Answer with Step-by-step explanation:
We are given that
[tex]r(t)=< t^2,\frac{2}{3}t^3,t >[/tex]
We have to find T,N and B at the given point t > (1,[tex]2/3[/tex],1)
[tex]r'(t)=<2t,2t^2,1>[/tex]
[tex]\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1[/tex]
[tex]T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{<2t,2t^2,1>}{2t^2+1}[/tex]
Now, substitute t=1
[tex]T(1)=\frac{<2,2,1>}{2+1}=\frac{1}{3}<2,2,1>[/tex]
[tex]T'(t)=\frac{-4t}{(2t^2+1)^2} <2t,2t^2,1>+\frac{1}{2t^2+1}<2,4t,0>[/tex]
[tex]T'(1)=-\frac{4}{9}<2,2,1>+\frac{1}{3} <2,4,0>[/tex]
[tex]T'(1)=\frac{1}{9}<-8+6,-8+12,-4+0>=<\frac{-2}{9},\frac{4}{9},\frac{-4}{9} >[/tex]
[tex]\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}[/tex]
[tex]N(1)=\frac{T'(1)}{\mid T'(1)\mid}[/tex]
[tex]N(1)=\frac{<\frac{-2}{9},\frac{4}{9},\frac{-4}{9}>}{\frac{2}{3}}=<\frac{-1}{3},\frac{2}{3},\frac{-2}{3}>[/tex]
[tex]N(1)=<\frac{-1}{3},\frac{2}{3},\frac{-2}{3}>[/tex]
[tex]B(1)=T(1)\times N(1)[/tex]
[tex]B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}[/tex]
[tex]B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})[/tex]
[tex]B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k[/tex]
[tex]B(1)=\frac{1}{3}<-2,1,2>[/tex]
