Consider the vector function r(t) =< 4t + 1, t2 − 3t, cos(t) >.
(a) Find the unit tangent vector T at (1, 0, 1)
I know that T = r'(t)/|r-(t)|
and r'(t) = < 4, 2t-3, -sin(t)>
I'm not sure why the book plugs in 0 for t.
Thanks in advance
Consider the vector function r(t) =< 4t + 1, t2 − 3t, cos(t) >.
(a) Find the unit tangent vector T at (1, 0, 1)
I know that T = r'(t)/|r-(t)|
and r'(t) = < 4, 2t-3, -sin(t)>
I'm not sure why the book plugs in 0 for t.
Thanks in advance
Ok $\displaystyle \bold{r}(t)=\left\langle4t+1,t^2-3t,\cos(t)\right\rangle$
So as you noted $\displaystyle \bold{T}(t)=\frac{\bold{T}'(t)}{|\bold{T}'(t)|}$
So we can compute this to be
$\displaystyle \bold{T}(t)=\frac{\langle4,2t-3,-\sin(t)\rangle}{\sqrt{16+\left(2t-3\right)^3+\sin^2(t)}}$
So $\displaystyle \bold{T}(1,0,1)=\frac{\langle4,-3,-\sin(1)\rangle}{\sqrt{16+9+\sin^2(1)}}$
$\displaystyle \bold r(t)=\left<4t+1,t^2-3t,\cos t\right>$ at the point $\displaystyle (1,0,1)$ implies that there is a t value such that $\displaystyle 4t+1=1,~t^2-3t=0,$ and $\displaystyle \cos(t)=1$
It should be a little clearer why they picked $\displaystyle t=0$...
--Chris