Chapter 6.3, Problem 29E

Solution

To find: a parametrization of the curve through the points $\left(3,4\right)$ and $\left(6,-3\right)$ .

The parametric equations are $x=3+3t\text{ and }y=4-7t\text{ where }0\le t\le 1$ .

Given data:

Line segment with the end-points $\left(3,4\right)$ and $\left(6,-3\right)$ .

Formula used:

Let $P\left(x,y\right)$ be an arbitrary point on the line through $A$ and $B$ . The vector $\overrightarrow{OP}$ is the tail to head vector sum of $\overrightarrow{OA}$ and $\overrightarrow{AP}$ where $\overrightarrow{AP}$ is a scalar multiple of $\overrightarrow{AB}$ . Let the scalar be $t$, then:

  $\begin{array}{l}\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}\\ \overrightarrow{OP}=\overrightarrow{OA}+t\cdot \overrightarrow{AB}\end{array}$

  $\overrightarrow{OA}$ and $\overrightarrow{OP}$ are vectors from origin $O$ as shown in Figure 1.

  

Calculation:

Calculate the parametric equations for the line through $A\left(3,4\right)$ and $B\left(6,-3\right)$ .

  $\begin{array}{l}\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}\\ \overrightarrow{OP}=\overrightarrow{OA}+t\cdot \overrightarrow{AB}\\ 〈x,y〉=〈3,4〉+t〈6-3,-3-4〉\\ 〈x,y〉=〈3,4〉+t〈3,-7〉\\ 〈x,y〉=〈3+3t,\text{ 4}-7t〉\end{array}$

Here, the vector equation is equivalent to the parametric equation $x=3+3t$ and $y=4-7t$

Put $t=0$ and compute $x$ and $y$ .

  $〈x,y〉=〈3,4〉$

Put $t=1$ and compute $x$ and $y$ .

  $〈x,y〉=〈6,-3〉$

It can be observed that $t=0$ produces point $A$ and $t=1$ produces point $B$ .

Therefore, a parametrization of the line segment is given by $x=3+3t,y=4-7t,0\le t\le 1$ .