91影视

Given $r\left(t\right)=\left(\sin \left(2t\right),\cos \left(2t\right),t\right)$.

The objective is to find the unit tangent vector and the principal unit normal vector at$t=\frac{\mathrm{蟺}}{4}$

For $r\left(t\right)=\left(\sin \left(2t\right),\cos \left(2t\right),t\right)$, the first derivative is given as

$r\text{'}\left(t\right)=\left(2\cos \left(2t\right),-2\sin \left(2t\right),1\right)$and the magnitude of first derivative is given as:

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

So the unit tangent vector is given as:

localid="1654165332485" $\begin{array}{rcl}T\left(t\right)& =& \frac{r\text{'}\left(t\right)}{鈭�r\text{'}\left(t\right)鈭�}\\ & =& \frac{\left(2\cos \left(2t\right),-2\sin \left(2t\right),1\right)}{\sqrt{5}}\end{array}$

And at $t=\frac{\mathrm{蟺}}{4}$we have:

localid="1654165516099" $\begin{array}{rcl}T\left(\frac{\mathrm{蟺}}{4}\right)& =& \frac{\left(2\cos \left(2脳{\displaystyle \frac{\mathrm{蟺}}{4}}\right),-2\sin \left(2脳{\displaystyle \frac{\mathrm{蟺}}{4}}\right),1\right)}{\sqrt{5}}\\ & =& \frac{\left(2\cos \left({\displaystyle \frac{\mathrm{蟺}}{2}}\right),-2\sin \left({\displaystyle \frac{\mathrm{蟺}}{2}}\right),1\right)}{\sqrt{5}}\\ & =& \frac{\left(0,-2,1\right)}{\sqrt{5}}\\ & =& \left(0,-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)\\ & =& \left(0,-\frac{2\sqrt{5}}{5},\frac{1}{\sqrt{5}}\right)\end{array}$

Now for the unit tangent vector $\begin{array}{rcl}T\left(t\right)& =& \frac{\left(2\cos \left(2t\right),-2\sin \left(2t\right),1\right)}{\sqrt{5}}\end{array}$, the first derivative is given as $T\text{'}\left(t\right)=\frac{1}{\sqrt{5}}\left(-4\sin \left(2t\right),-4\cos \left(2t\right),0\right)$and the magnitude of first derivative of unit tangent vector is:

role="math" localid="1654165430231" $\begin{array}{rcl}& 鈭�& T\left(t\right)鈭�=鈭�\frac{1}{\sqrt{5}}\left(-4\sin \left(2t\right),-4\cos \left(2t\right),0\right)鈭�\\ & =& \frac{1}{\sqrt{5}}\sqrt{16\left({\sin }^2\left(2t\right)+{\cos }^2\left(2t\right)\right)}\\ & =& \frac{4}{\sqrt{5}}\end{array}$

So the principal unit normal vector is given as:

$\begin{array}{rcl}N\left(t\right)& =& \frac{T\text{'}\left(t\right)}{鈭�T\text{'}\left(t\right)鈭�}\\ & =& \frac{\frac{1}{\sqrt{5}}\left(-4\sin \left(2t\right),-4\cos \left(2t\right),0\right)}{{\displaystyle \frac{4}{\sqrt{5}}}}\end{array}$

And at $t=\frac{\mathrm{蟺}}{4}$we have

$\begin{array}{rcl}N\left(\frac{\mathrm{蟺}}{4}\right)& =& \frac{1}{4}\left(-4\sin \left(\frac{2\mathrm{蟺}}{4}\right),-4\cos \left(\frac{2\mathrm{蟺}}{4}\right),0\right)\\ & =& \frac{1}{4}\left(-4,0,0\right)\\ & =& \left(-1,0,0\right)\end{array}$