Question

consider the helix r(t) = <cos(3t), sin(3t), -1t>. compute the following at t = π/6: a. the unit tangent vector t = <, , > b. the unit normal vector n = <, , > c. the unit binormal vector b = <, , > d. the curvature κ =

Explanation:

Step1: Find the derivative of $\mathbf{r}(t)$

First, find $\mathbf{r}'(t)$. Given $\mathbf{r}(t)=\langle\cos(3t),\sin(3t), - t
angle$, then $\mathbf{r}'(t)=\langle - 3\sin(3t),3\cos(3t),-1
angle$.

Step2: Calculate the magnitude of $\mathbf{r}'(t)$

The magnitude of $\mathbf{r}'(t)$ is $\|\mathbf{r}'(t)\|=\sqrt{(-3\sin(3t))^{2}+(3\cos(3t))^{2}+(-1)^{2}}=\sqrt{9\sin^{2}(3t)+9\cos^{2}(3t) + 1}=\sqrt{9(\sin^{2}(3t)+\cos^{2}(3t))+1}=\sqrt{9 + 1}=\sqrt{10}$.

Step3: Find the unit - tangent vector $\mathbf{T}(t)$

The unit - tangent vector $\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}=\langle-\frac{3}{\sqrt{10}}\sin(3t),\frac{3}{\sqrt{10}}\cos(3t),-\frac{1}{\sqrt{10}}
angle$.
When $t = \frac{\pi}{6}$, $\sin(3t)=\sin(\frac{\pi}{2}) = 1$ and $\cos(3t)=\cos(\frac{\pi}{2})=0$. So $\mathbf{T}(\frac{\pi}{6})=\langle-\frac{3}{\sqrt{10}},0,-\frac{1}{\sqrt{10}}
angle$.

Step4: Find the derivative of $\mathbf{T}(t)$

$\mathbf{T}'(t)=\langle-\frac{9}{\sqrt{10}}\cos(3t),-\frac{9}{\sqrt{10}}\sin(3t),0
angle$.

Step5: Calculate the magnitude of $\mathbf{T}'(t)$

$\|\mathbf{T}'(t)\|=\sqrt{(-\frac{9}{\sqrt{10}}\cos(3t))^{2}+(-\frac{9}{\sqrt{10}}\sin(3t))^{2}+0^{2}}=\frac{9}{\sqrt{10}}$.

Step6: Find the unit - normal vector $\mathbf{N}(t)$

$\mathbf{N}(t)=\frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|}=\langle-\cos(3t),-\sin(3t),0
angle$.
When $t=\frac{\pi}{6}$, $\mathbf{N}(\frac{\pi}{6})=\langle0, - 1,0
angle$.

Step7: Calculate the binormal vector $\mathbf{B}(t)$

$\mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t)$. Using the cross - product formula for two vectors $\mathbf{a}=\langle a_1,a_2,a_3
angle$ and $\mathbf{b}=\langle b_1,b_2,b_3
angle$: $\mathbf{a}\times\mathbf{b}=\langle a_2b_3 - a_3b_2,a_3b_1 - a_1b_3,a_1b_2 - a_2b_1
angle$.
$\mathbf{T}(\frac{\pi}{6})\times\mathbf{N}(\frac{\pi}{6})=

$$\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-\frac{3}{\sqrt{10}}&0&-\frac{1}{\sqrt{10}}\\0&- 1&0\end{vmatrix}$$

=\langle-\frac{1}{\sqrt{10}},0,\frac{3}{\sqrt{10}}
angle$.

Step8: Calculate the curvature $\kappa(t)$

The curvature $\kappa(t)=\frac{\|\mathbf{T}'(t)\|}{\|\mathbf{r}'(t)\|}$. Since $\|\mathbf{T}'(t)\|=\frac{9}{\sqrt{10}}$ and $\|\mathbf{r}'(t)\|=\sqrt{10}$, then $\kappa(t)=\frac{9}{10}$.

Answer:

A. $\mathbf{T}(\frac{\pi}{6})=\langle-\frac{3}{\sqrt{10}},0,-\frac{1}{\sqrt{10}}
angle$
B. $\mathbf{N}(\frac{\pi}{6})=\langle0, - 1,0
angle$
C. $\mathbf{B}(\frac{\pi}{6})=\langle-\frac{1}{\sqrt{10}},0,\frac{3}{\sqrt{10}}
angle$
D. $\kappa=\frac{9}{10}$