Question

Help with a parametric equation

Answer 1

?

Answer 2

You forgot to post the question

Answer 3

HERE it is

Answer 4

Hi tyler

Answer 5

Hey Rid

Answer 6

what is ferry?

Answer 7

I don't have word on this computer. Can you just write the problem?

Answer 8

A ferris wheel

Answer 9

ok I will type it

Answer 10

I dont know what ferry is or how ferry looks like so cant help HERO will!

Answer 11

A child is sitting on a ferris wheel of diameter of 10 meters making one revolution every 2 minutes Find the speed of the child

Answer 12

A ferris wheel is a ride at the amusement park!!!!!

Answer 13

Thanks for having so much confidence in me. I think you put 2 at 12 oclock, .5 at 3'clock

1 at 6 o'clock and 1.5 at 9 oclock, but use x^2 + y^2 = 5^2

somehow you have to write that in terms of t

Good luck

Answer 14

totally didn't understand that

Answer 15

I don't blame you

Answer 16

ok I have the answer but maybe u can explain me y that is the answer

Answer 17

Dont you just find the circumfrence of the ferris wheel and then use \[speed = \frac{d}{t}\] That will give you a speed in meters per minitue

Answer 18

It's actually a lot easier to not use parametric equations for this (just do distance divided by time, but anyway...). Here's a possible set of parametric equations, with the center of the ferris wheel being the origin.\[x=10\sin \theta\]\[y=10\cos \theta\]\(\theta\) is of course dependent on \(t\), which I'll leave you to figure out. :P Then find \(\frac{dy}{dx}=\frac{dy/dy}{dx/dt}\).

Answer 19

Just a guess lol

Answer 20

can't we just solve this like a physics problem?\[ v=r \omega=5{2\pi \over2}=5\pi \]the answer is in m/min, you could convert to m/s, but I guess that's not the goal here.

Answer 21

Look at the way the book solved it

Answer 22

I see, they want you to practice solving these problems in a particular way. Annoying, but reasonable.

Answer 23

What is the answer?? I dont have word on this computer

Answer 24

5 pi m/min

Answer 25

ok I will save it as a pdf

Answer 26

my answer is correct above, but it's not done the right way.

Answer 27

Yeah thats what i got too Turing.. Then i got \[\frac{5\pi}{60}\space m/\sec\]

Answer 28

yup, but the answer on the sheet in in m/min

Answer 29

Yeah thats what i got.. 15.7

Answer 30

I totally don't understand what the book did. can someone explain

Answer 31

@tyler but did you do the problem that way?
@rld let me see...

Answer 32

I have no idea.. Is this Calc 2?? or 1 or not calc at all lol

Answer 33

No i found the circumfrence of the ferris wheel and then used s = d/t

Answer 34

It is calc 1

Answer 35

Ahh because im in Calc1 and i havent seen this stuff before lol

Answer 36

I think we skipped it or somethin lol

Answer 37

ya every book does it in a diff order you will come by t by the end of calc 3 forsure

Answer 38

I did this in calc 2 myself.
I'm not sure I can explain it better that the book. What step do you get lost on?

Answer 39

the simple step!!!!!

Answer 40

why is it 10pi

Answer 41

where do you see 10pi?
the answer is 5pi
which line are you looking at?

Answer 42

oh you mean in the geometric method?

Answer 43

ya but look at the first answer and it brings in pi. wwhere does pi get in?

Answer 44

yes

Answer 45

The circumference of the circle is the distance the child travels in one revolution: \[C=2\pi r=\pi d=10\pi\]So we just used the old-fashioned circumference formula :)

Answer 46

LOL. I am so stupid

Answer 47

It happens when you focus on other things. I forgot how to complete the square just the other day, lol!. Don't worry about it :D

Answer 48

ok wait

Answer 49

Now I want to understand part B

Answer 50

Yes, that's a little trickier to explain. Start with the parameterizations, do you understand that?\[x=5\cos(\omega t)\]\[y=5\sin(\omega t)\]

Answer 51

how did they get thoose equations?

Answer 52

in general from trig you should know that
\[x=r \cos \theta \]\[y=r \sin \theta\]does that ring a bell?

Answer 53

ya i forgot all abt that . i gotta review my trig totally forgot e/t

Answer 54

You still have to be able to put it in terms of t

Answer 55

how did they get thoose equations?

Answer 56

but what is the r in front?

Answer 57

radius

Answer 58

here let me derive that from the definitions of sine and cosine...

Answer 59

ok. thanks

Answer 60

\[\sin \theta={y \over r}\to y=r \sin \theta\]\[\cos \theta={x \over r}\to x=r \cos \theta\]make sense now?

Answer 61

ya that helped ;-)

Answer 62

S/t is wrong with this site!!!!

Answer 63

it is so solw i gotta refresh every second

Answer 64

yes it has been repeating posts for me too lately

Answer 65

ok so now we need an expression for theta. do you see why it is \[\theta=\omega t\]or do you know what omega represents?

Answer 66

If I could figure out how to open Microsoft Office in safe mode, I would just keep it in safe mode

Answer 67

remind me of what omega stands for

Answer 68

i don't know what wt is

Answer 69

Never ever seen it b4

Answer 70

It is the angular velocity.
The analogue of this formula for distance is\[ d=vt\]for angles it is\[\theta=\omega t\]That is, the angle covered is the angular velocity times time. Good so far?

Answer 71

nope cuz i have never seen this b4 let me try to understand it

Answer 72

it is nuts my book doesn't mention it at all

Answer 73

here's another way to look at it, if you remember the arc of a circle that subtends angle theta is \[s=r \theta\]since our distance covered in this problem is in fact an arc we can write it as\[d=r \theta=r \omega t\]dividing by r gives\[\theta=\omega t\]If it's not in your book it's because they think you should know this already. I don't know why, I didn't learn that in calc, I learned it in physics.

Answer 74

*the length of an arc that subtends angle theta

Answer 75

if you don't know that arc formula you should blame your trig teacher, not your calc book

Answer 76

i never took a specific trig course

Answer 77

Well, that will be a problem for you in parametric equations because converting from cartesian coordinates requires these formulas all the time.

Answer 78

when did u take a trig course?

Answer 79

high school

Answer 80

like in college?

Answer 81

oh well in canada it works different

Answer 82

every year u take a math course that includes a chapter or two of trig

Answer 83

I refresh my memory on things like identities a lot, because they are easy to forget. I'll post a nice trig review site, hold on...

Answer 84

plz

Answer 85

http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/AlgebraTrigIntro.aspx

scroll down and check out the trig section when you get the chance

Answer 86

It's not a complete course, just a review, so you should look to deepen your understanding further

Answer 87

ok thanks but can u expalin what anguar velocity is in words?

Answer 88

what level of math r u in?

Answer 89

Well, if something goes around a circle in, say 2 minutes like our problem, it covers an angle of 2pi radians. Angular velocity is angle divided by time:\[\omega={\theta \over t}={2\pi \over 2}=\pi rad/\min\]So that is the angular velocity for our problem.

(like I said the analogue for distance is "speed is distance divided by time")

Answer 90

I'm in mulitvariable calculus, but I have done differential equations on my own to some extent.

Answer 91

what is the diff btwn taht and calc 3 lets say

Answer 92

Nothing really, just depends how you organize it. Where I am there are only two calculus classes, single-variable and multi.

Answer 93

oh i see

Answer 94

k that was clear but