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Exercises $13-15$ concern the subdivision of a Bézier curve shown in Figure $7 .$ Let $\mathbf{x}(t)$ be the Bézier curve, with control points $\mathbf{p}_{0}, \ldots, \mathbf{p}_{3},$ and let $\mathbf{y}(t)$ and $\mathbf{z}(t)$ be the subdividing Bézier curves as in the text, with control points $\mathbf{q}_{0}, \ldots, \mathbf{q}_{3}$ and $\mathbf{r}_{0}, \ldots, \mathbf{r}_{3}$ respectively.a. Justify each equal sign:$$3\left(\mathbf{r}_{3}-\mathbf{r}_{2}\right)=\mathbf{z}^{\prime}(1)=.5 \mathbf{x}^{\prime}(1)=\frac{3}{2}\left(\mathbf{p}_{3}-\mathbf{p}_{2}\right)$$b. Show that $r_{2}$ is the midpoint of the segment from $\mathbf{p}_{2}$ to $\mathbf{p}_{3}$c. Justify each equal sign: $3\left(\mathbf{r}_{1}-\mathbf{r}_{0}\right)=\mathbf{z}^{\prime}(0)=.5 \mathbf{x}^{\prime}(.5)$d. Use part (c) to show that $8 \mathbf{r}_{1}=-\mathbf{p}_{0} \mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3}+$ 8 $\mathbf{r}_{0} .$e. Use part $(\mathrm{d}),$ equation $(8),$ and part (a) to show that $\mathbf{r}_{1}$ isthe midpoint of the segment from $\mathbf{r}_{2}$ to the midpoint of the segment from $\mathbf{p}_{1}$ to $\mathbf{p}_{2} .$ That is, $\mathbf{r}_{1}=\frac{1 {2}\left[\mathbf{r}_{2}+\frac{1}{2}\left(\mathbf{p}_{1}+\mathbf{p}_{2}\right)\right]$

A. $3\left(\mathrm{r}_{3}-\mathrm{r}_{2}\right)=\mathrm{z}^{\prime}(1)=0.5 \mathrm{x}^{\prime}(1)=\frac{3}{2}\left(\mathrm{p}_{3}-\mathrm{p}_{2}\right)$B. $3\left(\mathbf{r}_{1}-\mathbf{r}_{0}\right)=\mathbf{z}^{\prime}(0)=0.5 \mathbf{x}^{\prime}(0.5)$C. $\mathbf{r}_{1}-\mathbf{r}_{0}=\frac{1}{8}\left(-\mathbf{p}_{0}-\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3}\right)$D. $8 \mathbf{r}_{1}=-\mathbf{p}_{0}-\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3}+8 \mathbf{r}_{0}$E. $\mathbf{r}_{1}=\frac{1}{2}\left[\mathbf{r}_{2}+\frac{1}{2}\left(\mathbf{p}_{1}+\mathbf{p}_{2}\right)\right]$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 6

Curves and Surfaces

Vectors

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and this problem want to find the Bessie a curve with the control points 143 12 6 15 and 74 Well, we can pretty much just plug that into the formula that were given in the equation. 89 and we just get that. The curve is given by X equals one plus 60 plus three T squared minus cube and why equals four plus 24 t minus 15 t squared minus 19. Q.

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