TrueType fonts, created by Apple Computer and Adobe Systems, use quadratic Bezier curves, while PostScript fonts, created by Microsoft, use cubic Bezier curves. The cubic curves provide more flexibility for typeface design, but it is important to Microsoft that every typeface using quadratic curves can be transformed into one that used cubic curves. Suppose that \({\mathop{\rm w}\nolimits} \left( t \right)\) is a quadratic curve, with control points \({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\) and \({{\mathop{\rm p}\nolimits} _2}\).

1. Find control points \({{\mathop{\rm r}\nolimits} _0},{{\mathop{\rm r}\nolimits} _1},{{\mathop{\rm r}\nolimits} _2},\), and \({{\mathop{\rm r}\nolimits} _3}\) such that the cubic Bezier curve \({\mathop{\rm x}\nolimits} \left( t \right)\) with these control points has the property that \({\mathop{\rm x}\nolimits} \left( t \right)\) and \({\mathop{\rm w}\nolimits} \left( t \right)\) have the same initial and terminal points and the same tangent vectors at \(t = 0\)and\(t = 1\). (See Exercise 16.)

2. Show that if \({\mathop{\rm x}\nolimits} \left( t \right)\) is constructed as in part (a), then \({\mathop{\rm x}\nolimits} \left( t \right) = {\mathop{\rm w}\nolimits} \left( t \right)\) for \(0 \le t \le 1\).

a)

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

It is given that the control points have the property that\({\mathop{\rm x}\nolimits} \left( t \right)\)and\({\mathop{\rm w}\nolimits} \left( t \right)\)have the same initial and terminal points and the same tangent vectors at\(t = 0\)and\(t = 1\).

Let\({\mathop{\rm w}\nolimits} \left( t \right)\)be a quadratic curve with control points\({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\)and\({{\mathop{\rm p}\nolimits} _2}\).

The standard parametric description of the curves for\(0 \le t \le 1\)as shown below

\({\mathop{\rm w}\nolimits} \left( t \right) = \left( {1 - 2t + {t^2}} \right){{\mathop{\rm p}\nolimits} _0} + 2t\left( {1 - t} \right){{\mathop{\rm p}\nolimits} _1} + {t^2}{{\mathop{\rm p}\nolimits} _2}\)

Differentiation of\({\mathop{\rm w}\nolimits} \left( t \right)\)and with\(t = 0\),\(t = 1\)as shown below

\(\begin{aligned}{}w'\left( t \right) = \left( { - 2 + 2t} \right){{\mathop{\rm p}\nolimits} _0} + \left( {2 - 4t} \right){{\mathop{\rm p}\nolimits} _1} + 2t{{\mathop{\rm p}\nolimits} _2}\\w'\left( 0 \right) = 2\left( {{{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}} \right)\\w'\left( 1 \right) = 2\left( {{{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _1}} \right)\end{aligned}\)

Compute the control points of \({{\mathop{\rm r}\nolimits} _0},{{\mathop{\rm r}\nolimits} _1},{{\mathop{\rm r}\nolimits} _2}\) and \({{\mathop{\rm r}\nolimits} _3}\) as shown below:

\({{\mathop{\rm r}\nolimits} _0} = {{\mathop{\rm p}\nolimits} _0}\)

\({{\mathop{\rm r}\nolimits} _3} = {{\mathop{\rm p}\nolimits} _2}\)

\(\begin{aligned}{}3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = w'\left( 0 \right)\left( {{\mathop{\rm since}\nolimits} \,\,\,{{\mathop{\rm r}\nolimits} _0} = {{\mathop{\rm p}\nolimits} _0}} \right)\\3{{\mathop{\rm r}\nolimits} _1} - 3{{\mathop{\rm p}\nolimits} _0} = 2\left( {{{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}} \right)\\3{{\mathop{\rm r}\nolimits} _1} = 2{{\mathop{\rm p}\nolimits} _1} - 2{{\mathop{\rm p}\nolimits} _0} + 3{{\mathop{\rm p}\nolimits} _0}\\{{\mathop{\rm r}\nolimits} _1} = \frac{{{{\mathop{\rm p}\nolimits} _0} + 2{{\mathop{\rm p}\nolimits} _1}}}{3}\end{aligned}\)

\(\begin{aligned}{}3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = w'\left( 1 \right)\left( {{\mathop{\rm since}\nolimits} \,\,\,{{\mathop{\rm r}\nolimits} _3} = {{\mathop{\rm p}\nolimits} _2}} \right)\\3{{\mathop{\rm p}\nolimits} _{ 2}} - 3{{\mathop{\rm r}\nolimits} _2} = 2\left( {{{\mathop{\rm p}\nolimits} _2} - {{\mathop{\rm p}\nolimits} _1}} \right)\\3{{\mathop{\rm r}\nolimits} _2} = - 2{{\mathop{\rm p}\nolimits} _2} + 2{{\mathop{\rm p}\nolimits} _1} + 3{{\mathop{\rm p}\nolimits} _2}\\{{\mathop{\rm r}\nolimits} _2} = \frac{{2{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}}}{3}\end{aligned}\)

Thus, the control points of \({{\mathop{\rm r}\nolimits} _0},{{\mathop{\rm r}\nolimits} _1},{{\mathop{\rm r}\nolimits} _2}\) and \({{\mathop{\rm r}\nolimits} _3}\) are \({{\mathop{\rm r}\nolimits} _0} = {{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm r}\nolimits} _1} = \frac{{{{\mathop{\rm p}\nolimits} _0} + 2{{\mathop{\rm p}\nolimits} _1}}}{3},{{\mathop{\rm r}\nolimits} _2} = \frac{{2{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}}}{3}\) and \({{\mathop{\rm r}\nolimits} _3} = {{\mathop{\rm p}\nolimits} _2}\).

Recall 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

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

Rewrite the equation (7) with \({{\mathop{\rm r}\nolimits} _i}\) in place of \({{\mathop{\rm p}\nolimits} _i}\) for \(i = 0,...,3\) as shown below

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

Substitute \({{\mathop{\rm r}\nolimits} _0} = {{\mathop{\rm p}\nolimits} _0}\) and \({{\mathop{\rm r}\nolimits} _3} = {{\mathop{\rm p}\nolimits} _2}\) as shown below:

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

Use the control points of \({{\mathop{\rm r}\nolimits} _1}\) and \({{\mathop{\rm r}\nolimits} _2}\) from part (a) as shown below:

\(\begin{aligned}{}{\mathop{\rm x}\nolimits} \left( t \right) &= \left( {1 - 3t + 3{t^2} - {t^3}} \right){{\mathop{\rm p}\nolimits} _0} + \left( {3t - 6{t^2} + 3{t^3}} \right)\left( {\frac{{{{\mathop{\rm p}\nolimits} _0} + 2{{\mathop{\rm p}\nolimits} _1}}}{3}} \right) + \left( {3{t^2} - 3{t^3}} \right)\left( {\frac{{2{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}}}{3}} \right) + {t^3}{{\mathop{\rm p}\nolimits} _2}\\ & = {{\mathop{\rm p}\nolimits} _0} - 3t{{\mathop{\rm p}\nolimits} _0} + 3{t^2}{{\mathop{\rm p}\nolimits} _0} - {t^3}{{\mathop{\rm p}\nolimits} _0} + 2{t^2}{{\mathop{\rm p}\nolimits} _1} + {t^2}{{\mathop{\rm p}\nolimits} _2} - 2{t^3}{{\mathop{\rm p}\nolimits} _1} - {t^3}{{\mathop{\rm p}\nolimits} _2} + t{{\mathop{\rm p}\nolimits} _0} + 2t{{\mathop{\rm p}\nolimits} _1} - 2{t^2}{{\mathop{\rm p}\nolimits} _0}\\ - 4{t^2}{{\mathop{\rm p}\nolimits} _1} + {t^3}{{\mathop{\rm p}\nolimits} _0} + 2{t^3}{{\mathop{\rm p}\nolimits} _1} + {t^3}{{\mathop{\rm p}\nolimits} _2}\\ & = {{\mathop{\rm p}\nolimits} _0} - 2t{{\mathop{\rm p}\nolimits} _0} + {t^2}{{\mathop{\rm p}\nolimits} _0} - 2{t^2}{{\mathop{\rm p}\nolimits} _1} + {t^2}{{\mathop{\rm p}\nolimits} _2} + 2t{{\mathop{\rm p}\nolimits} _1}\\ & = \left( {1 - 2t + {t^2}} \right){{\mathop{\rm p}\nolimits} _0} + \left( {2t - 2{t^2}} \right){{\mathop{\rm p}\nolimits} _1} + {t^2}{{\mathop{\rm p}\nolimits} _2}\\ & = {\left( {1 - t} \right)^2}{{\mathop{\rm p}\nolimits} _0} + 2t\left( {1 - t} \right){{\mathop{\rm p}\nolimits} _1} + {t^2}{{\mathop{\rm p}\nolimits} _2}\\ & = {\mathop{\rm w}\nolimits} \left( t \right)\end{aligned}\)

Thus, it is proved that \({\mathop{\rm x}\nolimits} \left( t \right) = {\mathop{\rm w}\nolimits} \left( t \right)\).