Parametric Equations and Curves — Question 10

Question 10

Problem

A cubic Bézier path is $\mathbf r(t)=(1-t)^3P_0+3(1-t)^2tP_1+3(1-t)t^2P_2+t^3P_3$. Explain, without expanding, why it begins at $P_0$ and ends at $P_3$.

Question 10 – Solution

Solution

1. The cubic Bézier path is $$\mathbf r(t)=(1-t)^3P_0+3(1-t)^2tP_1 +3(1-t)t^2P_2+t^3P_3.$$

2. Evaluate the path at $t=0$: $$\begin{aligned} \mathbf r(0) &=(1-0)^3P_0+3(1-0)^2(0)P_1\\ &\quad+3(1-0)(0)^2P_2+(0)^3P_3\\ &=P_0. \end{aligned}$$ Every term containing a factor of $t$ vanishes, leaving only $P_0$.

3. Evaluate the path at $t=1$: $$\begin{aligned} \mathbf r(1) &=(1-1)^3P_0+3(1-1)^2(1)P_1\\ &\quad+3(1-1)(1)^2P_2+(1)^3P_3\\ &=P_3. \end{aligned}$$ Every term containing a factor of $1-t$ vanishes, leaving only $P_3$.

4. Therefore, $$\boxed{\mathbf r(0)=P_0 \qquad\text{and}\qquad \mathbf r(1)=P_3}.$$ The path begins at the first control point and ends at the last control point.