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Expert-verified Found in: Page 437 ### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384 # 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 derivative $${\bf{x}}'\left( t \right)$$ is $${\bf{x}}'\left( t \right) = \frac{1}{6}\left[ { - 3{{\left( {1 - t} \right)}^2}{{\bf{p}}_o} + \left( { - 12t + 9{t^2}} \right){{\bf{p}}_1} + \left( { - 9{t^2} + 6t + 3} \right){{\bf{p}}_2} + 3{t^2}{{\bf{p}}_3}} \right]$$. The relation with control points is $${\bf{x}}'\left( 0 \right) = \frac{1}{2}\left( {{{\bf{p}}_2} - {{\bf{p}}_o}} \right)$$ and $${\bf{x}}'\left( 1 \right) = \frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)$$. The tangent vector at a 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 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 $$\frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)$$.

b) The derivative is. The relation with control points is and .

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

## Find the derivative of the B-Spline curve

The B-spline curve is 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:

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

## Find $$x'\left( 0 \right)$$, and $$x'\left( 1 \right)$$

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 $$\frac{1}{2}\left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)$$.

## Find  x''(0), and X''(1)

The x''(0) is calculated as,

## Draw the figure

The line segment that points in the direction of x''(0) is shown below:  ### Want to see more solutions like these? 