Solution:

$P(t)=P_{0}+\left(P_{1}-P_{0}\right) t$

$\begin{aligned} x(t)=P_{x} &=x_{0}+\left(x_{1}-x_{0}\right) t \\ &=1+(4-1) t \\ P_{x} &=1+3 t \end{aligned}$

$\begin{aligned} y(t)=P_{y} &=y_{0}+\left(y_{1}-y_{0}\right) t \\ &=2+(3-2) t \\ P_{y} &=2+t \end{aligned}$

Slope $=\frac{d_{y}}{d_{x}}=\frac{1}{3}$

Tangent Vector,

$\begin{aligned} P^{1} &=\left[x^{1}(t) \quad y^{1}(t)\right] \\ &=[3 \quad 1] \\ \therefore \vec{V} &=3 i+j \end{aligned}$

Parametric Representation of a circle:

1) Center of circle is at origin $0 \leq \theta \leq 2 \pi$

$x=r \cos \theta$

$y=r \sin \theta$

2) Center of circle is not at origin

$x=x_{c}+r \cos \theta$

$y=y_{c}+r \sin \theta$

$x_{i}=r \cos \theta i, y_{i}=r \sin \theta i$

$x_{i}+1=r \cos (\theta i+\delta \theta)$

$y_{i}+1=r \sin (\theta i+\delta \theta)$

$x_{i}+1=r[\cos \theta i \cdot \cos \delta \theta-\sin \theta i \cdot \sin \delta \theta]$

$\therefore x_{i}+1=x_{i} \cos \delta \theta-y_{i} \sin \delta \theta ]$

$\begin{aligned} y_{i}+1 &=r[\sin \theta i . \cos \delta \theta+\cos \theta i . \sin \delta \theta] \\ &=y_{i} \cos \delta \theta+x_{i} \sin \delta \theta \\ \therefore y_{i}+1&=x_{i} \sin \delta \theta-y_{i} \cos \delta \theta \end{aligned}$