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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

# In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let $${\mathop{\rm x}\nolimits} \left( t \right)$$ be the Bezier curve, with control points $${{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}$$, and let $${\mathop{\rm y}\nolimits} \left( t \right)$$ and $${\mathop{\rm z}\nolimits} \left( t \right)$$ be the subdividing Bezier curves as in the text, with control points $${{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}$$ and $${{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}$$, respectively.14.a. Justify each equal sign:$$3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)$$b. Show that $${{\mathop{\rm r}\nolimits} _2}$$ is the midpoint of the segment from $${{\mathop{\rm p}\nolimits} _2}$$ to $${{\mathop{\rm p}\nolimits} _3}$$.c. Justify each equal sign: $$3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right)$$.d. Use part (c) to show that $$8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}$$.e. Use part (d) equation (8), and part (a) to show that $${{\mathop{\rm r}\nolimits} _1}$$ is the midpoint of the segment from $${{\mathop{\rm r}\nolimits} _2}$$ to the midpoint of the segment from $${{\mathop{\rm p}\nolimits} _1}$$ to $${{\mathop{\rm p}\nolimits} _2}$$. That is, $${{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)$$.

1. The control points for $${\mathop{\rm z}\nolimits} \left( t \right)$$ with $$t = 1$$ is $$3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)$$.
2. It is proved that $${{\mathop{\rm r}\nolimits} _2}$$ is the midpoint of the segment from $${{\mathop{\rm p}\nolimits} _2}$$ to $${{\mathop{\rm p}\nolimits} _3}$$.
3. The control points for $${\mathop{\rm z}\nolimits} \left( t \right)$$ is $$3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right) = \frac{3}{8}\left( { - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)$$.
4. It is proved that $$8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}$$.
5. It is proved that $${{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)$$.
See the step by step solution

## Step 1: Explain each equal sign $$3\left( {{r_3} - {r_2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{p_3} - {p_2}} \right)$$

a)

Let $${\mathop{\rm x}\nolimits} \left( t \right)$$ be the Bezier curve with control points $${{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}$$ and let $${\mathop{\rm y}\nolimits} \left( t \right)$$ and $${\mathop{\rm z}\nolimits} \left( t \right)$$ be the subdividing Bezier curve with the control points $${{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}$$ and $${{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}$$.

For the Bezier curve in (7) as shown below:

$$x'\left( t \right) = \left( { - 3 + 6t - 3{t^2}} \right){{\mathop{\rm p}\nolimits} _0} + \left( {3 - 12t + 9{t^2}} \right){{\mathop{\rm p}\nolimits} _1} + \left( {6t - 9{t^2}} \right){{\mathop{\rm p}\nolimits} _2} + 3{t^2}{{\mathop{\rm p}\nolimits} _3}$$

In particular, for $$0 \le t \le 1$$;

$$x'\left( 0 \right) = 3\left( {{{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}} \right)\,\,\,\,\,{\mathop{\rm and}\nolimits} \,\,\,\,\,\,x'\left( 1 \right) = 3\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,...{\rm{ }}(9)$$

From $$x'\left( t \right)$$ compute $$x'\left( {.5} \right)$$ as shown below:

$$x'\left( {.5} \right) = \frac{3}{4}\left( { - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)$$…(10)

Recall the chain rule for the derivatives as shown below

$$y'\left( t \right) = .5x'\left( {.5t} \right)y'\left( t \right)\,\,\,\,{\mathop{\rm and}\nolimits} \,\,\,z'\left( t \right) = .5x'\left( {.5 + .5t} \right)\,\,\,\,{\mathop{\rm for}\nolimits} 0 \le t \le 1$$…(11)

From (9) with $$z'\left( 1 \right)$$ in place of $$x'\left( 1 \right)$$, from (11) with $$t = 1$$, and from (10), the control points for $${\mathop{\rm z}\nolimits} \left( t \right)$$ satisfy

$$3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)$$

Thus, the control points for $${\mathop{\rm z}\nolimits} \left( t \right)$$ with $$t = 1$$ is $$3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)$$.

## Step 2: Show that $${{\mathop{\rm r}\nolimits} _2}$$ is the midpoint of the segment from $${{\mathop{\rm p}\nolimits} _2}$$ to $${{\mathop{\rm p}\nolimits} _3}$$

b)

Use the part (a) as shown below:

\begin{aligned}{}3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) &= \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\\{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2} &= \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\\{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2} &= \frac{1}{2}{{\mathop{\rm p}\nolimits} _3} - \frac{1}{2}{{\mathop{\rm p}\nolimits} _2}\\{{\mathop{\rm r}\nolimits} _2} &= {{\mathop{\rm r}\nolimits} _3} - \frac{1}{2}{{\mathop{\rm p}\nolimits} _3} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _2}\end{aligned}

Substitute $${{\mathop{\rm r}\nolimits} _3} = {{\mathop{\rm p}\nolimits} _3}$$ in the obtained equation as shown below:

\begin{aligned}{}{{\mathop{\rm r}\nolimits} _2} &= {{\mathop{\rm p}\nolimits} _3} - \frac{1}{2}{{\mathop{\rm p}\nolimits} _3} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _2}\\ &= \frac{1}{2}{{\mathop{\rm p}\nolimits} _2} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _3}\\ &= \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)\end{aligned}

Thus, it is proved that $${{\mathop{\rm r}\nolimits} _2}$$ is the midpoint of the segment from $${{\mathop{\rm p}\nolimits} _2}$$ to $${{\mathop{\rm p}\nolimits} _3}$$.

## Step 3: Explain each equal sign $$3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right)$$

c)

From (9) with $$z'\left( 0 \right)$$ in place of $$x'\left( 0 \right)$$, from (11) with $$t = 0$$, and from (9), the control points for $${\mathop{\rm z}\nolimits} \left( t \right)$$ satisfyingas shown below:

$$3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right) = \frac{3}{8}\left( { - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)$$

Thus, the control points for $${\mathop{\rm z}\nolimits} \left( t \right)$$ is $$3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right) = \frac{3}{8}\left( { - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)$$.

## Step 4: Show that $$8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}$$

d)

Use the part (c) to show that $$8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}$$ as shown below:

$$3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = \frac{3}{8}\left( { - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)$$

Left-multiply the above equation by $$\frac{8}{3}$$ as shown below:

\begin{aligned}{}\frac{8}{3} \times 3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) &= \frac{8}{3} \times \frac{3}{8}\left( { - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)\\8\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) &= - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}\\8{{\mathop{\rm r}\nolimits} _1} - 8{{\mathop{\rm r}\nolimits} _0} &= - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}\\8{{\mathop{\rm r}\nolimits} _1} &= - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}\end{aligned}

Thus, it is proved that $$8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}$$.

## Step 5: Show that $${{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)$$

e)

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 as shown below:

$${\mathop{\rm x}\nolimits} \left( t \right) = \left( {1 - 3t + 3{t^2} - {t^3}} \right){{\mathop{\rm p}\nolimits} _0} + \left( {3t - 6{t^2} + 3{t^3}} \right){{\mathop{\rm p}\nolimits} _1} + \left( {3{t^2} - 3{t^3}} \right){{\mathop{\rm p}\nolimits} _2} + {t^3}{{\mathop{\rm p}\nolimits} _3}$$……(7)

The new control points $${{\mathop{\rm q}\nolimits} _3}$$ and $${{\mathop{\rm r}\nolimits} _0}$$ are given as shown below:

\begin{aligned}{}{{\mathop{\rm r}\nolimits} _0} &= {{\mathop{\rm q}\nolimits} _3}\\ &= {\mathop{\rm x}\nolimits} \left( {.5} \right)\\{{\mathop{\rm r}\nolimits} _0} &= \frac{1}{8}\left( {{{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm p}\nolimits} _1} + 3{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)\end{aligned} ……(8)

$$8{{\mathop{\rm r}\nolimits} _0} = {{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm p}\nolimits} _1} + 3{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}$$

Use equation (8) to substitute for 8$${{\mathop{\rm r}\nolimits} _0}$$ in part (d) as shown below:

\begin{aligned}{}8{{\mathop{\rm r}\nolimits} _1} &= - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + {{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm p}\nolimits} _1} + 3{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}\\ &= 2{{\mathop{\rm p}\nolimits} _1} + 4{{\mathop{\rm p}\nolimits} _2} + 2{{\mathop{\rm p}\nolimits} _3}\end{aligned}

Multiply the above equation by $$\frac{1}{8}$$ as shown below:

$${{\mathop{\rm r}\nolimits} _1} = \frac{1}{4}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _2} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _3}$$

Substitute $$\frac{1}{2}{{\mathop{\rm p}\nolimits} _2} = \frac{1}{4}{{\mathop{\rm p}\nolimits} _2} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _2}$$ in the above equation as shown below:

\begin{aligned}{}{{\mathop{\rm r}\nolimits} _1} &= \frac{1}{4}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _2} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _2} + \frac{1}{4}{{\mathop{\rm p}\nolimits} _3}\\ &= \frac{1}{2}\left( {\frac{1}{2}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)} \right)\end{aligned}

Use part (b) in the above equation as shown below:

\begin{aligned}{}{{\mathop{\rm r}\nolimits} _1} &= \frac{1}{2}\left( {\frac{1}{2}{{\mathop{\rm p}\nolimits} _1} + \frac{1}{2}{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm r}\nolimits} _2}} \right)\\ &= \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\end{aligned}

Thus, it is proved that $${{\mathop{\rm r}\nolimits} _1}$$ is the midpoint of the segment from $${{\mathop{\rm r}\nolimits} _2}$$ to the midpoint of the segment from $${{\mathop{\rm p}\nolimits} _1}$$ to $${{\mathop{\rm p}\nolimits} _2}$$ is $${{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)$$.