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Chapter 7: Q7E (page 395)

Let \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) be B茅zier 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

Expert verified

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

Step by step solution

01

Step 1:Describe the given information

It is given that\({\bf{x}}\left( t \right)\)and\({\bf{y}}\left( t \right)\)are the B茅zier 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 \({\mathbf{y}}''(0) = 6\left( {{{\mathbf{p}}_3} - {{\mathbf{p}}_4}} \right) + 6\left( {{{\mathbf{p}}_5} - {{\mathbf{p}}_4}} \right)\) .

02

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 \({{\bf{p}}_3}\).

03

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:

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

04

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

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

05

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 \({{\bf{p}}_3}\).

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Most popular questions from this chapter

Question: Repeat Exercise 15 for the following SVD of a \({\bf{3 \times 4}}\) matrix \(A\):

\(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{.86}}}&{ - {\bf{.11}}}&{ - {\bf{.50}}}\\{{\bf{.31}}}&{{\bf{.68}}}&{ - {\bf{.67}}}\\{{\bf{.41}}}&{ - {\bf{.73}}}&{ - {\bf{5}}{\bf{.5}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\bf{12}}{\bf{.48}}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{{\bf{6}}{\bf{.34}}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right){\bf{ \times }}\left( {\begin{array}{*{20}{c}}{{\bf{.66}}}&{ - {\bf{.03}}}&{ - {\bf{.35}}}&{{\bf{.66}}}\\{ - {\bf{1}}{\bf{.3}}}&{ - {\bf{.90}}}&{ - {\bf{.39}}}&{ - {\bf{.13}}}\\{{\bf{.65}}}&{{\bf{.08}}}&{ - {\bf{.16}}}&{ - {\bf{.73}}}\\{ - {\bf{.34}}}&{{\bf{.42}}}&{ - {\bf{8}}{\bf{.4}}}&{ - {\bf{0}}{\bf{.8}}}\end{array}} \right)\)

Question: 3. Let A be an \(n \times n\) symmetric matrix of rank r. Explain why the spectral decomposition of A represents A as the sum of r rank 1 matrices.

Question 8: Use Exercise 7 to show that if A is positive definite, then A has a LU factorization, \(A = LU\), where U has positive pivots on its diagonal. (The converse is true, too).

Determine which of the matrices in Exercises 1鈥�6 are symmetric.

\(\left( {\begin{array}{{}{}}1&2&2&1\\2&2&2&1\\2&2&1&2\end{array}} \right)\)

Question: 13. Exercises 12鈥�14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Suppose the equation\(A{\rm{x}} = {\rm{b}}\)is consistent, and let\({{\rm{x}}^ + } = {A^ + }{\rm{b}}\). By Exercise 23 in Section 6.3, there is exactly one vector\({\rm{p}}\)in Row\(A\)such that\(A{\rm{p}} = {\rm{b}}\). The following steps prove that\({{\rm{x}}^ + } = {\rm{p}}\)and\({{\rm{x}}^ + }\)is the minimum length solution of\(A{\rm{x}} = {\rm{b}}\).

  1. Show that \({{\rm{x}}^ + }\) is in Row \(A\). (Hint: Write \({\rm{b}}\) as \(A{\rm{x}}\) for some \({\rm{x}}\), and use Exercise 12.)
  2. Show that\({{\rm{x}}^ + }\)is a solution of\(A{\rm{x}} = {\rm{b}}\).
  3. Show that if \({\rm{u}}\) is any solution of \(A{\rm{x}} = {\rm{b}}\), then \(\left\| {{{\rm{x}}^ + }} \right\| \le \left\| {\rm{u}} \right\|\), with equality only if \({\rm{u}} = {{\rm{x}}^ + }\).
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