Question

Construct a bezier curve of order 3 and with 4 polygon vertices a (1, 1) b(2, 3) c(4,3) and d(6, 4).

Answer 1

This problem can be solved in the following way: 
(Please feel free to raise questions in case of any confusions)

Answer 2

Answer:

$P_1=P_3=(6,4)$

Step-by-step explanation:

Bezier curves can be generated under the control of other points. Approximate tangents by using control points are used to generate curve. The Bezier curve can be represented mathematically as –

$\sum_{k=0}^{n}P_iB_{i}^{n}(t)$

Where pi is the set of points and Bni(t)Bin(t) represents the Bernstein polynomials which are given by –

$B_{i}^{n}(t)=_{i}^{n}(1-t)^{n-i}t^{i}$

Where n is the polynomial degree, i is the index, and t is the variable.

The equation for the Bezier curve is given

as:-

$P(U)=(1-U)^{3}P_{1}+3U(1-U)^{^{2}}P_{2}+3U^{^{2}}(1-U)P_{3}+U^{3}P_{4}$

for $0\leq u\leq 1$ where P(U)

is the point on the curve $P_1,P_2,P_3,P_4$

Let us take $U=0,\frac{1}{4} ,\frac{1}{2} ,\frac{3}{4}$

Therefore,$P(0)=(1,1)$

$P(1/4)=(1-1/4)x^{3} P_1+3\frac{1}{4} (1-1/4)x^{2} P_2+3(1/4)x^{2} (1-1/4)P_3+(1/4)x^{3} P_4$

$\Rightarrow \frac{27}{6^{2}}(1,1)+\frac{27}{64}(2,3)+\frac{9}{64}(4,3)+\frac{1}{6^{4}}(6,4)$

$\Rightarrow [\frac{27}{64}\times 1+\frac{27}{64}\times 2+\frac{9}{64}\times 6,\frac{27}{64}\times 1+\frac{27}{64}\times 2+\frac{9}{64}\times 3+\frac{1}{64}\times 4]$

$\Rightarrow [\frac{123}{64},\frac{139}{64}]$

$=(1.9218,2.1718)$

$P(\frac{1}{2})=(1-\frac{1}{2}^{3})P_1+3\frac{1}{2}(1-\frac{1}{2})^{2}P_2+3\frac{1}{2}^{4}(1-\frac{1}{2})P_3+\frac{1}{2}^{3}P_4$

$\Rightarrow \frac{1}{8}(1,1)+\frac{3}{8}(2,3)+\frac{3}{8}(4,3)+\frac{1}{8}(6,4)$

$\Rightarrow [\frac{1}{8}\times 1+\frac{3}{8}\times 2+\frac{3}{8}\times 4+\frac{1}{8}\times 6,\frac{1}{8}\times 1+\frac{3}{8}\times 3+\frac{3}{8}\times 3+\frac{1}{8}\times 4]$

$\Rightarrow [\frac{25}{8},\frac{23}{8}]$

$=(3.125,2.375)$

Therefore, $P(\frac{3}{4})=(1-\frac{3}{4})^{3}P_1+3\frac{3}{4}(1-\frac{3}{4})^{2}P_2+3(\frac{3}{4})^{2}(1-\frac{3}{4})P_3+(\frac{3}{4})^{2}P_4$

$\Rightarrow \frac{1}{64}P_1+\frac{9}{64}P_2+\frac{27}{64}P_3+\frac{27}{64}P_4$

$\Rightarrow \frac{1}{64}(1,1)+\frac{9}{64}(2,3)+\frac{27}{64}(4,3)+\frac{27}{64}(6,4)$

$\Rightarrow [\frac{1}{64}\times 1+\frac{9}{64}\times 2+\frac{27}{64}\times 4+\frac{27}{64}\times 6,\frac{1}{64}\times 1+\frac{9}{64}\times 3+\frac{27}{64}\times 3+\frac{27}{64}\times 4]$

$\Rightarrow [\frac{289}{64},\frac{217}{64}]$

$=(4.5156,3.3906)$

$P_1=P_3=(6,4)$