Proving a parametric equation intersects itself at a certain point.

[ocarina_noob]

What's the best way to go about doing these types of questions? For example:

Prove that the graph of the relation x= sin 2t, y=sin t, t is [0, 2pi] intersects itself at the origin.

Would you get a Cartesian equation and let y or x=0? How do you prove that it intersects itself?

Someone please solve that for me!

Oh, by the way, nice to meet you guys. I'm going to posting very often in this section.

[Will T]

Hey,
as for solving it,

x = 2sin(t)cos(t)
y = sin(t)
dividing top by bottom
x/y = 2cost

draw a triangle, find cos(t) given y=sin(t)
then re-substitute to eliminate the t, and complete the square to solve.

You should get 1 relation comprised of 4 functions (sketch it on your CAS)
If I've done it correctly, they look like:
with alternating plus or minus sings.
Then substitute t = 0, t = pi/4 and t = pi/2 into the original equations to get the path, and it is clear it goes through the point 0,0 multiple times on its circuit.

[abeybaby]

you could say:
intersect at the origin when x=y=0:
sin2t=0            and                 sint=0
t=0,pi/2,pi,3pi/2,2pi             t=0,pi,2pi
therefore it goes through the origin 3 times, once at the start, once at the end, and once in the middle at t=pi

[ocarina_noob]

Thanks guys.

[Planck's constant]

The simplest way is to prove that there exist distinct t's, say t1 and t2 (both in [0,2 pi]) when the x coordinates of the path are equal AND the y coordinates of the path are equal (and then show that those t's correspond to the origin). Therefore,

sin2t1 = sin2t2 (1)

and

sint1 = sint2 (2)

Thus from (1)

2sint1cost1 = 2sint2cost2 (3)

Substitute (2) into (3) and simplify

sint1cost1 = sint1cost2
sint1(cost1-cost2) = 0 (4)

(4) implies

sint1 = 0 (=sint2)

or

cost1 = cost2 => t1 - t2 = 2 n*pi

The rest is straightforward.

[BubbleWrapMan]

Hm, I would think you'd also have to show that the velocity vectors aren't parallel at those times. For example, if a particle was travelling along the circle x^2 + (y-1)^2 = 1, it would pass through the origin multiple times, but the path wouldn't intersect itself, since it's the same section of the path that it returns to.