Chapter 3.5, Problem 1E

Find the one-piece BĂ©zier curve $\left(x\left(t\right),y\left(t\right)\right)$ defined by the given four points.

(a) $\left(0,0\right)$, $\left(0,2\right)$, $\left(2,0\right)$, $\left(1,0\right)$

(b) $\left(1,1\right)$, $\left(0,0\right)$, $\left(-2,0\right)$, $\left(-2,1\right)$

(c) $\left(1,2\right)$, $\left(1,3\right)$, $\left(2,3\right)$, $\left(2,2\right)$

To determine

To find:The one- piece Bezier curve defined by the given four points.

Answer to Problem 1E

The Bezier curve defined by the given points is;

   $\begin{array}{l}x(t)=6t^2-5t^3\\ y(t)=6t-12t^2+6t^3\end{array}$

Explanation of Solution

Given:

Four points are $(0,0),(0,2),(2,0),(1,0).$

Formula used:

Bezier curve equations are

   $\begin{array}{l}x(t)=x_1+b_{x}t+c_x{t}^2+d_y{t}^3\\ y(t)=y_1+b_{y}t+c_y{t}^2+d_y{t}^3\\ \\ \text{co-efficients are}\\ \\ b_x=3(x_2-x_1)\\ c_x=3(x_2-x_1)-b_x\\ d_x=x_4-x_1-b_x-c_x\\ b_y=3(y_2-y_1)\\ c_y=3(y_3-y_2)-b_x\\ d_y=y_4-y_1-b_y-c_y\end{array}$

Calculation:

From the given data,

We have,

   $\begin{array}{l}(x_1,y_1)=(0,0)\\ (x_2,y_2)=(0,2)\\ (x_3,y_3)=(2,0),\\ (x_4,y_4)=(1,0)\end{array}$

The coefficients of Bezier curve are determined as follows:

   $\begin{array}{l}{b}_x=3(x_2-x_1)=3(0-0)=0\\ c_x=3(x_2-x_1)-b_x=3(2-0)-0=6\\ d_x=x_4-x_1-b_x-c_x=1-0-0-6=-5\end{array}$

   $\begin{array}{l}{b}_y=3(y_2-y_1)=3(2-0)=6\\ c_y=3(y_3-y_2)-b_x=3(0-2)-6=-12\\ d_y=y_4-y_1-b_y-c_y=0-0-6+12=6\end{array}$

   $\begin{array}{l}\Rightarrow x(t)=x_1+b_{x}t+c_x{t}^2+d_y{t}^3\\ =0+0+6t^2-5t^3\\ =6t^2-5t^3\\ \Rightarrow y(t)=y_1+b_{y}t+c_y{t}^2+d_y{t}^3\\ =0+6t-12t^2+6t^3+0\\ =6t-12t^2+6t^3\end{array}$

Therefore, the Bezier curve equations are

   $\begin{array}{l}x(t)=6t^2-5t^3\\ y(t)=6t-12t^2+6t^3\end{array}$

To determine

To find:The one- piece Bezier curve defined by the given four points.

Answer to Problem 1E

The Bezier curve defined by the equations

   $\begin{array}{l}x(t)=1-3t-3t^2+3t^3\\ y(t)=1-3t+3t^2\end{array}$

Explanation of Solution

Given:

Four points are $(1,1),(0,0),(-2,0),(-2,1).$

Formula used:

Bezier curve equations are

   $\begin{array}{l}x(t)=x_1+b_{x}t+c_x{t}^2+d_y{t}^3\\ y(t)=y_1+b_{y}t+c_y{t}^2+d_y{t}^3\\ \\ co-efficientsare\\ \\ b_x=3(x_2-x_1)\\ c_x=3(x_2-x_1)-b_x\\ d_x=x_4-x_1-b_x-c_x\\ b_y=3(y_2-y_1)\\ c_y=3(y_3-y_2)-b_x\\ d_y=y_4-y_1-b_y-c_y\end{array}$

Calculation:

From the given data,

We have,

   $\begin{array}{l}(x_1,y_1)=(1,1\\ (x_2,y_2)=(0,0)\\ (x_3,y_3)=(-2,0),\\ (x_4,y_4)=(-2,1)\end{array}$

The coefficients of Bezier curve are determined as follows:

   $\begin{array}{l}{b}_x=3(x_2-x_1)=3(0-1)=-3\\ c_x=3(x_2-x_1)-b_x=3(-2-0)-0=-3\\ d_x=x_4-x_1-b_x-c_x=-2-1+3+3=3\end{array}$

   $\begin{array}{l}{b}_y=3(y_2-y_1)=3(0-1)=-3\\ c_y=3(y_3-y_2)-b_x=3(0-0)+3=3\\ d_y=y_4-y_1-b_y-c_y=1-1+3-3=0\end{array}$

   $\begin{array}{l}\Rightarrow x(t)=x_1+b_{x}t+c_x{t}^2+d_y{t}^3\\ =1+(-3)t+(-3)t^2+3t^3\\ =1-3t-3t^2+3t^3\\ \Rightarrow y(t)=y_1+b_{y}t+c_y{t}^2+d_y{t}^3\\ =1-3t+3t^2+0\\ =1-3t+3t^2\end{array}$

Therefore, the Bezier curve equations are

   $\begin{array}{l}x(t)=1-3t-3t^2+3t^3\\ y(t)=1-3t+3t^2\end{array}$

To determine

To find:The one- piece Bezier curve defined by the given four points.

Answer to Problem 1E

The Bezier curve defined by the equations

   $\begin{array}{l}x(t)=1+3t^2-2t^3\\ y(t)=2+3t-3t^2\end{array}$

Explanation of Solution

Given:

Four points are $(1,2),(1,3),(2,3),(2,2).$

Formula used:

Bezier curve equations are

   $\begin{array}{l}x(t)=x_1+b_{x}t+c_x{t}^2+d_y{t}^3\\ y(t)=y_1+b_{y}t+c_y{t}^2+d_y{t}^3\\ \\ co-efficientsare\\ \\ b_x=3(x_2-x_1)\\ c_x=3(x_2-x_1)-b_x\\ d_x=x_4-x_1-b_x-c_x\\ b_y=3(y_2-y_1)\\ c_y=3(y_3-y_2)-b_x\\ d_y=y_4-y_1-b_y-c_y\end{array}$

Calculation:

From the given data,

We have,

   $\begin{array}{l}(x_1,y_1)=(1,2)\\ (x_2,y_2)=(1,3)\\ (x_3,y_3)=(2,3),\\ (x_4,y_4)=(2,2)\end{array}$

The coefficients of Bezier curve are determined as follows:

   $\begin{array}{l}{b}_x=3(x_2-x_1)=3(1-1)=0\\ c_x=3(x_2-x_1)-b_x=3(2-1)-0=3\\ d_x=x_4-x_1-b_x-c_x=2-1-0-3=-2\end{array}$

   $\begin{array}{l}{b}_y=3(y_2-y_1)=3(3-2)=3\\ c_y=3(y_3-y_2)-b_x=3(3-3)-3=-3\\ d_y=y_4-y_1-b_y-c_y=2-2+3-3=0\end{array}$

   $\begin{array}{l}\Rightarrow x(t)=x_1+b_{x}t+c_x{t}^2+d_y{t}^3\\ =1+(0)t+(3)t^2+(-2)t^3\\ =1+3t^2-2t^3\\ \Rightarrow y(t)=y_1+b_{y}t+c_y{t}^2+d_y{t}^3\\ =2+3t-3t^2\end{array}$

Therefore, the Bezier curve equations are

   $\begin{array}{l}x(t)=1+3t^2-2t^3\\ y(t)=2+3t-3t^2\end{array}$