Find the tangential and normal components of the acceleration vector for the curve
$$ \vec r (t) = 〈 2t,t^2 ,\ln t 〉$$
I found this problem in a notebook and I did a part, but I was stuck and I don't know if I was doing it correctly. If you can help me it would be great. Thanks!
$$r'(t)= v(t)=<2,2t,1/t>$$ $$v'(t)=<0,2,-1/t^2>$$ $$\Vert r'(t)\Vert= \sqrt{4+4t^2+1/t^2}=2t+1/t. $$ $$<-4/t,-2/t^2,4>$$ $$k(t)=\frac{4t^3+2t}{(2t^2+1)^3}$$ $$a_T=<0,2,-1/t^2>$$ $$a_N=<4,4t^2,1/t^2>$$
I've come here and I do not know if it's this way or how?Sorry for the mistakes I'm starting to use the page.
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$\begingroup$Outline:
Position: $$ \vec r (t) = < 2t, t^2 , \ln t >$$
Velocity: $$ \vec v(t)= < 2,2t,1/t>$$
Acceleration: $$\vec a (t)=<0,2,-1/t^2>$$
Speed: $$ v(t) = \Vert \vec v(t)\Vert= \sqrt{4+4t^2+1/t^2}=2t+1/t $$
The unit vector in the direction of the velocity is $$ \vec u(t) = \frac{\vec v(t)}{ v(t)} = \frac{1}{2t+1/t} < 2,2t,1/t> $$
The tangential component of the acceleration $a_T(t)$ can be found by $a_T(t) = \vec a(t) \cdot \vec u (t)$
The normal component of the acceleration, $\vec a_N$ can then be found using $\vec a(t) = a_T (t) \vec u_T(t) + a_N (t) \vec u_N(t)$.
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