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Chapter 8: Q16E (page 437)

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 \(x'\left( 1 \right)\).

Short Answer

Expert verified

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

Step by step solution

01

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

The cubic Bezier curve is determined by four control points.

Let\({\mathop{\rm x}\nolimits} \left( t \right)\)be the cubic Bezier curve determined by the control points\({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2},\)and\({{\mathop{\rm p}\nolimits} _3}\).

The midpoint of the original curve\({\mathop{\rm x}\nolimits} \left( t \right)\)occurs at\({\mathop{\rm x}\nolimits} \left( {.5} \right)\)when\({\mathop{\rm x}\nolimits} \left( t \right)\)has the standard parameterization

\({\mathop{\rm x}\nolimits} \left( t \right) = \left( {1 - 3t + 3{t^2} - {t^3}} \right){{\mathop{\rm p}\nolimits} _0} + \left( {3t - 6{t^2} + 3{t^3}} \right){{\mathop{\rm p}\nolimits} _1} + \left( {3{t^2} - 3{t^3}} \right){{\mathop{\rm p}\nolimits} _2} + {t^3}{{\mathop{\rm p}\nolimits} _3}\)

02

Cubic Bezier curve

Differentiation of\({\mathop{\rm x}\nolimits} \left( t \right)\)as shown below:

\(\begin{aligned}{}x'\left( t \right) &= \left( { - 3 + 6t - 3{t^2}} \right){{\mathop{\rm p}\nolimits} _0} + \left( {3 - 12t + 9{t^2}} \right){{\mathop{\rm p}\nolimits} _1} + \left( {6t - 9{t^2}} \right){{\mathop{\rm p}\nolimits} _2} + 3{t^2}{{\mathop{\rm p}\nolimits} _3}\\x'\left( 0 \right) & = - 3{{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm p}\nolimits} _1} & = 3\left( {{{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}} \right)\\x'\left( 1 \right) & = - 3{{\mathop{\rm p}\nolimits} _2} + 3{{\mathop{\rm p}\nolimits} _3} & = 3\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\end{aligned}\)

This demonstrates that the tangent vector\(x'\left( 0 \right)\)points in the direction from\({{\mathop{\rm p}\nolimits} _0}\)to\({{\mathop{\rm p}\nolimits} _1}\)and is 3 times the length of\({{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}\).

Similarly,\(x'\left( 1 \right)\)points in the direction from\({{\mathop{\rm p}\nolimits} _2}\)to\({{\mathop{\rm p}\nolimits} _3}\)and is three times the length of\({{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}\).

Thus, the cubic Bezier curve is determined by \({\mathop{\rm x}\nolimits} \left( 0 \right),x'\left( 0 \right),{\mathop{\rm x}\nolimits} \left( 1 \right),\)and\(x'\left( 1 \right)\).

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Most popular questions from this chapter

Question 2: Given points \({{\mathop{\rm p}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}0\\{ - 1}\end{array}} \right),{\rm{ }}{{\mathop{\rm p}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right),\) and \({{\mathop{\rm p}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\) in \({\mathbb{R}^{\bf{2}}}\), let \(S = {\mathop{\rm conv}\nolimits} \left\{ {{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2},{{\mathop{\rm p}\nolimits} _3}} \right\}\). For each linear functional \(f\), find the maximum value \(m\) of \(f\), find the maximum value \(m\) of \(f\) on the set \(S\), and find all points x in \(S\) at which \(f\left( {\mathop{\rm x}\nolimits} \right) = m\).

a. \(f\left( {{x_1},{x_2}} \right) = {x_1} + {x_2}\)

b. \(f\left( {{x_1},{x_2}} \right) = {x_1} - {x_2}\)

c. \(f\left( {{x_1},{x_2}} \right) = - 2{x_1} + {x_2}\)

Question: 24. Repeat Exercise 23 for \({v_1} = \left( \begin{array}{l}1\\2\end{array} \right)\), \({v_2} = \left( \begin{array}{l}5\\1\end{array} \right)\), \({v_3} = \left( \begin{array}{l}4\\4\end{array} \right)\) and \(p = \left( \begin{array}{l}2\\3\end{array} \right)\).

Question: Let \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\), \({\bf{n}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{5}}\\{ - {\bf{1}}}\end{array}} \right)\), \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{0}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{4}}\\{\bf{0}}\\{\bf{4}}\end{array}} \right)\), and let H be the hyperplane in\({\mathbb{R}^{\bf{4}}}\) with normal n and passing through p. Which of the points \({{\bf{v}}_{\bf{1}}}\), \({{\bf{v}}_{\bf{2}}}\), and \({{\bf{v}}_{\bf{3}}}\) are on the same side of H as the origin, and which are not?

A quartic Bézier curve is determined by five control points,

\({{\bf{p}}_{\bf{o}}}{\bf{,}}\,{\rm{ }}{{\bf{p}}_{\bf{1}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{2}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{3}}}\)and \({{\bf{p}}_4}\):

\({\bf{x}}\left( t \right) = {\left( {1 - t} \right)^4}{{\bf{p}}_0} + 4t{\left( {1 - t} \right)^3}{{\bf{p}}_1} + 6{t^2}{\left( {1 - t} \right)^2}{{\bf{p}}_2} + 4{t^3}\left( {1 - t} \right){{\bf{p}}_3} + {t^4}{{\bf{p}}_4}\)for \(0 \le t \le 1\)

Construct the quartic basis matrix \({M_B}\) for \({\bf{x}}\left( t \right)\).

Question: Suppose that the solutions of an equation \(A{\bf{x}} = {\bf{b}}\) are all of the form \({\bf{x}} = {x_{\bf{3}}}{\bf{u}} + {\bf{p}}\), where \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{4}}\end{array}} \right)\). Find points \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) such that the solution set of \(A{\bf{x}} = {\bf{b}}\) is \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).
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