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D.E.V Project

The Wheelgoes Roundand 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 revolutionsper 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 oftime. 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 orderfor the car to travel one (1) kilometer.

Click the speaker tohear the question.

The Wheel goes Round and Round! Solution

Ocm (touching the pavement)and where Point X begins.

52cm abovethe pavementis where pointX is farthestfrom thepavement andis also thediameter of thecar wheel.

26cm is the radiusof the car wheel,which is wherepoint X is half thedistance betweenwhere X istouching thepavement (0cm),and where X isfarthest from thepavement (52cm).

0cm is where point X is touching thepavement, X's closest point to thepavement.

Car Wheel

radius

Point X

52cm

0cm

26cm

Period:20 revolutions per minute

20 revolutions per 60 seconds

60 secs secs20 revolutions rev

In order to find the period whengiven the amount of revolutions ina certain amount of time, youmust divide the amount of time bythe number of revolutions. In thiscase there were 20 revolutions inone minute, one minute is also 60seconds. So 60 seconds dividedby 20 revolutions is 3seconds/revolutions ( / = over)which is the Period. 3 = Period

Period = 3

The Wheel goes Round and Round! Solution

= 3

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

The Wheel goes Round and Round! A) SolutionPoint X's journey on the Car's w heel

0

10

20

30

40

50

60

0 2 4 6 8

Height of Point X

Period

Period of the Graph is 3 meaning that one complete revolution will end at 3seconds. Although, the period in seconds of the complete two revolutionsstarting at zero and moving to the right are, 0.75 secs, 1.5 secs, 2.25 secs, 3secs, 3.75 secs, 4.5 secs, 5.25 secs, 6 secs. In order to sketch two completerevolutions the graph is just basically duplicated from (0 to 3) to (3 to 6). A sineand cosine graph duplicates itself every period.

The sinusoidal axis is located at 26, the reason being that 26 is the radius ofthe wheel, meaning it is have the distance between 0 and 52. (expressed bythe solid red line).

(Seconds)

(cm)

Maximum of the graph is 52,due to the fact that the highestpoint on the wheel was 52cmand the minimum of the graphwas 0, for 0cm (Touching thepavement).

The Amplitude of this graph is26, as it is the distance fromeither the max or min value.

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 thosecharts are filled the equation is just pieced together by the numbers in thechart.

Sine Cosine A = 26 A = -26

B = B =

C = C = 0

D = 26 D = 26

Sine Equation:

h=26sin[ (t- )] +26

Cosine Equation:

h=-26cos[ t] + 26

3

4!

2

3

!

2

3

!

2

3

!

3

4

2

3

!

A= Amplitude (The distance from either the Max or Min value, in this case it is26, 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 equationdoes not begin at a maximum or minimum on y=0. In this case the Cosineequation has a minimum value on y =0, although the Sine equation does havea phase shift, forward 0.75 or seconds so that y=0 shifts forward 0.75 orseconds so the Sine equation begins at a maximum. Although if y=0 is shiftedforward 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.

4

3

4

3

The Wheel goes Round and Round! B) Solution

The Wheel goes Round and Round! C) SolutionC) 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 tobe over 20cm, one with the sine equation and one with the cosine equation. Iwill begin by showing you the sine equation method.

2 320 26sin 26

3 4t

!" #$ %= & +' () *

+ ,- .

You must begin this problem byplugging in the appropriatevalues into the equation.

2 36 26sin

3 4t

!" #$ %& = &' () *

+ ,- .

I brought over the 26 from theright side of the equation to theleft side to add the two valuestogether. 20+(-26)=-6.

The Wheel goes Round and Round! C) Solution

I took the arch sine of (-0.2307) on the left side, so thatthe sin on the right side of theequation would reduce.

( )1 2 3sin 0.2307 sin

3 4t

!"# $ %& '# = #( )* +

, -. /

2 3

3 40.2329

2 2

3 3

t!

! !

" #$ %&' () *& + ,- .

=

I then after I received thearch sine of -0.2307, dividedeach side by 2 pi over 3, sothat the right side wouldreduce and I would have todivide the arch sine of(-0.2307), which is (-0.2329)by 2 pi over 3.

2 326sin

3 46

26 26

t!" #$ %

&' () *& + ,- .=

Next I am dividing each side ofthe equation by 26, the rightside would reduce, and the leftside becomes -0.2307.

30.0124

4t! + =

The Wheel goes Round and Round! C) Solution

-0.2329 divided by 2 pi over 3equals -0.0124. Then I broughtover the -3/4 on the right side tothe left, so I can isolate t andderive a value for t.

0.7376 t=t = 0.7376, which is the amountof time that it takes point X(0cm) to reach 20 cm above thepavement.

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

The Wheel goes Round and Round! D) Solution

D) How long is one revolutionof this car wheel.

One (1) complete revolution is the circumference of the wheel. So tofigure 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 findthe circumference of the wheel.

C = πd

C = π (52cm)

C ≈ 163.3628cm

One revolution of the car wheel measures 163.3628 centimeters.

E) How many complete revolutions would the wheel have to take to reach one(1) kilometer (Km).

C ≈ 163.3628Km = 1000m

m = 100cm

If one (1) kilometer is (1000) one thousand meters, and one meter is (100) onehundred centimeters, then C ≈ 1.633628m. You must change it to meters sothat you can divide a kilometer by the circumference of the wheel to figure outhow many revolutions the wheel must take.

# of revolutions =

# 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 highestnumber which is 613.

1000m1.603628m

The Wheel goes Round and Round! E) Solution

The NextScrooge!

Question #2

The Next Scrooge!Bobby wants to invest his savings at a bank. Right now hekeeps his money at home, but he wants his money to gaininterest, 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 thebank that would increase his savings the most frominterest. The Royal Bank of Canada has offered Bobby aninterest rate of 8.0%, bi-annually. The Toronto DominionBank 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 tofind out which bank would increase his savings the mostover two years. So, are you going to help Bobby?


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

Before I solve the problem I am going to explain to you what each letter isand 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 caseit 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 theinterest rate, in this case it is 2, for two years.

( )1nt

r

nPA = +

Royal Bank of Canada

( )4

$5631.00 1 0.04A = +

I began this problem by taking theCompound interest formula and insertingthe appropriate values into the equationthat were given in the question.

I multiplied the exponents together so Ican work with the one exponent and Idivided the interest rate by the amount ofcompounding periods.

I began working out the equation byadding the 1 and 0.04 and then I workedout 1.04 to the exponent of 4 to equal1.1699.I worked out the last little bit of theequation by multiplying $5631.00 by1.1699 to equal $6587.47.

$6587.47 $5631.00 $956.47! =

So, over two years of$5631.00 beingcompounded bi-annually at8%, Bobby collected$956.47 worth of interest atRBC.

( )2*2

0.08

2$5631.00 1A = +

( )$5631.00 1.1699A =

$6587.47A =

Toronto Dominion BankI began this problem by taking theCompound interest formula and insertingthe appropriate values into the equationthat were given in the question.



$5978.75 $5631.00 $347.75! =

So, over two years of$5631.00 beingcompounded monthly at3%, Bobby collected$347.75 worth of interest atTD.

2*12

0.03$5631.00 1

12A

! "= +# $

% &

( )24

$5631.00 1 0.0025A = +

( )$5631.00 1.0618A =

$5978.75A =

Scotia Bank2*4

0.05$5631.00 1

4A

! "= +# $

% &

( )8

$5631.00 1 .0125A = +

( )$5631.00 1.1045A =

$6219.36A =

$6219.36 $5631.00 $588.36! =

I began this problem by taking theCompound interest formula and insertingthe appropriate values into the equationthat were given in the question.



So, over two years of$5631.00 beingcompounded quarterly at5%, Bobby collected$347.75 worth of interest atScotia Bank.

The Next Scrooge! SolutionSo, 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 thatBobby would be making $956.47 of interest after two years. Whereas if hechose the TD Bank, he would only be making $347.75 and if he chose theScotia Bank he would have only made $588.36 of interest.

The Bank that Bobby had invested in:

The RBC

The Planet ofZorbia

Question #3

The Planet of ZorbiaDear Human,

The Planet of Zorbia is situated 10 light years away from earth. There arecreatures inhabiting that planet, those creatures are named Zorbians. Theleader of the planet Zorbia is calling on humans to help the planet Zorbiaanalyze its population and population rates. There are 4600000 zorbians livingon the planet Zorbia. How many Zorbians would there be on the planet Zorbiain 25 years, if the rate of growth is 2.72% a year. Zorbians also want to knowanother piece of information regarding their planet. There are two large citieson the planet of Zorbia, Xora and Yorp. The leader of Zorbia wants to pick aCapital City for the Planet, although he wants the city with the highest growthrate, so that the city will always grow and stay the largest city on the planet ofZorbia. Xora is a city located near the equator of the planet Zorbia and has apopulation of 256,230, but 7 years ago it had a population of 227,351. Yorp onthe other hand is situated in the Northern hemisphere of the planet Zorbia andhas 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 andcities of Zorbia, so the rest is up to you.

Your favorite Zorbian,

Xavi Zorb


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

P: Is the final population, in this case it is the population we are trying to findthat is 25 years away.

Po: Is the current population, in this case it is the population we are startingwith, 4,600,000.

Model: Is 1 + 0.0272, because it is the full population (1, which is 100%) plusthe 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 finalpopulation, in this case it is 25 (25 years).

mode( l) yPoP =

The Planet of Zorbia Solution

254600000(1 0.0272)P = +You begin this problem byplugging in the appropriate valuesinto the equation that we wereusing from the previous slide.

254600000(1.0272)P =

You then add the two modelnumbers together to achieve1.0272.

4600000(1.9560)P =

You then calculate (1.0272) to thepower of 25, to give you the valueof (1.9560), so you are able tomultiply 4,600,000, to achieve thevalue of P.

8,997,700P =After you multiply 4,600,000 by(1.9560), you achieve the totalpopulation.

The Planet of Zorbia SolutionFor the (First Part) of the second question, the city of Xora has a population of256,230, but 7 years ago it had a population of 227,351. What is its growthrate? 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.

7256,230 227,351( )m=You must begin this problem byplugging in the appropriatevalues into the equation.

7256,230 227,351( )

227,351 227,351

m=

You must divide 227,351 fromeach side, so that the m^7becomes isolated, so then youare able to solve for m.

ln1.1270 7 lnm=

The Planet of Zorbia SolutionYou must then add Ln to each side,so that m can later be isolated, whenyou add the Ln, the exponent on them, in this case 7 moves in front ofthe Ln symbol.

You must then multiply each side by1/7 (1 over 7) so that the 7 on theright side reduces, and then youmust multiply 1.1270 by Ln, and thenmultiply by 1/7 to equal m.

0.0171e m=

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

0.0171 lnm=



7256,230 227,351( )

227,351 227,351

m=


log1.1270 7 logm=

You must then add Log to eachside, so that m can later beisolated, when you add the Log,the exponent on the m, in thiscase 7 moves in front of the Logsymbol.

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

The Planet of Zorbia SolutionYou must then multiply each side by1/7 (1 over 7) so that the 7 on theright side reduces, and then youmust multiply 1.1270 by log, andthen multiply by 1/7 to equal m.

Due to the fact that we are using thelog method, 0.0074 is put to theexponent of base 10 (log), whichequals m. You must calculate10^0.0074 in order to receive agrowth rate of 0.0171 or 1.71%.

0.0074 logm=

0.007410 m=

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

The Planet of Zorbia SolutionFor the (Second Part) of the second question, the city of Yorp has a populationof 239,973, but 7 years ago it had a population of 208,659. What is its growthrate? 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.


7239,973 208,659( )

208,659 208,659

m=



ln1.1501 7 lnm=

You must then add Ln to each side,so that m can later be isolated, whenyou add the Ln, the exponent on them, in this case 7 moves in front ofthe Ln symbol.

0.0200 lnm=You must then multiply each side by1/7 (1 over 7) so that the 7 on theright side reduces, and then youmust multiply 1.1501 by Ln, and thenmultiply by 1/7 to equal m.

0.0200e m=

Due to the fact that we are using thee method, 0.0200 is put to theexponent of e, which equals m.0.0200 is the actual percentage ofgrowth (2.00%).

The Planet of Zorbia SolutionHere is the log version on how to solve this equation:

You must begin this problem byplugging in the appropriatevalues into the equation.


You must then add Log to eachside, so that m can later beisolated, when you add the Log,the exponent on the m, in thiscase 7 moves in front of the Logsymbol.

7239,973 208,659( )m=

7239,973 208,659( )

208,659 208,659

m=

log1.1501 7 logm=


0.0087 logm=You must then multiply each side by1/7 (1 over 7) so that the 7 on theright side reduces, and then youmust multiply 1.1501 by log, andthen multiply by 1/7 to equal m.

Due to the fact that we are using thelog method, 0.0087 is put to theexponent of base 10 (log), whichequals m. You must calculate10^0.0087 in order to receive agrowth rate of 0.0200 or 2.00%.

0.008710 m=

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

The Planet of Zorbia SolutionNow that you have found the answers to the questions that the leader of Zorbiahas asked of you, you are ready to present your answers.

For the first question, the population of the planet Zorbia in 25 years at a2.72% 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

Question #4

Trig Mania

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

2

2 4 4

1 1

2csc 1 cos 1 cos

1 2sin cos sin

! ! !

! ! !

+" +=

" "


Trig Mania! Solution

We will work on the right side of theidentity. 1 over 1-cosθ + 1 over 1+cosθbecomes 1+cosθ + 1-cosθ over 1-2sin²θ so that there is a commondenominator.

22

2 4 4

1 cos 1 cos

2csc 1 cos

1 2sin cos sin

! !

! !

! ! !

+ + "

"=" "

22

2 4 4

2

2csc sin

1 2sin cos sin

! !

! ! !=

" "

The two cosθ’s reduce and 1+1=2.1-cos²θ is a trigonometric identity soit becomes sin²θ. It is now 2 over2sin²θ

2 2

2 2 2 2 2

2csc 2csc

1 2sin (cos sin )(cos sin )

! !

! ! ! ! !=

" " +

2 over sin²θ afterdividing is nicelyconverted into2csc²θ. Cos θ – sinθ is nicely multipliedout to (cos²θ -sin²θ)(cos²θ+sin²θ)

4 4

2 2

2 2 2

2csc 2csc

1 2sin (cos sin )(1)

! !

! ! !=

" "

Cos²θ + sin²θ becomes 1, it is atrigonometric identity.



2 2

2 2 2

2csc 2csc

1 2sin (1 sin sin )(1)

! !

! ! !=

" " "

Cos²θ becomes 1 - sin²θ, it isanother trigonometric identity.

2 2

2 2

2csc 2csc

1 2sin (1 2sin )(1)

! !

! !=

" "

The two -sin²θ are addedtogether to become -2sin²θ.

2 2

2 2

2csc 2csc

1 2sin 1 2sin

! !

! !=

" "

(1-2sin²θ) is multiplied byone, obviously nothingchanges. Now both sides areequal to one another. Do notforget Q.E.D..

Q.E.D.


MYTHOUGHTSON D.E.V.

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 postthem onto the DEV blog. Although, it worked out for the best and I am herefinally done this project that was like hiking over Mount Everest. I would notsay 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 likeme will probably agree. All in all I liked this project, because it expanded yourimagination in how to create fun and well thought out questions, and how toexplain those question in the easiest way possible. So personally, I believe thisproject is good to have and this should be used for every 40s math course inDMCI. It makes you appreciate how much time teachers spend in creatingquestions for students to do in class.

Dinoppc40sw07

dev project

Technology

toronto dominion

compound interest

highest growth

population

trig mania

trigonometric

growth rate

compounding