Calc-Vector Question?

Question

A particle moves in the plane so that its position at any time t>=0 is given by:
x=cos(t) y=sin(2t)

Eliminate the parameter and find an equation in terms of x and y for the path of the particle.
Multiple choice:
A) y=x(1-x^2)^.5
B) y=2x(1-x)^.5
C) y=2x(1-x^2)^.5
D) y=2x(1-x^2)

I got y=2xsin(t) but I can't get rid of the "t". I've spent the last 10 minutes looking at a list of trig identities and none of them are doing me any good. I remember doing a problem like this before but I can't remember how to do it. So, if you could tell me how to do it I would appreciate it.

Thanks.

Answer 1

x=cos(t) y=sin(2t)

sin(2t) = 2 sin(t) cos(t)
sin²(2t) = 4 sin²(t) cos²(t)
4x² -y² = 4cos²(t) - 4sin²(t) cos²(t)
= 4cos²(t) { 1 -1 sin²(t) }

using the identity cos²(t) +sin²(t) =1
we get
4x² -y² = 4cos²(t) cos²(t) = 4x^4

y^2 = 4x^2 -4x^4 = 4x^2(1 -x^2)
y^2 = 4x^2(1 -x^2)
y = 2x sqrt(1 -x^2)

Answer is C

Answer 2

y = 2 sin t cos t
x = cos t

y = (2 sin t) (x)

cos t = x / 1 thus:-
sin t = (1 - x²)^(1/2) / 1 = (1 - x²)^(1/2)

y = (2) (1 - x²)^(1/2) (x)
y = (2x) (1 - x²)^(1/2)
OPTION C)