Find unit vector, normal vector, binomial vector, and curvature at the given point

Question

Suppose r(t) = (√2t, e^t, e^-t).

(a) Find the unit tangent vector. T(t), at (0,1,1).

(b) Find the principal unit normal vector, N(t), at (0.1.1).

(c) Find the binormal vector. B(t). at (0.1.1).

(d) Find the curvature at (0, 1, 1).

Answer 1

r(t) = < √2t, e^t, e^-t >

r' (t ) = < √2, e^t, -e^-t > , || r' (t ) || = e^t + e^-t and the point (0 , 1, 1 ) can be obtained when t = 0

Therefore r' (0 ) = < √2 , 1, -1 > and || r' (0 ) || =2

T ( 0 ) = r' (0 ) / || r' (0 ) || = (1/2) < √2 , 1, -1 > a unit vector.

T (t ) = r' (t ) / || r' (t ) || and T ' (t ) = (d / dt ) { < √2, e^t, -e^-t > [ 1/ (e^t + e^-t ) ] }

T ' (t ) = <0, e^t, e^-t > [ 1/ (e^t + e^-t ) ] − ( e^t -e^t ) < √2, e^t, -e^-t > [ 1/ (e^t + e^-t )² ] and

T ' (0 ) =(1/2)< 0 , 1, 1 > ⇒ || T ' (0 ) || = 1 /√2

Then N (0) = ( 1 /√2 ) < 0, 1 ,1 > another unit vector.

Hence B (0 ) = T ( 0 ) x N (0) = [ (1/2) < √2 , 1, -1 > ] x [ ( 1 /√2 ) < 0, 1 ,1 > ] = ( 1 / 2√2 ) < 2, − √2, √2 >

B (0 ) = ( 1 / 2√2 ) < 2, − √2, √2 >.

Curvature

κ (t ) = || r' (t ) x r' ' (t ) || / [ || | r' (t ) || ]³

κ (0) = { || < 2 , −√2, √2 || } /{ || < √2, 1, − 1 > || }³ = (2√2 ) / 2³ = √2 / 4