Question

Graphig, Finding position vectors, and velocity vectors.

Answer 1

\[r(\theta)=2+\cos(2\pi t)\]hm... not so sure here

Answer 2

thinking on it...\[r^2=x^2+y^2\]\[x=r\cos\theta\]\[y=r\sin\theta\]

Answer 3

\[r=2+\cos(2\pi t)\]\[\theta=\pi t\]so\[x(t)=(2+\cos(2\pi t))\cos(\pi t)\]\[y(t)=(2+\cos(2\pi t))\sin(\pi t)\]

Answer 4

I bet you can simplify that a bit with some trig formulas or whatever

Answer 5

I don't know what they want you to do exactly, graph it?

Answer 6

I just checked wolfram and it didn't give me any simplified solutions.

Answer 7

no, I know

Answer 8

very ugly "orbit" too
http://www.wolframalpha.com/input/?i=polar%20plot%20r%3D2%2Bcos(2pi*t)&t=crmtb01

Answer 9

well that's just from t=0 to t=2

Answer 10

Wolfram gives from t=o to t=1

Answer 11

oh here we go
http://www.wolframalpha.com/input/?i=parametric+plot+x%3D%282%2Bcos%282pi*t%29%29%28cos%28pi*t%29%29%2Cy%3D%282%2Bcos%282pi*t%29%29%28sin%28pi*t%29%29
0 to 2 will give a full orbit (because theta goes from 0 to 2pi), so it's just once around the loop

Answer 12

Oh that makes sense

Answer 13

what else do they want us to say about it?

you could find the intercepts by hand by finding when x=0 and y=0 to find the y and x-intercepts respectively

Answer 14

that would give you a rough idea of the graph without wolfram

Answer 15

but we already used wolfram, so... what else is there to say here?

Answer 16

the position vector is just\[\langle x(t),y(t)\rangle\]and I wrote the expressions for x and y above
the velocity vector is the derivative of the position vector with respect to time

Answer 17

So I take the derivative of x and then y? It wants me to calculate the velocity at t=.5 and t =1

Answer 18

yes, so take the derivative
position:\[\vec s(t)=\langle x(t),y(t)\rangle\]velocity:\[\vec v(t)=\langle x'(t),y'(t)\rangle\]then plug in the values

Answer 19

do they want the velocity as a vector or do the want the magnitude?

Answer 20

It wants velocity and speed

Answer 21

velocity is then just plugging in the numbers and leaving it in vector form

speed will be the magnitude of velocity, which is\[\|\vec v(t)\|=\sqrt{(x')^2+(y')^2}\]

Answer 22

I don't think we have learned how to take derivatives this big. This is what wolfram gives me http://www.wolframalpha.com/input/?i=%282%2Bcos%282pi*t%29%29%28cos%28pi*t%29%29

Answer 23

yeah, the derivative is not that hard (or, well, shouldn't be by this point in calc)
it just involves the product and chain rules
you only have the derivative of x(t) there though, you need y(t) as well

Answer 24

Alright thanks. I think I can do take it from here.

Answer 25

okey-doke
good luck!