Chapter 0, Problem 51RE

To determine

To find: the parameterization for the curve.

Answer to Problem 51RE

  $x=t,y=\frac{5}{3}t+\frac{5}{3},t\le 2$

Explanation of Solution

Given information:

The curve is:

The ray with initial point $\left(2,5\right)$ and that passes through $\left(-1,0\right)$ .

Formula Used:

The slope of a line segment containing two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

The equation of the line segment containing the point $\left(x_1,y_1\right)$ and with slope m is:

  $y-y_1=m\left(x-x_1\right)$

It is given that the ray with initial point $\left(2,5\right)$ and that passes through $\left(-1,0\right)$ . So, the slope of this line segment is:

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{0-5}{-1-2}\\ =\frac{-5}{-3}\\ =\frac{5}{3}\end{array}$

Then, the equation of this line segment is:

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-5=\frac{5}{3}\left(x-2\right)\\ y-5=\frac{5}{3}x-\frac{10}{3}\end{array}$

  $\begin{array}{c}y=\frac{5}{3}x-\frac{10}{3}+5\\ y=\frac{5}{3}x+\frac{5}{3}\end{array}$

Since, the initial point of this ray is $\left(2,5\right)$ and it passes through $\left(-1,0\right)$ , so the domain of this function is $x\le 2$ .

To find the parameterization of the Cartesian equation, let $x=t$ .

Now, find y in terms of t ,

  $\begin{array}{c}y=\frac{5}{3}x+\frac{5}{3}\\ y=\frac{5}{3}t+\frac{5}{3}\end{array}$

So, a possible parameterization is:

  $x=t,y=\frac{5}{3}t+\frac{5}{3}$

Now, domain is given to be $x\le 2$ , substituting $x=t$ gives the interval $t\le 2$ for parameterization.

Thus, the parameterization for the curve is:

  $x=t,y=\frac{5}{3}t+\frac{5}{3},t\le 2$