# 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 quick Google search gives this formula from Wikipedia. It is just a matter of putting $P_0=(0,0), P_1=(2,2), P_2=(4,0)$. – cutculus May 27 '14 at 7:03
• I think this en.wikipedia.org/wiki/De_Casteljau%27s_algorithm should help you. – pointer May 27 '14 at 7:24
• I asked for someone to walk me through this for a reason. – Requiem May 27 '14 at 12:19

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 $(1.2, 0.84)$.