determine unit outward normal vector on a curve It is necessary for me to find unit outward normal vector for the curve:
$$\gamma=(x(t),y(t))$$
where
$$x(t)=(0.6)\cos(t)-(0.3)\cos(3t)$$
and 
$$y(t)=(0.7)\sin(t)+(0.07)\sin(7t)+(0.1)\sin(3t)$$
I know how to find unit outward  normal vector  for this: using
$$T=\frac{\gamma'(t)}{||\gamma(t)||},\;\text{ so }\,N=\frac{T'(t)}{||T(t)||}$$
but my problem is that I do not have $t$. Just I have $x(t)$ and $y(t)$.
How could I find $t$ or $N$ without need to $t$.
Is there any command in MATLAB or MAPLE to this?
 A: $$
\gamma'(t) = (-\sin(t) + 1.5 \sin(3t), \cos(t)+7\cos(7t)+3\cos(3t))\\
\| \gamma'(t) \| = \sqrt{(-\sin(t) + 1.5 \sin(3t))^2, (\cos(t)+7\cos(7t)+3\cos(3t))^2}
$$
In general, for a vector $(a, b)$ in the plane, $(-b, a)$ is perpendicular to it. 
So your normal vector is 
$$
N(t) = \frac{\pm (-\cos(t)-7\cos(7t)-3\cos(3t), -\sin(t) + 1.5 \sin(3t))}{\| \gamma'(t) \|}
$$
To choose the sign, you may want to make it point in the direction that has a positive dot product with $T'$; to do so by differentiating the quotient that defines $T$ would be a pain in the neck, but fortunately, you can merely make your nromal vector point in the direction that has positive dot product with $\gamma''$, since $T'$ ends up being a linear combination of this and a vector in the $T$ direction, which will not affect the dot product. So:
Compute
$$
s = (-\cos(t)-7\cos(7t)-3\cos(3t), -\sin(t) + 1.5 \sin(3t)) \cdot 
(-\cos(t)+4.5\cos(3t), -\sin(t)-49\sin(7t)-9\sin(3t))
$$
and if $s$ is positive, select the "+" choice in the $\pm$ formula I gave you above. 
If $s$ changes sign as a function of $t$, then the means that the curve has an inflection, and there's no way to continuously choose the normal vector to be on the "concave" side of the curve. 
A: This problem [restated as : Given a point, $(x_0, y_0)$ on the given curve, compute the unit normal to the curve at $(x_0, y_0)$.] cannot be solved. Why? Because there might be two distinct normal lines to the curve at $(x_0, y_0)$. The following picture shows this:

At points where the plot intersects itself, there are two distinct normals, so any unambiguous formula in terms of $(x_0, y_0)$ will give at most one...which means it's wrong, (esp. if the other one is the one you wanted). 
Pretty plot, though. 
A: Since your curve is in the plane, you can find the normal vector without differentiating.  If the components of the unit tangent vector are $(a,b)$ then the components of a unit normal are $(-b,a)$.  "Outward" doesn't really make sense until you have a closed curve, and involves global considerations.
