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e to the i pi
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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!
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!