p. 812 # 25. Find the curvature of the path given parametrically for r(t) = t i + \frac{1}{2}t^{2} j + \frac{1}{3}t^{3} k and find the tangential and normal components of acceleration.

r(t) = t i + \frac{1}{2}t^{2} j + \frac{1}{3}t^{3} k
v = r′(t) = i + t j + t^{2} k
a = r′′(t) =  j + 2t k
speed = \frac{ds}{dt} = ||v|| = \sqrt{t^{4} + t^{2} + 1}
a_{T} = \frac{d^{2}s}{dt^{2}} = ||v||′ = \frac{\frac{1}{2}(4t^{3} + 2t)}{\sqrt{t^{4} + t^{2} + 1}} = \frac{2t^{3} + t}{\sqrt{t^{4} + t^{2} + 1}}

T = ( i + t j + t^{2} k)/\sqrt{t^{4} + t^{2} + 1}
\frac{d^{2}s}{dt^{2}}T = \frac{2t^{3} + t}{t^{4} + t^{2} + 1} ( i + t j + t^{2} k)
k (\frac{ds}{dt})^{2}N = a − \frac{d^{2}s}{dt^{2}}T

= \frac{(−2t^{3} − t) i + (−t^{4} + 1) j + (t^{3} + 2t) k}{t^{4} + 2t + 1}

a_{N} = k (\frac{ds}{dt})^{2} = ||k (\frac{ds}{dt})^{2}N||

= \frac{\sqrt{(−2t^{3} − t)^{2} + (−t^{4} + 1)^{2} + (t^{3} + 2t)^{2}}}{t^{4} + t^{2} + 1}

= \frac{\sqrt{t^{8} + 5t^{6} + 6t^{4} + 5t^{2} + 1}}{t^{4} + t^{2} + 1}

= \frac{\sqrt{(t^{4} + t^{2} + 1)(t^{4} + 4t^{2} + 1)}}{t^{4} + 4t^{2} + 1}

= \begin{matrix}√ & \overline{t^{4} + 4t^{2} + 1} \\ √ & \overline{t^{4} + t^{2} + 1} \end{matrix}

k = \frac{a_{N}}{(ds/dt)^{2}} = \frac{(t^{4} + 4t^{2} + 1)^{\frac{1}{2}}}{(t^{4} + t^{2} + 1)^{\frac{3}{2}}}