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 $\mathbf T(t)$.

$\mathbf r(t) = \langle \sin(t), 1+t, \cos(t) \rangle \implies \dfrac{d\mathbf r}{dt} = \langle \cos(t), 1, -\sin(t) \rangle \implies \dfrac{d\mathbf r}{dt}(0) = \langle 1, 1, 0 \rangle$

$\|\langle1,1,0\rangle\| = \sqrt{1^2+1^2+0^2} = \sqrt2$

$\implies\mathbf T(0) = \dfrac{\langle1,1,0\rangle}{\sqrt2} = \boxed{\left\langle \dfrac1{\sqrt2}, \dfrac1{\sqrt2}, 0\right\rangle}$