Chapter 0.4, Problem 18E

a.

To determine

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

Answer to Problem 18E

  

Initial point is $\left(3,0\right)$ and terminal point is $\left(0,2\right)$ .

Explanation of Solution

Given information:

The parametric equations:

  $x=3-3t,y=2t,\text{ 0}\le t\le 1$

Use the graphing calculator to graph the given curve.

First step is to set the mode to parametric.

For that press the MODE key.

Then Scroll down to FUNC.

Then move it to right till PAR

Press ENTER key.

  

Now, go back to main window by quitting (to do that press and then ).

Now, press the key

Now enter the given parametric equations as shown below:

  

Press the WINDOW key.

Enter values

  

Press GRAPH (here observe the direction in which the graph is being traced.)

  

This is the required graph and the red arrow gives the direction in which it is traced.

To find the initial point, substitute $t=0$ in the given parametric equation.

  $\begin{array}{c}x=3-3t\\ =3-3\left(0\right)\\ =3-0\\ =3\end{array}$

  $\begin{array}{c}y=2t\\ =2\left(0\right)\\ =0\end{array}$

So, the initial point is $\left(3,0\right)$ .

To find the terminal point, substitute $t=1$ in the given parametric equation.

  $\begin{array}{c}x=3-3t\\ =3-3\left(1\right)\\ =3-3\\ =0\end{array}$

  $\begin{array}{c}y=2t\\ =2\left(1\right)\\ =2\end{array}$

So, the terminal point is $\left(0,2\right)$ .

b.

To determine

To find: Cartesian equation for the curve that contains the parameterized curve. Also explain what portion of the Cartesian equation is traced by the parameterized curve.

Answer to Problem 18E

  $y=-\frac{2}{3}x+2$

The parameterized curve traces the Cartesian curve for $\text{0}\le x\le 3$ .

Explanation of Solution

Given information:

The parametric equations:

  $x=3-3t,y=2t,\text{ 0}\le t\le 1$

Solve $y=2t$ for t ,

  $\begin{array}{c}y=2t\\ t=\frac{y}{2}\end{array}$

Substitute $t=\frac{y}{2}$ in $x=3-3t$ ,

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

Thus, the required Cartesian equation is $y=-\frac{2}{3}x+2$ .

From the graph in part (a), it is clear that the parameterized curve traces the Cartesian curve for $\text{0}\le x\le 3$ .