Kotlin Multiplatform & Compose Multiplatform - Starter kit for pragmatics
DEV
1. Problem #1
You are at a museum with an antique bicycle exhibit that allows visitors to pedal the bikes. There
is a penny farthing with the highest point of the front wheel at 4 feet off the ground and the
bottom of the wheel is on a 6 inch stand off the ground to allow you to pedal. The boarding point
is 45 counterclockwise from the top of the wheel. The ramp leading to the boarding point is at an
incline of 20 . When you start pedaling, you pedal at a rate of 5 mph clockwise.
Complete the following:
-write a sine or cosine equation
-find the amplitude
-find the period
-find the vertical shift
-find the horizontal shift
-find the “b” value
-find the height of the boarding point
-find the length of the ramp
-draw a graph for your equation
2. How to find the amplitude:
In order to find the amplitude, you must take the highest point on the wheel which is 4 feet tall and take away the 6
inches that the wheel is off the ground, giving you 42 inches. This is the diameter, and you need the radius for the
amplitude so divide the diameter by 2. The work below shows how to find the radius:
Highest point of the wheel=4ft=48 inches
48in-6 inch stand=42in diameter
=radius of 21in
3. How to find the period and “b” value:
The period is given to you in miles per hour but you need seconds per revolution so you need
to convert it. Change rate into inches by multiplying it by 5280 feet in a mile and then 12
inches in a foot.
This gives you a rate of inches per hour so divide it by 3600 seconds, this is the number of
seconds in one hour.
This just gives you a rate, you want the time it takes for the wheel to make one complete
revolution. To do this, use the equation distance=rate*time. You have the rate and in order
to find the distance, you must find the circumference of the circle. C=2πr so take the radius
and plug it into the equation.
Rearrange the distance equation to make t=d/r
This is the period of the wheel.
To find the “b” value, use the equation Period=2π/b. Rearrange the equation to find the “b”
value
4. How to find the vertical shift
The vertical shift is quite simple, you just have to take the amplitude which is 21in
and add the 6 inches from the stand.
21in+6in=27in
5. How to find the horizontal shift:
option #1
It takes 1.499 seconds to go around the wheel. Find the fraction of the circle that
the boarding shift is away from the bottom of the wheel.
45 from the top
half of the circle=180
180-45=135 from the bottom of the circle
which is of the whole circle.
Multiply that fraction by the time it takes to go around the circle and that is the
horizontal shift.
6. How to find the horizontal shift:
option #2
You can also find the horizontal shift by finding the arc length from the bottom of
the circle to the boarding point. Use the fraction of the arc and multiply it by the
circumference to find the length of the arc.
This is the distance of the arc so use the distance formula to find the time it takes to
travel this length. D=rt
This is the horizontal shift, which is the same value that you found the other way.
7. Write an equation:
The standard equation for a cosine wave is: y=k+a*cosb(X-h) or y=k+a*sinb(X-h)
for a sin wave. You have all the information found so now just plug it in! k is the
vertical shift which is 27in. A is the amplitude which is 21. B is the horizontal
stretch and equals . Lastly, h equals the horizontal shift which is .5621.
When writing the equation, don’t forget that the cos is negative because the
graph is starting at the bottom and positive cos graphs start at the top of the
wheel. Also, the horizontal shift is negative because you are shifted right when
you start pedaling. So the equation is:
8. How to find the height of the boarding point
Option #1: Use the equation that you found, type it into your graphing calculator,
look at time=0. This is the boarding height which should equal 41.84in.
Option #2:Use right triangle trigonometry. You know the angle from the top of the
circle to the boarding point, and you know the length from the center of the
circle to the boarding point which is the radius. Use the cosine of that angle to
find X.
21*cos(45)=14.84in. Then add the radius and stand height to
that to get the height of the boarding point.
14.849+21+6=41.84in
9. To find the ramp length:
Use right triangle trigonometry again to find the length of the ramp. You now know
the height of the boarding point and the angle of incline so use a sin equation
Rearrange to get:
Don’t forget to make sure that your calculator is in degrees!
10. Now graph the equation:
To graph the equation, first look at the
vertical shift which was 27 inches and
mark the midline on the graph. Next, use
the amplitude of 21 to help you
determine the max and min of the graph.
Since the boarding point was not at the
bottom of the wheel, there is a
horizontal shift to the left. Just move the
start of the graph (which is at the min
because it is a –cos graph) to the left
.5621 seconds. Finally, use the period
that you found earlier to graph the
horizontal stretch. The period was 1.499
seconds so one full revolution should
occur every 1.499 seconds.
12. The first step is to look and see if there are any identities that you can see right away.
The equations in red are Pythagorean identities. The identities are sec²θ-1=tan²θ and cot²θ+1=
csc²θ which can be substituted with tan²θ and csc²θ to give you:
Keep looking for different identities to simplify the equation until you have reached an equation that
is identical on both sides.
The red is an example of the half angle formula cos²= 1/2(1+cos2θ)
13. The red represents a double angle formula where 2cos²θ-1=cos2θ after substituting
cos2θ, you get:
There is only one more step until the equation is solved
The double angle formula for tan is highlighted in red. tan2θ=(2tanθ)/(1-tan²θ) which
leaves the final equation of:
14. Problem #3
A ferris wheel reaches a height of 431 feet. The lowest point of the wheel is
5 feet from the ground. The boarding point is located 160 counter clockwise
from the top of the wheel. The wheel spins counter clockwise at a rate of 9
miles per hour. The ramp to the boarding point has an incline of 5 .
Find:
-amplitude
-period
-vertical shift
-horizontal shift (in seconds)
-height of the boarding point
-length of the ramp
15. How to find the amplitude:
• The amplitude is the measurement from the midline to the highest
point on the wheel
• First, subtract the lowest point of 5 feet from the highest point of
431,giving you the total height of the ferris wheel, 426 feet.
431-5=426
• Then, divide the wheel’s height by 2, giving you the amplitude of
213 feet
426/2=213
16. How to find the vertical shift:
• Add the amplitude of 213 feet to the lowest point of the
ferris wheel of 5 feet
213+5=218
• This show how much the midline had shifted from the
ground
17. How to find the period:
• First, find the circumference of the ferris wheel by using the
equation c=2πr.
2π213=1338.32 feet
• Then convert the rate of 9 miles per hour into feet. There is
5,280 feet in a mile so you need to multiply 9 by 5,280.
9 5,280=47,520 feet per hour
• To convert the rate into feet per second, you need to divide
by 60 two times.
47,520/60=792 feet per minute, 792/60=13.2 feet per second
• To find the time it takes to make one period, you need to divide
the circumference by the rate
1,338.32/13.2=101.39 seconds, giving you the period
18. How to find the horizontal shift:
• The boarding point had shifted π/9, which is 1/18 of the wheel
• You need to multiply the period by (1/18)
101.39 (1/18)=5.63 seconds
• Because the period was already converted into seconds, you
do not need to convert the horizontal shift
19. How to find the height of the boarding point:
• To find the height of the boarding point, you need to find the b value
• You need to divide the period by 2π
2π/101.39=.062
• You then plug in the values you’ve found into the equation
y=k+a*cosb(X-h)
y=-213cos.062(x-5.63)+218
• By plugging in 0 for x, you will find the boarding point when x is 0
y=17.845 feet
20. How to find the length of the ramp:
• Use soh cah toa find the length of the ramp
• Use the equation sin5=17.845/X
• To get x out from the bottom, multiply both sides by x
xsin5=17.845
• Then divide both sides by sin5
x=17.845/sin5
x=204.75 feet
24. Find where sin √3/2 is on the unit circle
π/3, 2π/3
Set the right side equal to the left side
Θ+5π/6=π/3
Subtract 5π/6 from both sides
Θ=-3π/6
Do on both sides
Θ+5π/6=2π/3
Θ=-π/6
25. Problem #5
• Cosecant of theta has the same values as sine of theta. These
values for theta are: 0, π/2, π, 3π/2, 2π and for y: 0, 1, 0, -1, 0.
• These are all the changes that need to be made
• Theta is multiplied by 1/3 so use 2π/b. 2π/1/3 then multiply
by the reciprocal of 1/3 which is just 3. Period is 6π
• Amplitude is the number being multiplied by the csc so for
this problem it is 4
• The phase shift is the number being added or subtracted to
the theta. The shift is always opposite of the symbol +/-. The
phase shift is -3π/4
• The vertical shift is either at the very beginning or end of the
equation which is being added or subtracted. The vertical
shift is +3 or just 3
Y= 4csc1/3(θ+3π/4)+3
26. Problem 5 continued
• First add the changes made to the y-values
• Multiply the values by 4 (amplitude)
• New values are: 0, 4, 0, -4, 0
• Add 3 to account for the vertical shift
• Final y-values are: 3, 7, 3, -1, 3
• Now onto the x-values
• To get the period, multiply every value by the reciprocal of
1/3 which is 3
• New values are: 0, 3π/2, 3π, 9π/2, 6π
• Next subtract 3π/4 from each value to account for phase
shift
• Do not forget to use common denominators
• Final x-values are: -3π/4, 3π/4, 9π/4, 15π/4, 21π/4
27. Problem 5 continued
When you go to graph, here are some things you should
know
• You can make the y-axis as spaced out as you want it
to be but you should have a maximum of at least 9
and a minimum of at least -3
• The x-axis can also be as spaced out as you want but
the spaces between each value should be consistent
• Do not forget to place the u-shapes (what I call them)
at the maximum and minimum of the sine graph
because the equation calls for cosecant
Look at that, boom you are done with my help. Good luck in
the future young padawan.
28. Problem 6: the last voyage
These problems are actually very easy if you pay attention to what you
are doing. We want cos(θ+2π/3) by itself so here are the steps to take
• Subtract 7 from both sides, on the left we need common
denominators so instead of 7 we subtract 35/5. We are left with
√2/5 on the left side and 2/5cos(Θ+2π/3)
• To get cosine by itself, multiply both sides by the reciprocal of 2/5
which is 5/2. This makes the equation 5√2 = cos(Θ+2π/3)
10
• 5/10 reduces to ½ so the left side is simplified to √2/2
Solve an equation for 0≤Θ<2π
35+√2= 7+ 2/5cos(Θ+2π/3)
5
29. Problem 6 continued
• Ignore the +2π/3 for a moment and see where cosine (x) is √2/2
• These 2 places are at π/4 and 7π/4
• Set Θ+ 2π/3 equal to each value previously found above
• 2 separate equations!
• Subtract the 2π/3 from Θ to get your final values
• Do not forget to get common denominators
• In the equation with π/4, the equation ends with a negative
value (-5π/12) so you need to add the period which is 2π in
order to place in between the range to which we are solving.
• The final two values should be 19π/12 and 13π/12
Bam, you are all done solving. Way to go young padawan!
30. Hannah’s Reflection:
The reason I chose the problems I did changed from last tri. Last tri, I chose the units I
thought I did very well in and understood, but this tri, I tried something different and chose a
problem from a unit that I was not very comfortable with. I was pretty confident with the
graphing unit but the identity unit was very confusing for me. I decided to do a problem from
that unit to give me extra practice for the exam on a topic that was harder for me. Since these
units were a little more difficult to understand and combine, I think it shows my understanding
of them that I actually was able to combine them and explain the solutions and steps I took to
find the answers.
This project is sometimes stressful especially since it is close to the end of the year, it is hard
to stay on track and many teachers give projects at the end of the year, but I am still thankful
for this project. It helps me review for the exam in a way that is more helpful than just staring
at problems to “study”. This project makes me go back to the units and actually look through
things and go over how to do problems which is very helpful for the exam. It also helps
because I try to find more than one way to solve the problems. It is a very valuable project
and Mr. Jackson should continue to use it. However, again, I still get caught up on the ways to
get bonus points with the presentation. I feel that I have to go crazy with different media tools
to get a good grade.
31. Amber’s Reflection:
I chose problems from the first two units because those are the units I
struggled with the most. I thought that practicing these units would help
my understanding. Teaching these problems helped me perfect the skills
even better and helped me to truly understand the material. In the ferris
wheel problem, I chose to add both a phase and vertical shift to make the
problem more complex. This made it possible for me to use everything I
have learned in one problem. This project was very helpful in my
understanding. It helped me review material from previous units and better
understand the material I was unclear about.
32. Hilary’s Reflection
During the DEV project, I created one problem from the measurement
unit and one from the graphing unit. I chose these two for my problem
set because I performed relatively well in both units and I had a full
understanding of how to create these problems. I also thought I could
explain my thought process throughout the problem so others could
understand. These problems provide an overview of my knowledge
because I did not take the easy way out. I created these problems
because they were not simple, get it done in 30 seconds problems. I
wanted to challenge myself into working on problems to possibly
expand my knowledge. As I entered this project, I want to at least score
into the knight category, which I believe I did. From this project I
learned that I am better with creating problems than I thought I was.
Making sure that the problem would actually solve was a major challenge
to work through. I also learned that I can solve problems similar to the ones
I used in the project without struggles. I realized that I am smarter than I
thought I was which happens to be a big confidence booster. I believe
this is a great project that tests your actual knowledge on various units
and it really helps educate oneself on the course of precalc trigonometry.
“The one exclusive sign of thorough knowledge is
the power of teaching” -Aristotle
