Question

Determine if the following trajectory lies on a circle...

I start with a vector I am down to 2(cos^2)t+4(sin^2)t+2sintcost-2sqrt(3)sintcost but am just bad enough at trig to not know what to do from here to find what the radius is this trajectory lies on...

Answer 1

The point \((x,y)\) lies on a circle radius \(r= \sqrt{x^2+y^2}\)

Answer 2

If we let the origin be the radius.

Answer 3

right, so when i square and sum the x- and y-component, I'm getting a big string of trig I don't know how to simplify

Answer 4

nope, it is down to r=2

Answer 5

perhaps it would help to show the step by step reduction...

Answer 6

(sin^2)t - 2(cos^2)t + 2sqrt(3)(cos^2)t-2sintcost is what i get when i find the mag of the original vector

Answer 7

@Canuckish that may be part of the problem, you might have simplified something incorrectly.

Answer 8

sorry that should be

***2***(sin^2)t - 2(cos^2)t + 2sqrt(3)(cos^2)t-2sintcost is what i get when i find the mag of the original vector.

I see pulling the 2 out of the sin^2-cos^2 and reducing that to 2, but what do I do with the rest of it?

Answer 9

well, sin^2 - cos^2 is not equal to 1, that is sin^2 +++ cos^2 = 1 :)

Answer 10

I think @OOOPS is writing out the derivation.

Answer 11

crap. okay -thanks

Answer 12

sorry - i'm terrible at trig

Answer 13

lol, you are on the right track, it's just some pesky algebra and trig mistakes. :) no worries.

Answer 14

\(x = sint +\sqrt3 cost\\x^2 =(sint+\sqrt3cost)^2= sin^2t+2\sqrt3sint cost +3 cos^2t\)
\(y= \sqrt3sint-cost \\y^2=(\sqrt3sint-cost)^2= 3sin^2t-2\sqrt3sint costt+cos^2t\)
at them together you get \(sin^2t+3cos^2t+3sin^2t+cos^2t=4sin^2t+4cos^2t=4\)
therefore r =2

Answer 15

An alternative approach :)
\[||v||^2 = (\sin{t} + \sqrt{3}\cos{t})^2 + (\sqrt{3} \sin{t} - \cos{t})^2\]
\[||v||^2 = \sin^2{t} + 2\sqrt{3}\sin{t}\cos{t} + 3 \cos^2{t} + 3\sin^2{t} - 2\sqrt{3}\sin{t}\cos{t} + \cos^2{t}\]
\[||v||^2 = 4\sin^2{t} + 4\cos^2{t}\]
\[||v||^2 = 4(\sin^2{t} + \cos^2{t}) = 4\]
\[||v|| = 2\]

Answer 16

yeah, it's clever. linear algebra approaching for vector problem is perfect.

Answer 17

Basically you prove that the magnitude is constant, thereby a circle. Hooray for math! :D

Answer 18

I was just hoping that the original problem wasn't going to be a shifted circle. -.-

Answer 19

Fortunately, it is just "determine if it is on a circle" and find the radius, not the center. hihihihi so we are ok

Answer 20

yeah, but as @wio pointed out, our formula kind of assumes that the origin is in fact the center. :)