## October 07, 2006

### Jho-Ann's Bicycle

The pedals on a bicycle have a maximum height of 30 cm above the ground and a minimum distance of 8 cm above the ground. Jho-Ann pedals at a rate of 20 cycles per minute.

a) What is the period, in seconds for this function?

b) At π = 0, Jho-Anns right foot is closest to the ground

i) Write 2 equations that represent the height of her right foot above the ground; 1 sine; 1 cosine

ii) For how long per cycle is Jho-Anns right foot 20 cm, or higher, above the ground.

1. period = 60seconds / 20 cycles
period = 3

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4. actually, i think this is the right answer:
a)

p = 2pi/(20/60)
p = 2pi/1 * 60/20
p = 120pi/20
p = 60pi

b) i) h(t) = 11sine1/30(x-15pi) + 19
h(t) = 11cos1/30(x- pi) + 19

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a)The period is 3 seconds because, if it takes 20 cycles per minute, that implies that ONE CYCLE occurs 1/20th of a minute. Divide 60 seconds by 20, and you get 3 seconds, hence, a period occurs every 3 seconds.

b) i)

h(t)=11sin[2π/3(t+3/4)]+19
h(t)=-11cos(2π/3t)+19

ii)I used my cosine equation to solves this question.

20=-11cos(2π/3t)+19
LET Θ=(2π/3t)

20=-11cosΘ+19
1=-11cosΘ
-1/11=cosΘ

We know that cosine is negatvie in quadrants II & III, so the related angle would be 2π-Θ.

Θ≈1.662

Now you solve for t.

1.662=2π/3t
1.662(3/2π)=t
0.7935≈t

Θ≈4.6214 [Related angle in Quadrant III]

4.6214=2π/3t
4.6214(3/2π)=t
2.2065≈t

Then, we take both answers and subtract them from each other to get how long her right pedal was in the air, at 20cm or higher.

2.2065-0.7935=1.4131

Keep in mind, I didn't round off those values with my calculator, I stored the full decimals, subtracted them and then rounded off my answer.

Jho-Ann's right foot is in the air 20cm or above for 1.4131 seconds.

8. I tried the question out on my own and was going to post it up but i saw that there where already a couple of answers, my answer was off a little but after re-doing the question, by following how jessica did it and storing the values in my calculator (which i forgot) my answer came out to be like her's. I was going to try out Jeffersons but i wasnt able to follow his math (No offense intended).

10. Trying the questions I did happen to get a few answers which matched up with what Jessica has. The other ones however were very different, when I looked over her answers was when I noticed that I had forgotten a few steps between the start and finish. As far as I can see (and after looking over what notes are in my notebook and on the blog several times) she's got the questions right.

11. I was looking over Jessicajill's answers and a few of the other peoples answers and managed to get A, B, and D of the equation, as well as the period, but had a hard time getting the rest of the other answers a little mixed up. I was reading over and checking some of the answers people made and it made me understand what I was missing.

12. I also got 3 seconds for my answer to a) and f(x)=11sin2pi/3(x)+19, f(x)=11cos2pi/3(x-1.5)+19, but when I punched it into my calculator the values in the table of values didnt look right...for the last question Im having a little bit of trouble I find it confusing....from what I can see though I think jessica has the right answer since everything else was the same as mine.

13. I made one mistake I think, with the first part of b. For my sin equation, it's suppose to be a subtract sign instead of an addition sign. So the equation will be this:

h(t)=11sin[2π/3(t-3/4)]+19

I think that's right now.

14. after trying the question.. i got the same thing as jessicajill. but then, i did make a few minor mistakes and saw where i got it wrong. so therefore, i think jessicajill got it right.

15. I do believe that jessica jill was right just by illustrating it on a graph that the period is 3 because within the 60 seconds their is 20 cycles which implies 1/3 of 60 seconds or period of 3 seconds. I found that A, D were both the same for finding my cosine/sine graph because between 30 and 8 is 19 so that proves that between one point to the sinusoidal axis is a distance of 11 which is my A and 19 which represents D for the vertical shift of the sinosuidal axis. 2π/3 affects the period to make it 3 seconds because when finding our period we reciprocate 2π/3 so our 2π cancels so were left with 3. I am questioned however as to why or how the phase shift is like because thats the only time i get stuck?!?

16. First of all, When I first looked at this question, I said to myself , this was going to take very long, so I decided to do it later on , and my answers to them , were , for question A, I got 3 seconds as did most people. For B(i) I got 11sin(2pi/3(x-1.5))+19, for the cos graph, I got -11cos(2pi/3(x))+19,and I kinda had trouble with the phaseshift, so I am not sure if my answer is right, ALso I would like to tell " anh " that I know why her calculator looked wrong, it was because she must of typed it wrong on the y=. For the last question however I got stuck, but I looked at jessicajill's comment and saw jessicajill's way of finding the last question, and I remembered I could check my answers on the calculator,which I did, and I ended up getting the same answer, so I believe jessicajill's answer for the last one is right.

17. Wow , never knew that i was going to pull this off, because i forgot my binder in my friend's car. I though i was going to be done, but good thing we have blogger.
let me see.
A). i got 3, i took 60sec/20cyc

i).
I got the same as jessicajill, as
h(t)=-11cos(2π/3t)+19
h(t)=11sin[2π/3(t+3/4)]+19

ii).
same thing as jessicajill

-1/11=cosΘ

but for Θ , didn't get the right answer. i did the wrong quadrants so i got a mix up. i put for that quadrant II & III was negative, i put Iv & II.Oops simple math mistake.

so all the other answers was wrong. Because if you mess up this part up ,your wrong through the whole thing so you got to be careful.

I have to agree with Jessicajill's answer because the step were simular to the notes that we learned and it made alot of sense that i can follow.

18. For question A I too got 3 seconds for the period. For part b the first time I tried it i got the same as anh in that f(x)=11sin2pi/3(x)+19 but my cosine was f(x)=11cos2pi/3(x-.25)+19. But after looking at what jessicajill got I realized that I didn't read the question properly in that it stated At π = 0, Jho-Anns right foot is closest to the ground. So i drew my graph wrong instead of starting the graph at 8 which is the minimum height at y=0 i drew it starting with 19 at y=0. After I realized my mistake I got the same as jessicajill. The last question I understood what it was asking for but I had trouble trying to get the answer. But after I looked at how jessicajill started answering the question I started to get what she was doing. I'll have trouble wording what i understood but what you are looking for is the intersection of y=20 on the graph which comes up to 2 points between .75(3/4) and 2.25(9/4). And then you do that equation thing looking for the angle i think is what you're looking for. then once you find the angle and its related angle you need to find it in seconds which you do by multiplying it by the reciprocal of the period. then you take away both times. That's what i understood and it made sense so yeah i think jessicajill got the answer right.

19. sorry it toke to long to be out have a vacation with the family to go camping for the long weekend, so here my answers for the assignment
A). i got 3 as period, so it took about 60sec/20cyl

i)h(t)=-11cos[(2π/3t)]+19
h(t)=11sin[2π/3(t+3/4)]+19

ii) 20=-11cos(2π/3t)+19
LET Θ=(2π/3t)
20=-11cosΘ+19
1=-11cosΘ
-1/11=cosΘ
so we can say that cosine is negitive so it will be in quadrant II and III. i didnt get thou getting how long was in the air at 20cm or higher but i get how it works out

20. i had lots of mistakes doing it and i did not explain my answers on my first comment so i'll just make a new one. what i almost did is just tried to follow everyone's explanation and how they ended up with such answers and compared them with the examples in my notes. i think jessicajill has the right answer.