the formula for curvature is |dT/dt|/|dr/dt| where T is the unit tangent
To see an explanation see Unit Tangent Vectors/Unit Normal Vectors
x = 2cos(t) and y = 3sin(t), where the parameter t satisfies 0 < t < 2pi.
This is only part of the problem but i'm having trouble with it, i need to find the curvature for this parametric equation at points (2,0) and (0,3). I was told to use CAS. Can anyone explain what CAS is and how to do this?
So so far i have
|r'(t)| = sqrt(13)
|T'(t)| = sqrt(4/13 + 9/13) = 1
But then i have nowhere to plug in the 2 given coordinantes, where did i go wrong?
...hmm i'm not very proficient with maple. Trying to get |T'(t)| i get
T'(t) = (-2cost/sqrt(4+5cos^2(t)),-3sint/sqrt(4+5cos^2(t)))
|T'(0)| = (sqrt(4sin^2(t)/sqrt(4+5cos^2(t)) + 9cos^2(t)/sqrt(4+5cos^2(t)) = sqrt(4/3+5/3) = sqrt(3)??
am i doing the algebra wrong?
kappa = |T'(t)|/|r'(t)|
so kappa = sqrt(3)/3 (wrong)