r(t) = <12t^3,18t^2,9t^4>
Find r'(t) and T(t) and evaluate T(1)
Any help is appreciated
You title this "curvature" but the problem is much simpler than calculating the cuvarture. r'(t)= < (12t^3)', (18t^2)', (9t^4)'>= <36t^2, 36t, 36t^3>. It's length is $\displaystyle 36\sqrt{t^4+ t^2+ t^6}= 36t\sqrt{t^4+ t^2+ 1}$ and the unit tangent vector, T(t) is <36t^2, 36t, 36t^3> divided by that: $\displaystyle T(t)= <t, 1,t^2>/\sqrt{t^4+ t^2+ 1}$. T(1) is, of course, that evaluated at t= 1.