Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

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 Vaia!

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

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 Bรฉzier 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 \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free