Question

Principal unit normal vectors: Find the principal unit normal vector for the given function at the specified value of t.

$\mathbf{r}\left(t\right)=⟨t,\alpha \sin \beta t,\alpha \cos \beta t⟩,t=0$

Short answer

The principal unit normal vector to r(t) is,

$⟨0,0,-1⟩$

Step-by-step solution

Step 1. Given Information 

We are given an function,

$\mathbf{r}\left(t\right)=⟨t,\alpha \sin \beta t,\alpha \cos \beta t⟩,t=0$

Step 2. Finding principal unit normal vector  

Principal unit normal vector:

If r(t) is a twice differentiable vector function, then the principal unit vector to r(t) is,

$N\left(t\right)=\frac{T^{\text{'}}\left(t\right)}{||T^{\text{'}}\left(t\right)||}$

Where $T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||}$

T(t) should be calculated first and then N(t) is to be evaluated.

We start by comparing $r^{\text{'}}\left(t\right)$first:

$$\begin{gathered} r^{\text{'}}\left(t\right)=⟨1,\alpha \beta \cos \beta t,-\alpha \beta \sin \beta t⟩ \\ \left|r^{\text{'}}\left(t\right)\right|=\sqrt{1+(\alpha \beta \cos \beta t{)}^2+1-(\alpha \beta \sin \beta t{)}^2} \\ =\sqrt{1+{\alpha }^2{\beta }^2} \end{gathered}$$

The unit tangent vector to r(t) is

$$\begin{gathered} T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{⟨1,\alpha \beta \cos \beta t,-\alpha \beta \sin \beta t⟩}{\sqrt{1+{\alpha }^2{\beta }^2}} \end{gathered}$$

Step 3. Finding principal unit normal vector.

Dividing $T^{\text{'}}\left(t\right)$by its magnitde,

The principal unit normal vector.

At $t=0,$

role="math" localid="1649675362152" $$\begin{gathered} N\left(t\right)=N\left(0\right) \\ =⟨0,-\sin 0,-\cos 0⟩ \\ =⟨0,0,-1⟩ \end{gathered}$$

Thus, the principle unit normal vector of r(t) is $⟨0,0,-1⟩$.