Find the unit normal $N$ of ${\bf r}=14 \mathrm{e}^{-10 t}\cos(t){\bf i}+14 \mathrm{e}^{-10 t}\sin(t){\bf j}$
The answer should be in vector form. 
I can't find an easy way to do this.
I end up with 
T being something complicated as in one term
$$\frac{\mathrm{e}^{-10 t}(\sin(t)+10\cos(t)}{\sqrt{101}\sqrt{\mathrm{e}^{-20 t}}}$$
 A: Hint. In order to find the unit normal of $\boldsymbol{r}(t)$, you should first obtain the unit tangent, which is
$$
\boldsymbol{\tau}(t)=\frac{1}{\|\boldsymbol{r}'(t)\|}\boldsymbol{r}'(t)=a(t)\,{\bf i}+b(t)\,\bf j,
$$
and then what you look for is going to be
$$
{\boldsymbol\nu}(t)=b(t)\,{\bf i}-a(t)\,\bf j.
$$
A: By letting $a(t)=14e^{-10t}$ we have 
$$r'(t)=a'(t)\begin{pmatrix}\cos(t)\\ \sin(t)\end{pmatrix}+a(t)\begin{pmatrix}-\sin(t)\\ \cos(t)\end{pmatrix}.$$
Rotating that vector counterclockwise by $\pi/2$ yields the normal vector
$$n(t)=a'(t)\begin{pmatrix}-\sin(t)\\ \cos(t)\end{pmatrix}+a(t)\begin{pmatrix}-\cos(t)\\ -\sin(t)\end{pmatrix}.$$
You'll easily verify that it's length is $\sqrt{a'^2(t)+a^2(t)}$.  Since $a'(t)=-10a(t)$ the length equals $a(t)\sqrt{101}$ and 
$$n(t)=a(t)\left(-10\begin{pmatrix}-\sin(t)\\ \cos(t)\end{pmatrix}+\begin{pmatrix}-\cos(t)\\ -\sin(t)\end{pmatrix}\right),$$
hence the unit length normal vector is 
$$\frac{1}{\sqrt{101}}\left(-10\begin{pmatrix}-\sin(t)\\ \cos(t)\end{pmatrix}+\begin{pmatrix}-\cos(t)\\ -\sin(t)\end{pmatrix}\right).$$
A: $$\mathbf{r}=\langle14e^{-10t}\cos(t),14e^{-10t}\sin(t)\rangle$$
Therefore, the magnitude $$||\mathbf{r}||=14e^{-10t}$$
By trignometric relations. Therefore, the unit vector is
$$N=\cos(t)\hat{i}+\sin(t)\hat{j}$$
This is just one of the unit vectors in cylindrical coordinates.
