Page: $P_{0}=(x, y)=(3,4), \quad P_{1}=(x, y)=(10,7)$
0

Solution:

$\begin{aligned} P_{x} &=\left[\begin{array}{ll}{3} & {10}\end{array}\right]\left[\begin{array}{cc}{-1} & {1} \\ {1} & {0}\end{array}\right]\left\{\frac{t}{1}\right\} \\ &=\left[\begin{array}{cc}{3} & {10}\end{array}\right]\left[\begin{array}{cc}{-t} & {1} \\ {t} & {0}\end{array}\right] \\ P_{x} &=3(-t+1)+10 \\ P_{x} &=7 t+3 \end{aligned}$

$P_{y}=\left[\begin{array}{ll}{4} & {7}\end{array}\right]\left[\begin{array}{cc}{-1} & {1} \\ {1} & {0}\end{array}\right]\left\{\frac{t}{1}\right\}$

$P_{y}=4(-t+1)+7 t$

$P_{y}=3 t+4$

$\begin{array}{|c|c|c|c|c|c|c|}\hline t \rightarrow & {0} & {0.2} & {0.4} & {0.6} & {0.8} & {1} \\ \hline P x & {3} & {4.4} & {5.8} & {7.2} & {8.6} & {10} \\ \hline P y & {4} & {4.6} & {5.2} & {5.8} & {6.4} & {7} \\ \hline\end{array}$

Nimitz Cubic curve:

enter image description here

Start point $\mathrm{t}=0$ , end point $\mathrm{t}=1$

Tangent vector at start point $=P_{0}^{1}=\left[P_{2}-P_{0}\right]$

Tangent vector at end point $=P_{1}^{1}=\left[P_{3}-P_{1}\right]$

$P(t)=\sum_{i=0}^{3} C_{i} t^{i}$

$P(t)=C_{0}+C_{1} t+C_{2} t^{2}+C_{3} t^{3}$

$P^{1}(t)=C_{1}+2 C_{2} t+3 C_{3} t^{2}$

At $t=0$

$P_{0}=C_{0}$

$P_{0}^{1}=C_{1}$

At $t=1$

$P_{1}=C_{0}+C_{1}+C_{2}+C_{3} \ldots (1)$

$P_{1}^{1}=C_{1}+2 C_{2}+3 C_{3} \ldots (2)$

Multiplying eqn 1 by 3 and Subtracting by 2

$3 P_{1}-P_{1}^{1}=3 C_{0}+3 C_{1}+3 C_{2}+3 C_{3}-C_{1}-2 C_{2}-3 C_{3}$

$3 P_{1}-P_{1}^{1}=3 C_{0}+2 C_{1}+C_{2}$

$\therefore C_{2}=3 P_{1}-P_{1}^{1}-3 P_{0}-2 P_{0}^{1}$

Similarly,

$\therefore C_{3}=3 P_{1}-P_{1}^{1}-3 P_{0}-2 P_{0}^{1}$

$2 P_{1}-P_{1}^{1}=2 C_{0}+2 C_{1}+2 C_{2}+2 C_{3}-C_{1}-2 C_{2}-3 C_{3}$

$2 P_{1}-P_{1}^{1}=2 C_{0}+C_{1}-C_{3}$

$C_{3}=2 C_{0}+C_{1}-2 P_{1}+P_{1}^{1}$

$\begin{aligned} P(t) &=C_{0}+C_{1} t+C_{2} t^{2}+C_{3} t^{3} \\ P(t) &=P_{0}+P_{0}^{1} t+\left[3 P_{1}-P_{1}^{1}-3 P_{0}-2 P_{0}^{1}\right] t^{2}+\left[2 P_{0}+P_{0}^{1}-2 P_{1}+P_{1}^{1}\right] t^{3} \\ &=P_{0}\left[2 t^{3}-3 t^{2}+1\right]+P_{1}\left[-2 t^{3}+3 t^{2}\right]+P_{0}^{1}\left[t^{3}-2 t^{2}+t\right]+P_{1}^{1}\left[t^{3}-t^{2}\right] \end{aligned}$

$P(t)=\left[P_{0} P_{1} P_{0}^{1} P_{1}^{1}\right]^{}\left[\begin{array}{ccc}{2 t^{3}} & {-3 t^{2}} & {1} \\ {-2 t^{3}} & {3 t^{2}} & {1} \\ {t^{3}} & {-2 t^{2}} & {t} \\ {t^{3}} & {-t^{2}} & {1} \end{array}\right]$

$P(t)=\left[\begin{array}{cccc}{2} & {-3} & {0} & {1} \\ {-2} & {3} & {0} & {0} \\ {1} & {-2} & {1} & {0} \\ {1} & {-1} & {0} & {0}\end{array}\right]\left[\begin{array}{l}{t^{3}} \\ {t^{2}} \\ {t} \\ {1}\end{array}\right]$

ADD COMMENTlink
modified 10 hours ago by gravatar for Snehal Bhise Snehal Bhise ♦♦ 10 written 8 days ago by gravatar for Ankit Pandey Ankit Pandey70
Please log in to add an answer.