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# System prepended metadata

title: GDV-Exercise 11

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# GDV-Exercise 11

## 11.1

### t=0.25:
![](https://i.imgur.com/4elk2da.png)

$\begin{pmatrix}0\\0\end{pmatrix} + 0.25 \cdot \begin{pmatrix}2\\4\end{pmatrix} = \begin{pmatrix}0.5\\1\end{pmatrix}$

$\begin{pmatrix}2\\4\end{pmatrix} + 0.25 \cdot \begin{pmatrix}5-2\\0-4\end{pmatrix} = \begin{pmatrix}2.75\\3\end{pmatrix}$

$\begin{pmatrix}5\\0\end{pmatrix} + 0.25 \cdot \begin{pmatrix}6-5\\2-0\end{pmatrix} = \begin{pmatrix}5.25\\0.5\end{pmatrix}$

Second order points:
$\begin{pmatrix}0.5\\1\end{pmatrix} + 0.25 \cdot \begin{pmatrix}2.75-0.5\\3-1\end{pmatrix} = \begin{pmatrix}1.0625\\1.5\end{pmatrix}$

$\begin{pmatrix}2.75\\3\end{pmatrix} + 0.25 \cdot \begin{pmatrix}5.25-2.75\\0.5-3\end{pmatrix} = \begin{pmatrix}3.375\\2.375\end{pmatrix}$

Final point:
$\begin{pmatrix}1.0625\\1.5\end{pmatrix} + 0.25 \cdot \begin{pmatrix}3.375-1.0625\\2.375-1.5\end{pmatrix} \approx \begin{pmatrix}1.64\\ 1.72\end{pmatrix}$


### t=0.5
![](https://i.imgur.com/oYLSxBa.png)


### t=0.75
$\begin{pmatrix}0\\0\end{pmatrix} + 0.75 \cdot \begin{pmatrix}2\\4\end{pmatrix} = \begin{pmatrix}1.5\\3\end{pmatrix}$

$\begin{pmatrix}2\\4\end{pmatrix} + 0.75 \cdot \begin{pmatrix}5-2\\0-4\end{pmatrix} = \begin{pmatrix}4.25\\1\end{pmatrix}$

$\begin{pmatrix}5\\0\end{pmatrix} + 0.75 \cdot \begin{pmatrix}6-5\\2-0\end{pmatrix} = \begin{pmatrix}5.75\\1.5\end{pmatrix}$

Second order points:
$\begin{pmatrix}1.5\\3\end{pmatrix} + 0.75 \cdot \begin{pmatrix}4.25-1.5\\1-3\end{pmatrix} = \begin{pmatrix}3.5625\\1.5\end{pmatrix}$

$\begin{pmatrix}4.25\\1\end{pmatrix} + 0.75 \cdot \begin{pmatrix}5.75-4.25\\1.5-1\end{pmatrix} = \begin{pmatrix}5.375\\1.375\end{pmatrix}$

Final point:
$\begin{pmatrix}3.5625\\1.5\end{pmatrix} + 0.75 \cdot \begin{pmatrix}5.375-3.5625\\1.375-1.5\end{pmatrix} \approx \begin{pmatrix}4.922\\ 1.406\end{pmatrix}$

![](https://0wo.in/file/mole-worm-pony/image_2023-01-30_230001438.png)


## 11.2

### a)
$P(t) = \sum_{i=0}^3 B^3_i(t) \cdot b_i$

$= B^3_0(t) b_0 + B^3_1(t) b_1 + B^3_2(t) b_2 + B^3_3(t) b_3$

$= \binom{3}{0}t^0(1-t)^3 + \binom{3}{1}t^1(1-t)^2 + \binom{3}{2}t^2(1-t)^1 + \binom{3}{3} t^3 \cdot (1-t)^0$


$= (1-t)^3 b_0 + 3t(1-t)^2 b_1 + 3 t^2 (1-t) b_2 + t^3 b_3$


$= (-t^3 + 3 t^2 - 3 t + 1)b_0 + (3 t^3 - 6 t^2 + 3 t)b_1 + (-3t^3 + 3t^2)b_2 + t^3b_3$

$= -b_0t^3 + 3b_1t^3 -3b_2t^3 + t^3b_3 + (3b_0 - 6b_1 + 3b_2)t^2$

$= (-b_0 + 3b_1 -3b_2 + b_3) t^3 + (3b_0 - 6b_1 + 3b_2)t^2 + (-3b_0 + 3b_1)t + b_0$

Derive:
$= 3(-b_0 + 3b_1 -3b_2 + b_3)t^2 + 2\cdot (3b_0 - 6b_1 + 3b_2)t + (-3b_0 + 3b_1)$
$= (-3b_0t^2 + 6b_0t - 3b_0) + (9b_1t^2-12b_1t + 3b_1) + (-9b_2t^2 + 6b_2t) + 3b_3t^2$
$= -3(t^2 + 2t - 1)b_0 + (9t^2-12t+3)b_1 + (-9t^2+6t) + 3t^2b_3$
$= -3(1-t)^2b_0 + (9t^2 - 12t + 3)b_1 + (-9t^2+6t) + 3t^2b_3$

Result:
$P'(t) = -3(1-t)^2b_0 + (9t^2 - 12t + 3)b_1 + (-9t^2+6t)b_2 + 3t^2b_3$


### b)
Derivative:
$P'(t) = -3(1-t)^2\begin{pmatrix}0\\0\end{pmatrix}  + (9t^2 - 12t + 3)\begin{pmatrix}2\\4\end{pmatrix}  + (-9t^2+6t)\begin{pmatrix}5\\0\end{pmatrix}  + 3t^2\begin{pmatrix}6\\2\end{pmatrix}$

<img src="https://i.imgur.com/AveVIDL.png" style="width: 500px;"/>

### c)



## 11.3
### a)
$B_0 = $
### b)
$\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ -3 & 3 & 0 & 0\\0 & 0 & -3 & 3\end{pmatrix} \cdot \begin{pmatrix}0 & 0\\2 & 4\\ 5 & 0 \\ 6 & 2\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 6 & 2\\ 6 & 12\\ 3 & 6\end{pmatrix}$

Parameters: $P_0 = (0, 0); P_1 = (6,2); P_2'=(6, 12); P_3'=(3,6)$




## 11.4
### a)
![](https://0wo.in/file/fish-spider-frog/P_20230130_171604.jpg)

### b) 
Catmull-Rom splines are continuous in $C^0$ and $C^1$, but not $C^2$ or higher.
### c)
NURBS are able to represent a perfect circle exactly (ref. lecture). 

