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Chapter 8: Q1E (page 437)

Suppose a B茅zier curve is translated to \({\bf{x}}\left( t \right) + {\bf{b}}\) . That is, for\(0 \le t \le 1\), the new curve is

\({\bf{x}}\left( t \right) = {\left( {1 - t} \right)^3}{{\bf{p}}_o} + 3t{\left( {1 - t} \right)^2}{{\bf{p}}_1} + 3{t^2}\left( {1 - t} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3} + {\bf{b}}\)

Show that this new curve is again a B茅zier curve.

Short Answer

Expert verified

It is shown that the new curve is also a B茅zier curve.

Step by step solution

01

Find the new control points

If the curve is shifted to\({\rm{x}}\left( t \right) + {\rm{b}}\),then the new control points become\({{\rm{p}}_o} + {\rm{b}},\,\,{{\rm{p}}_1} + {\rm{b}},\,\,{{\rm{p}}_2} + {\rm{b}},\,\,{{\rm{p}}_3} + {\rm{b}}\).

02

Find the new Bezier curve

The new Bezier curve is represented as shown below:

\(\begin{array}{c}{\bf{y}}\left( t \right) = {\left( {1 - t} \right)^3}\left( {{{\bf{p}}_o} + {\bf{b}}} \right) + 3t\left( {1 - {t^2}} \right)\left( {{{\bf{p}}_1} + {\bf{b}}} \right) + 3{t^2}\left( {1 - t} \right)\left( {{{\bf{p}}_2} + {\bf{b}}} \right) + {t^3}\left( {{{\bf{p}}_3} + {\bf{b}}} \right)\\ = {\left( {1 - t} \right)^3}{{\bf{p}}_o} + 3t\left( {1 - {t^2}} \right){{\bf{p}}_1} + 3{t^2}\left( {1 - t} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3} + {\left( {1 - t} \right)^3}{\bf{b}} + 3t\left( {1 - {t^2}} \right){\bf{b}}\\ + 3{t^2}\left( {1 - t} \right){\bf{b}} + {t^3}{\bf{b}}\end{array}\)

03

Verify the sum of weight is equal to 1 or not

The weighted sum of linear combination must be equal to 1.

\(\begin{array}{c}{\left( {1 - t} \right)^3} + 3t\left( {1 - {t^2}} \right) + 3{t^2}\left( {1 - t} \right) + {t^3} = 1 - {t^3} - 3t + 3{t^2} + 3t - 3{t^2} + {t^3}\\ = 1\end{array}\)

04

 Draw a conclusion

As \({\left( {1 - t} \right)^3} + 3t\left( {1 - {t^2}} \right) + 3{t^2}\left( {1 - t} \right) + {t^3} = 1\), for all \(t\). This implies that \({\rm{y}}\left( t \right) = {\rm{x}}\left( t \right) + {\rm{b}}\) is also true for all \(t\).

Thus, translation by \(b\)maps a B茅zier curve into a new B茅zier curve.

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Most popular questions from this chapter

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: 30. Prove that the convex hull of a bounded set is bounded.

Question: 29. Prove that the open ball \(B\left( {{\rm{p}},\delta } \right) = \left\{ {{\rm{x:}}\left\| {{\rm{x - p}}} \right\| < \delta } \right\}\)is a convex set. (Hint: Use the Triangle Inequality).

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).

22. If \(A \subset B\), then \(affA \subset aff B\).

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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