Find the tangent and normal lines to the curve $\gamma(t)=(2\cos(t)-\cos(2t), 2\sin(t)-\sin(2t))$ at $t=\frac{\pi}{4}$ The normal line to a curve in the plane at a point $\mathbf p$ is the straight line passing through $\mathbf p$ perpendicular to the tangent line at $\mathbf p$. Find the tangent and normal lines to the curve $\gamma(t)=(2\cos(t)-\cos(2t), 2\sin(t)-\sin(2t))$ at $t=\frac{\pi}{4}$
What I have so far:
$$\gamma(\frac{\pi}{4})=(\sqrt{2} , \sqrt{2} -1)$$
and then determine $\dot\gamma$:
$$\dot\gamma(t)=(-2\sin(t)+2\sin(2t), 2\cos(t)-2\cos(2t)$$
$$\dot\gamma(\frac{\pi}{4})=\sqrt{2}(\sqrt{2}-1,1)$$
$\dot\gamma(\frac{\pi}{4})$ gives you the slope of the tangent line, $\mathbf m$, and from this you can determine the slope of the normal line:
$$-\frac{1}{\mathbf m} = -\frac{1}{\sqrt{2}}(1-\frac{1}{\sqrt{2}},1)$$
This is as far as I got as I can't fathom how to put all this information together to find the tangent and normal lines from here.
 A: HINT: 
I would say that
$x=2cos(t)-cos(2t), y=2sin(t)-sin(2t), t=\pi/4 \Rightarrow P(\sqrt{2},\sqrt{2}-1)$
$dx = -2sin(t)+2sin(2t), dy = 2cos(t)-2cos(2t)$
$\Rightarrow m=\frac{dy}{dx}_{[t=\pi/4]}=\frac{\sqrt{2}}{2-\sqrt{2}}, m'=-\frac{1}{m}=-\frac{2-\sqrt{2}}{\sqrt{2}}$
So
Tangent line: equation of straight line passing through $P(\sqrt{2},\sqrt{2}-1)$, slope of the line $m =\frac{\sqrt{2}}{2-\sqrt{2}}$,
Normal line: equation of straight line passing through $P(\sqrt{2},\sqrt{2}-1)$, slope of the line $m'=-\frac{2-\sqrt{2}}{\sqrt{2}}$.
A: [Notice that $\gamma^{\prime}(\frac{\pi}{4})$ gives a tangent vector to the curve instead of the slope of a line.]
If you let $\vec{v}=\gamma^{\prime}(\frac{\pi}{4})=\langle2-\sqrt{2}, \sqrt{2}\rangle$, you can write the tangent line in the form
$\vec{r}=\langle x,y\rangle=\vec{p}+t\vec{v}=\langle\sqrt{2},\sqrt{2}-1\rangle+t\langle2-\sqrt{2},\sqrt{2}\rangle$, and this gives
$x=\sqrt{2}+t(2-\sqrt{2})$ and $y=\sqrt{2}-1+t\sqrt{2}$.
(You can write this in the form $y=mx+b$ by solving for $t$ in the 1st equation and then substituting into the 2nd equation.)

Similarly, if we take a vector $\vec{w}=\langle\sqrt{2}, \sqrt{2}-2\rangle$ which is orthogonal to $\vec{v}$,
we can write the normal line in the form
$\vec{r}=\langle x,y\rangle=\vec{p}+t\vec{w}=\langle\sqrt{2},\sqrt{2}-1\rangle+t\langle\sqrt{2},\sqrt{2}-2\rangle$, and this gives
$x=\sqrt{2}+t\sqrt{2}$ and $y=\sqrt{2}-1+t(\sqrt{2}-2)$.
A: I think you have solved it completely.

Normal is perpendicular to Tangent.
