Word Problems
In order to solve problems modelled with Sinusoidals more effectively, we typically need to write the equation using \( \cos\) or \( \sin \). These variables include :
- Period (or Frequency)
- Axis of Symmetry
- Amplitude
- Phase Shift
To identify the Period, we can identify 2 points with the same altitude (for example 2 max or min points). We can then subtract the value of Point 2 by that of Point 1. Sometimes the Frequency is given. Take the inverse to calculate the Period.
To identify the Axis, we would need to calculate the average of the max and min points.
To identify the Amplitude, we would need to calculate the average of the difference between the max and min.
The word problem might not use the words max or min; instead look for the top and bottom.
To identify the Phase Shift, first decide if you will write the equation in terms of \(\cos\) or \(\sin\). Then, find the 'starting' \(x\)-value. Remember \(\cos\) starts at the max and then decreases and \(\sin\) starts at the axis then increases.
Objects travelling in a circle are often modelled using sinusodials. The diameter of the circle is the difference between the max and min.
Review the basic characteristics of sinusoidals and writing sinusoidal equations before attempting word problems.
Once you have the equation, you will need to solve for either \(x\) or \(y\)-values. To solve for \(y\), plug in an \(x\)-value. Make sure you use the correct units! To solve for \(x\)- values, plug in \(y\) and isolate for the trig ratio. Remember when you take the inverse there could be multiple answers. Draw out the CAST diagram! Finally, double check your values are valid before stating the final answer!
A weight attached to the end of a spring is bouncing up and down. As it bounces, its distance from the floor varies sinusoidally with time. You start a stopwatch. When the stopwatch reads \(0.3\) seconds, the weight reaches a high point \(60 \; [\text{cm}]\) above the floor. The next low point, \(40\; [\text{cm}]\) above the floor, occurs at \(1.8\) seconds (Don't assume that at time zero the weight is at the minimum!)
- Sketch a graph of this sinusoidal function.
- Write an equation expressing distance from the floor in terms of the number of seconds the stopwatch reads.
- What was the distance from the floor when you started the watch?
- Predict the time at which the weight was \(59 \; [\text{cm}]\) above the floor for the first and second times.
First, we can identify the max and min points as \(1.8\) and \(0.3\) respectively.
Next, we need to determine the period and axis in order to accurately draw the graph.
We can identify half the period's length by subtracting the time at the min point from that of the max point. We can than multiply this value by \(2\) to get the full period length:
\(\cfrac{1}{2}\text{period} = \text{max} - \text{min}\)
\(\cfrac{1}{2}\text{period} = 1.8 - 0.3\)
\(\cfrac{1}{2}\text{period} = 1.5\)
\(\left(\cfrac{1}{2}\right)(2)\text{period} = (1.5)(2)\)
\(\text{period} = 3\;[\text{s}]\)
We can determine the axis by finding the average of the maximum and minimum points:
\(\text{axis} =\cfrac{\text{max}+\text{min}}{2}\)
\(\text{axis} =\cfrac{60+40}{2}\)
\(\text{axis} =\cfrac{100}{2}\)
\(\text{axis} = 50\)
Using this information, we can create a table of values:
| Time (s) | 0.3 | 1.05 | 1.8 | 2.55 | 3.3 |
|---|---|---|---|---|---|
| Distance from Floor (cm) | 60 | 50 | 40 | 50 | 60 |
We can now draw our graph:
ii. As a reference, our formula needs to be in the form \(f(x) = |a|\sin(k(x+d))+c\).
First, we can find the amplitude by subtracting the min (\(40\)) from the min (\(60\)) and dividing by \(2\):
\(\text{amplitude} = \cfrac{\text{max} - \text{min}}{2}\)
\(\text{amplitude} = \cfrac{60 - 40}{2}\)
\(\text{amplitude} = \cfrac{20}{2}\)
\(\text{amplitude} = 10\)
Next, we can find \(k\) by dividing \(360°\) by the period length (\(3\)):
\(k = \cfrac{360°}{\text{period}}\)
\(k = \cfrac{360°}{3}\)
\(k = 120°\)
Then, we can determine \(d\) by finding the first horizontal point in the cycle (meaning axis point that increases to the maximum). By looking at our graph, we can determine that this point lies at \(2.55\).
We have already determined the axis (or \(c\)) in Part i. As a result, we can put our formula together:
\(f(x) = |a|\sin(k(x+d))+c\)
\(f(x) = 10\sin(120°(x-2.55))+50\)
Therefore, we can determine that our formula is \(\boldsymbol{f(x) = 10\sin(120°(x-2.55))+50}\).
iii. In order to determine the initial distance from the ground, we must set \(x=0\). Since our graph represents that of a cosine function, we will use \(\cos\) in place of \(\sin\). We will also use \(0.3\) in place of \(2.55\) to represent the phase shift:
\(f(x) = 10\cos(120°(x-0.3))+50\)
\(f(x) = 10\cos(120°(0-0.3))+50\)
\(f(x) = 10\cos(120°(-0.3))+50\)
\(f(x) = 10\cos(-36)+50\)
\(f(x) = 10(0.809016994)+50\)
\(f(x) = 8.090169944 + 50\)
\(f(x) = 58.1\;[\text{cm}]\)
Therefore, we can determine that the spring's initial height was \(\boldsymbol{58.1\;[\textbf{cm}]}\).
iv. In order to determine the times that the spring was \(59\;[\text{cm}]\) above the ground, we will once again use the cosine formula. We will also represent \(120°(x-0.3) \) as \(\theta\).
\(59 = 10\cos\theta + 50\)
\(10\cos\theta = 59 - 50\)
\(10\cos\theta = 9\)
\(\cfrac{\cancel{10}\cos\theta}{\cancel{10}} = \cfrac{9}{10}\)
\(\cos\theta = 0.9\)
We can find the inverse of \(0.9\) in order to calculate \(\theta\):
\(\theta = \cos(0.9)^{-1}\)
\(\theta = 26°\)
We can now set \(26°\) equal to \(120°(x-0.3)\) and solve for \(x\) in order to determine the actual time:
\(26° = 120°(x-0.3)\)
\(\cfrac{26°}{120°} = \cfrac{120°(x-0.3)}{120°}\)
\(0.2166 = x-0.3\)
\(x = 0.2166 + 0.3 \)
\(x = 0.5\;[\text{s}]\)
In order to determine the second time, we can use the Unit Circle. According to the Unit Circle, cosine produces positive results for Quadrant \(1\) (which we've already covered) and Quadrant \(4\).
We can determine the angle in Quadrant \(4\) by subtracting \(360°\) by \(26°\) since it contains the same reference angle:
\(\theta' = 360° - 26°\)
\(\theta' = 334°\)
We can now use this new angle to determine what the second time will be. We can do so by setting \(334°\) equal to \(120°(x-0.3)\) and solving for \(x\):
\(334° = 120°(x-0.3)\)
\(\cfrac{334°}{120°} = \cfrac{120°(x-0.3)}{120°}\)
\(2.7833 = x-0.3\)
\(x = 2.7833 + 0.3\)
\(x \approx 3.1\;[\text{s}]\)
Therefore, we can determine the first and second times the spring was \(59\;[\text{cm}]\) above the ground were \(\boldsymbol{0.5\;[\textbf{s}]}\) and \(\boldsymbol{3.1\;[\textbf{s}]}\) respectively.
You are on a 8-seat Ferris wheel at an unknown height when the ride starts. It takes you \(24\) seconds to reach the top of the wheel \(26\;[\text{m}]\) above the ground. The loading platform is \(4\;[\text{m}]\) high. The wheel makes a complete revolution in \(60\) seconds.
- Create a diagram/sketch to show your location on the ride and the key features of the wheel.
- Find the sinusoidal equation that models your ride on the ferris wheel.
- How high above the ground will your seat be after \(90\) seconds?
- Find the \(2\) times within \(1\) cycle when the height is at \(24\;[\text{m}]\)
i. First, we can determine the full period length as \(60\). We can also determine the max and min heights as \(26\) and \(4\) respectively.
Next, we need to determine the period and axis in order to accurately draw the graph.
Although we have the full period length, we can determine the amount of time it takes to go from the max to min by dividing this value by \(2\):
\(\cfrac{1}{2}\text{period} = 30\;[\text{s}]\)
Therefore, we can determine that it takes \(30\;[\text{s}]\) to go from the max to min (and vice versa).
Since it took us \(24\;[\text{s}]\) to reach the top of the ferris wheel, we are also able to determine that we didn't start the ride at the minimum height (\(4\;[m]\)); that would've meant beginning the ride at \(-6\;[s]\), which is impossible.
Next, we need to determine the axis. We can do so by finding the average of the maximum and minimum points:
\(\text{axis} =\cfrac{\text{max}+\text{min}}{2}\)
\(\text{axis} =\cfrac{26 + 4}{2}\)
\(\text{axis} =\cfrac{30}{2}\)
\(\text{axis} = 15 \; [\text{m}]\)
Using this information, we can create a table of values:
| Time (s) | -6 | 9 | 24 | 39 | 54 | 69 | 84 |
|---|---|---|---|---|---|---|---|
| Height (m) | 4 | 15 | 26 | 15 | 4 | 15 | 26 |
We can now sketch our graph as such:
ii. As a reference our formula needs to be in the form \(f(x) = |a|\sin(k(x+d))+c\).
First, we can find the amplitude (\(a\)) by subtracting the maximum by the minimum and dividing by \(2\):
\(\text{amplitude} =\cfrac{\text{max}-\text{min}}{2}\)
\(\text{amplitude} =\cfrac{26-4}{2}\)
\(\text{amplitude} =\cfrac{22}{2}\)
\(\text{amplitude} = 11\)
Next, we can find \(k\) by dividing \(360°\) by the period length (\(60\)):
\(k = \cfrac{360°}{k}\)
\(k = \cfrac{360°}{60}\)
\(k = 6°\)
Then, we can determine \(d\) by finding the first horizontal (or axis) point in the cycle. By looking at our graph, we can determine that this point lies at \(9\).
We have already determined the axis \(c\) in the previous part. As a result, we can put our formula together:
\(f(x) = |a|\sin(k(x+d))+c\)
\(f(x) = 11\sin(6°(x-9))+15\)
Therefore, we can determine that our formula is \(\boldsymbol{f(x) = 10\sin(120°(x-2.55))+50}\).
iii. In order to determine your height after \(90\;[\text{s}]\), we must set \(x=0\):
\(f(x) = 11\sin(6°(x-9))+15\)
\(f(x) = 11\sin(6°(90-9))+15\)
\(f(x) = 11\sin(6°(81))+15\)
\(f(x) = 11\sin(486)+15\)
\(f(x) = 11(0.809016994)+15\)
\(f(x) = 8.899186938+15\)
\(f(x) = 23.9\;[\text{m}]\)
Therefore, we can determine that you will be \(\boldsymbol{23.9\;[\textbf{m}]}\) above the ground at \(90\;[\text{s}]\).
iv. In order to determine the two times in the cycle that the height was at \(24\;[\text{m}]\), we will set it equal to the \(\sin\) formula, \(f(x) = 11\sin(6°(x-9))+15\) and solve for \(\theta]\). To make things more efficient, we will represent \(6°(x-9)\) as \(\theta\):
\(24 = 11\sin\theta + 15\)
\(11\sin\theta = 24 - 15\)
\(11\sin\theta = 9\)
\( \cfrac{\cancel{11}\sin\theta}{\cancel{11}} = \cfrac{9}{11}\)
\(\sin\theta = 0.818181818\)
We can find the inverse of \(0.818181818\) to find \(\theta\):
\(\theta = \sin(0.818181818)^{-1}\)
\(\theta \approx 54.9°\)
We can now set \(54.9°\) equal to \(6°(x-9)\) and solve for \(x\) in order to determine the actual time:
\(54.90319877° = 6°(x-9)\)
\(\cfrac{\cancel{6°}(x-9)}{\cancel{6°}} = \cfrac{54.90319877°}{6°}\)
\(9.150533129 = x-9\)
\(x = 9.150533129 + 9\)
\(x \approx 18.15\;[\text{s}]\)
In order to determine the second time, we can use the Unit Circle. According to the Unit Circle, sin produces positive results for Quadrant \(1\) (which we've already covered) and Quadrant \(2\). We can determine the angle in Quadrant \(2\) by subtracting \(180°\) by \(54.9°\):
\(\theta' = 180° - 54.9°\)
\(\theta' = 125.1°\)
We can now use this new angle to determine what the second time will be. We can do so by setting the angle, \(125.1°\), equal to \(6°(x-9)\) and solving for \(x\):
\(125.1° = 6°(x-9)\)
\(\cfrac{125.1°}{6°} = \cfrac{\cancel{6°}(x-9)}{\cancel{6°}}\)
\(20.85 = x-9\)
\(x = 20.85 + 9\)
\(x = 29.85\;[\text{s}]\)
Therefore, we can determine that the two times in the cycle that the height is at \(24\;[\text{m}]\) are at \(\boldsymbol{18.15\;[\textbf{s}]}\) and \(\boldsymbol{29.85\;[\textbf{s}]}\).
You are at Risser's Beach, N.S. to search for interesting shells. At \(2\colon00 \;[\text{p.m.}]\) on June 19, the tide is in (ie the water is at its deepest). At that time, you find that the depth of the water at the end of the breakwater is \(15\;[\text{m}]\). At \(8\colon00 \;[\text{p.m.}]\) the same day when the tide is out, you find that the depth of the water is \(11\;[\text{m}]\). Assume that the depth of the water varies sinusoidally with time.
- Derive an equation expressing depth in terms of the number of hours that have elapsed since \(12\colon00 \;[\text{p.m.}]\) on June 19.
- Use your mathematical model to predict the depth of the water at \(7\colon00 \;[\text{a.m.}]\) on June 20.
- At what time will the first low tide occur on June 20?
- What is the earliest time on June 20 that the water will be at \(12.7\;[\text{m}]\) deep?
i. As a reference, our formula needs to be in the form \(f(x) = |a|\sin(k(x+d))+c\).
First, we can find the amplitude by subtracting the maximum (\(15\)) by the minimum (\(11\)) and then dividing by \(2\):
\(\text{amplitude} = \cfrac{\text{max} - \text{min}}{2}\)
\(\text{amplitude} = \cfrac{15 - 11}{2}\)
\(\text{amplitude} = \cfrac{4}{2}\)
\(\text{amplitude} = 2\)
Next, we can determine the period by taking into account the amount of time it takes to go from the max to the min. In this instance, it takes 6 hours from max (\(2:00\;[\text{p.m.}]\)) to min (\(8:00\;[\text{p.m.}]\)). We can then double this amount to find the full period length:
\(\cfrac{1}{2}\text{period} = 6\)
\(\text{period} = (6)(2)\)
\(\text{period} = 12 \; [h]\)
Then, we can determine the axis by finding the sum of the max and min points and dividing the value by \(2\):
\(\text{axis} = \cfrac{\text{max} + \text{min}}{2}\)
\(\text{axis} = \cfrac{15 + 11}{2}\)
\(\text{axis} = \cfrac{26}{2}\)
\(\text{axis} = 13\)
Assuming that the cycle begins at \(12:00\;[\text{p.m.}]\) (\(0\)), this means that this cycle ends at \(12:00\;[\text{a.m.}]\) (\(12\)).
Now, we can find \(k\) by dividing \(360°\) by the period (\(12\)):
\(k = \cfrac{360°}{\text{period}}\)
\(k = \cfrac{360°}{12}\)
\(k = 30\)
We can create a graph to make it easier to determine the rest of the values for our formula. First, we can create a table of values to help draw our graph:
| Time (h) | 2 | 5 | 8 | 11 | 14 |
|---|---|---|---|---|---|
| Depth (m) | 15 | 13 | 11 | 13 | 15 |
From looking at the graph we can determine the phase shift, \(d\), as \(11\).
We can put our formula together:
\(f(x) = a\sin(k(x+d))+c\)
\(f(x) = 2\sin(30°(x-11))+13\)
Therefore, we can determine that our formula is \(\boldsymbol{f(x) = 2\sin(30°(x-11))+13}\).
ii. In order to predict the water depth at \(\text{7:00}\;[am]\), we need to determine what to substitute \(x\) with:
\(x = 12:00\;[\text{a.m.}] + 7\)
\(x = 12 + 7\)
\(x = 19 \; [\text{h}]\)
We can now substitute \(x\) with \(19\) to predict the water depth:
\(f(19) = 2\sin(30°(19-11))+13\)
\(f(19) = 2\sin(30°(8))+13\)
\(f(19) = 2\sin(240)+13\)
\(f(19) = 2(-0.866025403)+13\)
\(f(19) = -1.732050808+13\)
\(f(19) = 11.27\;[\text{m}]\)
Therefore, we can determine that the depth of the water at \(7:00\;[\text{a.m.}]\) is \(\boldsymbol{11.27\;[\textbf{m}]}\).
iii. We can determine when the first low tide of June 20th will occur by setting the formula equal to \(11\) (the min) and solving for \(x\):
\(f(x) = 2\sin(30°(x-11))+13\)
\(11 = 2\sin(30°(x-11))+13\)
\(2\sin(30°(x-11)) = 11-13\)
\(2\sin(30°(x-11)) = -2\)
\(\cfrac{\cancel{2}\sin(30°(x-11))}{\cancel{2}} = \cfrac{-\cancel{2}}{\cancel{2}}\)
\(\sin(30°(x-11)) = -1\)
We can find the inverse of \(-1\) and simplify the rest of the expression to find \(x\):
\(30°(x-11) = \sin(-1)⁻¹\)
\(\cfrac{\cancel{30°}(x-11)}{\cancel{30}} = \cfrac{-90}{30}\)
\(x - 11 = -3\)
\(x = -3 + 11\)
\(x = 8\; [\text{h}]\)
We can now add \(8\) to the period (\(12\)) to get our final time:
\(T = 12 + 8\)
\(T = 20 = 8\colon00\;[\text{a.m.}]\)
Therefore, the first low tide of June 20th will occur at \(\boldsymbol{8\colon00\;[\textbf{a.m.}]}\).
iv. In order to determine the earliest time on June 20th that the water will be \(12.7\;[\text{m}]\) deep, we will use the sin formula, \(f(x) = 2\sin(30°(x-11))+13\). To make things more efficient, we will represent \(30°(x-11)\) as \(\theta\):
\(12.7 = 2\sinθ + 13\)
\(2\sin\theta = 12.7 -13\)
\(2\sin\theta = -0.3\)
\( \cfrac{\cancel{2}\sin\theta}{\cancel{2}} = \cfrac{-0.3}{2}\)
\(\sin\theta = -0.15\)
We can find the inverse of \(-0.15\) and add it to \(360\) to determine \(\theta\):
\(\theta = \sin(-0.15)^{-1}\)
\(\theta = -8.626926559°\)
\(\theta = 360° - 8.626926559°\)
\(\theta = 351.4^{\circ}\)
We can now set \(351.4°\) equal to \(30°(x-11)\) in order to determine the actual time:
\(351.4° = 30°(x-11)\)
\(\cfrac{351.4°}{30°} = \cfrac{\cancel{30°}(x-11)}{\cancel{30°}}\)
\(11.71 = x-11\)
\(x = 11.71 + 11\)
\(x = 22.71\)
We can determine the time by converting the number and decimal to hour and minute respectively for June 20:
\(\text{Hour} = (22)(1\;[\text{h}]) = 10 \colon 00\;[\text{p.m.}]\)
\(\text{Minute} = (0.71)(60\;[\text{min}]) = 42.6\;[\text{min}] \approx 43[\text{min}]\)
\(x = \text{June} \; 20, \; 10 \colon 43\;[\text{a.m.}]\)
We will round the time to the nearest full hour:
In order to determine additional time, we can use the Unit Circle. According to the Unit Circle, \(\sin\) produces negative results for Quadrants \(3\) and \(4\). As we've already taken Quadrant \(4\) into account , we can determine the second angle in Quadrant \(3\) by adding \(180°\) to \(8.6°\):
\(\theta = 180° + 8.6°\)
\(\theta = 188.6°\)
We can now use this new angle to determine what the second time will be. We can deteremine what this time will be by setting \(188.6°\) equal to \(30°(x-11)\) and solving for \(x\):
\(188.6° = 30°(x-11)\)
\(\cfrac{188.6°}{30°} = \cfrac{\cancel{30°}(x-11)}{\cancel{30°}}\)
\(6.286666667 = x - 11\)
\(x = 6.286666667 + 11\)
\(x \approx 17.3\)
We can determine the time by converting the number and decimal to hour and minute respectively for June 20:
\(\text{Hour} = (17)(1\;[\text{h}]) = 5 \colon 00\;[\text{a.m.}]\)
\(\text{Minute} = (0.3)(60\;[\text{min}]) = 18\;[\text{min}]\)
\(x = \text{June} \; 20 \; 5 \colon 00\;[\text{a.m.}] + 18\;[\text{min}]\)
\(x = \text{June} \; 20,\; 5 \colon 18\;[\text{a.m.}]\)
Finally, we can compare both times to determine which comes earlier:
Therefore, we can determine that the earliest time on June 20 that the water will be \(12.7\;[\text{m}]\) deep is \(\boldsymbol{5 \colon 18\;[\textbf{a.m.}]}\).