Question

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

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

Short answer

Ans: The principal unit normal vector of $\mathbf{r}\left(t\right)=⟨\alpha \sin \beta t,\alpha \cos \beta t⟩$at $t=0$is$⟨0,-1⟩$

Step-by-step solution

Step 1. Given information:

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

Step 2. Simplifying the principal unit normal vector:

The principal unit normal vector of $r\left(t\right)$denoted by $N\left(t\right)$is given by

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

$T\left(t\right)$should be calculated first and then $N\left(t\right)$is to be evaluated.

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

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

The unit tangent vector

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

To find $N\left(t\right)$we divide $T^{\text{'}}\left(t\right)$by its magnitude.

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

Step 3. Finding the principal unit normal vector:

The principal unit normal vector

$$\begin{gathered} N\left(t\right)=\frac{T^{\text{'}}\left(t\right)}{||T^{\text{'}}\left(t\right)||} \\ =\frac{⟨-\beta \sin \beta t,-\beta \cos \beta t⟩}{\beta } \\ =⟨-\sin \beta t,-\cos \beta t⟩ \end{gathered}$$

At $t=0,N\left(t\right)=N\left(0\right)=⟨-\sin 0,-\cos 0⟩=⟨0,-1⟩$

Thus the principle unit normal vector of $r\left(t\right)$at $t=0$is $⟨0,-1⟩$