Applications of sine and cosine functions

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+12​sinπ6​t 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$12​sinnt 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$12​sinπ6​t component is equal to zero in our function $d\left(t\right)=2+\frac{1}{2}\sin\frac{\pi}{6}t$d(t)=2+12​sinπ6​t. 

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$12​sinπ6​t=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$12​sinπ6​t=0 at $\frac{\pi}{6}t=0,\pi,2\pi,3\pi$π6​t=0,π,2π,3π$,$,...etc.

Since the question asked for two values of $t$t, we solve for $\frac{\pi}{6}t=0$π6​t=0 and $\frac{\pi}{6}t=\pi$π6​t=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+12​sinπ6​t	$=$=	$1.75$1.75
$\frac{1}{2}\sin\frac{\pi}{6}t$12​sinπ6​t	$=$=	$-0.25$−0.25
$\sin\frac{\pi}{6}t$sinπ6​t	$=$=	$-0.5$−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,$\frac{\pi}{6}t=\frac{7\pi}{6}$π6​t=7π6​ and $\frac{\pi}{6}t=\frac{11\pi}{6}$π6​t=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)=−60cos2π3​t+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 $-60\cos\frac{2\pi}{3}t+61=91$−60cos2π3​t+61=91 and solve for $t$t.

$-60\cos\frac{2\pi}{3}t+61$−60cos2π3​t+61	$=$=	$91$91
$-60\cos\frac{2\pi}{3}t$−60cos2π3​t	$=$=	$30$30
$\cos\frac{2\pi}{3}t$cos2π3​t	$=$=	$-\frac{1}{2}$−12​
$\frac{2\pi}{3}t$2π3​t	$=$=	$\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.

Worked Examples

question 1

question 2

question 3