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Chapter 8: Q11E (page 437)

Question: Find an example of a closed convex set S in \({\mathbb{R}^{\bf{2}}}\) such that its profile P is nonempty but \({\bf{conv}}\,P \ne S\).

Short Answer

Expert verified

A rectangle without lower and right boundaries represents a bounded convex set S in \({\mathbb{R}^2}\) such that profile P is nonempty but \({\rm{conv}}\,\,P \ne S\).

Step by step solution

01

Find the example of a closed convex set

One example of such a closed convex set is a rectangle that has all parts except lower and right boundary as shown below:

02

Create the convex hull

The convex hull can be created by outlining a triangular region with three of its vertices.

It represents one example of a bounded convexset S in \({\mathbb{R}^2}\) such that its profile P in non empty but \({\rm{conv}}\,\,P \ne S\).

So, a rectangle without lower and right boundary represents a bounded convex set S in \({\mathbb{R}^2}\) such that profile P is nonempty but \({\rm{conv}}\,\,P \ne S\).

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Most popular questions from this chapter

In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

15. Sometimes only one-half of a Bezier curve needs further subdividing. For example, subdivision of the 鈥渓eft鈥� side is accomplished with parts (a) and (c) of Exercise 13 and equation (8). When both halves of the curve \({\mathop{\rm x}\nolimits} \left( t \right)\) are divided, it is possible to organize calculations efficiently to calculate both left and right control points concurrently, without using equation (8) directly.

a. Show that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

b. Use part (a) to show that \({{\mathop{\rm q}\nolimits} _3}\) (which equals \({{\mathop{\rm r}\nolimits} _0}\)) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({{\mathop{\rm r}\nolimits} _1}\).

c. Using part (b) and the results of Exercises 13 and 14, write an algorithm that computes the control points for both \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) in an efficient manner. The only operations needed are sums and division by 2.

Explain why a cubic Bezier curve is completely determined by \({\mathop{\rm x}\nolimits} \left( 0 \right)\), \(x'\left( 0 \right)\), \({\mathop{\rm x}\nolimits} \left( 1 \right)\), and \(x'\left( 1 \right)\).

Let \({\bf{x}}\left( t \right)\) be a cubic B茅zier curve determined by points \({{\bf{p}}_o}\), \({{\bf{p}}_1}\), \({{\bf{p}}_2}\), and \({{\bf{p}}_3}\).

a. Compute the tangent vector \({\bf{x}}'\left( t \right)\). Determine how \({\bf{x}}'\left( 0 \right)\) and \({\bf{x}}'\left( 1 \right)\) are related to the control points, and give geometric descriptions of the directions of these tangent vectors. Is it possible to have \({\bf{x}}'\left( 1 \right) = 0\)?

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 \({{\bf{p}}_1}\) as the origin of the coordinate system.]

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

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