Page: Plot the Hermitz Cubic Curve having end points $P_{0}(1,3)$ and $P_{1}(7,2) .$ The tangent vector for end $P_{0}$ is defined by a line joining $P_{0}$ and another point $P_{2}(10,8)$
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whereas the tangent vector for end $P_{1}$ is defined by line joining $P_{1}$ and another point $P_{3}(6,0)$

Solution:

$P_{0}=\left[\begin{array}{ll}{1} & {3}\end{array}\right]$

$P_{1}=\left[\begin{array}{ll}{7} & {2}\end{array}\right]$

$P_{2}=\left[\begin{array}{ll}{10} & {8}\end{array}\right]$

$P_{3}=\left[\begin{array}{ll}{6} & {0}\end{array}\right]$

Tangent vector,

$P_{0}^{1}=\left[P_{2}-P_{0}\right]$

$P_{0}^{1}=[9 \quad 5]$

$P_{1}^{1}=\left[P_{3}-P_{1}\right]$

$P_{1}^{1}=[-1 \quad -2]$

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

For X -direction

$P_{x}=1\left[2 t^{3}-3 t^{2}+1\right]+7\left[-2 t^{3}+3 t^{2}\right]+9\left[t^{3}-2 t^{2}+t\right]-1\left[t^{3}-t^{2}\right]$

$P_{x}=2 t^{3}-3 t^{2}+1-14 t^{3}+21 t^{2}+9 t^{3}-18 t^{2}+9 t-t^{3}+t^{2}$

$\therefore P_{x}=-4 t^{3}+t^{2}+9 t+1$

For Y-direction

$P_{y}=3\left[2 t^{3}-3 t^{2}+1\right]+2\left[-2 t^{3}+3 t^{2}\right]+5\left[t^{3}-2 t^{2}+t\right]-2\left[t^{3}-t^{2}\right]$

$P_{y}=6 t^{3}-9 t^{2}+3-4 t^{3}+6 t^{2}+5 t^{3}-10 t^{2}+5 t-2 t^{3}+2 t^{2}$

$\therefore P_{y}=5 t^{3}-11 t^{2}+5 t+3$

$\begin{array}{|c|c|c|c|c|c|c|}\hline t \rightarrow & {0} & {0.2} & {0.4} & {0.6} & {0.8} & {1} \\ \hline P_{x} & {1} & {2.808} & {4.504} & {5.896} & {6.792} & {7} \\ \hline P_y & {3} & {3.6} & {3.56} & {3.12} & {2.52} & {2} \\ \hline\end{array}$

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Bezier Curve:

$P(t)=\sum_{i=0}^{n} \mathrm{P}_{\mathrm{i}} \mathrm{B}_{\mathrm{i}}, \mathrm{n}(\mathrm{t})$

Where, $\mathrm{B}_{\mathrm{i}}, \mathrm{n}(\mathrm{t})=\mathrm{C}_{\mathrm{n}, \mathrm{i}} \mathrm{t}^{\mathrm{i}}(1-\mathrm{t})^{\mathrm{n}-\mathrm{i}}$

$\mathrm{C}_{\mathrm{n}, \mathrm{i}}=\frac{\mathrm{n} !}{\mathrm{i} !(\mathrm{n}-\mathrm{i}) !}$

Cubic Curve:

$\mathrm{N}=3 |$ Points $P_{0}, P_{1}, P_{2}, P_{3}$ $P(t)=P_{0} B_{0,3}(t)+P_{1} B_{1,3}(t)+P_{2} B_{2,3}(t)+P_{3} B_{3,3}(t)$ $\quad=P_{0} C_{0,3}(1-t)^{3}+P_{1} C_{1,3} t(1-t)^{2}+P_{2} C_{2,3} t^{2}(1-t)+P_{3} C_{3,3} t^{3}$

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

Use formula, $(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}$ for above equation

Quadratic Curve:

$P(t)=P_{0}(1-t)^{2}+2 P_{1}(1-t) t+P_{2} t^{2}$

$(1-t) \downarrow$

(t) $\uparrow$

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

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modified 10 hours ago by gravatar for Snehal Bhise Snehal Bhise ♦♦ 10 written 8 days ago by gravatar for Ankit Pandey Ankit Pandey70
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