Question

attach file I need help all number.Thank

Answer 1

By rectangular shape to model the paddle wheel I assume they just mean the outer line it traces as it moves, which is a circle. The equation of a circle is\[r^2=x^2+y^2\]so you should be able to get that rather quickly.

Answer 2

After that for part b you will use the equations\[x=r\cos\theta\]\[y=r\sin\theta\]but you sill need to find a function for theta in terms of t, so we have to find the angular rate of motion, which is given by\[\omega=\frac{2\pi}{T}\]where T is the period (the time it takes a point on the wheel to make a revolution)which will convert your formulas to\[x=r\cos(\omega t)\]\[y=r\sin(\omega t)\]so they will then be in parametric form.

Answer 3

This is all about the first problem of course.

Answer 4

Now for part c if you have the rest this should be no problem. We simply need to remember that the time t will we have is in seconds, so remembering that 60sec is one minute we plug that into our final equation for t.

Answer 5

\[a).... x^2+y^2=6^2\]?

Answer 6

seems like it.

Answer 7

\[b).... x=6cost , y= 6 sint ,.... 0\le t \le2\pi?\]

Answer 8

c) ?

Answer 9

never mind, you will keep the equations separate.
no we need omega as I mentioned above, for now they should be in terms of sin(theta)

Answer 10

and we don't need boundaries yet...

Answer 11

can you help me c?

Answer 12

first you need b

Answer 13

then c is easy

Answer 14

show me please!

Answer 15

well you are on the right track above, but as I said, for now they should be in terms of theta\[x=\cos\theta\]\[y=\sin\theta\]and we need to convert theta to a function of time with the formula I posted above...

Answer 16

those should have an r in them obviously^

Answer 17

\[x=r\cos\theta\]\[y=r\sin\theta\]

Answer 18

\[6\cos\frac{1}{2}+6\frac{\sqrt{3}}{2}=36 ?\]

Answer 19

that's not a functions of t...

Answer 20

the conversions I said to do were from\[\theta\to\omega t\]where\[\omega=\frac{2\pi}{T}\]is the "angular velocity" and where T is the period it takes to make one revolution. Those value are given in the problem.

Answer 21

This is the equivalent of \[v=d/t\]for linear motion, only for angles instead.

Answer 22

Actually one condition to add: in standard notation we will have to add a phase shift of \[\frac{\pi}{2}\]to account for the fact that we are starting at the top of the wheel.

Answer 23

so the form will be\[y=r\cos(\omega t+\frac{\pi}{2})\]\[x=r\sin(\omega t+\frac{\pi}{2})\]now use the formula I gave you to find omega.

Answer 24

y and x are switched above, sorry

Answer 25

\[x=r\cos(\omega t+\frac{\pi}{2})\]\[y=r\sin(\omega t+\frac{\pi}{2})\]

Answer 26

r=36
t= second
w= ?

Answer 27

No, r is just the radius no need to square it.
t will stay a variable, it is our parameter, so don't worry about that.
and omega is given, again, by\[\omega=\frac{2\pi}{T}\]where T is the period it takes to make one revolution. What is T ?

Answer 28

\[T=\frac{2\pi}{w}?\]

Answer 29

no, T is a given number in the problem. It says that that it takes 2 seconds for the wheel to make one revolution.

Answer 30

so plug that in and you will get omega

Answer 31

one solution is 2pi ?

Answer 32

2 seconds is 2pi?

Answer 33

Yes, so what is one second?

Answer 34

one seconds is 1pi?

Answer 35

exactly.
so that is our omega, because omega=(angle)/(unit time).
The unit time for us is one second.

Answer 36

I now need to note that this should actually be\[\omega=-\pi\]because we are going clockwise, which is negative in conventional trigonometric notation.

So now that we have omega and our phase shift to put us at the to, what do you get?

Answer 37

I don't know, sorry

Answer 38

using pi/2 to put us at the top, as I mentioned earlier we have\[x=\cos(\omega t+\frac{\pi}{2})\]\[y=\sin(\omega t+\frac{\pi}{2})\]and we just found out that \[\omega=-\pi\]so what are our equations when we plug in omega?

Answer 39

x = (-pi*t + pi/2)?

Answer 40

cosine of...

Answer 41

sorry I missing cos

Answer 42

if you are surprised that x=rcos(theta) then I think you have been missing class for a while :/

Answer 43

thank

Answer 44

sorry, forgot r:\[x=6\cos(\frac{\pi}{2}-\pi t)\]\[y=6\sin(\frac{\pi}{2}-\pi t)\]

Answer 45

Happy chrismast and new year

Answer 46

likewise :)