/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q24E Given control points \({{\rm{p}}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Q24E (page 437)

Given control points \({{\rm{p}}_{\rm{o}}}\) , \({{\rm{p}}_{\rm{1}}}\) , \({{\rm{p}}_{\rm{2}}}\) and \({{\rm{p}}_{\rm{3}}}\) in \({\mathbb{R}^n}\) , let \({g_1}\left( t \right)\)for \(0 \le t \le 1\) be the quadratic B茅zier curve from Exercise 23 determined by \({{\rm{p}}_{\rm{o}}}\) , \({{\rm{p}}_{\rm{1}}}\) , and \({{\rm{p}}_{\rm{2}}}\), and let \({g_2}\left( t \right)\)be defined similarly for \({{\rm{p}}_{\rm{1}}}\) , \({{\rm{p}}_{\rm{2}}}\) and \({{\rm{p}}_{\rm{3}}}\). For \(0 \le t \le 1\), define \(h\left( t \right) = \left( {1 - t} \right){g_1}\left( t \right) + t{g_2}\left( t \right)\). Show that the graph of \(h\left( t \right)\)lies in the convex hull of the four control points. This curve is called a cubic B茅zier curve, and its definition here is one step in an algorithm for constructing B茅zier curves (discussed later in Section 8.6). A B茅zier curve of degree \(k\) is determined by \(k + 1\) control points, and its graph lies in the convex hull of these control points.

Short Answer

Expert verified

\({\bf{h}}\left( t \right) = \left( {1 - 3t + 3{t^2} - {t^3}} \right){{\bf{p}}_{\bf{o}}} + \left( {3t - 6{t^2} + 3{t^3}} \right){{\bf{p}}_{\bf{1}}} + \left( {3{t^2} - 3{t^3}} \right){{\bf{p}}_{\bf{2}}} + {t^3}{{\bf{p}}_{\bf{3}}}\)

It is shown that \({\bf{h}}\left( t \right)\)is the convex combination of \(\left\{ {{{\bf{p}}_{\bf{o}}}{\bf{,}}\,{{\bf{p}}_{\bf{1}}}{\bf{,}}\,{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\).

Step by step solution

01

Use the definition of affine hull

Assume that \({{\bf{p}}_{\bf{o}}}{\bf{,}}\,{{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}} \in {{\bf{g}}_{\bf{o}}}\left( t \right)\) and \({{\bf{p}}_{\bf{1}}}{\bf{,}}\,{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}} \in {{\bf{g}}_{\bf{1}}}\left( t \right)\).

The affine hull of distinct points \({v_1}\) and \({v_2}\) is \(y = \left( {1 - t} \right){v_1} + t{v_2}\).

Similarly, the affine hull of \({{\bf{g}}_{\bf{o}}}\left( t \right)\)and \({{\bf{g}}_{\bf{1}}}\left( t \right)\)is \({\bf{h}}\left( t \right) = \left( {1 - t} \right){{\bf{g}}_{\bf{o}}}\left( t \right) + t{{\bf{g}}_{\bf{1}}}\left( t \right)\).

02

 Apply the affine hull for \({g_o}\left( t \right)\), and\({g_1}\left( t \right)\)

\(\begin{aligned}{}{\bf{h}}\left( t \right) = \left( {1 - t} \right)\left( {{{\left( {1 - t} \right)}^2}{{\bf{p}}_{\bf{o}}} + 2t\left( {1 - t} \right){p_1} + {t^2}{p_2}} \right) + t\left( {{{\left( {1 - t} \right)}^2}{{\bf{p}}_{\bf{1}}} + 2t\left( {1 - t} \right){{\bf{p}}_{\bf{2}}} + {t^2}{{\bf{p}}_{\bf{3}}}} \right)\\ = {\left( {1 - t} \right)^3}{{\bf{p}}_{\bf{o}}} + 2t\left( {1 - 2t + {t^2}} \right){{\bf{p}}_{\bf{1}}} + \left( {{t^2} - {t^3}} \right){{\bf{p}}_{\bf{2}}} + t\left( {1 - 2t + {t^2}} \right){{\bf{p}}_{\bf{1}}} + 2{t^2}\left( {1 - t} \right){{\bf{p}}_{\bf{2}}} + {t^3}{{\bf{p}}_{\bf{3}}}\\ = \left( {1 - 3t + 3{t^2} - {t^3}} \right){{\bf{p}}_{\bf{o}}} + \left( {2t - 4{t^2} + 2{t^3}} \right){{\bf{p}}_{\bf{1}}} + \left( {{t^2} - {t^3}} \right){{\bf{p}}_{\bf{2}}} + \left( {t - 2{t^2} + {t^3}} \right){{\bf{p}}_{\bf{1}}}\\ + \left( {2{t^2} - 2{t^3}} \right){{\bf{p}}_{\bf{2}}} + {t^3}{{\bf{p}}_{\bf{3}}}\\ = \left( {1 - 3t + 3{t^2} - {t^3}} \right){{\bf{p}}_{\bf{o}}} + \left( {3t - 6{t^2} + 3{t^3}} \right){{\bf{p}}_{\bf{1}}} + \left( {3{t^2} - 3{t^3}} \right){p_2} + {t^3}{{\bf{p}}_{\bf{3}}}\end{aligned}\)

03

Use the concept that weights in linear combination sum to 1

A point \(y\) in\({\mathbb{R}^n}\) is an affine combination of \({v_1},.......,{v_p}\)in \({\mathbb{R}^n}\), if \(\overline y = {c_1}{\overline v _1} + ...... + {c_p}{\overline v _p}\) such that \({c_1} + ...... + {c_p} = 1\).

So, the sum of the weight of \(h\left( t \right)\) should be 1; that is, \(\left( {1 - 3t + 3{t^2} - {t^3}} \right) + \left( {3t - 6{t^2} + 3{t^3}} \right) + \left( {3{t^2} - 3{t^3}} \right) + {t^3} = 1\).

04

Find the range of weight when \(0 \le t \le 1\).

The weight sum is \(\left( {1 - 3t + 3{t^2} - {t^3}} \right) + \left( {3t - 6{t^2} + 3{t^3}} \right) + \left( {3{t^2} - 3{t^3}} \right) + {t^3} = 1\). This sum also varies between 0 and 1 if \(0 \le t \le 1\).

This implies, it is convex combination of \(\left\{ {{{\bf{p}}_{\bf{o}}}{\bf{,}}\,{{\bf{p}}_{\bf{1}}}{\bf{,}}\,{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.]

In Exercises 1-6, determine if the set of points is affinely dependent. (See Practice Problem 2.) If so, construct an affine dependence relation for the points.

3.\(\left( {\begin{aligned}{{}}1\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 2}\\{ - 4}\\8\end{aligned}} \right),\left( {\begin{aligned}{{}}2\\{ - 1}\\{11}\end{aligned}} \right),\left( {\begin{aligned}{{}}0\\{15}\\{ - 9}\end{aligned}} \right)\)

Question:In Exercises 21 and 22, mark each statement True or False. Justify each answer.

21. a. A linear transformation from\(\mathbb{R}\)to\({\mathbb{R}^n}\)is called a linear functional.

b. If\(f\)is a linear functional defined on\({\mathbb{R}^n}\), then there exists a real number\(k\)such that\(f\left( x \right) = kx\)for all\(x\)in\({\mathbb{R}^n}\).

c. If a hyper plane strictly separates sets\(A\)and\(B\), then\(A \cap B = \emptyset \)

d. If\(A\)and\(B\)are closed convex sets and\(A \cap B = \emptyset \), then there exists a hyper plane that strictly separate\(A\)and\(B\).

Question: 16. Let \({\rm{v}} \in {\mathbb{R}^n}\)and let \(k \in \mathbb{R}\). Prove that \(S = \left\{ {{\rm{x}} \in {\mathbb{R}^n}:{\rm{x}} \cdot {\rm{v}} = k} \right\}\)is an affine subset of \({\mathbb{R}^n}\).

Question: In Exercise 7, let Hbe the hyperplane through the listed points. (a) Find a vector n that is normal to the hyperplane. (b) Find a linear functional f and a real number d such that \(H = \left( {f:d} \right)\).

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{4}}\\{\bf{1}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{ - {\bf{2}}}\\{\bf{5}}\end{array}} \right)\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.