Find a formula for the curvature of the curve:
x=(e^t + e^(-t))/2
y=(e^t - e^(-t))/2
Write an equation of the osculating circle when t=0.
curvature=|x'y'' - x''y'|/(x'^2 + y'^2)^(3/2)
The Attempt at a Solution
First, wouldn't the formula for curvature be:
(e^(2t) + e^(-2t))/2 * sqrt(e^(2t) + e^(-2t))
? But for some reason, my teacher marked a big "X" across this and wrote question marks.
EDIT: Nevermind about the first part of the question - I just should have simplified it more.
Now, for the equation of the osculating circle when t=0, I'd get the center of the circle by doing: <x_center,y_center>=<x(0)-K*y'(0)/sqrt(x'(0)^2+y'(0)^2),y(0)+K*x'(0)/sqrt(x'(0)^2+y'(0)^2)>