Find the unit normal find the unit normal $\bf \hat{N}$ of
$${\bf r}=6 \mathrm{e}^{-14 t}\cos(t){\bf i}+6 \mathrm{e}^{-14 t}\sin(t){\bf j}$$
The answer should be in vector form. Use t as parameter. Write $e^x$ for exponentials.
Have been working with this a long time now but cant get the right answer.
My answer is 
$$(-((e^{-14t})(\cos(t)-14\sin(t)))/((\sqrt{12})\sqrt{e^{-28t}}), (-((e^{-14t})(\sin(t)+14\cos(t)))/((\sqrt{12})\sqrt{(e^{-28t}})),0)$$
but it aint right. Thx for help!
 A: Here is the long computation...
Compute $v(t) = r'(t)$:
$$\begin{aligned}
v(t) = \left\langle -84e^{-14t}\cos(t)- 6e^{-14t}\sin(t), -84e^{-14t}\sin(t) + 6e^{-14t}\cos(t) \right\rangle
\end{aligned}$$
By the definition of unit tangent vector,
$$\begin{aligned}
T(t) &= \dfrac{1}{\| v(t) \|}\cdot v(t)\\
&= \dfrac{1}{\sqrt{(-84e^{-14t}\cos(t)- 6e^{-14t}\sin(t))^2 + (-84e^{-14t}\sin(t) + 6e^{-14t}\cos(t))^2}} \cdot \left\langle -84e^{-14t}\cos(t)- 6e^{-14t}\sin(t), -84e^{-14t}\sin(t) + 6e^{-14t}\cos(t) \right\rangle\\
&= \dfrac{1}{6\sqrt{197e^{-28t}}} \cdot \left\langle -84e^{-14t}\cos(t)- 6e^{-14t}\sin(t), -84e^{-14t}\sin(t) + 6e^{-14t}\cos(t) \right\rangle
\end{aligned}$$
Use this form $N(t) = \dfrac{T'(t)}{\|T'(t) \|}$ and see if you get the different answer.
A: Let $s=\sin(t)$, $c=\cos(t)$, $a=6\exp(-14t)$, $r=r(t)$.  We have $r'=(a'c-as,a's+ac)$. Now observe that $a'=-14a$, hence $r'=a(-14c-s,-14s+c)$.  From here $\|r'\|^2=197a^2$.  Finally the unit normal is
$$\frac{1}{\sqrt{197}}(-14c-s,-14s+c).$$
