Question

Binormal vectors and osculating planes: Find the binormal vector and equation of the osculating plane for the given function at the specified value of t.

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

Short answer

The binomial vector is $⟨0,0,-1⟩$.

The equation of the osculating plane is $z=0$.

Step-by-step solution

Step 1. Given Information  

We are given,

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

Step 2. Finding the binormal vector. 

Finding the binormal vector,

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

At $t=0$,

role="math" localid="1649754634630" $$\begin{gathered} T\left(t\right)=T\left(0\right) \\ =⟨\cos 0,-\sin 0⟩ \\ =⟨1,0⟩ \\ {T}^{\text{'}}\left(t\right)=⟨-\beta \sin \beta t,-\beta \cos \beta t⟩ \\ ||T^{\text{'}}\left(t\right)||=\sqrt{(-\beta \sin \beta {)}^2+(-\beta \cos \beta t{)}^2} \\ =\beta \\ N\left(t\right)=\frac{T^{\text{'}}\left(t\right)}{||T^{\text{'}}\left(t\right)||} \\ =\frac{⟨-\beta \sin \beta t,-\beta \cos \beta t⟩}{\beta } \\ N\left(t\right)=⟨-\sin \beta t,-\cos \beta t⟩ \end{gathered}$$

Step 3. Finding the binormal vector. 

At $t=0$,

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

Now,

The binormal vector is given by,

$$\begin{gathered} B\left(0\right)=T\left(0\right)\times N\left(0\right) \\ =⟨1,0⟩\times ⟨0,-1⟩ \\ =\left|\begin{array}{ccc}i& j& k\\ 1& 0& 0\\ 0& -1& 0\end{array}\right| \\ =k(-1) \\ =-k \\ =⟨0,0,-1⟩ \end{gathered}$$

Hence, the binormal vector is $⟨0,0,-1⟩$.

Step 4. Finding the equation of the osculating plane  

The equation of the osculating at $r(t=0)$is defined by,

$$\begin{gathered} B\left(0\right)·⟨x-x\left(0\right),y-y\left(0\right),z-z\left(0\right)⟩=0 \\ ⟨0,0,-1⟩·⟨x-0,y-\alpha ,z⟩=0 \\ -z=0 \\ z=0 \end{gathered}$$

Hence, the equation is $z=0$.