Computing the coordinates of a Bezier Curve I just started messing with Bezier Curves over the past couple days and I'm trying to get some of the basics down.  I have this problem.  
Consider a quadratic Bezier curve with control points (0, 0), (2, 2), and (4, 0). 
What are the coordinates of the curve at t = 0.3?

How would I go about solving this?  Would I just use the quadratic Bezier curve formula and go from there?  If anyone could walk me through this and explain, that would be much appreciated.  
 A: The curve point $\mathbf{C}(t)$ at parameter value $t$ is given by the standard formula
$$
\mathbf{C}(t) = (1-t)^2\mathbf{P}_0 + 2t(1-t)\mathbf{P}_1 + t^2\mathbf{P}_2
$$
In our case, we have $\mathbf{P}_0 = (0,0)$, $\mathbf{P}_1 = (2,2)$, $\mathbf{P}_2 = (4,0)$, and we're interested in the parameter value $t = 0.3$. Plugging all these into the formula, we get
\begin{align*}
\mathbf{C}(0.3) &= (0.7)^2(0,0) + 2(0.3)(0.7)(2,2) + (0.3)^2(4,0) \\
 &= (0.49)(0,0) + (0.42)(2,2) + (0.09)(4,0) \\
 &= (1.2, 0.84)
\end{align*}
Note that $0.49+0.42+0.09=1$. This is significant.
You can get the same answer by using de Casteljau's algorithm. This says that
$$
\mathbf{C}(t) = (1-t)\big[(1-t)\mathbf{P}_0 + t\mathbf{P}_1\big] + 
                    t\big[(1-t)\mathbf{P}_1 + t\mathbf{P}_2\big]
$$
which has a nice geometric interpretation. Plugging $t=0.3$ into this, we get
\begin{align*}
\mathbf{C}(0.3) &= (0.7)\big[0.7\mathbf{P}_0 + 0.3\mathbf{P}_1\big] + 
                  (0.3)\big[0.7\mathbf{P}_1 + 0.3\mathbf{P}_2\big]  \\
&= (0.7)(0.6,0.6) + (0.3)(2.6, 1.4)  \\
&=  (1.2, 0.84)            
\end{align*}
You'll probably learn something if you draw a picture showing the original points $\mathbf{P}_0 = (0,0)$, $\mathbf{P}_1 = (2,2)$, $\mathbf{P}_2 = (4,0)$, the intermediate de Casteljau points $(0.6,0.6)$, and $(2.6, 1.4)$, and the final point $(1.2, 0.84)$.
