Question

Find the unit tangent vector T(t) to the curve r(t) = [sin(t), 1 + t, cos(t)] when t = 0.

Answer 1

Compute the derivative of
`\mathbf r(t)` at
`t=0` - this will be the tangent vector - then normalize it by dividing it by its magnitude to get the unit tangent vector
`\mathbf T(t)`.

`\mathbf r(t) = \langle \sin(t), 1+t, \cos(t) \rangle \implies (d\mathbf r)/(dt) = \langle \cos(t), 1, -\sin(t) \rangle \implies (d\mathbf r)/(dt)(0) = \langle 1, 1, 0 \rangle`

`\|\langle1,1,0\rangle\| = √(1^2+1^2+0^2) = \sqrt2`

`\implies\mathbf T(0) = (\langle1,1,0\rangle)/(\sqrt2) = \boxed{\left\langle \frac1{\sqrt2}, \frac1{\sqrt2}, 0\right\rangle}`