[Math] How to derive the equation for a bézier curve

bezier-curvederivativesgraphing-functions

So, I remember a while back there was a maths competition and we were given a curve that we needed to write an equation for. I just skipped the question since I didn't even know where to begin. I remember it was one among the last few questions of the paper and it was worth a lot of points.

I don't really remember what the curve looked like; it was something spirally, but I can't recall it to save my life right now.

So, I drew this curve in Inkscape (it's a Bézier curve. Or a few of them linked together, according to Wikipedia. If it's required I will post the whole path). And I would like to write the equation for it (with someone's help, obviously).

The Bézier Curve

I was always a bit bad with curves, graphs and lines, but I want to understand them better. So, I was hoping someone could explain the process of deriving the equation for a curve.

P.S: I'd like it if you could use another curve (it can be something simpler, but try avoiding something overly complicated) so I can crack this one on my own, but if you feel like using this curve as an example I won't mind.

EDIT

So have been browsing the internet, read a few Wikipedia entries about Bazier curves, and I understand how they're drawn (mostly the GIFs helped, haha), but I am still stumped when it comes to mathematically representing a Bézier curve. Also, I will add this image, which is the path and its control points (at the end of the blue lines; I didn't paint them in):

The Path

And also, the contents of the .tex file for the shape.

%LaTeX with PSTricks extensions
%%Creator: 0.48.2
%%Please note this file requires PSTricks extensions
\psset{xunit=.5pt,yunit=.5pt,runit=.5pt}
\begin{pspicture}(451.46875,34.25392151)
	{
	\newrgbcolor{curcolor}{1 0 0}
	\pscustom[linewidth=3,linecolor=curcolor]
	{
	\newpath
	\moveto(450.48448,1.10834551)
	\curveto(404.89404,41.45133951)(333.34998,42.21654151)(281.90128,9.03018551)
	\curveto(258.09407,-6.32636849)(228.42388,9.91159551)(202.75741,15.38398551)
	\curveto(145.68728,27.55199551)(85.852286,40.32786151)(28.08402514,26.23698551)
	\curveto(18.5710181,23.91656551)(9.403556,20.24334551)(0.681686,15.78116551)
	}
	}
	\end{pspicture}

Thanks!

Best Answer

Linear Bézier curve is simply a line given by parametric equation $R(t) = A+t(AB)$ , A being initial point and B being final point.

For Quadratic Bézier curve, take a look at the following picture.
enter image description here
Let the point between $P_1$ and $P_0$ be $Q_1$ and $P_1$ and $P_2$ be $Q_2$. Let our path be traced by $Q_0$. Then from above figure. $$ \frac{P_0Q_1}{P_0P_1} = \frac{P_1Q_2}{P_1P_2} = \frac{Q_1Q_0}{Q_1Q2} = t \text{ (say)} $$ $$Q_1 = P_0 + t(P_0P_1), Q_2 = P_1 + t(P_1P_2)$$ So we have $$Q_0 = Q_1 + t(Q_1Q_2) = P_0 + t(P_0P_1) + t(P_1 + t(P_1P_2) - (P_0 + t(P_0P_1)))$$ Have a look at more elaborate article on Wikipedia.

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