Now that we're familiar with the properties of sine and cosine functions, we can apply them to real-world situations. Many phenomena in the world around us change periodically such as pendulums, springs, rotors and wheels. In general, when solving problems involving applications like these, the first step is translating the information provided into either an equation or graph, and answering the relevant questions that we might be interested in.
Worked example
Peter is riding forwards on a unicycle. When he gets on to start riding, the pedals are horizontally inline. The height of the back pedal in centimetres is given by the equation $y=20\sin\theta+40$y=20sinθ+40 where $\theta$θ is the measure of the angle that the back pedal has rotated.
a) What is the maximum and minimum height that the back pedal reaches?
Think: The maximum height of the back pedal occurs whenever $\sin\theta=1$sinθ=1 and the minimum height occurs whenever $\sin\theta=-1$sinθ=−1.
Do: Let's substitute $\sin\theta=1$sinθ=1 and $\sin\theta=-1$sinθ=−1 into the equation.
| $y$y | $=$= | $20\sin\theta+40$20sinθ+40 | (writing down the equation) |
| $y$y | $=$= | $20\times1+40$20×1+40 | (making the substitution) |
| $y$y | $=$= | $60$60 cm | (simplifying the expression) |
So the maximum height of the back pedal is $60$60 centimetres.
| $y$y | $=$= | $20\sin\theta+40$20sinθ+40 | (writing down the equation) |
| $y$y | $=$= | $20\times\left(-1\right)+40$20×(−1)+40 | (making the substitution) |
| $y$y | $=$= | $20$20 cm | (simplifying the expression) |
So the minimum height of the back pedal is $20$20 centimetres.
b) How high is the back pedal from the ground when Peter gets on?
Think: Peter gets on when the pedals are horizontally level and no angle has been formed. In other words, Peter gets on when $\theta=0^\circ$θ=0°.
Do: Let's substitute $\theta=0^\circ$θ=0° into the equation.
| $y$y | $=$= | $20\sin\theta+40$20sinθ+40 | (writing down the equation) |
| $y$y | $=$= | $20\sin0^\circ+40$20sin0°+40 | (making the substitution) |
| $y$y | $=$= | $20\times0+40$20×0+40 | (simplifying the $\sin$sin function) |
| $y$y | $=$= | $40$40 cm | (simplifying the expression) |
So the back pedal is $40$40 centimetres above ground when Peter gets on.
c) Graph the function given by the equation $y=20\sin\theta+40$y=20sinθ+40.
Think: We can draw the function $y=20\sin\theta+40$y=20sinθ+40 by first considering $y=\sin\theta$y=sinθ.
Do: If we take the graph of $y=\sin\theta$y=sinθ, stretch the amplitude and vertically translate the curve up, we get our desired function.
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| The graph of $y=20\sin\theta+40$y=20sinθ+40 |
d) How high is the back pedal from the ground when it has rotated through $150^\circ$150°?
Think: We wish to find the value of $y$y when $\theta=150^\circ$θ=150°. We can see graphically that this looks like $50$50 centimetres, but we can show this algebraically as well.
Do: Let's substitute $\theta=150^\circ$θ=150° into the equation.
| $y$y | $=$= | $20\sin\theta+40$20sinθ+40 | (writing down the equation) |
| $y$y | $=$= | $20\sin150^\circ+40$20sin150°+40 | (making the substitution) |
| $y$y | $=$= | $20\times\frac{1}{2}+40$20×12+40 | (simplifying the $\sin$sin function) |
| $y$y | $=$= | $50$50 cm | (simplifying the expression) |
So we can confirm that the back pedal is $50$50 centimetres above the ground after the pedal has rotated $150^\circ$150°.
e) How many degrees has the back pedal first rotated through when it is $30$30 cm off the ground?
Think: We can see from the graph of $y=20\sin\theta+40$y=20sinθ+40 and $y=30$y=30 that the pedal has rotated somewhere between $180^\circ$180° and $240^\circ$240° - perhaps the midpoint of $210^\circ$210°. Again, we can confirm this algebraically.
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| The graph of $y=20\sin\theta+40$y=20sinθ+40 (green) and $y=30$y=30 (blue) |
Do: Let's substitute $y=30$y=30 into the equation.
| $y$y | $=$= | $20\sin\theta+40$20sinθ+40 | (writing down the equation) |
| $30$30 | $=$= | $20\sin\theta+40$20sinθ+40 | (making the substitution) |
| $-10$−10 | $=$= | $20\sin\theta$20sinθ | (subtracting $40$40 from both sides) |
| $-\frac{1}{2}$−12 | $=$= | $\sin\theta$sinθ | (dividing by $20$20 on both sides) |
| $\theta$θ | $=$= | $\sin^{-1}\left(-\frac{1}{2}\right)$sin−1(−12) | (taking the inverse $\sin$sin of both sides) |
| $\theta$θ | $=$= | $210^\circ$210° | (writing down the answer in degrees) |
So after the back pedal rotates $210^\circ$210°, it is $30$30 centimetres above the ground.
f) Graph the height of the front pedal as a function over $\theta$θ.
Think: Whenever the back pedal is closest to the ground, the front pedal is furthest from the ground and visa versa. From this, the peaks and troughs of the back and front pedal should occur alternately.
Do: We draw the two functions alongside one another below.
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| The height of the back pedal (green) and the front pedal (blue) as functions of $\theta$θ. |
Practice questions
question 1
A projectile is launched at an angle with measure $\theta$θ with an initial velocity of $a$a m/s. Let $x$x be the horizontal distance covered by the projectile in metres and $t$t be the number of seconds the projectile spends airborne.
The horizontal distance is related to the launch angle through the equation $x=at\cos\theta$x=atcosθ. If the initial velocity of the projectile is $20$20 m/s and we wish to determine the horizontal distance covered by the projectile after $3$3 seconds, form an equation that relates $x$x and $\theta$θ.
Find the value of $x$x if the launch angle of the projectile is $45^\circ$45°. Give your answer in exact form.
Find the value of $\theta$θ if the projectile travels a horizontal distance of $57$57 metres. Give your answer to the nearest degree.
question 2
A bridge has length $70$70 metres and is joined by two arms at its centre. The two arms lift up when a boat wishes to pass underneath. Let $\theta$θ be the angle of elevation of the two arms.
The height of the bridge above the water when the two arms lie flat is $50$50 metres. Find $y$y, the height of the bridge in metres above the water when the two arms are lifted at an angle of elevation $\theta$θ.
Find the value of $y$y if the angle of elevation of the bridge arms is $60^\circ$60°. Give your answer to the nearest metre if necessary.
Find the value of $\theta$θ if the height of the bridge from the water is $67$67 metres. Give your answer to the nearest degree if necessary.
question 3
A large tent is pitched on a levelled surface at the points $P$P and $R$R. An adjustable pole fixed at the midpoint of $P$P and $R$R holds the tent up at $Q$Q. The arms of the tent, $PQ$PQ and $QR$QR form an angle with measure $\theta$θ.
Let $h$h represent the height of the tent in metres. If the length of $QR$QR is $15$15 metres, form an equation for $h$h in terms of $\frac{\theta}{2}$θ2.
Complete the table of values and give your answers to two decimal places if necessary.
$\theta$θ
$0^\circ$0° $15^\circ$15° $30^\circ$30° $60^\circ$60° $90^\circ$90° $120^\circ$120° $150^\circ$150° $180^\circ$180° $h$h
$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Draw the height of the tent as a function of the angle $\theta$θ.
Loading Graph...From part (c), what is the maximum height of the tent?
Hence what are the possible values of the height of the tent?
$\editable{}\le h\le\editable{}$≤h≤
Phase shift
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^\circ$1° 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 $360$360 seconds, or $6$6 minutes. 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.
Thirty 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 $h=\cos\left(t-30^\circ\right)+1$h=cos(t−30°)+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 $30$30 seconds corresponds to a horizontal translation of the graph of $h=\cos t+1$h=cost+1 to the graph of $h=\cos\left(t-30^\circ\right)+1$h=cos(t−30°)+1.
Graph of the functions $h=\cos t+1$h=cost+1 and $h=\cos\left(t-30^\circ\right)+1$h=cos(t−30°)+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 $15^\circ$15° per second. The motion of the point on the first disc is then described by the equation $h=\cos\left(15t\right)+1$h=cos(15t)+1, and the point on the second disc has the equation $h=\cos\left(15\left(t-30\right)\right)+1$h=cos(15(t−30))+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{360^{\circ}}{15^{\circ}\text{ per second}}=24$360∘15∘ per second=24 seconds.
Graph of the functions $h=\cos\left(15t\right)+1$h=cos(15t)+1 and $h=\cos\left(15\left(t-30\right)\right)+1$h=cos(15(t−30))+1.
Worked example
The height of a point on a disc from the scenario above is described by the equation $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t−7))+10. The graph of the function is shown below.
Graph of the function $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t−7))+10.
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 $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t−7))+10.
| $h$h | $=$= | $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t−7))+10 | |
| $=$= | $10\sin\left(20\left(0-7\right)\right)+10$10sin(20(0−7))+10 | (Substitute the value of $t$t) | |
| $=$= | $10\sin\left(20\times\left(-7\right)\right)+10$10sin(20×(−7))+10 | (Simplify the product) | |
| $=$= | $10\sin\left(-140^\circ\right)+10$10sin(−140°)+10 | (Evaluate the multiplication) | |
| $=$= | $3.57$3.57 cm (2 d.p.) | (Evaluate the expression) |
The initial height of the point is $3.57$3.57 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 $\sin\left(20\left(t-7\right)\right)=1$sin(20(t−7))=1.
Do: By substituting $\sin\left(20\left(t-7\right)\right)=1$sin(20(t−7))=1 into the equation $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t−7))+10 we see that the maximum height is $h=10\times1+10=20$h=10×1+10=20 cm above the surface.
c. Find the time when the point is first at a height of $15$15 cm.
Think: We have a value of $h$h, and we want to find the corresponding value of $t$t.
Do: Substitute $h=15$h=15 into the equation and rearrange to solve for $t$t.
| $h$h | $=$= | $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t−7))+10 | |
| $15$15 | $=$= | $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t−7))+10 | (Substitute the value of $h$h) |
| $5$5 | $=$= | $10\sin\left(20\left(t-7\right)\right)$10sin(20(t−7)) | (Subtract $10$10 from both sides) |
| $\frac{1}{2}$12 | $=$= | $\sin\left(20\left(t-7\right)\right)$sin(20(t−7)) | (Divide both sides by $10$10) |
| $30$30 | $=$= | $20\left(t-7\right)$20(t−7) | (Take the inverse $\sin$sin of both sides) |
| $\frac{3}{2}$32 | $=$= | $t-7$t−7 | (Divide both sides by $20$20) |
| $t$t | $=$= | $8.5$8.5 seconds | (Add $7$7 to both sides) |
The point on the disk will first reach a height of $15$15 cm when it has rolled for $8.5$8.5 seconds.
Practice question
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?