Question

Find the unit tangent T, the unit normal N and the curvature K for the curve r(t) =<2 sint, 5t, 2 cost>.

Answer 1

$\vec{r}(t) = \lbrace 2\sin t, 5t, 2\cos t \rbrace.$

Let us take derivatives with respect to t:

$\vec{r}'(t) = \lbrace 2\cos t, 5, -2\sin t \rbrace, \;\; \vec{r}''(t) = \lbrace -2\sin t, 0, -2\cos t \rbrace.$

The tangent is $\vec{\tau}(t) = \vec{r}'(t) = \lbrace 2\cos t, 5, -2\sin t \rbrace.$ We should normalize this vector (norm is $\sqrt{4\cos^2t + 25 + 4\sin^2t} = \sqrt{29}$ ), so the unit tangent is $\vec{T} =\left \lbrace \dfrac{2\cos t}{\sqrt{29}}, \dfrac{5}{\sqrt{29}}, \dfrac{-2\sin t}{\sqrt{29}} \right\rbrace$ .

The normal is $\vec{\beta}(t) = \vec{r}'(t)\times (\vec{r}'(t)\times\vec{r}''(t) ) .$

$\vec{r}'(t)\times\vec{r}''(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2\cos t & 5 & -2\sin t \\ -2\sin t & 0 & -2\cos t \end{vmatrix} = \lbrace -10\cos t, 4 , 10\sin t \rbrace .$

$\vec{r}'(t)\times (\vec{r}'(t)\times\vec{r}''(t) ) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2\cos t & 5 & -2\sin t \\ -10\cos t & 4 & 10\sin t \end{vmatrix} = \lbrace 58\cos t, 0, 58\sin t \rbrace.$

The unit normal is $\vec{N} = \lbrace \cos t, 0, \sin t \rbrace.$

The curvature is $k = \dfrac{||\vec{r}'(t)\times\vec{r}''(t)||}{||\vec{r}'(t)||^3} = \dfrac{\sqrt{116}}{\sqrt{29}^3} = \dfrac{2}{29}.$