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Mathematics 20 Online
OpenStudy (anonymous):

Describe the motion of a particle with position (x,y) as t varies in the given interval.

OpenStudy (anonymous):

\[x=5sint~~~y=2cost~~~~~-\pi \le t \le 5\pi\]

OpenStudy (anonymous):

@ganeshie8

ganeshie8 (ganeshie8):

from the first glance it looks like an ellipse, what do u think ?

OpenStudy (anonymous):

I understand it's an ellipse but I don't understand why it moves clockwise around the ellipse 3 times, maybe just the drawing I don't understand haha.

ganeshie8 (ganeshie8):

period of sint is 2pi so one complete revolution = 2pi radians, yes ?

ganeshie8 (ganeshie8):

|dw:1409902062553:dw|

ganeshie8 (ganeshie8):

we don't know orientation yet, but we can see that [-pi, 5pi] has 3 full periods in it : (5pi - -pi)/(2pi) = 6pi/2pi = 3

ganeshie8 (ganeshie8):

so it makes 3 full revolutions starting at SOME point and moving in SOME direction we need to find these starting point and direction

OpenStudy (anonymous):

Ah right right, thank you so much :P!

ganeshie8 (ganeshie8):

at t = -pi, (x,y) = (0, -2) : |dw:1409902254176:dw|

ganeshie8 (ganeshie8):

thats the starting position, you need to figure out whether the particle moves in clockwise or counterclockwise

ganeshie8 (ganeshie8):

we need to plot atleast TWO more points to know the orientation unambiguously

OpenStudy (anonymous):

Yeah I got it, as t goes from - pi to 5 pi, (0,-2) is the start point and moves clockwise around 3 times. Thanks Ganeshie ^.^

ganeshie8 (ganeshie8):

:) yes and we need to plot the points within the same period.. otherwise there will be a clash

ganeshie8 (ganeshie8):

or else if we recall transformations in xy plane : (x, y) = (cos, sin) : counterclockwise rotation (x, y) = (sin, cos) : clockwise rotation

ganeshie8 (ganeshie8):

that tells immediately the orientation of particle

OpenStudy (anonymous):

Yup, can I ask one more just to confirm?

ganeshie8 (ganeshie8):

yeah sure il try

OpenStudy (anonymous):

\[x=3+2cost~~~y=1+2sint~~~~~~\pi/2 \le t \le 3\pi/2\] for this what I thought of was x = h+rcost y = k+sint which gives an equation of a circle so I got (x-3)^2+(y-1)^2=4

OpenStudy (anonymous):

start point (3,3) moving counterclockwise

ganeshie8 (ganeshie8):

looks perfect ! but do you really get a full circle ?

OpenStudy (anonymous):

No it should be half of a circle I think

ganeshie8 (ganeshie8):

you should draw it

OpenStudy (anonymous):

y=k+rsint* yeah I'm drawing it now haha

OpenStudy (anonymous):

one half of a circle

ganeshie8 (ganeshie8):

diagram

ganeshie8 (ganeshie8):

diagram tells everything - startpoint, endpoint and orientation

ganeshie8 (ganeshie8):

|dw:1409903747363:dw|

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