Chapter 10.6, Problem 30E

To determine

To find: the rectangular equations of the curve and graph the equation by using arrows to indicate orientation.

Answer to Problem 30E

  $y^2=4\left(\frac{1}{8}\right)\left(x-\left(-1\right)\right)$

Counter-clockwise

Explanation of Solution

Given information:

The parametric equations are as follows:

  $\begin{array}{l}x=2t-1\mathrm{...}\left(1\right)\\ y=\sqrt{t}\left[\text{where}0\le t\le 4\right]\mathrm{...}\left(2\right)\end{array}$

Calculation:

Solve equation (1) to evaluate $t$ as follows:

  $\begin{array}{c}x=2t-1\\ 2t-1=x\left[\text{Rewriting the equation}\right]\\ 2t=x+1\left[\text{Add 1 on each side}\right]\\ t=\frac{x+1}{2}\left[\text{Divide each side by 2}\right]\end{array}$

Now substitute $t$ in equation (1) and evaluate the equation as shown below.

  $\begin{array}{c}y=\sqrt{t}\\ y=\sqrt{\frac{x+1}{2}}\left[\text{Put the value of }t\right]\\ y^2=\frac{x+1}{2}\left[\text{Squaring each side}\right]\\ y^2=4\left(\frac{1}{8}\right)\left(x-\left(-1\right)\right)\left[\text{Multiply and divide each side by 8}\right]\end{array}$

This equation represents the parabola, ${\left(y-k\right)}^2=4p\left(x-h\right)$ . The center $\left(h,k\right)$ is $\left(-1,0\right)$ and focal length $\left|p\right|$ is $\frac{1}{8}$ . The curve and its orientation for the given interval are shown below.

  

Thus, the orientation is counter-clockwise.