Let \({\bf{x}}\left( t \right)\) be a B-spline in Exercise 2, with control points \({{\bf{p}}_o}\), \({{\bf{p}}_1}\) , \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\).

a. Compute the tangent vector \({\bf{x}}'\left( t \right)\) and determine how the derivatives \({\bf{x}}'\left( 0 \right)\) and \({\bf{x}}'\left( 1 \right)\) are related to the control points. Give geometric descriptions of the directions of these tangent vectors. Explore what happens when both \({\bf{x}}'\left( 0 \right)\)and \({\bf{x}}'\left( 1 \right)\)equal 0. Justify your assertions.

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 \({{\bf{p}}_2}\) as the origin of the coordinate system.]

The B-spline curveis given as\({\bf{x}}\left( t \right) = \frac{1}{6}\left[ {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\).

Its first derivative is shown below:

\(\begin{array}{l}{\bf{x}}\left( t \right) = \frac{1}{6}\left[ {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\\{\bf{x}}'\left( t \right) = \frac{1}{6}\left[ { - 3{{\left( {1 - t} \right)}^2}{{\bf{p}}_o} + \left( { - 12t + 9{t^2}} \right){{\bf{p}}_1} + \left( { - 9{t^2} + 6t + 3} \right){{\bf{p}}_2} + 3{t^2}{{\bf{p}}_3}} \right]\end{array}\)

The \({\rm{x'}}\left( 0 \right)\) is calculated as shown below:

\(\begin{array}{c}{\bf{x}}'\left( 0 \right) = \frac{1}{6}\left[ { - 3{{\left( {1 - \left( 0 \right)} \right)}^2}{{\bf{p}}_o} + \left( { - 12\left( 0 \right) + 9{{\left( 0 \right)}^2}} \right){{\bf{p}}_1} + \left( { - 9{{\left( 0 \right)}^2} + 6\left( 0 \right) + 3} \right){{\bf{p}}_2} + 3{{\left( 0 \right)}^2}{{\bf{p}}_3}} \right]\\ = \frac{1}{6}\left( { - 3{{\bf{p}}_o} + 3{{\bf{p}}_{_2}}} \right)\\ = \frac{1}{2}\left( {{{\bf{p}}_2} - {{\bf{p}}_o}} \right)\end{array}\)

The tangent vector at point\({{\bf{p}}_o}\)is directed from\({{\bf{p}}_o}\)to\({{\bf{p}}_2}\)and its length is twice the length of\(\frac{1}{2}\left( {{{\bf{p}}_2} - {{\bf{p}}_o}} \right)\).

The \({\rm{x'}}\left( 1 \right)\)is calculated as shown below:

\(\begin{array}{c}{\bf{x}}'\left( 1 \right) = \frac{1}{6}\left[ { - 3{{\left( {1 - \left( 1 \right)} \right)}^2}{{\bf{p}}_o} + \left( { - 12\left( 1 \right) + 9{{\left( 1 \right)}^2}} \right){{\bf{p}}_1} + \left( { - 9{{\left( 1 \right)}^2} + 6\left( 1 \right) + 3} \right){{\bf{p}}_2} + 3{{\left( 1 \right)}^2}{{\bf{p}}_3}} \right]\\ = \frac{1}{6}\left[ { - 3{{\bf{p}}_1} + 0{{\bf{p}}_2} + 3{{\bf{p}}_3}} \right]\\ = \frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)\end{array}\)

The tangent vector at point \({{\bf{p}}_1}\) is directed from \({{\bf{p}}_1}\) to \({{\bf{p}}_3}\) and its length is twice the length of \(\frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)\).

The x''(0) is calculated as,

The line segment that points in the direction of x''(0) is shown below: