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The Wheel
goes Round
and Round!
Question #1

The Wheel goes Round and Round!
A wheel of a car has a radius of 26cm, and rotates at a rate of 20 revolutions
per minute. The wheel of the car is constantly touching the pavement.

A) Point X is situated on the wheel of the car and is touching the pavement.
Sketch a graph of point X making two complete revolutions as a function of
time. The graph begins at t = 0 seconds.

B) Write a sine and cosine equation for the function.

C) Determine one time, when point X 20cm above the pavement.

D) How long, in centimeters is one full revolution of the car wheel.

E) How many complete revolutions would the car wheel have to make, in order
for the car to travel one (1) kilometer.

Click the speaker to
hear the question.

The Wheel goes Round and Round! Solution

26cm is the radius
52cm above 52cm of the car wheel,
the pavement which is where
is where point point X is half the
Car Wheel
X is farthest distance between
from the 26cm where X is
pavement and touching the
radius
is also the pavement (0cm),
diameter of the and where X is
car wheel. farthest from the
0cm Point X pavement (52cm).

0cm is where point X is touching the
Ocm (touching the pavement)
pavement, X's closest point to the
and where Point X begins.
pavement.

The Wheel goes Round and Round! Solution
In order to find the period when
given the amount of revolutions in Period:
a certain amount of time, you 20 revolutions per minute
must divide the amount of time by
the number of revolutions. In this 20 revolutions per 60 seconds
case there were 20 revolutions in
one minute, one minute is also 60 60 secs secs
=3
seconds. So 60 seconds divided 20 revolutions rev
by 20 revolutions is 3
Period = 3
seconds/revolutions ( / = over)
which is the Period.
3 = Period

Now that we have figured out the diagram for the wheel and found the period
of the function. The revolution of the wheel being the function, we now have
enough information to answer part A) of the question. Part A) asks that we
sketch a graph of the function point X situated on the wheel, for two complete
revolutions.

The Wheel goes Round and Round! A) Solution
Maximum of the graph is 52,
Point X's journey on the Car's w heel

due to the fact that the highest
60
point on the wheel was 52cm
50

and the minimum of the graph
40
Height
of Point X
was 0, for 0cm (Touching the
30
(cm)
20
pavement).
10

The Amplitude of this graph is
0
0 2 4 6 8
26, as it is the distance from
Period (Seconds)
either the max or min value.
Period of the Graph is 3 meaning that one complete revolution will end at 3
seconds. Although, the period in seconds of the complete two revolutions
starting at zero and moving to the right are, 0.75 secs, 1.5 secs, 2.25 secs, 3
secs, 3.75 secs, 4.5 secs, 5.25 secs, 6 secs. In order to sketch two complete
revolutions the graph is just basically duplicated from (0 to 3) to (3 to 6). A sine
and cosine graph duplicates itself every period.
The sinusoidal axis is located at 26, the reason being that 26 is the radius of
the wheel, meaning it is have the distance between 0 and 52. (expressed by
the solid red line).

The Wheel goes Round and Round! B) Solution
B) Write a Sine And Cosine equation for the function of Point X.
To begin, we must Create charts for both Sine and Cosine so that once those
charts are filled the equation is just pieced together by the numbers in the
chart.

Sine Cosine
A = 26 A = -26
2! 2!
B= 3 B=
3
C=!3 C= 0
4

D = 26 D = 26

Sine Equation: Cosine Equation:
3
h=-26cos[ 23 t] + 26
2! !
h=26sin[ (t- )] +26
4
3

The Wheel goes Round and Round! B) Solution

A= Amplitude (The distance from either the Max or Min value, in this case it is
26, as it is 26 from the sinusoidal axis to either the max (52) or min (0) value).

B= Period Determiner (2Π divided by the period of the graph, in this case 3)

C= Phase Shift ( It is the horizontal shift that occurs if a cosine or sine equation
does not begin at a maximum or minimum on y=0. In this case the Cosine
equation has a minimum value on y =0, although the Sine equation does have
a phase shift, forward 0.75 or 3 seconds so that y=0 shifts forward 0.75 or 3
4 4
seconds so the Sine equation begins at a maximum. Although if y=0 is shifted
forward C becomes negative.

D= Vertical Shift ( Number on the sinusoidal axis, as it is technically the new x-
axis, it is half way between the minimum and maximum, in this case it is 26.

The Wheel goes Round and Round! C) Solution
C) Determine one time, when Point X is 20cm above the pavement.

There are two ways of trying to figure out when Point x on the wheel is going to
be over 20cm, one with the sine equation and one with the cosine equation. I
will begin by showing you the sine equation method.

" 2! $ 3 %# You must begin this problem by
20 = 26sin ) ' t & ( * + 26 plugging in the appropriate
-3 + 4 ,. values into the equation.

I brought over the 26 from the
" 2! $ 3 %# right side of the equation to the
&6 = 26sin ) ' t & (* left side to add the two values
-3 + 4 ,. together. 20+(-26)=-6.

Next I am dividing each side of
" 2! $ 3 % #
26sin ) ' t & ( * the equation by 26, the right
- 3 + 4 ,.
&6 side would reduce, and the left
side becomes -0.2307.
=
26 26
I took the arch sine of (-
$ 2! & 3 '%
#1 0.2307) on the left side, so that
sin " (#0.2307 ) = sin * ( t # )+
.3 , 4 -/ the sin on the right side of the
equation would reduce.
I then after I received the
" 2! 3 %#
$ arch sine of -0.2307, divided
' t & (* each side by 2 pi over 3, so
)3 4 ,.
&0.2329 - + that the right side would
= reduce and I would have to
2! 2! divide the arch sine of
3 3 (-0.2307), which is (-0.2329)
by 2 pi over 3.


-0.2329 divided by 2 pi over 3
3 equals -0.0124. Then I brought
!0.0124 + = t over the -3/4 on the right side to

4 the left, so I can isolate t and
derive a value for t.

t = 0.7376, which is the amount

0.7376 = t of time that it takes point X
(0cm) to reach 20 cm above the
pavement.

It takes 0.7376 seconds for Point X (starts at 0cm) to reach 20 cm above the
pavement.

The Wheel goes Round and Round! D) Solution

D) How long is one revolution
of this car wheel.

One (1) complete revolution is the circumference of the wheel. So to
figure out this question we will use the circumference formula.

C = πd
d = 2r = 2(26cm) = 52cm
Now that we have found the diameter of the car wheel, we can now find
the circumference of the wheel.

C = πd

C = π (52cm)
C ≈ 163.3628cm
One revolution of the car wheel measures 163.3628 centimeters.

The Wheel goes Round and Round! E) Solution
E) How many complete revolutions would the wheel have to take to reach one
(1) kilometer (Km).

C ≈ 163.3628
Km = 1000m
m = 100cm

If one (1) kilometer is (1000) one thousand meters, and one meter is (100) one
hundred centimeters, then C ≈ 1.633628m. You must change it to meters so
that you can divide a kilometer by the circumference of the wheel to figure out
how many revolutions the wheel must take.
1000m
# of revolutions = 1.603628m

# of revolutions = 612.1345

That is not the answer, as the question asks for the # of Complete revolutions,
so the # of complete revolutions is 613, as you must round to the next highest
number which is 613.

The Next
Scrooge!
Question #2

The Next Scrooge!
Bobby wants to invest his savings at a bank. Right now he
keeps his money at home, but he wants his money to gain
interest, so he can increase his savings. Bobby has (3)
three banks to choose from, the Royal Bank of Canada
(RBC), the Toronto Dominion bank (TD), and Scotia Bank.
Bobby has $5631.00 to invest. Bobby wants to choose the
bank that would increase his savings the most from
interest. The Royal Bank of Canada has offered Bobby an
interest rate of 8.0%, bi-annually. The Toronto Dominion
Bank has offered Bobby an interest rate of 3.0%, monthly.
The Scotia Bank has offered Bobby an interest rate of 5%,
quarterly. Bobby is asking for your help, he wants you to
find out which bank would increase his savings the most Click the speaker to
over two years. So, are you going to help Bobby? hear the question.

The Next Scrooge! Solution
We will begin with the Royal Bank of Canada, but to begin we must use the
Compound interest formula, which is :

tn
A = P (1 + )
r
n
Before I solve the problem I am going to explain to you what each letter is
and what number it is going to be in accordance to the question.
A: The amount of money you are going to receive after two years.
P: The amount of money (Principle) you begin with, in this case $5631.00.
r: The amount of interest the bank is joining to be giving you, in this case
it is 0.08 (8%).
n: The amount of compounding periods in a year, in this case it is two (Bi-
annually/twice a year).
t: The amount of years the principle is going to be compounded by the
interest rate, in this case it is 2, for two years.

Royal Bank of Canada
I began this problem by taking the
0.08 2*2
A = $5631.00 (1 + ) Compound interest formula and inserting
the appropriate values into the equation
2
that were given in the question.
I multiplied the exponents together so I
4
A = $5631.00 (1 + 0.04 ) can work with the one exponent and I
divided the interest rate by the amount of
compounding periods.
I began working out the equation by
A = $5631.00 (1.1699 ) adding the 1 and 0.04 and then I worked
out 1.04 to the exponent of 4 to equal
1.1699.
I worked out the last little bit of the
A = $6587.47 equation by multiplying $5631.00 by
1.1699 to equal $6587.47.
So, over two years of
$5631.00 being
$6587.47 ! $5631.00 = $956.47 compounded bi-annually at
8%, Bobby collected
$956.47 worth of interest at
RBC.

Toronto Dominion Bank
2*12 I began this problem by taking the
! 0.03 " Compound interest formula and inserting
A = $5631.00 #1 + $ the appropriate values into the equation
12 &
% that were given in the question.
I multiplied the exponents together so I
24 can work with the one exponent and I
A = $5631.00 (1 + 0.0025) divided the interest rate by the amount of

1.0618.
1.0618 to equal $5978.75.
$5631.00 being
$5978.75 ! $5631.00 = $347.75 compounded monthly at
3%, Bobby collected
TD.

Scotia Bank
I began this problem by taking the
2*4
! 0.05 " Compound interest formula and inserting
A = $5631.00 #1 + $ the appropriate values into the equation
4&
% that were given in the question.

8 I multiplied the exponents together so I
A = $5631.00 (1 + .0125 ) can work with the one exponent and I
divided the interest rate by the amount of

1.1045.
1.1045 to equal $6219.36.
$5631.00 being
$6219.36 ! $5631.00 = $588.36 compounded quarterly at
5%, Bobby collected
Scotia Bank.

The Next Scrooge! Solution
So, now that we have found the amount of interest Bobby could be
making at each bank, which bank should Bobby invest his money at?
Well, it is obviously the Royal Bank of Canada. Due to the fact that
Bobby would be making $956.47 of interest after two years. Whereas if he
chose the TD Bank, he would only be making $347.75 and if he chose the
Scotia Bank he would have only made $588.36 of interest.
The Bank that Bobby had invested in:

The RBC

The Planet of
Zorbia
Question #3

The Planet of Zorbia
Dear Human,
The Planet of Zorbia is situated 10 light years away from earth. There are
creatures inhabiting that planet, those creatures are named Zorbians. The
leader of the planet Zorbia is calling on humans to help the planet Zorbia
analyze its population and population rates. There are 4600000 zorbians living
on the planet Zorbia. How many Zorbians would there be on the planet Zorbia
in 25 years, if the rate of growth is 2.72% a year. Zorbians also want to know
another piece of information regarding their planet. There are two large cities
on the planet of Zorbia, Xora and Yorp. The leader of Zorbia wants to pick a
Capital City for the Planet, although he wants the city with the highest growth
rate, so that the city will always grow and stay the largest city on the planet of
Zorbia. Xora is a city located near the equator of the planet Zorbia and has a
population of 256,230, but 7 years ago it had a population of 227,351. Yorp on
the other hand is situated in the Northern hemisphere of the planet Zorbia and
has a population of 239,973, but 7 years ago it had a population of 208,659.
The leader of Zorbia has given all the information that he has on the planet and
cities of Zorbia, so the rest is up to you.
Your favorite Zorbian,
hear the question.
Xavi Zorb

The Planet of Zorbia Solution
For the first question, the population of the Planet Zorbia is 4,600,000 and has
a growth rate of 2.72% a year. The question is how many Zorbians would there
be in 25 years? For this question we are going to be using this equation, as
this equation represents population growth.

y
P = Po(mod el)
P: Is the final population, in this case it is the population we are trying to find
that is 25 years away.
Po: Is the current population, in this case it is the population we are starting
with, 4,600,000.
Model: Is 1 + 0.0272, because it is the full population (1, which is 100%) plus
the annual growth of (0.0272, which is 2.72%).
Y: Is the amount of years that are to be elapsed in order to get the final
population, in this case it is 25 (25 years).

You begin this problem by
25 plugging in the appropriate values
P = 4600000(1 + 0.0272) into the equation that we were
using from the previous slide.

You then calculate (1.0272) to the
25
P = 4600000(1.0272) power of 25, to give you the value
of (1.9560), so you are able to
multiply 4,600,000, to achieve the
value of P.

You then add the two model
P = 4600000(1.9560) numbers together to achieve
1.0272.
After you multiply 4,600,000 by
P = 8,997, 700 (1.9560), you achieve the total
population.

For the (First Part) of the second question, the city of Xora has a population of
256,230, but 7 years ago it had a population of 227,351. What is its growth
rate? We are using the same equation that we used to solve the first question,
and the answer will be showed in log and in e, I will do e first.

You must begin this problem by
7
256, 230 = 227,351(m) plugging in the appropriate
values into the equation.

7
256, 230 227,351(m) You must divide 227,351 from
each side, so that the m^7
= becomes isolated, so then you
227,351 227,351 are able to solve for m.

You must then add Ln to each side,
so that m can later be isolated, when
ln1.1270 = 7 ln m you add the Ln, the exponent on the
m, in this case 7 moves in front of
the Ln symbol.

You must then multiply each side by
1/7 (1 over 7) so that the 7 on the
0.0171 = ln m right side reduces, and then you
must multiply 1.1270 by Ln, and then
multiply by 1/7 to equal m.

Due to the fact that we are using the
e method, 0.0171 is put to the
0.0171
e =m exponent of e, which equals m.
0.0171 is the actual percentage of
growth (1.71%).

Here is the log version on how to solve this equation:
7 plugging in the appropriate
256, 230 = 227,351(m) values into the equation.

You must divide 227,351 from
7
256, 230 227,351(m) each side, so that the m^7
227,351 227,351 are able to solve for m.

You must then add Log to each
side, so that m can later be
log1.1270 = 7 log m isolated, when you add the Log,
the exponent on the m, in this
case 7 moves in front of the Log
symbol.


0.0074 = log m 1/7 (1 over 7) so that the 7 on the
right side reduces, and then you
must multiply 1.1270 by log, and
then multiply by 1/7 to equal m.

log method, 0.0074 is put to the
0.0074
10 =m exponent of base 10 (log), which
equals m. You must calculate
10^0.0074 in order to receive a
growth rate of 0.0171 or 1.71%.

The city of Xora has a population growth rate of 0.0171 or 1.71%

For the (Second Part) of the second question, the city of Yorp has a population
of 239,973, but 7 years ago it had a population of 208,659. What is its growth
rate? We are using the same equation that we used to solve the first question,
and the answer will be showed in log and in e, I will do e first.

7
239,973 = 208, 659(m) plugging in the appropriate

7
239,973 208, 659(m) You must divide 208,659 from
each side, so that the m^7
208, 659 208, 659 are able to solve for m.

You must then add Ln to each side,
so that m can later be isolated, when
ln1.1501 = 7 ln m you add the Ln, the exponent on the
m, in this case 7 moves in front of
the Ln symbol.


0.0200 = ln m 1/7 (1 over 7) so that the 7 on the
must multiply 1.1501 by Ln, and then
multiply by 1/7 to equal m.

e method, 0.0200 is put to the
0.0200
e =m exponent of e, which equals m.
0.0200 is the actual percentage of
growth (2.00%).

Here is the log version on how to solve this equation:

7 You must begin this problem by
239,973 = 208, 659(m) plugging in the appropriate

You must divide 208,659 from
7
239, 973 208, 659(m) each side, so that the m^7
208, 659 208, 659 are able to solve for m.

You must then add Log to each
side, so that m can later be
log1.1501 = 7 log m isolated, when you add the Log,
the exponent on the m, in this
case 7 moves in front of the Log
symbol.


0.0087 = log m 1/7 (1 over 7) so that the 7 on the
must multiply 1.1501 by log, and
then multiply by 1/7 to equal m.

log method, 0.0087 is put to the
0.0087
10 =m exponent of base 10 (log), which
equals m. You must calculate
10^0.0087 in order to receive a
growth rate of 0.0200 or 2.00%.

The city of Yorp has a population growth rate of 0.0200 or 2.00%

Now that you have found the answers to the questions that the leader of Zorbia
has asked of you, you are ready to present your answers.

For the first question, the population of the planet Zorbia in 25 years at a
2.72% growth rate, is going to be: 8,997,700 (Zorbians).

For the second question, the growth rate for Xora and Yorp is:

Xora: 1.71% Yorp: 2.00%
So the capital city of Zorbia should be Yorp, as it has the highest growth rate.

Trig Mania

This trigonometric problem, is a problem that spans the better part of the
trigonometric identities unit. It showcases the multiple identity formulas and
also requires careful though in how to workout the problem as there are many
ways of doing so. So to not keep you waiting here is the question:
hear the question.

1 1
+
2
2csc ! 1 " cos ! 1 + cos !
=
2 4 4
1 " 2sin ! cos ! " sin !

Trig Mania! Solution

1 + cos ! + 1 " cos ! We will work on the right side of the
2csc2 ! identity. 1 over 1-cosθ + 1 over 1+cosθ
1 " cos 2 ! becomes 1+cosθ + 1-cosθ over 1-
=
2
cos 4 ! " sin 4 !
1 " 2sin ! 2sin²θ so that there is a common
denominator.

2 The two cosθ’s reduce and 1+1=2.
2
2 csc ! 1-cos²θ is a trigonometric identity so
2
sin !
= it becomes sin²θ. It is now 2 over
2 4 4
1 " 2sin ! cos ! " sin ! 2sin²θ


2csc2 ! 2csc2 ! Cos²θ + sin²θ becomes 1, it is a
= trigonometric identity.
2 2 2
1 " 2sin ! (cos ! " sin ! )(1)

2 over sin²θ after
dividing is nicely
2csc2 ! 2csc2 ! converted into
= 4
2csc²θ. Cos4 θ – sin
2 2 2 2 2
1 " 2sin ! (cos ! " sin ! )(cos ! + sin ! ) θ is nicely multiplied
out to (cos²θ -
sin²θ)(cos²θ+sin²θ)


2 2
2csc ! 2csc ! Cos²θ becomes 1 - sin²θ, it is
= another trigonometric identity.
1 " 2sin ! (1 " sin 2 ! " sin 2 ! )(1)
2

2 2
2 csc ! 2 csc ! The two -sin²θ are added
= together to become -2sin²θ.
2 2
1 " 2sin ! (1 " 2sin ! )(1)


2 2 (1-2sin²θ) is multiplied by
2csc ! 2csc ! one, obviously nothing
= changes. Now both sides are
2 2
1 " 2sin ! 1 " 2sin ! equal to one another. Do not
forget Q.E.D..

Q.E.D.

MY THOUGHTS ON D.E.V.
Well, I am finally done this project. This was somewhat of a challenge
for me, especially when it came to posting these slides on the DEV blog,
because I have never imported slides to slideshare and bliptv and then post
them onto the DEV blog. Although, it worked out for the best and I am here
finally done this project that was like hiking over Mount Everest. I would not
say it was a very hard project, in terms of making the questions and solutions.
The hardest part was the posting to the blog, as computer illiterate people like
me will probably agree. All in all I liked this project, because it expanded your
imagination in how to create fun and well thought out questions, and how to
explain those question in the easiest way possible. So personally, I believe this
project is good to have and this should be used for every 40s math course in
DMCI. It makes you appreciate how much time teachers spend in creating
questions for students to do in class.

Dinoppc40sw07

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