Chapter 1.4, Problem 18E

(a)

To determine

To find: The graph of the parametric equations, initial and terminal points. Also, indicate the direction in which the curve is traced.

Answer to Problem 18E

The graph of the parametric equations is shown in figure (1), initial point is $\left(3,0\right)$ and terminal point is $\left(0,2\right)$.

Explanation of Solution

Given information: The equations are $x=3-3t$ and $y=2t$ , $0\le t\le 1$.

Calculation:

Use the following step to graph the parametric equations by graphing calculator.

Step 1: First press the “ON” button graphical calculator.

Step 2: Press the $\boxed{\text{MODE}}$ button, under FUNC and choose PAR option.

Step 3: Press $\boxed{\text{Y=}}$ , and set the parametric equations equal to ${\text{X}}_{\text{1T}}$ and ${\text{Y}}_{\text{1T}}$.

Step 4: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the parametric equations and the window can be set to $\left[-2,4\right]$ by $\left[-1,3\right]$ as shown below.

  

Figure (1)

If $t=0$ , then $x=3$ and $y=0$ . So, the initial point is $\left(3,0\right)$.

If $t=1$ , then $x=0$ and $y=2$ . So, the terminal point is $\left(0,2\right)$.

Therefore, the graph of the parametric equations is shown in figure (1), initial point is $\left(3,0\right)$ and terminal point is $\left(0,2\right)$.

(b)

To determine

To find: The Cartesian equation for a curve that contains the parameterized curve and the portion of the graph of the Cartesian equation that is traced by the parameterized curve.

Answer to Problem 18E

The Cartesian equation for a curve that contains the parameterized curve is $y=-\frac{2}{3}x+2$ and portion traced by the parameterized curve is the segment from $\left(3,0\right)$ to $\left(0,2\right)$.

Explanation of Solution

Given information: The equations are $x=3-3t$ and $y=2t$ , $0\le t\le 1$.

Calculation:

Rewrite the equation $x=3-3t$ as $t=\frac{3-x}{3}$.

Substitute $\frac{3-x}{3}$ for $t$ in the equation $y=2t$.

  $\begin{array}{c}y=2\left(\frac{3-x}{3}\right)\\ =\frac{6}{3}-\frac{2}{3}x\\ =-\frac{2}{3}x+2\end{array}$

As shown in the graph, the portion traced by the parameterized curve is the segment from $\left(3,0\right)$ to $\left(0,2\right)$.

Therefore, the Cartesian equation for a curve that contains the parameterized curve is $y=-\frac{2}{3}x+2$ and portion traced by the parameterized curve is the segment from $\left(3,0\right)$ to $\left(0,2\right)$.