Question

Let \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) be Bézier curves from Exercise 5, and suppose the combined curve has \({C^2}\)continuity (which includes \({C^1}\) continuity) at \({{\bf{p}}_3}\) . Set \({\mathbf{x}}''\left( 1 \right) = {\mathbf{y}}''\left( 0 \right)\) and show that \({{\bf{p}}_5}\) is completely determined by \({{\bf{p}}_1}\) , \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\) . Thus, the points \({{\bf{p}}_0},...,\,{{\bf{p}}_3}\)and the \({C^2}\) condition determine all but one of the control points for \({\bf{y}}\left( t \right)\).

Short answer

It is shown that \({{\bf{p}}_5}\) is dependent on \({{\bf{p}}_1}\),\({{\bf{p}}_2}\)and \({{\bf{p}}_3}\).

Step-by-step solution

Step 1:Describe the given information

It is given that\({\bf{x}}\left( t \right)\)and\({\bf{y}}\left( t \right)\)are the Bézier curves their combined curve has\({C^2}\)continuity including\({C^1}\)continuityat\({{\bf{p}}_3}\).

It has already been shown that \({\mathbf{x}}''(0) = 6\left( {{{\mathbf{p}}_o} - {{\mathbf{p}}_1}} \right) + 6\left( {{{\mathbf{p}}_2} - {{\mathbf{p}}_1}} \right)\), \({\mathbf{x}}''(1) = 6\left( {{{\mathbf{p}}_1} - {{\mathbf{p}}_2}} \right) + 6\left( {{{\mathbf{p}}_3} - {{\mathbf{p}}_2}} \right)\) and if x''(0) is the control point of\({\bf{y}}\left( t \right)\), then \({\mathbf{y}}''(0) = 6\left( {{{\mathbf{p}}_3} - {{\mathbf{p}}_4}} \right) + 6\left( {{{\mathbf{p}}_5} - {{\mathbf{p}}_4}} \right)\) .

Step 2:Set \({\mathbf{x}}''\left( 1 \right) = {\mathbf{y}}''\left( 0 \right)\)

Find the relation as shown below:

\(\begin{array}{c}\left( {{{\bf{p}}_1} - {{\bf{p}}_2}} \right) + 6\left( {{{\bf{p}}_3} - {{\bf{p}}_2}} \right) = 6\left( {{{\bf{p}}_3} - {{\bf{p}}_4}} \right) + 6\left( {{{\bf{p}}_5} - {{\bf{p}}_4}} \right)\\\left( {{{\bf{p}}_1} - {{\bf{p}}_2}} \right) + \left( {{{\bf{p}}_3} - {{\bf{p}}_2}} \right) = \left( {{{\bf{p}}_3} - {{\bf{p}}_4}} \right) + \left( {{{\bf{p}}_5} - {{\bf{p}}_4}} \right)\end{array}\)

We can write \({{\bf{p}}_3} = \frac{1}{2}\left( {{{\bf{p}}_2} + {{\bf{p}}_4}} \right)\)or\({{\bf{p}}_4} = {{\bf{p}}_3} + {{\bf{p}}_3} - {{\bf{p}}_2}\) as the curves have \({C^1}\) continuity at \({{\bf{p}}_3}\).

Plug in  \({{\bf{p}}_3} = \frac{1}{2}\left( {{{\bf{p}}_2} + {{\bf{p}}_4}} \right)\) into the resulting equation

The equation becomes as shown below:

\(\begin{array}{c}\left( {{{\bf{p}}_1} - {{\bf{p}}_2}} \right) + \left( {\frac{1}{2}\left( {{{\bf{p}}_2} + {{\bf{p}}_4}} \right) - {{\bf{p}}_2}} \right) = \left( {\frac{1}{2}\left( {{{\bf{p}}_2} + {{\bf{p}}_4}} \right) - {{\bf{p}}_4}} \right) + \left( {{{\bf{p}}_5} - {{\bf{p}}_4}} \right)\\\left( {{{\bf{p}}_1} - {{\bf{p}}_2}} \right) + 2\left( {{{\bf{p}}_3} - {{\bf{p}}_2}} \right) + {{\bf{p}}_4} = {{\bf{p}}_5}\end{array}\)

Plugin  \({{\bf{p}}_4} = {{\bf{p}}_3} + {{\bf{p}}_3} - {{\bf{p}}_2}\) into the resulting equation 

\(\begin{array}{c}\left( {{{\bf{p}}_1} - {{\bf{p}}_2}} \right) + 2\left( {{{\bf{p}}_3} - {{\bf{p}}_2}} \right) + {{\bf{p}}_3} + {{\bf{p}}_3} - {{\bf{p}}_2} = {{\bf{p}}_5}\\{{\bf{p}}_5} = {{\bf{p}}_3} + \left( {{{\bf{p}}_1} - {{\bf{p}}_2}} \right) + 3\left( {{{\bf{p}}_3} - {{\bf{p}}_2}} \right)\end{array}\)

Draw a conclusion

As \({{\bf{p}}_4} = {{\bf{p}}_3} + {{\bf{p}}_3} - {{\bf{p}}_2}\) and \({{\bf{p}}_5} = {{\bf{p}}_3} + \left( {{{\bf{p}}_1} - {{\bf{p}}_2}} \right) + 3\left( {{{\bf{p}}_3} - {{\bf{p}}_2}} \right)\), it is stated that \({{\bf{p}}_4}\)and \({{\bf{p}}_5}\)depends on \({{\bf{p}}_1}\),\({{\bf{p}}_2}\)and \({{\bf{p}}_3}\).