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.

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

We are given,

Finding the binormal vector,

$$\begin{gathered} \mathbf{r}\left(t\right)=⟨t,\alpha \sin \beta t,\alpha \cos \beta t⟩,t=0 \\ \mathbf{r}\left(t\right)=⟨t,\alpha \sin \beta t,\alpha \cos \beta t⟩,t=0 \\ ||r^{\text{'}}\left(t\right)||=\sqrt{1+{\alpha }^2{\beta }^2\left({\cos }^2\beta t+{\sin }^2\beta t\right)} \\ =\sqrt{1+{\alpha }^2{\beta }^2} \\ T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{1}{\sqrt{1+{\alpha }^2{\beta }^2}}⟨1,\alpha \beta \cos \beta t,-\alpha \beta \sin \beta t⟩ \end{gathered}$$

At $t=0$,

$$\begin{gathered} T\left(t\right)=T\left(0\right) \\ =\frac{1}{\sqrt{1+{\alpha }^2{\beta }^2}}⟨1,\alpha \beta ,0⟩ \end{gathered}$$

Thus, the unit tangent vector is $\frac{1}{\sqrt{1+{\alpha }^2{\beta }^2}}⟨1,\alpha \beta ,0⟩$.

By using the Quotient Rule for derivatives,

Thus, the principal unit normal vector to r(t) is $⟨0,0,-1⟩$,

The binormal vector is given by,

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

Hence, the equation is $-\alpha \beta x+y=0$.