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, \({{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).

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

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 \({{\mathop{\rm p}\nolimits} _3}\).

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

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 \(8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}\).

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 \({{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).