Recent content by Mikeylikesit182

So it seems at this point to find where the graph would intersect itself, we would ask what value for t would give us 2 sets of identical x,y coordinates?

(x-2)^2 = (pi^2)(cos^2(t)) (y-2t)^2 = (pi^2)(sin^2(t)) added = (x-2)^2 + (y-2t)^2 = (pi^2)(cos^2(t)) + (pi^2)(sin^2(t)) (x-2)^2 + (y-2t)^2 = (pi^2)[(cos^2(t)) + (sin^2(t))] (x-2)^2 + (y-2t)^2 = (pi^2) [(x-2)^2]/(pi^2) + [(y-2t)^2]/(pi^2) = 1 It seems to look like an ellipse adjusted upwards by...

Thank you for the help! I can't figure what that equation would be ( I can see one involving arccos and substitution but I think that would be a dead end.) I thought maybe finding the derivative dy/dx for the parametric eqn and realizing that at the point of intersection the 2 slopes would be...

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...