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Chapter 8: Q10E (page 437)
Question 10: Find an example of a closed convex set \(S\) in \({\mathbb{R}^2}\) such that its profile P is nonempty but \({\mathop{\rm conv}\nolimits} P \ne S\).
P is not empty but \({\mathop{\rm conv}\nolimits} P \ne S\) when it is a ray.
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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.
13. a. Use equation (12) to show that \({{\mathop{\rm q}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\).
b. Use equation (13) to show that \(8{{\mathop{\rm q}\nolimits} _2} = 8{{\mathop{\rm q}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + {{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _3}\).
c. Use part (b), equation (8), and part (a) to show that \({{\mathop{\rm q}\nolimits} _2}\) to the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _1}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\). That is, \({{\mathop{\rm q}\nolimits} _2} = \frac{1}{2}\left( {{{\mathop{\rm q}\nolimits} _1} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).
Question: In Exercises 11 and 12, mark each statement True or False. Justify each answer.
11.a. The cubic Bezier curve is based on four control points.
b. Given a quadratic Bezier curve \({\mathop{\rm x}\nolimits} \left( t \right)\) with control points \({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\) and \({{\mathop{\rm p}\nolimits} _2}\), the directed line segment \({{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}\) (from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\)) is the tangent vector to the curve at \({{\mathop{\rm p}\nolimits} _0}\).
c. When two quadratic Bezier curves with control points \(\left\{ {{{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2}} \right\}\) and \(\left\{ {{{\mathop{\rm p}\nolimits} _2},{{\mathop{\rm p}\nolimits} _3},{{\mathop{\rm p}\nolimits} _4}} \right\}\) are joined at \({{\mathop{\rm p}\nolimits} _2}\), the combined Bezier curve will have \({C^1}\) continuity at \({{\mathop{\rm p}\nolimits} _2}\)if\({{\mathop{\rm p}\nolimits} _2}\) is the midpoint of the line segment between \({{\mathop{\rm p}\nolimits} _1}\) and \({{\mathop{\rm p}\nolimits} _3}\).
Question: In Exercise 9, 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 \(H = \left( {f:d} \right)\).
9. \(\left( {\begin{array}{*{20}{c}}1\\0\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\3\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\2\\2\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}1\\1\\1\\1\end{array}} \right)\)
Question 4: Repeat Exercise 2 where \(m\) is the minimum value of \(f\) on \(S\) instead of the maximum value.
The conditions for affine dependence are stronger than those for linear dependence, so an affinely dependent set is automatically linearly dependent. Also, a linearly independent set cannot be affinely dependent and therefore must be affinely independent. Construct two linearly dependent indexed sets\({S_{\bf{1}}}\)and\({S_{\bf{2}}}\)in\({\mathbb{R}^2}\)such that\({S_{\bf{1}}}\)is affinely dependent and\({S_{\bf{2}}}\)is affinely independent. In each case, the set should contain either one, two, or three nonzero points.
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