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Magazine Chapter 8: Q-8.6-3E (page 437)
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.]
Short Answer
The derivative \({\bf{x}}'\left( t \right)\) is \({\bf{x}}'\left( t \right) = \left( { - 3 + 6t - 3{t^2}} \right){{\bf{p}}_o} + \left( {3 - 12t + 9{t^2}} \right){{\bf{p}}_1} + \left( {6t - 9{t^2}} \right){{\bf{p}}_2} + 3{t^2}{{\bf{p}}_3}\). The relation with control points is \({\bf{x}}'\left( 0 \right) = 3\left( {{{\bf{p}}_o} - {{\bf{p}}_1}} \right)\)and\({\bf{x}}'\left( 1 \right) = 3\left( {{{\bf{p}}_3} - {{\bf{p}}_2}} \right)\).The tangent vector at point\({{\bf{p}}_o}\)is directed from\({{\bf{p}}_o}\)to\({{\bf{p}}_1}\)and its length is three times the length of\({{\bf{p}}_o}{{\bf{p}}_1}\). The tangent vector at point\({{\bf{p}}_1}\)is directed from\({{\bf{p}}_2}\)to\({{\bf{p}}_3}\)and its length is three times the length of\({{\bf{p}}_2}{{\bf{p}}_3}\).
b) The derivative is
. The relation with control points is and.The line segment that points in the direction of is shown below:
Step by step solution
Find the derivative of Beizer curve
The Beizer curveis given as\({\bf{x}}\left( t \right) = {\left( {1 - t} \right)^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2}} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right)} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}\).
Its first derivative is shown below:
\(\begin{array}{l}{\bf{x}}\left( t \right) = {\left( {1 - t} \right)^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2}} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right)} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}\\{\bf{x}}'\left( t \right) = {\left( { - 3 + 6t - 3{t^2}} \right)^3}{{\bf{p}}_o} + \left( {3 - 12t + 9{t^2}} \right){{\bf{p}}_1} + \left( {6t - 9{t^2}} \right){{\bf{p}}_2} + 3{t^2}{{\bf{p}}_3}\end{array}\)
Find \(x'\left( 0 \right)\), and \(x'\left( 1 \right)\)
The value of \({\rm{x'}}\left( 0 \right)\) is calculated as shown below:
\(\begin{array}{c}{\bf{x}}'\left( t \right) = {\left( { - 3 + 6t - 3{t^2}} \right)^3}{{\bf{p}}_o} + \left( {3 - 12t + 9{t^2}} \right){{\bf{p}}_1} + \left( {6t - 9{t^2}} \right){{\bf{p}}_2} + 3{t^2}{{\bf{p}}_3}\\{\bf{x}}'\left( 0 \right) = {\left( { - 3 + 6\left( 0 \right) - 3{{\left( 0 \right)}^2}} \right)^3}{{\bf{p}}_o} + \left( {3 - 12\left( 0 \right) + 9{{\left( 0 \right)}^2}} \right){{\bf{p}}_1} + \left( {6\left( 0 \right) - 9{{\left( 0 \right)}^2}} \right){{\bf{p}}_2} + 3{\left( 0 \right)^2}{{\bf{p}}_3}\\ = - 3{{\bf{p}}_o} + 3{{\bf{p}}_1}\\ = 3\left( {{{\bf{p}}_o} - {{\bf{p}}_1}} \right)\end{array}\)
The tangent vector at point\({{\bf{p}}_o}\)is directed from\({{\bf{p}}_o}\)to\({{\bf{p}}_1}\)and its length is three times the length of\({{\bf{p}}_o}{{\bf{p}}_1}\).
The \({\rm{x'}}\left( 1 \right)\)is calculated as shown below:
\(\begin{array}{c}{\bf{x}}'\left( 1 \right) = {\left( { - 3 + 6\left( 1 \right) - 3{{\left( 0 \right)}^2}} \right)^3}{{\bf{p}}_o} + \left( {3 - 12\left( 1 \right) + 9{{\left( 1 \right)}^2}} \right){{\bf{p}}_1} + \left( {6\left( 1 \right) - 9{{\left( 1 \right)}^2}} \right){{\bf{p}}_2} + 3{\left( 1 \right)^2}{{\bf{p}}_3}\\ = - 3{{\bf{p}}_2} + 3{{\bf{p}}_3}\\ = 3\left( {{{\bf{p}}_3} - {{\bf{p}}_2}} \right)\end{array}\)
The tangent vector at point \({{\bf{p}}_1}\) is directed from \({{\bf{p}}_2}\) to \({{\bf{p}}_3}\) and its length is three times the length of \({{\bf{p}}_2}{{\bf{p}}_3}\).
Find the second derivative of Beizer curve
The second derivative of the Beizer curve is calculated as shown below:
Find , and
The is calculated as shown below:
The is calculated as shown below:
The tangent vector at point \({{\bf{p}}_2}\) lies along the line starting from \({{\bf{p}}_0}\) to \({{\bf{p}}_1}\) and \({{\bf{p}}_3}\) lies along the line starting from \({{\bf{p}}_1}\) to \({{\bf{p}}_2}\).
Draw the figure
The line segment that points in the direction of is shown below:
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