Converting Position Vector vs Time to Cartesian Coordinates

[e to the i pi]

1. The position vector of a particle at time t ≥ 0 is given by r = sin(t)*i + cos(2t)*j. Find the cartesian equation for the path of the particle.



2. I was told that the answer is:
y = 1 - 2x^2
But I don't know how to obtain that solution.



3. r = sin(t)*i + cos(2t)*j
At first I thought I would merely plug in the values:
x = i, y = j and √(x^2 + y^2) = r, but that wasn't working out:
√(x^2 + y^2) = x sin(t) + y cos(2t)
x^2 + y^2 = x^2(sin(t))^2 + 2xysin(t)cos(2t) + y^2(cos(2t))^2
Solve with CAS Calculator:
t = -pi/2
Substitute that back in:
x^2 + y^2 = x^2(-1)^2 + 2xy(-1)(0) + y^2(-1)^2
x^2 + y^2 = x^2 + y^2
0 = 0
Now I am lost. Please help me!

 

[dynamicsolo]

Use the "double-angle formula" for cosine -- there are three forms of it, but one of the expressions for y = cos(2t) will have a simple connection to x = sin t in the vector equation for r .

(The double-angle formulas for sine and cosine from trigonometry are ones that it are useful to know. They see a lot of application...)

 

[e to the i pi]

cos(2t) = 1 - 2(sin(t))^2
So r = sin(t)*i + (1 - 2(sin(t))^2)*j
But I'm still confused because I don't know how to get rid of the i's and j's...
Wait I think I get it:
x = sin(t)
y = 1 - 2(sin(t))^2
y = 1 - 2x^2
Is that right?

 

[SteamKing]

Let x = sin (t). What is y in terms of x?

 

[e to the i pi]

Thanks a lot for your help, dynamicsolo and SteamKing. Now I know how to solve these types of problems for the future.
I let x = the value that is being multiplied by i and y = the value that is being multiplied by j and then ignore the r.
Then I express y in terms of the value being multiplied by i and then make the substitution to get y in terms of x.
Thank you once again!

 

[dynamicsolo]

Wait I think I get it:
x = sin(t)
y = 1 - 2(sin(t))^2
y = 1 - 2x^2
Is that right?

Yes. You don't have to "get rid of" the i and j: they are just the unit vectors connected with the x- and y-components of the vector equation. We would write

r(t) = x(t) i + y(t) j = sin(t) i + cos(2t) j .

You are just being asked to look for the equation relating y(t) to x(t) .