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.

15. Sometimes only one-half of a Bezier curve needs further subdividing. For example, subdivision of the “left” side is accomplished with parts (a) and (c) of Exercise 13 and equation (8). When both halves of the curve \({\mathop{\rm x}\nolimits} \left( t \right)\) are divided, it is possible to organize calculations efficiently to calculate both left and right control points concurrently, without using equation (8) directly.

a. Show that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

b. Use part (a) to show that \({{\mathop{\rm q}\nolimits} _3}\) (which equals \({{\mathop{\rm r}\nolimits} _0}\)) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({{\mathop{\rm r}\nolimits} _1}\).

c. Using part (b) and the results of Exercises 13 and 14, write an algorithm that computes the control points for both \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) in an efficient manner. The only operations needed are sums and division by 2.

a)

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 equation (11) with \(t = 1\) in \(y'\left( t \right)\) and \(t = 0\) in \(z'\left( t \right)\) as shown below:

\(\begin{aligned}{}y'\left( 1 \right) &= .5x'\left( {.5} \right)\\ &= z'\left( 0 \right)\end{aligned}\)

Thus, it is proved that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

It is noted that \(y'\left( 1 \right) = 3\left( {{{\mathop{\rm q}\nolimits} _3} - {{\mathop{\rm q}\nolimits} _2}} \right)\). The following equation follows from equation (9), for \({\mathop{\rm y}\nolimits} \left( t \right)\) its control points in place of \({\mathop{\rm x}\nolimits} \left( t \right)\) and its control points. Likewise, for \({\mathop{\rm z}\nolimits} \left( t \right)\) and its control points, \(z'\left( 0 \right) = 3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right)\).

From part (a) as shown below:

\(3\left( {{{\mathop{\rm q}\nolimits} _3} - {{\mathop{\rm q}\nolimits} _2}} \right) = 3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right)\)

Substitute \({{\mathop{\rm r}\nolimits} _0} = {{\mathop{\rm q}\nolimits} _3}\) in the above equation as shown below:

\(\begin{aligned}{}3\left( {{{\mathop{\rm q}\nolimits} _3} - {{\mathop{\rm q}\nolimits} _2}} \right) &= 3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm q}\nolimits} _3}} \right)\\{{\mathop{\rm q}\nolimits} _3} - {{\mathop{\rm q}\nolimits} _2} &= {{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm q}\nolimits} _3}\\2{{\mathop{\rm q}\nolimits} _3} &= {{\mathop{\rm r}\nolimits} _1} + {{\mathop{\rm q}\nolimits} _2}\\{{\mathop{\rm q}\nolimits} _3} &= \frac{{\left( {{{\mathop{\rm q}\nolimits} _2} + {{\mathop{\rm r}\nolimits} _1}} \right)}}{2}\end{aligned}\)

Thus, it is proved that \({{\mathop{\rm q}\nolimits} _3}\) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({r_1}\).

The new control points are related to the original control points by simple formulas are\({{\mathop{\rm q}\nolimits} _0} = {{\mathop{\rm p}\nolimits} _0}\)and\({{\mathop{\rm r}\nolimits} _3} = {{\mathop{\rm p}\nolimits} _3}\).

Recall the result from exercise 13 as shown below:

\({{\mathop{\rm q}\nolimits} _1}\)is the midpoint of the segment from\({{\mathop{\rm p}\nolimits} _0}\)to\({{\mathop{\rm p}\nolimits} _1}\)is\({{\mathop{\rm q}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _0}} \right)\).

\({{\mathop{\rm q}\nolimits} _2}\)is the midpoint of the segment from\({{\mathop{\rm q}\nolimits} _1}\)to the midpoint of the segment from

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

Recall the result from Exercise 14 as shown below:

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

Here\({{\mathop{\rm r}\nolimits} _2}\)is the midpoint of the segment from\({{\mathop{\rm p}\nolimits} _2}\)to\({{\mathop{\rm p}\nolimits} _3}\)is\({{\mathop{\rm r}\nolimits} _2} = \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)\).

Write the algorithm that computes the control points for both\({\mathop{\rm y}\nolimits} \left( t \right)\)and\({\mathop{\rm z}\nolimits} \left( t \right)\)as shown below

Set the formulas\({{\mathop{\rm q}\nolimits} _0} = {{\mathop{\rm p}\nolimits} _0}\)and\({{\mathop{\rm r}\nolimits} _3} = {{\mathop{\rm p}\nolimits} _3}\).

Compute\({{\mathop{\rm q}\nolimits} _1}\)is the midpoint of the segment from\({{\mathop{\rm p}\nolimits} _0}\)to\({{\mathop{\rm p}\nolimits} _1}\)is\({{\mathop{\rm q}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _0}} \right)\)and\({{\mathop{\rm r}\nolimits} _2}\)is the midpoint of the segment from\({{\mathop{\rm p}\nolimits} _2}\)to\({{\mathop{\rm p}\nolimits} _3}\)is\({{\mathop{\rm r}\nolimits} _2} = \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3}} \right)\).

Consider\({\mathop{\rm m}\nolimits} = \frac{{{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}}}{2}\)

Compute\({{\mathop{\rm q}\nolimits} _2}\)is the midpoint of the segment from\({{\mathop{\rm q}\nolimits} _1}\)to\({\mathop{\rm m}\nolimits} \)is\({{\mathop{\rm q}\nolimits} _2} = \frac{1}{2}\left( {{{\mathop{\rm q}\nolimits} _1} + {\mathop{\rm m}\nolimits} } \right)\).

Compute\({{\mathop{\rm r}\nolimits} _1}\)is the midpoint of the segment from\({{\mathop{\rm r}\nolimits} _2}\)to\({\mathop{\rm m}\nolimits} \)is\({{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + {\mathop{\rm m}\nolimits} } \right)\).

Compute\({{\mathop{\rm q}\nolimits} _3}\)is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({r_1}\) is \({{\mathop{\rm q}\nolimits} _3} = \frac{{\left( {{{\mathop{\rm q}\nolimits} _2} + {{\mathop{\rm r}\nolimits} _1}} \right)}}{2}\).

Set the control points \({{\mathop{\rm r}\nolimits} _0} = {{\mathop{\rm q}\nolimits} _3}\).