91Ó°ÊÓ

We have been given the vector valued function $r\left(t\right)=⟨\sin 2t,\cos 2t,tâŸ\copyright$.

The objective is to find the unit tangent vector, the principal unit normal vector, the binormal vector, and the equation of the osculating plane at$t=\frac{\mathrm{Ï€}}{2}$.

For the vector valued function $r\left(t\right)=⟨\sin 2t,\cos 2t,tâŸ\copyright$, the first derivative is given as- role="math" localid="1654247791574" and so its magnitude is given as:

$\begin{array}{rcl}||r\text{'}\left(t\right)||& =& \sqrt{4{\cos }^2\left(2t\right)+4{\sin }^2\left(2t\right)+1}\\ & =& \sqrt{4\left({\cos }^2\left(2t\right)+{\sin }^2\left(2t\right)\right)+1}\\ & =& \sqrt{4+1}\\ & =& \sqrt{5}\end{array}$

Now the unit tangent vector is given as:

role="math" localid="1654248032465"

And at $t=\frac{\mathrm{Ï€}}{2}$we have

For $T\left(t\right)=\begin{array}{rcl}& & ⟨\frac{2}{\sqrt{5}}\cos 2t,-\frac{2}{\sqrt{5}}\sin 2t,\frac{1}{\sqrt{5}}⟩\end{array}$the first derivative is given as: role="math" localid="1654248202393" and its magnitude is:

$\begin{array}{rcl}||T\text{'}\left(t\right)||& =& \sqrt{\frac{16}{5}{\sin }^2\left(2t\right)+\frac{16}{5}{\cos }^2\left(2t\right)+0}\\ & =& \sqrt{\frac{16}{5}\left({\sin }^2\left(2t\right)+{\cos }^2\left(2t\right)\right)}\\ & =& \sqrt{\frac{16}{5}}\\ & =& \frac{4}{\sqrt{5}}\end{array}$

So the principal normal unit vector is given as:

role="math"

At $t=\frac{\mathrm{Ï€}}{2}$its value is:

The binomial vector is given as:

The equation of the osculating plane is given as: