Calculus
posted by James .
Find an equation of the tangent line to the curve for the given value of t.
x=t^22t, y=t^2+2t when t=1

dx/dt = 2t2
dy/dt = 2t+2
when t=1
slope = dy/dx = (dy/dt) / (dx/dt) = 4/0 , ahhh, a vertical line
when t=1
x = 12 = 1
y = 1 + 2 = 3
so a vertical line through (1,2) is
x = 1