Find the eqn of Bezier curve and determine co-ordinate of points on the curve for t = 0, 0.25, 0.5, 0.75 ,1

Solution:

No. of points = 4 = k

Degree of curve $=(\mathrm{n})=\mathrm{k}-1=3$

$\therefore$ Cubic Curve

$P(t)=P_{0}(1-t)^{3}+3 P_{1}(1-t)^{2} t+3 P_{2}(1-t) t^{2}+P_{3} t^{3}$

$P_{0}=\left[\begin{array}{ccc}{2} & {2} & {0}\end{array}\right]$

$P_{1}=\left[\begin{array}{ccc}{2} & {3} & {0}\end{array}\right]$

$P_{2}=\left[\begin{array}{ccc}{2} & {3} & {0}\end{array}\right]$

$P_{3}=\left[\begin{array}{lll}{3} & {2} & {0}\end{array}\right]$

$\begin{aligned} P_{x} &=2(1-t)^{3}+3(2)(1-t)^{2} t+3(3)(1-t) t^{2}+3 t^{3} \\ &=2(1-t)^{3}+6(1-t)^{2} t+9(1-t) t^{2}+3 t^{3} \\ &=2\left(1-3 t+3 t^{2}-t^{3}\right)+6\left(1-2 t+t^{2}\right) t+9 t^{2}-9 t^{3}+3 t^{3} \\ &=2-6 t+6 t^{2}-2 t^{3}+ 6 t-12 t^{2}+6 t^{3}+9 t^{2}-9 t^{3}+3 t^{3} \\ \therefore & P_{x}=2+3 t^{2}-2 t^{3} \end{aligned}$

$\begin{aligned} P_{y} &=2(1-t)^{3}+3(3)(1-t)^{2} t+3(3)(1-t) t^{2}+2 t^{3} \\ &=2\left[1-3 t+3 t^{2}-t^{3}\right]+9\left[1-2 t+t^{2}\right] t+9\left[t^{2}-t^{3}\right]+2 t^{3} \\ &=2-6 t+6 t^{2}-2 t^{3}+\left[9-18 t+9 t^{2}\right] t+9 t^{2}-9 t^{3}+2 t^{3} \\ P_{y} &=2-6 t+6 t^{2}-2 t^{3}+9 t-18 t^{2}+9 t^{3}+9 t^{2}-7 t^{3} \\ \therefore & P_{y}=2+3 t-3 t^{2} \end{aligned}$

$\begin{array}{|c|c|c|c|c|c|}\hline t \rightarrow & {0} & {0.25} & {0.5} & {0.6} & {0.75} & {1} \\ \hline P_{x} & {2} & {2.15625} & {2.5} & {5.896} & {} & {3} \\ \hline P_y & {2} & {} & {3.56} & {2.75} & {} & {2} \\ \hline\end{array}$