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Q-8.6-4E

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Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Let \({\bf{x}}\left( t \right)\) be a B-spline in Exercise 2, with control points \({{\bf{p}}_o}\), \({{\bf{p}}_1}\) , \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\).

a. Compute the tangent vector \({\bf{x}}'\left( t \right)\) and determine how the derivatives \({\bf{x}}'\left( 0 \right)\) and \({\bf{x}}'\left( 1 \right)\) are related to the control points. Give geometric descriptions of the directions of these tangent vectors. Explore what happens when both \({\bf{x}}'\left( 0 \right)\)and \({\bf{x}}'\left( 1 \right)\)equal 0. Justify your assertions.

b. Compute the second derivative and determine how and are related to the control points. Draw a figure based on Figure 10, and construct a line segment that points in the direction of . [Hint: Use \({{\bf{p}}_2}\) as the origin of the coordinate system.]

The derivative \({\bf{x}}'\left( t \right)\) is \({\bf{x}}'\left( t \right) = \frac{1}{6}\left[ { - 3{{\left( {1 - t} \right)}^2}{{\bf{p}}_o} + \left( { - 12t + 9{t^2}} \right){{\bf{p}}_1} + \left( { - 9{t^2} + 6t + 3} \right){{\bf{p}}_2} + 3{t^2}{{\bf{p}}_3}} \right]\). The relation with control points is \({\bf{x}}'\left( 0 \right) = \frac{1}{2}\left( {{{\bf{p}}_2} - {{\bf{p}}_o}} \right)\) and \({\bf{x}}'\left( 1 \right) = \frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)\). The tangent vector at a point \({{\bf{p}}_o}\) is directed from \({{\bf{p}}_o}\)to \({{\bf{p}}_2}\) and its length is twice the length of \(\frac{1}{2}\left( {{{\bf{p}}_2} - {{\bf{p}}_o}} \right)\). The tangent vector at point \({{\bf{p}}_1}\) is directed from \({{\bf{p}}_1}\)to\({{\bf{p}}_3}\) and its length is twice the length of \(\frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)\).

b) The derivative is. The relation with control points is and .

The line segment that points in the direction of x''(0) is shown below:

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Find the derivative of the B-Spline curve

The B-spline curve is given as\({\bf{x}}\left( t \right) = \frac{1}{6}\left[ {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\).

Its first derivative is shown below:

\(\begin{array}{l}{\bf{x}}\left( t \right) = \frac{1}{6}\left[ {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\\{\bf{x}}'\left( t \right) = \frac{1}{6}\left[ { - 3{{\left( {1 - t} \right)}^2}{{\bf{p}}_o} + \left( { - 12t + 9{t^2}} \right){{\bf{p}}_1} + \left( { - 9{t^2} + 6t + 3} \right){{\bf{p}}_2} + 3{t^2}{{\bf{p}}_3}} \right]\end{array}\)

Find \(x'\left( 0 \right)\), and \(x'\left( 1 \right)\) 

The \({\rm{x'}}\left( 0 \right)\) is calculated as shown below:

\(\begin{array}{c}{\bf{x}}'\left( 0 \right) = \frac{1}{6}\left[ { - 3{{\left( {1 - \left( 0 \right)} \right)}^2}{{\bf{p}}_o} + \left( { - 12\left( 0 \right) + 9{{\left( 0 \right)}^2}} \right){{\bf{p}}_1} + \left( { - 9{{\left( 0 \right)}^2} + 6\left( 0 \right) + 3} \right){{\bf{p}}_2} + 3{{\left( 0 \right)}^2}{{\bf{p}}_3}} \right]\\ = \frac{1}{6}\left( { - 3{{\bf{p}}_o} + 3{{\bf{p}}_{_2}}} \right)\\ = \frac{1}{2}\left( {{{\bf{p}}_2} - {{\bf{p}}_o}} \right)\end{array}\)

The tangent vector at point \({{\bf{p}}_o}\) is directed from \({{\bf{p}}_o}\)to \({{\bf{p}}_2}\) and its length is twice the length of \(\frac{1}{2}\left( {{{\bf{p}}_2} - {{\bf{p}}_o}} \right)\).

The \({\rm{x'}}\left( 1 \right)\) is calculated as shown below:

\(\begin{array}{c}{\bf{x}}'\left( 1 \right) = \frac{1}{6}\left[ { - 3{{\left( {1 - \left( 1 \right)} \right)}^2}{{\bf{p}}_o} + \left( { - 12\left( 1 \right) + 9{{\left( 1 \right)}^2}} \right){{\bf{p}}_1} + \left( { - 9{{\left( 1 \right)}^2} + 6\left( 1 \right) + 3} \right){{\bf{p}}_2} + 3{{\left( 1 \right)}^2}{{\bf{p}}_3}} \right]\\ = \frac{1}{6}\left[ { - 3{{\bf{p}}_1} + 0{{\bf{p}}_2} + 3{{\bf{p}}_3}} \right]\\ = \frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)\end{array}\)

The tangent vector at point \({{\bf{p}}_1}\) is directed from \({{\bf{p}}_1}\) to \({{\bf{p}}_3}\) and its length is twice the length of \(\frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)\).

Find  x''(0), and X''(1)   

The x''(0) is calculated as,

Draw the figure

The line segment that points in the direction of x''(0) is shown below:

Most popular questions for Math Textbooks

TrueType fonts, created by Apple Computer and Adobe Systems, use quadratic Bezier curves, while PostScript fonts, created by Microsoft, use cubic Bezier curves. The cubic curves provide more flexibility for typeface design, but it is important to Microsoft that every typeface using quadratic curves can be transformed into one that used cubic curves. Suppose that \({\mathop{\rm w}\nolimits} \left( t \right)\) is a quadratic curve, with control points \({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\) and \({{\mathop{\rm p}\nolimits} _2}\).

  1. Find control points \({{\mathop{\rm r}\nolimits} _0},{{\mathop{\rm r}\nolimits} _1},{{\mathop{\rm r}\nolimits} _2},\), and \({{\mathop{\rm r}\nolimits} _3}\) such that the cubic Bezier curve \({\mathop{\rm x}\nolimits} \left( t \right)\) with these control points has the property that \({\mathop{\rm x}\nolimits} \left( t \right)\) and \({\mathop{\rm w}\nolimits} \left( t \right)\) have the same initial and terminal points and the same tangent vectors at \(t = 0\)and\(t = 1\). (See Exercise 16.)

  1. Show that if \({\mathop{\rm x}\nolimits} \left( t \right)\) is constructed as in part (a), then \({\mathop{\rm x}\nolimits} \left( t \right) = {\mathop{\rm w}\nolimits} \left( t \right)\) for \(0 \le t \le 1\).

Question: 19. Let \(S\) be an affine subset of \({\mathbb{R}^n}\) , suppose \(f:{\mathbb{R}^n} \to {\mathbb{R}^m}\)is a linear transformation, and let \(f\left( S \right)\) denote the set of images \(\left\{ {f\left( {\rm{x}} \right):{\rm{x}} \in S} \right\}\). Prove that \(f\left( S \right)\)is an affine subset of \({\mathbb{R}^m}\).

In Exercises 1-4, write y as an affine combination of the other point listed, if possible.

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{\bf{0}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{6}}}\\{\bf{7}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}\\{\bf{3}}\\{\bf{1}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{ - {\bf{4}}}\end{aligned}} \right)\)

In Exercises 7 and 8, find the barycentric coordinates of p with respect to the affinely independent set of points that precedes it.

7. \(\left( {\begin{array}{{}}1\\{ - 1}\\2\\1\end{array}} \right),\left( {\begin{array}{{}}2\\1\\0\\1\end{array}} \right),\left( {\begin{array}{{}}1\\2\\{ - 2}\\0\end{array}} \right)\), \({\mathop{\rm p}\nolimits} = \left( {\begin{array}{{}}5\\4\\{ - 2}\\2\end{array}} \right)\)

In Exercises 21-26, prove the given statement about subsets A and B of \({\mathbb{R}^n}\), or provide the required example in \({\mathbb{R}^2}\). A proof for an exercise may use results from earlier exercises (as well as theorems already available in the text).

25. \({\mathop{\rm aff}\nolimits} \left( {A \cap B} \right) \subset \left( {{\mathop{\rm aff}\nolimits} A \cap {\mathop{\rm aff}\nolimits} B} \right)\)
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