pi}{3}t$2π3t | $=$= | $\frac{2\pi}{3}$2π3,$\frac{4\pi}{3}$4π3,... | Recalling our quadrants and exact values for sine and cosine angles. |
| $t$t | $=$= |
$1$1,$2$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.
Phase Shifts
Behaviour in the real world that involves rotation or repetitive motion can be modelled using trigonometric equations like $y=\sin x$y=sinx and $y=\cos x$y=cosx.
We can develop more accurate models by transforming these equations. In this lesson we will focus on phase shifts of the form $y=\sin\left(x-c\right)$y=sin(x−c), and also look at other transformations.
Exploration
A small disc sitting on a flat surface begins to roll slowly with a constant angular velocity of $1$1 radian per second when a fan is switched on.
As the disc rolls, the height $h$h in centimetres of a point on the disc is described by the equation $h=\cos t+1$h=cost+1, where $t$t is in seconds.
A graph of the equation is shown below. Notice that the period of the equation is $2\pi$2π seconds, or just over $6$6 seconds. This is the time it takes the point on the disc to complete one full revolution.
Graph of the function $h=\cos t+1$h=cost+1.
Two seconds after the first disc starts rolling along the surface, a second disc of the same size is placed in front of the fan. It begins to roll in a similar way, and the height of a point on its surface is described by the equation $h=\cos\left(t-2\right)+1$h=cos(t−2)+1.
The second disc has the same behaviour as the first disc, except it has been translated in time. The graph below shows that the phase shift of $2$2 seconds corresponds to a horizontal translation of the graph of $h=\cos t+1$h=cost+1 to the graph of $h=\cos\left(t-2\right)+1$h=cos(t−2)+1.
Graph of the functions $h=\cos t+1$h=cost+1 and $h=\cos\left(t-2\right)+1$h=cos(t−2)+1.
Suppose now that we want the discs to roll a bit faster. If we crank up the speed of the fan, we can increase the angular velocity of the disc to $4$4 radians per second. The motion of the point on the first disc is then described by the equation $h=\cos\left(4t\right)+1$h=cos(4t)+1, and the point on the second disc has the equation $h=\cos\left(4\left(t-2\right)\right)+1$h=cos(4(t−2))+1.
In the image below we see the consequence of increasing the rolling speed is to decrease the period of each equation. The new period is $\frac{2\pi\text{ radians}}{4\text{ radians per second}}=\frac{\pi}{2}$2π radians4 radians per second=π2 seconds, or about $1.57$1.57 seconds.
Graph of the functions $h=\cos\left(4t\right)+1$h=cos(4t)+1 and $h=\cos\left(4\left(t-2\right)\right)+1$h=cos(4(t−2))+1.
Worked example
The height of a point on a disc from the scenario above 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 question
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?