- #1
Mikeylikesit182
- 4
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x = 2 − π cos t
y = 2t − π sin t
−π ≤ t < π
I understand how to eliminate parameter using sin^2 + cos^2 = 1.
I can't figure out how to deal with the "2t" in the y equation, if you solve for sin(t) and square, you get
((2t-y)/π )^2 which leaves the parameter. Is there a way to get it into the argument of sin?
I need this to then figure out intersection points and tangents later. Thanks in advance for any insight!
y = 2t − π sin t
−π ≤ t < π
I understand how to eliminate parameter using sin^2 + cos^2 = 1.
I can't figure out how to deal with the "2t" in the y equation, if you solve for sin(t) and square, you get
((2t-y)/π )^2 which leaves the parameter. Is there a way to get it into the argument of sin?
I need this to then figure out intersection points and tangents later. Thanks in advance for any insight!