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# Dino's DEV Project

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### Dino's DEV Project

1. 1. D.E.V Project
2. 2. The Wheel goes Round and Round! Question #1
3. 3. 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.
4. 4. The Wheel goes Round and Round! Solution Ocm (touching the pavement) and where Point X begins. 52cm above the pavement is where point X is farthest from the pavement and is also the diameter of the car wheel. 26cm is the radius of the car wheel, which is where point X is half the distance between where X is touching the pavement (0cm), and where X is farthest from the pavement (52cm). 0cm is where point X is touching the pavement , X's closest point to the pavement. Car Wheel radius Point X 52cm 0cm 26cm
5. 5. Period: 20 revolutions per minute 20 revolutions per 60 seconds 60 secs secs 20 revolutions rev In order to find the period when given the amount of revolutions in a certain amount of time, you must divide the amount of time by the number of revolutions. In this case there were 20 revolutions in one minute, one minute is also 60 seconds. So 60 seconds divided by 20 revolutions is 3 seconds/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 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.
6. 6. The Wheel goes Round and Round! A) Solution 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). (Seconds) (cm) Maximum of the graph is 52, due to the fact that the highest point on the wheel was 52cm and the minimum of the graph was 0, for 0cm (Touching the pavement). The Amplitude of this graph is 26, as it is the distance from either the max or min value.
7. 7. 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 B = B = C = C = 0 D = 26 D = 26 Sine Equation: h=26sin[ (t- )] +26 Cosine Equation: h=-26cos[ t] + 26
8. 8. 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 (2 Π 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 seconds so that y=0 shifts forward 0.75 or 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! B) Solution
9. 9. 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. You must begin this problem by plugging in the appropriate values into the equation. I brought over the 26 from the right side of the equation to the left side to add the two values together. 20+(-26)=-6.
10. 10. The Wheel goes Round and Round! C) Solution I took the arch sine of (-0.2307) on the left side, so that the sin on the right side of the equation would reduce. I then after I received the arch sine of -0.2307, divided each side by 2 pi over 3, so that the right side would reduce and I would have to divide the arch sine of (-0.2307), which is (-0.2329) by 2 pi over 3. Next I am dividing each side of the equation by 26, the right side would reduce, and the left side becomes -0.2307.
11. 11. The Wheel goes Round and Round! C) Solution -0.2329 divided by 2 pi over 3 equals -0.0124. Then I brought over the -3/4 on the right side to the left, so I can isolate t and derive a value for t. t = 0.7376, which is the amount 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.
12. 12. 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 = Π d 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 = Π d C = Π (52cm) C ≈ 163.3628cm One revolution of the car wheel measures 163.3628 centimeters.
13. 13. 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. # 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 highest number which is 613. 1000m 1.603628m The Wheel goes Round and Round! E) Solution
14. 14. The Next Scrooge! Question #2
15. 15. 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 over two years. So, are you going to help Bobby? Click the speaker to hear the question.
16. 16. 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 : 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.
17. 17. Royal Bank of Canada I began this problem by taking the Compound interest formula and inserting the appropriate values into the equation that were given in the question. I multiplied the exponents together so I 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 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 equation by multiplying \$5631.00 by 1.1699 to equal \$6587.47. So, over two years of \$5631.00 being compounded bi-annually at 8%, Bobby collected \$956.47 worth of interest at RBC.
18. 18. Toronto Dominion Bank I began this problem by taking the Compound interest formula and inserting the appropriate values into the equation that were given in the question. I multiplied the exponents together so I 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 adding the 1 and 0.0025 and then I worked out 1.0025 to the exponent of 24 to equal 1.0618. I worked out the last little bit of the equation by multiplying \$5631.00 by 1.0618 to equal \$5978.75. So, over two years of \$5631.00 being compounded monthly at 3%, Bobby collected \$347.75 worth of interest at TD.
19. 19. Scotia Bank I began this problem by taking the Compound interest formula and inserting the appropriate values into the equation that were given in the question. I multiplied the exponents together so I 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 adding the 1 and 0.0125 and then I worked out 1.0125 to the exponent of 8 to equal 1.1045. I worked out the last little bit of the equation by multiplying \$5631.00 by 1.1045 to equal \$6219.36. So, over two years of \$5631.00 being compounded quarterly at 5%, Bobby collected \$347.75 worth of interest at Scotia Bank.
20. 20. 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
21. 21. The Planet of Zorbia Question #3
22. 22. 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, Xavi Zorb Click the speaker to hear the question.
23. 23. 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. 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).
24. 24. The Planet of Zorbia Solution You begin this problem by plugging in the appropriate values into the equation that we were using from the previous slide. You then add the two model numbers together to achieve 1.0272. You then calculate (1.0272) to the 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. After you multiply 4,600,000 by (1.9560), you achieve the total population.
25. 25. The Planet of Zorbia Solution 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 plugging in the appropriate values into the equation. You must divide 227,351 from each side, so that the m^7 becomes isolated, so then you are able to solve for m.
26. 26. The Planet of Zorbia Solution You must then add Ln to each side, so that m can later be isolated, when 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 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 exponent of e, which equals m. 0.0171 is the actual percentage of growth (1.71%).
27. 27. The Planet of Zorbia Solution You must begin this problem by plugging in the appropriate values into the equation. You must divide 227,351 from each side, so that the m^7 becomes isolated, so then you are able to solve for m. You must then add Log to each side, so that m can later be isolated, when you add the Log , the exponent on the m, in this case 7 moves in front of the Log symbol. Here is the log version on how to solve this equation:
28. 28. The Planet of Zorbia Solution You must then multiply each side by 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. Due to the fact that we are using the log method, 0.0074 is put to the 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%
29. 29. The Planet of Zorbia Solution 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. You must begin this problem by plugging in the appropriate values into the equation. You must divide 208,659 from each side, so that the m^7 becomes isolated, so then you are able to solve for m.
30. 30. The Planet of Zorbia Solution You must then add Ln to each side, so that m can later be isolated, when 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 right side reduces, and then you must multiply 1.1501 by Ln, and then multiply by 1/7 to equal m. Due to the fact that we are using the e method, 0.0200 is put to the exponent of e, which equals m. 0.0200 is the actual percentage of growth (2.00%).
31. 31. The Planet of Zorbia Solution Here is the log version on how to solve this equation: You must begin this problem by plugging in the appropriate values into the equation. You must divide 208,659 from each side, so that the m^7 becomes isolated, so then you are able to solve for m. You must then add Log to each side, so that m can later be isolated, when you add the Log , the exponent on the m, in this case 7 moves in front of the Log symbol.
32. 32. The Planet of Zorbia Solution You must then multiply each side by 1/7 (1 over 7) so that the 7 on the right side reduces, and then you must multiply 1.1501 by log, and then multiply by 1/7 to equal m. Due to the fact that we are using the log method, 0.0087 is put to the 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%
33. 33. The Planet of Zorbia Solution 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.
34. 34. Trig Mania Question #4
35. 35. 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: Click the speaker to hear the question.
36. 36. Trig Mania! Solution We will work on the right side of the identity. 1 over 1-cos θ + 1 over 1+cos θ becomes 1+cos θ + 1-cos θ over 1-2sin ² θ so that there is a common denominator. The two cos θ ’s reduce and 1+1=2. 1-cos² θ is a trigonometric identity so it becomes sin² θ . It is now 2 over 2sin² θ
37. 37. 2 over sin² θ after dividing is nicely converted into 2csc² θ . Cos θ – sin θ is nicely multiplied out to (cos² θ -sin² θ )(cos² θ +sin² θ ) Cos² θ + sin² θ becomes 1, it is a trigonometric identity. Trig Mania! Solution
38. 38. Trig Mania! Solution Cos² θ becomes 1 - sin² θ , it is another trigonometric identity. The two -sin² θ are added together to become -2sin² θ .
39. 39. (1-2sin² θ ) is multiplied by one, obviously nothing changes. Now both sides are equal to one another. Do not forget Q.E.D.. Q.E.D. Trig Mania! Solution
40. 40. MY THOUGHTS ON D.E.V.
41. 41. 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